\begin{document}

\title[Instanton moduli spaces and $\scW$-algebras]
{Instanton moduli spaces and $\scW$-algebras

}
\author[A.~Braverman]{Alexander Braverman}
\address{Department of Mathematics, Brown University, 151 Thayer st,
Providence RI 02912, USA}
\email{braval@math.brown.edu}
\author[M.~Finkelberg]{Michael Finkelberg}
\address{National Research University Higher School of Economics,
Department of Mathematics, 20 Myasnitskaya st, Moscow 101000 Russia}
\email{fnklberg@gmail.com}
\author[H.~Nakajima]{Hiraku Nakajima}
\address{Research Institute for Mathematical Sciences,
Kyoto University, Kyoto 606-8502,
Japan}
\email{nakajima@kurims.kyoto-u.ac.jp}

\dedicatory{In memory of Kentaro Nagao}

\subjclass[2000]{Primary 14C05; Secondary 14D21, 14J60}
\maketitle

\begin{NB}
\begin{verbatim}
\newcommand{\Uh}[2][G]{{\mathcal U}_{#1}^{#2}}
\end{verbatim}

\verb+\Uh{d}+ yields $\Uh{d}$.
\end{NB}

\setcounter{tocdepth}{1}
\tableofcontents

\begin{NB}
\subsection*{To do list}

\begin{itemize}
\item Complete \secref{sec:whittaker-state}. Give the proof of
  Conjectures~\ref{c1} : {\color{red}$\checkmark$}
\item Write introduction : SB {\color{red}$\checkmark$}
\item Check the proof of \lemref{lem:Fock} : SB,MF
\item Give the reference to Laumon's theorem at line \ref{Laumon} : SB {\color{red}$\checkmark$}
\item Give the reference to the dimension estimate at lines
  \ref{dim_est1}, \ref{dim_est2} : HN {\color{red}$\checkmark$}
\item Generalize the description of irreducible components of
  $\Uh[P,0]d$ in \propref{prop:Un} to arbitrary $G$. This is not
  necessary now, but may be useful for type $B_2$, $G_2$ : ALL {\color{red}$\checkmark$}
\item Give the proof of \lemref{lem:inter=pair} : ALL {\color{red}$\checkmark$}
\item Study the additional condition (2) in \lemref{lem:cochar} : SB,MF
\item Check the computation \eqref{eq:Vira} : SB,MF
\item Check the computation \eqref{eq:92} : SB,MF
\item Check the assertion \eqref{eq:32} : SB,MF
\item Check \subsecref{sec:highest-weight} : SB,MF
\item Check \subsecref{sec:kac-shapovalov-form} : SB,MF
\item Use character formula to complete the proof of
  \propref{prop:lambda}, \propref{prop:span} : Either SB or MF, as it
  depends on \cite{BFG}. {\color{red}$\checkmark$}
\item Fill the gap in the proof of \propref{prop:diff_proof} : ALL {\color{red}$\checkmark$}
\item Give the proof of \propref{prop:SpanP} : ALL {\color{red}$\checkmark$}
\item Give the proof of \lemref{lem:geomind} : ALL {\color{red}$\checkmark$}
\item Check HN's comments in \secref{sec:append-exactn} : SB {\color{red}$\checkmark$}
\item Write the proof of the embedding of $\scW_\bA(\g)$ into the
  $\scW$-algebra of Levi : SB {\color{red}$\checkmark$}
\end{itemize}
\end{NB}

\input{introduction}


\section{Preliminaries}\label{sec:preliminaries}

A basic reference to results in this section is \cite{BFG}, where
\cite{AHS,Lecture} are quoted occasionally.

\subsection{Instanton number}\label{subsec:instnumber}

We define an instanton number of a $G$-bundle $\mathcal F$ over
$\proj^2$. It is explained in, for example, \cite{AHS}.

Since it is related to our assumption that $G$ is simply-laced, we
briefly recall the definition.

The instanton number is the characteristic class associated with an
invariant bilinear form $(\ ,\ )$ on the Lie algebra $\g$ of
$G$. Since we assume $G$ is simple, the bilinear form is unique up to
scalar. We normalize it so that the square length of the highest root
$\theta$ is $2$.
\index{( , )@$(\ ,\ )$}

When $G = SL(r)$, it is nothing but the second Chern class of the
associated complex vector bundle.

For an embedding $SL(2)\to G$ corresponding to a root $\alpha$, we can
induce a $G$-bundle $\mathcal F$ from an $SL(2)$-bundle $\mathcal
F_{SL(2)}$. Then the corresponding instanton numbers are related by
\begin{equation}
  d(\mathcal F) = d(\mathcal F_{SL(2)})\times \frac2{(\alpha,\alpha)}.
\end{equation}
Since we assume $G$ is simply-laced, we have $(\alpha,\alpha) = 2$ for
any root $\alpha$. Thus the instanton number is preserved under the
induction.

\subsection{Moduli of framed \texorpdfstring{$G$}{G}-bundles}

Let $\Bun{d}$ be the moduli space of $G$-bundles with trivialization
at $\linf$ of instanton number $d$ as before.
\begin{NB}
  If the following is explained in Introduction, it will be deleted.
\end{NB}
We often call them {\it framed $G$-bundles}.

The tangent space of $\Bun{d}$\index{Bun@$\Bun{d}$} at $\mathcal F$ is
equal to the cohomology group $H^1(\proj^2,\g_{\mathcal F}(-\linf))$,
where $\g_{\mathcal F}$ is the vector bundle associated with $\mathcal
F$ by the adjoint representation $G\to GL(\g)$ (\cite[3.5]{BFG}).

Other degree cohomology groups vanish, and hence the dimension of
$H^1$ is given by the Riemann-Roch formula.

It is equal to $2dh^\vee$ (\cite{AHS}). Here
$h^\vee$\index{hcheck@$h^\vee$ (dual Coxeter number)} is the dual
Coxeter number of $G$, appearing as the ratio of the Killing form and
our normalized inner product $(\ ,\ )$.

It is known that $\Bun{d}$ is connected, and hence irreducible
(\cite[Prop.~2.25]{BFG}).

It is also known that $\Bun{d}$ is a holomorphic symplectic
manifold. Here the symplectic form is given by the isomorphism
\begin{equation}
  H^1(\proj^2,\g_{\mathcal F}(-2\linf))
  \xrightarrow{\cong}
  H^1(\proj^2,\g^*_{\mathcal F}(-\linf)),
\end{equation}
where $\g\cong\g^*$ is induced by the invariant bilinear form, and
$\shfO_{\proj^2}(-\linf)\to\shfO_{\proj^2}(-2\linf)$ is given by the
multiplication by the coordinate $z_0$ corresponding to $\linf$.

The tangent space $T_{\mathcal F}\Bun{d}\cong H^1(\proj^2,\g_{\mathcal
  F}(-\linf))$ is isomorphic also to $H^1(\proj^2,\g_{\mathcal
  F}(-2\linf))$ and the above isomorphism can be regarded as
$T_{\mathcal F}\Bun{d}\to T_{\mathcal F}^*\Bun{d}$. It is
nondegenerate and closed. (See \cite[Ch.~2, 3]{Lecture} for
$G=SL(r)$. General cases can be deduced from the $SL(r)$-case by a
faithful embedding $G\to SL(r)$.)

\subsection{Stratification}

Let $\Uh{d}$ be the Uhlenbeck space for $G$.

It has a stratification
\begin{equation}
  \label{eq:strat}
  \Uh{d} = \bigsqcup \BunGl{d_1},\quad
  \BunGl{d_1} = \Bun{d_1}\times S_\lambda\AA^2,
  \index{BunGlambda@$\BunGl{d_1}$}
\end{equation}
where the sum runs over pairs of integers $d_1$ and partitions
$\lambda$ with $d_1+|\lambda| = d$. Here
$S_\lambda\AA^2$\index{SlambdaA@$S_\lambda\AA^2$} is a stratum of the
symmetric product $S^{|\lambda|}\AA^2$, consisting of configurations
of points whose multiplicities are given by $\lambda$, that is
\begin{equation}
  \label{eq:Sl}
  S_\lambda\AA^2 = \left\{
    \sum \lambda_i x_i\in S^{|\lambda|}\AA^2 \middle|\,
      \text{$x_i\neq x_j$ for $i\neq j$}\right\}
\end{equation}
for $\lambda = (\lambda_1\ge\lambda_2\ge\cdots)$. We have
\begin{equation}
  \dim \BunGl{d_1} = 2(d_1 h^\vee + l(\lambda)).
\end{equation}

Let $\UhGl{d_1}$ be the closure of $\BunGl{d_1}$. We
have a finite morphism
\begin{equation}\label{eq:finite}
  \Uh{d_1}\times\overline{S_\lambda\AA^2}\to \UhGl{d_1},
\end{equation}
extending the identification $\Bun{d_1}\times S_\lambda\AA^2 =
\BunGl{d_1}$, where $\overline{S_\lambda\AA^2}$ is the closure of
$S_\lambda\AA^2$ in $S^{|\lambda|}\AA^2$.

\subsection{Factorization}\label{sec:factorization}

For any projection $a\colon\AA^2 \to \AA^1$\index{a@$a$ (projection)}
we have a natural map $\pi^d_{a,G}\colon \Uh{d} \to
S^d\AA^1$\index{paidpg@$\pi^d_{p,G}$ (factorization morphism)}. See
\cite[\S6.4]{BFG}. It is equivariant under $\GG =
G\times\CC^*\times\CC^*$: it is purely invariant under $G$. We also
change the projection $a$ according to the $\CC^*\times\CC^*$-action.

Let us explain a few properties. Let $\mathcal F\in\Bun{d}$. It is a
principal $G$-bundle over $\proj^2$ trivialized at $\linf$, but can be
also considered as a $G$-bundle over $\proj^1\times\proj^1$
trivialized at the union of two lines $\{\infty\}\times\proj^1$ and
$\proj^1\times\{\infty\}$. We extend $a$ to
$\proj^1\times\proj^1\to\proj^1$.

Then $\pi^d_{a,G}(\mathcal F)$ measures how the restriction of
$\mathcal F$ to a projective line $a^{-1}(x)$ differs from the trivial
$G$-bundle for $x\in\proj^1$. If $x$ is disjoint from
$\pi^d_{a,G}(\mathcal F)$, then $\mathcal F|_{a^{-1}(x)}$ is a trivial
$G$-bundle. If not, the coefficient of $x$ in $\pi^d_{a,G}(\mathcal
F)$ counts non-triviality with an appropriate multiplicity. (See
\cite[\S4]{BFG}.)

On the stratum $\Bun{d_1}\times S_{\lambda}\AA^2$, $\pi^d_{a,G}$ is
given as the sum of $\pi^{d_1}_{a,G}$ and the natural morphism
$S_{\lambda}\AA^2\to S^{|\lambda|}\AA^1$ induced from $a$. This
property comes from the definition of the Uhlenbeck as a space of
quasi-maps. (See \cite[\S\S1,2]{BFG}.)

For type $A$, it is given as follows in terms of the ADHM description
$(B_1,B_2,I,J)$ (see \cite[Ch.~2]{Lecture}): let $B_a$ be the linear
combination of $B_1$, $B_2$ corresponding to the projection
$a\colon\AA^2\to \AA^1$. Then $\pi^d_{a,G}$ is the characteristic
polynomial of $B_a$. (See \cite[Lem.~5.9]{BFG}.)

Moreover, most importantly, this map enjoys the factorization
property, which says the following. Let us write $d = d_1+d_2$ with
$d_1, d_2 > 0$. Let $(S^{d_1}\AA^1\times S^{d_2}\AA^1)_0$ be the open
subset of $S^{d_1}\AA^1\times S^{d_2}\AA^1$ where the first divisor is
disjoint from the second divisor. Then we have a natural isomorphism
\begin{equation}
    \label{eq:4}
    \Uh{d}\times_{S^d\AA^1} (S^{d_1}\AA^1\times S^{d_2}\AA^1)_0
    \cong (\pi^{d_1}_{a,G} \times \pi^{d_2}_{a,G})^{-1}(
    (S^{d_1}\AA^1\times S^{d_2}\AA^1)_0).
\end{equation}
See \cite[Prop.~6.5]{BFG}. We call $\pi^d_{a,G}$ the {\it
  factorization morphism}. Often we are going to make statements
about $\Uh{d}$ and we are going to prove them by induction on $d$;
\eqref{eq:4} will usually allow us to say that the inductive step is
trivial away from the preimage under $\pi^d_{a,G}$ of the main
diagonal in $S^d\AA^1$. In this case we are going to say that (the
generic part of) the induction step ``follows by the factorization
argument''.

\begin{NB}
\subsection{Parabolic subgroups}

Let us fix a Borel subgroup $B$ and a maximal torus $T$. Then we have
simple roots $\alpha_i\in X^*(T) = \Hom(T,\GG_m)$. Let $I$ denote the
set of simple roots. Let us choose a one parameter subgroup
$\lambda\in X_*(T) = \Hom(\GG_m,T)$ from the fundamental chamber,
i.e., $\langle \lambda, \alpha_i\rangle > 0$ for any $i\in I$. Then we
recover $B$ and $T$ from $\lambda$ as
\begin{equation}
  B = \left\{ g\in G\, \middle|\,\text{$\lim_{t\to 0}
  \lambda(t) g\lambda(t)^{-1}$ exists}\right\},
\quad
  T = G^{\lambda(\GG_m)}.
\end{equation}

We identify $G/T$ as the space of one parameter subgroups
\begin{equation}
  G/T \ni h \mapsto \{ h\lambda h^{-1}\colon \GG_m\to G \}.
\end{equation}
Let
\begin{equation}
  \begin{split}
    & B^h \defeq \left\{ g\in G\, \middle|\, \text{$\lim_{t\to 0}
  (h\lambda(t) h^{-1}) g (h\lambda(t) h^{-1})^{-1}$
  exists}\right\} = h B h^{-1},
\\
    & T^h \defeq G^{h\lambda(\GG_m) h^{-1}} = h T h^{-1}.
  \end{split}
\end{equation}
All pairs $(B^h, T^h)$ of Borel subgroups and tori appear in this
way. Therefore $G/T$ can be considered as the space of pairs.

Choose a subset $J\subset I$. Then we choose $\lambda_J\in X_*(T)$ in
so that
\begin{equation}
  \begin{split}
  & \langle \lambda_J, \alpha_i\rangle > 0 \quad\text{for $i\notin J$},
\\
  & \langle \lambda_J, \alpha_i\rangle = 0 \quad\text{for $i\in J$}.
  \end{split}
\end{equation}
Then we define
\begin{equation}
  \begin{split}
    & P_J \defeq \left\{ g\in G\, \middle|\, \text{$\lim_{t\to 0}
  \lambda_J(t) g\lambda_J(t)^{-1}$ exists}\right\},
\\
    & L_J \defeq G^{\lambda_J(\GG_m)}.
  \end{split}
\end{equation}
It is a parabolic subgroup and its Levi subgroup corresponding to
$J$. (Note $P_\emptyset = B$, $L_\emptyset = T$, $P_I = L_I = G$.)

We can define $P_J^h$, $L_J^h$ in the same way as above. The space of
pairs $(P, L)$ of parabolic subgroups and its Levi subgroups is equal
to $G/L_J$.
\begin{NB2}
  This is not true. Misha's message on Oct. 30: Consider $G=SL(3)$ and
  $L$ is of type $(2,1)$. The standard parabolic $P$ and opposite
  parabolic $P_-$ is not conjugate. Let us take $w_0$, the longest
  element of the Weyl group. We have a parabolic subgroup $P^{w_0}$,
  but its Levi subgroup is of type $(1,2)$.
\end{NB2}

Let $B^\text{op}$ be the opposite Borel subgroup. It is given by the
longest element $w_0$ in the Weyl group $W = N(T)/T$ as
$B^{w_0}$. Note that $w_0$ induces an involutive diagram automorphism
$\varpi$ of $I$ by
\begin{equation}
  w_0(\alpha_i) = - \alpha_{\varpi(i)}.
\end{equation}
For example, for type $A_\ell$, it is the reflection with respect to
the center in the Dynkin diagram. In the McKay correspondence, $I$ is
the set of nontrivial irreducible representations of a finite subgroup
$\Gamma$ of $SL_2(\CC)$. Then $\varpi$ is given by the dual representation.

We have a projection $G/T\to G/L_J$. If we regard $w_0$ as an element
in $G/L_J$, then $L_J^{w_0} = L_{\varpi(J)}$. This probably means that
what is written after \eqref{eq:decomp2} is {\it wrong}. We have
decomposition
\begin{equation}
  U^B_{T,G} = U^{P_i}_{T,L_i} \oplus (U^B_{T,L_i})^\perp.
\end{equation}
If we move this by $w_0$, we get
\begin{equation}
  U^{B^w_0}_{T,G} = U^{P_i^{w_0}}_{T,L_{\varpi(i)}} \oplus
  (U^{P_i^{w_0}}_{T,L_{\varpi(i)}})^\perp.
\end{equation}

Therefore the inner product using the trivialization of the local
system should not be a standard one. It should be given by
\begin{equation}
  d(x, w_0 y) \qquad\text{for $x,y\in\h$}
\end{equation}
under $U^d_{T,G}\cong\h$.
\end{NB}

\input{local}

\section{Hyperbolic restriction on Uhlenbeck spaces}\label{sec:hyper-uhlenbeck}

This section is of technical nature, but will play a quite important
role later. Feigin-Frenkel realized the $\scW$-algebra $\scW_k(\g)$
\index{Wkg@$\scW_k(\g)$|textit} in the Heisenberg algebra $\Heis(\h)$
associated with the Cartan subalgebra $\h$ of $\g$.  (See
\cite[Ch.~15]{F-BZ}.)

We will realize this picture in a geometric way. In \cite{MO}
Maulik-Okounkov achieved it by {\it stable envelops\/} which relate the
cohomology group of Gieseker space to that of the fixed point set with
respect to a torus.

The former is a module over $\scW_k(\g)$ and the latter is a Heisenberg
module.

In \cite{SV} Schiffmann-Vasserot also related two cohomology groups by
a different method.

We will take a similar approach, but we need to use a sheaf theoretic
language, as Uhlenbeck space is singular. We use the {\it hyperbolic
  restriction functor} in \subsecref{sec:hyperb-restr-1}, and combine
it with the theory of stable envelops. This study was initiated by the
third author \cite{tensor2}. A new and main result here is
\thmref{thm:perverse}, which says that perversity is preserved under
the hyperbolic restriction in our situation.

\begin{NB}
    Sasha, please put an explanation on the result of
    Mirkovi\'c-Vilonen \cite{MV2}.

    Sasha says that there is no formal relation. I agree, but at least
    that they proved that a hyperbolic restriction preserves the
    perversity in their situation.
\end{NB}

We fix a pair $T\subset B$ of a maximal torus $T$ and a Borel subgroup
$B$, and consider only parabolic subgroups $P$ containing $B$, except
we occasionally use opposite parabolic subgroups $P_-$ until
\subsecref{sec:autg-invariance}. In \subsecref{sec:autg-invariance},
we consider other parabolic subgroups also.

\subsection{A category of semisimple perverse sheaves}

Let $\IC(\BunGl{d},\rho)$ denote the intersection cohomology (IC)
complexes, where $\rho$ is a simple local system on $\BunGl{d} =
\Bun{d}\times S_\lambda\AA^2$ corresponding to an irreducible
representation of $S_{n_1}\times S_{n_2}\times\cdots$ via the covering
\begin{equation}
  (\AA^2)^{n_1}\times (\AA^2)^{n_2}\times
  \cdots \setminus\text{diagonal}
  \to
  S_{\lambda}\AA^2,
\end{equation}
where $\lambda = (1^{n_1} 2^{n_2}\cdots)$. (Recall $S_\lambda\AA^2$ is
a stratum of $S^{|\lambda|}\AA^2$, see \eqref{eq:Sl}.)

\begin{Definition}
Let $\Perv(\Uh{d})$\index{perveU@$\Perv(\Uh{d})$} be the additive
subcategory of the abelian category of semisimple perverse sheaves on
$\Uh{d}$, consisting of finite direct sum of $\IC(\BunGl{d},\rho)$.
\end{Definition}

By abuse of notation, we use the same notation $\IC(\BunGl{d},\rho)$
even if $\rho$ is a reducible representation of $S_{n_1}\times
S_{n_2}\times\cdots$. It is the direct sum of the corresponding
simple IC sheaves.

If $\rho$ is the trivial rank $1$ local system, we omit $\rho$ from
the notation and denote the corresponding IC complex by
$\IC(\BunGl{d})$, or $\IC(\UhGl{d})$.

Furthermore, we omit $\lambda$ from the notation when it is the empty
partition $\emptyset$. Therefore $\IC(\Uh{d})$ means
$\IC(\Bun[G,\emptyset]{d})$.

Objects in $\Perv(\Uh{d})$ naturally have structures of equivariant
perverse sheaves in the sense of \cite{BL} with respect to the group
action $\GG = G\times\CC^*\times\CC^*$ on $\Uh{d}$. We often view
$\Perv(\Uh{d})$ as the subcategory of equivariant perverse sheaves.

\subsection{Fixed points}\label{subsec:fixed}

Let $P$ be a parabolic subgroup of $G$ with a Levi subgroup $L$. Let
$A = Z(L)^0$ denote the connected center of $L$.
\begin{NB}
  $Z(L)$ is not connected in general. Misha's message on Oct. 30:
  consider $G = SL(4)$ and the Levi subgroup of type $(2,2)$. Then the
  center consists of diagonal matrices of the form $(x,x,y,y)$ with
  $x^2 y^2 = 1$. Therefore we have components correspond to $xy = \pm
  1$.
\end{NB}
Let $\BunL{d}$ denote the moduli space of $L$-bundles on $\proj^2$
with trivialization at $\linf$ of `instanton number $d$'. The latter
expression makes sense, since the notion of instanton number, defined
as in \subsecref{subsec:instnumber}, corresponds to a choice of a
bilinear form on the coweight lattice, which is the same for $G$ and
for $L$.

Suppose that $\mathcal F\in \Bun{d}$ is fixed by the $A$-action. It
means that bundle automorphisms at $\linf$ parametrized by $A$
extend to the whole space $\proj^2$. The extensions are
unique. Therefore the structure group $G$ of $\mathcal F$ reduces to
the centralizer of $A$, which is $L$. Hence $(\Bun{d})^{A} =
\BunL{d}$.

Let us consider the fixed point subvariety
\begin{equation}
    \UhL{d} = (\Uh{d})^{A} \index{UhL@$\UhL{d}$}
\end{equation}
in the Uhlenbeck space.

Then we have an induced stratification
\begin{equation}
    \UhL{d} = \bigsqcup_{d_1+d_2=d, \lambda \vdash d_2} \BunLl{d_1},
    \quad \BunLl{d_1} = \BunL{d_1} \times S_\lambda\AA^2.
\end{equation}

Strictly speaking, our $\UhL{d}$ depends on the choice of the
embedding $L\to G$, therefore should be denoted, say by
$\Uh[L,G]{d}$. We think that there is no fear of confusion.

Note that $[L,L]$ is again semi-simple and simply-connected.
\begin{NB}
  Do you have a reference ? I googled and find

  \url{http://mathoverflow.net/questions/126516/simply-connected-algebraic-groups-and-reductive-subgroups-of-maximal-rank}.

It is referred to Borel-Tits. Which paper ?

And I give some detail on the relation between $\Uh[{[L,L]}]{d}$ and
$\UhL{d}$. Please check it.
\end{NB}
(See \cite[Cor.~4.4]{Borel-Tits}.)

Suppose that we have only one simple factor. Since we assume $G$ is
simply-laced, $[L,L]$ is also. The instanton number is the same for
$G$ and $[L,L]$. Otherwise we define the instanton number for $[L,L]$
by the invariant form on $\operatorname{Lie}([L,L])$ induced from one
on $\g$.

Since we only have trivial framed $L/[L,L]$-bundles as $\proj^2$ is
simply-connected, we have
\begin{equation}\label{eq:BunLL}
  \BunL{d_1} = \Bun[{[L, L]}]{d_1}.
\end{equation}

Since $[L,L]$ is a subgroup of $G$, we have the induced closed
embedding $\Uh[{[L,L]}]{d}\to \Uh{d}$ (see \cite[Lem.~6.2]{BFG}),
which clearly factors as
\begin{equation}
  \Uh[{[L,L]}]{d}\to \UhL{d}.
\end{equation}
By \eqref{eq:BunLL}, this map is bijective.
\begin{NB}
  We do not know $\UhL{d}$ is normal.
\end{NB}
Since both spaces are closed subschemes of $\Uh{d}$, we have

\begin{Proposition}\label{prop:LL}
  The morphism $\Uh[{[L,L]}]{d}\to \UhL{d} = (\Uh{d})^A$ is a
  homeomorphism between the underlying topological spaces.
\end{Proposition}

We are interested in perverse sheaves on $\UhL{d}$, hence we only need
underlying topological spaces. Hence we may identify $\UhL{d}$ and
$\Uh[{[L,L]}]{d}$.

We define the category $\Perv(\UhL{d})$ in the same way as
$\Perv(\Uh{d})$.

\begin{Example}\label{ex:torus_fixed}
  The case when $L$ is a maximal torus $T$ is most important. We have
\begin{equation}
  \Uh[T]{d} = S^d\AA^2 = \bigsqcup_{\lambda\vdash d} S_\lambda\AA^2,
\end{equation}
as we do not have nontrivial framed $T$-bundles.
\end{Example}

\subsection{Polarization}\label{sec:polarization}

Following \cite[\S3.3.2]{MO}, we introduce the notion of a {\it
  polarization\/} of a normal bundle of the smooth part of a fixed
point component.

Let us give a definition in a general situation. Suppose a torus $A$
acts on a holomorphic symplectic manifold $X$, preserving the
symplectic structure. Let $Z$ be a connected component of $X^A$ and
$N_Z$ be its normal bundle in $X$. Consider $A$-weights of a fiber of
$N_Z$.

Let $e(N_Z)|_{H^*_A(\mathrm{pt})}$ be the $H^*_A(\mathrm{pt})$-part of
the Euler class of the normal bundle, namely the product of all
$A$-weights of a fiber of $N_Z$.

Since $A$ preserves the symplectic form, $Z$ is a symplectic
submanifold, and weights of $N_Z$ appear in the pairs
$(\alpha_i,-\alpha_i)$. Hence
\begin{equation}\label{eq:e^2}
  (-1)^{(\codim Z)/2} e(N_Z)|_{H^*_A(\mathrm{pt})} = \prod \alpha_i^2
\end{equation}
is a perfect square. A choice of a square root
$\pol$\index{delta@$\pol$ (polarization)} of \eqref{eq:e^2} is called a
{\it polarization\/} of $Z$ in $X$.

In the next subsection we consider attractors and repellents. We have
a polarization $\pol_{\text{rep}}$ given by product of weights in
repellent directions. However this will not be a right choice to save
signs. Our choice of the polarization $\pol$, which follows
\cite[Ex.~3.3.3]{MO}, will be explained in \subsecref{sec:Gifixed} for
Gieseker spaces, and in \subsecref{sec:anotherbase2} for Uhlenbeck
spaces. Then we understand $\pol = \pm 1$, depending on whether it is
the same as or the opposite to $\pol_{\text{rep}}$, in other words we
identify $\pol$ with $\pol/\pol_{\text{rep}}$, as
$\delta_{\text{rep}}$ is clear from the context.

Note that a polarization does not make sense unless the variety $X$ is
smooth. Therefore we restrict the normal bundle to $Z\cap\Bun{d} =
Z\cap\Bun[L]d$ and consider a polarization there for Uhlenbeck spaces.

However a fixed point component $Z$, in general, does not intersect
with $\Bun{d}$. Say $Z\cap\Bun{d} = \emptyset$ if $L = T$. We do not
consider a polarization of $Z$ in this case, and smooth cases are
enough for our purpose.

\subsection{Definition of hyperbolic restriction functor}\label{sec:hyperb-restr}

We now return to the situation when $X = \Uh{d}$. We choose a
parabolic subgroup $P$ with a Levi subgroup $L$ as before.

We consider the setting in
\S\S\ref{sec:attr-repell},\ref{sec:hyperb-restr-1} with $A =
Z(L)^0$. Then \eqref{eq:29} is the hyperplane arrangements induced by
roots:
\begin{equation}
  \fa_\RR \setminus \bigcup_{\alpha} \{ \left.\alpha\right|_{\fa_\RR} = 0 \},
\end{equation}
where the union runs over all positive roots $\alpha$ which do not vanish
on $\fa_\RR$. The chambers are in one to one correspondence to the parabolic
subgroups containing $L$ as their Levi (associated parabolics).

Therefore the fixed $P$ determines a `positive' chamber.

We denote the corresponding attracting and repelling sets $\mathcal
A_X$, $\mathcal R_X$ by $\UhP{d}$\index{UhP@$\UhP{d}$} and
$\UhPm{d}$. Often we are going to drop the instanton number $d$ from
the notation, when there is no fear of confusion. We let
$i$\index{i@$i$ (inclusion $\UhL{d}\to\UhP{d}$)} and $p$\index{p@$p$
  (projection $\UhP{d}\to\UhL{d}$)} denote the corresponding maps from
$\UhL{}$ to $\UhP{}$ and from $\UhP{}$ to $\UhL{}$. Also we denote by
$j$\index{j@$j$ (inclusion $\UhP{d}\to \Uh{d}$)} the embedding of
$\UhP{}$ to $\Uh{}$. We shall sometimes also use similar maps $i_-$,
$j_-$ and $p_-$ where $\UhP{}$ is replaced with $\UhPm{}$. We have
diagrams
\begin{equation}
    \label{eq:1}
    \UhL{} \overset{p}{\underset{i}{\leftrightarrows}}
    \UhP{} \overset{j}{\rightarrow} \Uh{},
\qquad
    \UhL{} \overset{p_-}{\underset{i_-}{\leftrightarrows}}
    \UhPm{} \overset{j_-}{\rightarrow} \Uh{}.
\end{equation}

\begin{Definition}
    We define the functor $\Phi_{L,G}$\index{UZphiLG@$\Phi_{L,G}$} by
    $i^* j^! = p_* j^!$.
\end{Definition}
\begin{NB}
  Sometimes, we are going to write $\Phi_{L,G}^d$ (when we want to fix
  $d$).

  Do we really need the notation $\Phi_{L,G}^d$ ? It seems to me that
  $P$-dependence is more important. So why do not we write
  $\Phi_{L,G}^P$ when we want to emphasize $P$ ?

  \begin{NB2}
    Oct. 24: Now it is comment out.
  \end{NB2}
\end{NB}

We apply it to weakly $A$-equivariant objects, in particular on
$\Perv(\Uh{d})$.


\noindent
{\bf Warning.} Of course, the functor $\Phi_{L,G}$ depends on $P$ and
not just on $L$. When we want to emphasize $P$, we write
$\Phi^P_{L,G}$. Otherwise $P$ is always chosen so that $P\supset B$
for the fixed Borel subgroup $B$.

Let us justify our notation $\UhP{}$ for the attracting set. We have a
one parameter subgroup $\lambda\colon\GG_m\to G$ such that
\begin{equation}\label{eq:PLoneparam}
  \begin{split}
  P & = \left\{ g\in G\, \middle|\, \text{$\lim_{t\to 0}
  \lambda(t) g\lambda(t)^{-1}$ exists}\right\},
\\
  L &= G^{\lambda(\mathbb G_m)}
  = \left\{ g\in G\, \middle|\, \text{$
  \lambda(t) g = g\lambda(t)$ for any $t\in\mathbb G_m$}\right\}.
  \end{split}
\end{equation}
Then we have
\begin{equation}
  \begin{split}
    \UhP{} & \defeq \left\{ x\in \Uh{}\, \middle|\, \text{$\lim_{t\to 0}
  \lambda(t)\cdot x$ exists}\right\},
\\
  \UhL{} &\defeq (\Uh{})^{\lambda(\mathbb G_m)}
  = \left\{ x\in \Uh{}\, \middle|\, \text{$
  \lambda(t)\cdot x = x$ for any $t\in\mathbb G_m$}\right\}.
  \end{split}
\end{equation}
We embed $G$ into $SL(N)$ and consider the corresponding space for $G
= SL(N)$. We use the ADHM description for $\Uh[SL(N)]{}$ to identify
it with the affine GIT quotient as in \cite[Ch.~3]{Lecture}. Then
$SL(N) = SL(W)$, and $\UhP{}$ coincides with the variety
$\pi(\mathfrak Z)$ studied in \cite[\S3]{Na-Tensor}. Here $\pi$ is
Gieseker-Uhlenbeck morphism, and $\mathfrak Z$ is the attracting set
in the Gieseker space, which will be denoted by $\Gi{P}$ later.

In \cite[Rem.~3.16]{Na-Tensor} it was remarked that $\mathfrak Z$
parametrizes framed torsion free sheaves having a filtration $E = E^0
\supset E^1 \supset \cdots \supset E^k \supset E^{k+1} = 0$.

If all $F^i = E^i/E^{i+1}$ are locally free, $E$ is a $P$-bundle. Thus
$\UhP{}$ contains a possibly empty open subset $p^{-1}(\BunL{})$
consisting of $P$-bundles.

Let us, however, note that $\UhP{}\cap \Bun{}$ is not entirely
consisting of $P$-bundles, hence larger than
$p^{-1}(\BunL{})$: Consider a short exact sequence
\begin{equation*}
   0 \to F^2 \to E \to F^1 = \mathcal I_x \to 0,
\end{equation*}
arising from the Koszul resolution of the skyscraper sheaf at a point
$x\in\AA^2$. Here $\mathcal I_x$ is the ideal sheaf for $x$. Then
$E\in\UhP{}\cap\Bun{}$, but $E$ is not a $P$-bundle as $F^1$ is not
locally free.

More detailed analysis will be given in the proof of \propref{prop:Un}.

\subsection{Associativity}\label{sec:ass}

\begin{Proposition}\label{prop:trans}
  Let $Q$ be another parabolic subgroup of $G$, contained in $P$ and
  let $M$ denote its Levi subgroup. Let $Q_L$ be the image of $Q$ in
  $L$ and we identify $M$ with the corresponding Levi group.

  Then we have a natural isomorphism of functors
    \begin{equation}
        \label{eq:5}
       \Phi_{M, L} \circ \Phi_{L,G} \cong \Phi_{M,G}.
    \end{equation}
\end{Proposition}

\begin{proof}
    It is enough to show that
    \begin{equation}
        \label{eq:2}
            \UhP{} \times_{\UhL{}} \Uh[{Q_L}]{} = \Uh[Q]{},
    \end{equation}
    as
    \begin{equation*}
      p'_* j^{\prime !} p_* j^!
      = p'_* p''_* j^{\prime\prime !} j^!
      = (p'\circ p'')_* (j\circ j'')^!
    \end{equation*}
    in the diagram
    \begin{equation}\label{eq:CDpj}
      \begin{CD}
        \Uh[Q]{} @>{j''}>> \UhP{} @>{j}>> \Uh{}
\\
        @V{p''}VV @VV{p}V @.
\\
        \Uh[Q_L]{} @>{j'}>> \UhL{} @.
\\
        @V{p'}VV @. @.
\\
        \Uh[M]{} @. @.
      \end{CD}
    \end{equation}

    The left hand side of \eqref{eq:2} is just equal to
    $p^{-1}(\Uh[{Q_L}]{})$. By embedding $G$ into $SL(N)$ we may
    assume that $G = SL(N)$.
    \begin{NB}
        $\UhP{}$ is defined via a one parameter subgroup, therefore it
        is respected under the embedding $G\to SL(N)$.
    \end{NB}
    In this case, we use the ADHM description
    to describe $\UhP{}$, $\Uh[Q]{}$, $\Uh[Q_L]{}$. By \cite[Proof of
    Lemma~3.6]{Na-Tensor}, they are consisting of data $(B_1,B_2,I,J)$
    such that $J F(B_1,B_2) I$ are in $P$, $Q$, $Q_L$ respectively,
    i.e., upper triangular in appropriate sense, for any products
    $F(B_1,B_2)$ of $B_1$, $B_2$ of arbitrary order. Now the assertion
    is clear.
    \begin{NB}
      Sep.15 : I give a proof via the ADHM.
    \end{NB}
\end{proof}

\subsection{Preservation of perversity}

The following is our first main result:

\begin{Theorem}\label{thm:perverse}
    $\Phi_{L,G}(\IC(\Uh{d}))$ is perverse \textup(and semi-simple,
    according to \cite[Theorem 2]{Braden}\textup). Moreover, the same is
    true for any perverse sheaf in $\Perv(\Uh{d})$.
\end{Theorem}

The proof will be given in \secref{sec:append-exactn}.

Let us remark that the result is easy to prove for type $A$, see
\cite[\S4.4, Lemma~3]{tensor2}. The argument goes back to an earlier
work by Varagnolo-Vasserot \cite{VV2}.

\begin{NB}
  Sep. 16 : The following subsection is moved from
  \subsecref{sec:anotherbase}, as the argument is already used.

  \begin{NB2}
    Oct. 23 : Don't we need the following subsection in the argument
    in Appendix ? It seems to me that you need it at the last
    moment. If so, why do not we move the following subsection to
    before this subsection ?
  \end{NB2}

  \begin{NB2}
    Answer is NO. It uses only the hyperbolic restriction of the
    trivial bundle on a smooth variety is trivial.
  \end{NB2}
\end{NB}

\subsection{Hyperbolic restriction on \texorpdfstring
{$\Bun[L]{d}$}{BunLd}}\label{sec:open}

Let us consider the restriction of $\Phi_{L,G}(\IC(\Uh{d}))$ to the
open subset $\Bun[L]{d}$ in this subsection.

For simplicity, suppose that $[L,L]$ has one simple factor so that the
instanton numbers of $L$-bundles are the same as those of
$[L,L]$-bundles. In particular, $\Bun[L]{d}$ is irreducible. Then
$\IC(\Uh[L]{d})$ is a simple perverse sheaf, and we study
\begin{equation}\label{eq:LG}
  \Hom_{\Perv(\UhL{d})}(\IC(\Uh[L]{d}),\Phi_{L,G}(\IC(\Uh{d}))).
\end{equation}

We restrict \eqref{eq:1} to the open subsets consisting of genuine
bundles:
\begin{equation}
    \label{eq:10}
    \BunL{d} \overset{p}{\underset{i}{\leftrightarrows}}
    p^{-1}(\BunL{d})
    \overset{j}{\rightarrow} \Bun{d}.
\end{equation}
Let us take $\mathcal F\in \BunL{d}$. Then the tangent space of
$\BunL{d}$ at $\mathcal F$ is $H^1(\proj^2, \mathfrak l_{\mathcal
  F}(-\linf))$, where $\mathfrak l$ is the Lie algebra of $L$. This is
the subspace of $H^1(\proj^2, \g_{\mathcal F}(-\linf)) = T_{\mathcal
  F}\Bun{d}$, consisting of $Z(L)^0$-fixed elements.

The normal bundle of $\BunL{d}$ in $\Bun{d}$ splits into the sum of
$H^1(\proj^2, \mathfrak n_{\mathcal F}(-\linf))$ and $H^1(\proj^2,
\mathfrak n^-_{\mathcal F}(-\linf))$, where $\mathfrak n$ is the nil
radical of $\mathfrak p = \operatorname{Lie}P$, and $\mathfrak n^-$ is
its opposite.

They correspond to attracting and repellent directions respectively.

Then $p^{-1}(\BunL{d})$ is a vector bundle over $\BunL{d}$, whose
fiber at $\mathcal F$ is $H^1(\proj^2, \mathfrak n_{\mathcal F}(-\linf))$. It
parametrizes framed $P$-bundles.
\begin{NB}
    For type $A$, this space parametrizes the extension of $E_2$ by
    $E_1$, hence is equal to $\Ext^1(E_1, E_2(-\linf))$. In general,
    it is a generalization of the Ext group, as above.
\end{NB}
The morphism $p$ is the projection and $i$ is the inclusion of the
zero section. Therefore we have the Thom isomorphism between
$i^*j^!(\cC_{\Bun{d}})$ and $\cC_{\BunL{d}}$ up to shift.

Note further that $\dim p^{-1}(\BunL{d})$ is the half of the sum of
dimensions of $\BunL{d}$ and $\Bun{d}$, as $H^1(\proj^2, \mathfrak
n_{\mathcal F}(-\linf))$ and $H^1(\proj^2, \mathfrak n^-_{\mathcal
  F}(-\linf))$ are dual to each other with respect to the symplectic
form.

Hence a shift is unnecessary, and the Thom isomorphism gives the
canonical identification $i^* j^! (\cC_{\Bun{d}})\cong
\cC_{\BunL{d}}$.

Therefore we normalize the canonical homomorphism
\begin{equation}\label{eq:1LG}
  1^d_{L,G}\in \Hom_{\Perv(\UhL{d})}(\IC(\UhL{d}),
  \Phi_{L,G}(\IC(\Uh{d})))
\end{equation}
so that it is equal to the Thom isomorphism on the open subset.
\index{1LG@$1^d_{L,G}$}

\begin{NB}
  Sep. 16 :
  This homomorphism is not a right one, as it must be corrected by a
  polarization.

  For $L = L_i$, the correction will be given by a bipartite coloring
  of the vertexes of Dynkin diagram, though I still need to check the
  detail.
  \begin{NB2}
    I checked the detail. It is explained in
    \subsecref{sec:anotherbase2}.
  \end{NB2}

  If this problem will be fixed, I may change my definition of
  $1^d_{L,G}$. But it applies only to the case $L = L_i$, so I
  probably will not.
\end{NB}

Note also that a homomorphism in \eqref{eq:LG} is determined by its
restriction to $\BunL{d}$, hence \eqref{eq:LG} is $1$-dimensional from
the above observation. And $1^d_{L,G}$ is its base.

If $[L,L]$ has more than one simple factors $G_1$, $G_2$, \dots,
$\Bun[L]d$ is not irreducible as it is isomorphic to
$\bigsqcup_{d_1+d_2+\dots=d} \Bun[G_1]{d_1}\times
\Bun[G_2]{d_2}\times\cdots$.

Then $\IC(\Uh[L]d)$ must be understood as the direct sum
\begin{equation}
  \bigoplus_{d_1+d_2+\dots=d} \IC(\Bun[G_1]{d_1}\times
  \Bun[G_2]{d_2}\times\cdots).
\end{equation}
In particular, \eqref{eq:LG} is not $1$-dimensional. But it does not
cause us any trouble. We have the canonical isomorphism for each
summand, and $1^d_{L,G}$ is understood as their sum.

\subsection{Space \texorpdfstring{$U^d$}{Ud} and its base}\label{sec:Ud}

We shall introduce the space $U^d$ of homomorphisms from
$\cC_{S_{(d)}\AA^2}$ to $\Phi_{L,G}(\IC(\Uh{d}))$ and study its
properties in this subsection. A part of computation is a byproduct of
the proof of \thmref{thm:perverse} (see \lemref{lem:Ud}).

The study of $U^d$ will be continued in the remainder of this section,
and also in the next section.

\begin{Definition}
For $d > 0$, we define a vector space
\begin{equation}\label{eq:Ud}
  \begin{split}
  U^d_{L,G} \equiv\index{ULG@$U^d_{L,G}$}
  U^d &\defeq \Hom_{\Perv(\UhL{d})}(\cC_{S_{(d)}\AA^2},\Phi_{L,G}(\IC(\Uh{d})))\\
  &= H^{-2}(S_{(d)}\AA^2,\xi^! \Phi_{L,G}(\IC(\Uh{d}))),
  \end{split}
\end{equation}
where $(d)$ is the partition of $d$ consisting of a single entry $d$,
and $\xi\colon S_{(d)}\AA^2\to \UhL{d}$ is the inclusion.
\index{SdA2@$S_{(d)}\AA^2$}
\end{Definition}

We use the notation $U^d$, when $L$, $G$ are clear from the context.
\begin{NB}
  Sep. 16 : I move the definition of $U^d_{L,G}$ here.
\end{NB}

Since the hyperbolic restriction $\Phi_{L,G}$ depends on $P$, the
space $U^d_{L,G}$ depends also on $P$. When we want to emphasize $P$,
we denote it by $U^{d,P}_{L,G}$\index{ULGdP@$U^{d,P}_{L,G}$} or simply
by $U^{d,P}$.

We have a natural evaluation homomorphism
\begin{equation}
    \label{eq:6}
    U^d\otimes \cC_{S_{(d)}\AA^2}\to \Phi_{L,G}(\IC(\Uh{d})),
\end{equation}
which gives the isotypical component of $\Phi_{L,G}(\IC(\Uh{d}))$
corresponding to the simple perverse sheaf $\cC_{S_{(d)}\AA^2}$.

By the factorization \secref{sec:factorization} together with the
Thom isomorphism $i^*j^!(\cC_{\Bun{d_1}})\cong \cC_{\BunL{d_1}}$,
\begin{NB}
  Sep. 16 : I add `Thom isomorphism'. This was missing in the previous
  version, but we certainly need it.

  We may introduce
  \begin{equation}
    W^d = \Hom_{\Perv(\UhL{d})}(\IC(\UhL{d}), \Phi_{L,G}(\IC(\Uh{d}))),
  \end{equation}
  and the canonical isomorphism is
  \begin{equation}
        \Phi_{L,G}(\IC(\Uh{d})) \cong \bigoplus
    \IC(\BunLl{d_1},\rho)\otimes W^{d_1}.
  \end{equation}
  I do not know which is better, at this moment, but I take an
  economical approach, fixing $W^{d_1}\cong \CC$.
\end{NB}
we get

\begin{Proposition}\label{prop:hyperbolic}
  We have the canonical isomorphism
\begin{equation}
    \label{eq:7}
    \Phi_{L,G}(\IC(\Uh{d})) \cong \bigoplus
    \IC(\BunLl{d_1},\rho).
\end{equation}
Here $\rho$ is the \textup(semisimple\textup) local system on
$\BunLl{d_1} = \BunL{d_1}\times S_\lambda\AA^2$ with $\lambda =
(1^{n_1} 2^{n_2}\cdots)$ corresponding to the representation of
$S_{n_1}\times S_{n_2}\times\cdots$ on $(U^1)^{\otimes n_1}\otimes
(U^2)^{\otimes n_2}\otimes\cdots$ given by permutation of factors.
\end{Proposition}

For example, the isotypical component for the intersection cohomology
complex $\IC(\Bun[L,\lambda]{d_1})$ for the trivial simple local system
is
\begin{equation}
    \label{eq:9}
    \Sym^{n_1} U^1\otimes \Sym^{n_2} U^2\otimes \cdots,
\end{equation}
where $\Sym$ denotes the symmetric power.

\begin{NB}
  The following will be used later, but the proof was not given
  before. Dec. 12.
\end{NB}

\begin{Lemma}\label{lem:Fock}
  Suppose $L=T$. We have
  \begin{equation}
    H^*(S^d\AA^2, \Phi_{T,G}(\IC(\Uh{d})))
    \cong \bigoplus_{|\lambda|=d} \Sym^{n_1}U^1\otimes\Sym^{n_2}U^2\otimes
    \cdots
  \end{equation}
where $\lambda = (1^{n_1}2^{n_2}\dots)$.
\end{Lemma}

\begin{proof}
  Since $L=T$, we have $\Uh[T]{d} = S^d\AA^2$. See
  Example~\ref{ex:torus_fixed}. Then the assertion means that only
  trivial representation of $S_{n_1}\times S_{n_2}\times\cdots$
  contribute to the global cohomology group.

  Let $U$ be an open subset of $(\AA^2)^{n_1}\times (\AA^2)^{n_2}
  \times\cdots$ consisting of pairwise disjoint $n_1$ ordered points,
  $n_2$ ordered points, and so on in $\AA^2$. Forgetting orderings, we
  get an $(S_{n_1}\times S_{n_2}\times\cdots)$-covering $p \colon U\to
  S_\lambda\AA^2$. The pushforward of the trivial rank $1$ system with
  respect to $p$ is the regular representation $\rho_{\mathrm{reg}}$
  of $(S_{n_1}\times S_{n_2}\times\cdots)$.

  Since $p$ extends to a finite morphism $(\AA^2)^{n_1}\times
  (\AA^2)^{n_2} \times\cdots\to \overline{S_\lambda\AA^2}$, we have
  $\IC(S_\lambda\AA^2,\rho_{\mathrm{reg}}) =
  p_*(\cC_{(\AA^2)^{n_1}\times (\AA^2)^{n_2} \times\cdots})$. By the
  K\"unneth theorem, the global cohomology group $H^*(\bullet)$ of the
  right hand side is $H^*((\AA^2)^{n_1})\otimes
  H^*((\AA^2)^{n_2})\otimes\cdots$. This is $1$-dimensional, and
  corresponds to the trivial isotypical component of
  $\rho_{\mathrm{reg}}$. Now the assertion follows.
\end{proof}

Let us continue the study of $U^d$.

Let us note that all of our spaces $\Uh{d}$, $\Uh[L]{d}$, $\Uh[P]{d}$
have trivial factors $\AA^2$ given by the center of instantons, or the
translation on the base space $\AA^2$ except $d=0$ where $\Uh{0} =
\Uh[L]{0} = \Uh[P]{0} = \mathrm{pt}$. We assume $d\neq 0$
hereafter. Let $\cUh{d}$ denote the centered Uhlenbeck space at the
origin, thus we have $\Uh{d} = \cUh{d}\times \AA^2$.

Let us compose factorization morphisms $\pi^d_{h,G}$, $\pi^d_{v,G}$
for the horizontal and vertical projections $h\colon\AA^2\to\AA^1$,
$v\colon\AA^2\to\AA^1$ with the sum map $\sigma\colon S^d\AA^1\to
\AA^1$. Then $\cUh{d} =
(\sigma\pi^d_{h,G}\times\sigma\pi^d_{v,G})^{-1}(0,0)$.

\index{UhcG@$\cUh{d}$ (centered Uhlenbeck space)}
We use the notation $\cUh[L]{d}$, $\cUh[P]{d}$ for $\Uh[L]{d}$,
$\Uh[P]{d}$ cases. The diagrams \eqref{eq:1} factor and induce the
diagrams for the centered spaces, and the factorization is compatible
with the hyperbolic restriction. Let us use the same notation for $i$,
$j$, $p$ for the centered spaces. Then we have
\begin{equation}
  U^{d} = H^0(\xi_0^! p_* j^!\IC(\cUh{d})),
\end{equation}
where $\xi_0$ is the inclusion of the single point $d\cdot 0$ in
$\cUh[L]{d}$.

Here $d\cdot 0$ is the point in $S_{(d)}\AA^2$, the origin with
multiplicity $d$.

By base change we get
\begin{equation}\label{eq:17}

  U^d 


   \cong H^{0}(p^{-1}(d\cdot 0), \tilde j^{!} \IC(\cUh{d})),

\end{equation}
where $\tilde j\colon p^{-1}(d\cdot 0)\to \cUh{d}$ is the
inclusion.
\begin{NB}
Consider the fiber square
\begin{equation}
    \begin{CD}
        p^{-1}(d\cdot 0) @>{\tilde \xi_0}>> \cUh[P]{d}
\\
       @V{p}VV @VV{p}V
\\
       \{ d\cdot 0\} @>{\xi_0}>> \cUh[L]{d}.
    \end{CD}
\end{equation}
We have $\xi_0^! p_* = p_* \tilde\xi_0^!$. Hence
\begin{equation}
    \begin{split}
    H^{0}(\xi_0^! p_* j^{!} \IC(\cUh{d}))
    &= H^{0}(p^{-1}(d\cdot 0),
    (j\circ \tilde\xi_0)^{!} \IC(\cUh{d}))
\\
   &= H^{0}(p^{-1}(d\cdot 0),
    \tilde j^{!} \IC(\cUh{d})).
    \end{split}
\end{equation}
\end{NB}

Note that the spaces of homomorphisms between objects in
$\Perv(\Uh[L]{d})$ are canonically isomorphic for equivariant category
with respect to $L\times\CC^*\times\CC^*$ and non-equivariant one.
(See \cite[1.16(a)]{Lu-cus2}.
\begin{NB}
  I cannot find a statement in \cite{BL}. Is this a standard result ?

  \begin{NB2}
    If we consider the corresponding statement for equivariant
    $D$-modules, it is obvious: Vector fields generated by Lie
    algebras determined the equivariant structure on the $D$-module
    for a {\it connected\/} group.
  \end{NB2}
\end{NB}
)

Therefore \eqref{eq:7} is an isomorphism in the equivariant derived
category, though we use the factorization, which is {\it not\/}
equivariant with respect to $\CC^*\times\CC^*$.

We have
\begin{Lemma}\label{lem:dimU}
    \begin{equation}
        \label{eq:8}
        \dim U^d = \rank G - \rank [L,L].
    \end{equation}
\end{Lemma}

\begin{proof}

  \begin{NB}
    The first sentence is comment out, as I changed the order of
    \propref{prop:hyperbolic} and \lemref{lem:dimU}.

  First, let us note that Proposition \ref{prop:hyperbolic} is
  independent of Lemma \ref{lem:dimU}. In addition,
  \end{NB}
  According to a theorem of Laumon~\cite{Laumon-chi}
  \begin{NB}
      \linelabel{Laumon}
  \end{NB}
  the Euler characteristic of the
  stalk of $\IC(\Uh{d})$ at a point of $S_{(d)}\AA^2$ is equal to the
  Euler characteristic of the stalk of $\Phi_{L,G}(\IC(\Uh{d}))$ at
  the same point; the former was computed in Theorem 7.10 in \cite{BFG}.
  \begin{NB}
    Put the reference of Laumon.
  \end{NB}

  \begin{NB}
    What do you mean ?

  Let us not prove Lemma \ref{lem:dimU}.
  \end{NB}

  First, let us give a proof in the case $L=T$. Then it is easy to see
  that Proposition \ref{prop:hyperbolic} implies that the stalk of
  $\Phi_{T,G}(\IC(\Uh{d}))$ at a point of $S_{(d)}\AA^2$ is isomorphic
  to $\Sym^d(\oplus_i U^i_{T,G})$, where we regard $\oplus_i
  U^i_{T,G}$ as a graded vector space (with the natural grading coming
  from $i$) and the super-script $d$ means degree $d$ with respect to
  that grading. Comparing it with Theorem 7.10 of \cite{BFG} (and
  using induction on $d$), we get $\dim U^d_{T,G}=\rank(G)$ for every
  $d$.

  Let us now consider the case of arbitrary $L$. Again, it is easy to
  deduce from Proposition \ref{prop:hyperbolic} that the stalk of
  $\Phi_{T,L}(\Phi_{L,G}(\IC(\Uh{d})))\simeq \Phi_{T,G}(\IC(\Uh{d}))$
  at a point of $S_{(d)}\AA^2$ is isomorphic to
$$
\bigoplus\limits_{d_1+d_2=d} \Sym^{d_1}(\oplus_i U^i_{T,L})\otimes \Sym^{d_2}(\oplus_j U^j_{L,G}),
$$
where the meaning of the super-scripts $d_1$ and $d_2$ is as above. In view of the preceding paragraph, we get
$\dim U^d_{L,G}=\rank(G)-\rank([L,L])$.
\end{proof}


The dimension estimate \corref{cor:dimest}
\begin{NB}
    \linelabel{dim_est1}
\end{NB}
\begin{NB}
  Reference to somewhere in the appendix. I must be careful, as the
  argument is not straight.
\end{NB}
and the argument in \cite[Prop.~3.10]{MV2}
implies that
\begin{equation}\label{eq:107}
  \begin{split}
    & H^{0}(p^{-1}(d\cdot 0), \tilde j^{!} \IC(\cUh{d}))
    \\
    \cong \; & H^{0}(p^{-1}(d\cdot 0)\cap\Bun{d}, \tilde
    j^{!} \IC(\cUh{d}))
    \\
    = \;& H_{[0]}(p^{-1}(d\cdot 0)\cap\Bun{d},\CC).
  \end{split}
\end{equation}
Here we use the degree shift convention of the Borel-Moore homology
group (see Convention~(\ref{conv2})), which is shift by $\dim\cUh{d} =
2dh^\vee-2$ in this case.

Let us set
\begin{equation}\label{247}
  \Uh[P,0]d \defeq p^{-1}(d\cdot 0). \index{UhP0@$\Uh[P,0]d$}
\end{equation}
The subscript $0$ stands for $d\cdot 0$, and this convention will be
also used later. More generally, we denote $p^{-1}(x)$ by $\Uh[P,x]d$ for
$x\in \UhL{d}$\index{UhPx@$\Uh[P,x]d$}.

Then $H_{[0]}(\Uh[P,0]d\cap\Bun{d},\CC)$ has a base given by
$(dh^\vee-1)$-dimensional irreducible components of
$\Uh[P,0]d\cap\Bun{d}$. The dimension estimate \corref{cor:dimest}
\begin{NB}
\linelabel{dim_est2}
\end{NB}
\begin{NB}
  Reference to somewhere in the appendix. I must be careful, as the
  argument is not straight.
\end{NB}
implies that $\Uh[P,0]d\cap\Bun{d'}$ ($d' < d$) is
lower-dimensional. Therefore

\begin{Lemma}\label{lem:Ud}
We have
\begin{equation}\label{eq:tu}
  U^d \cong H_{[0]}(\Uh[P,0]d).
\end{equation}
This space has a base given by $(dh^\vee-1)$-dimensional irreducible
components of $\Uh[P,0]d$.
\end{Lemma}

\subsection{Irreducible components}\label{sec:irred-comp}

Let us describe $(dh^\vee - 1)$-dimensional irreducible components of
$\Uh[P,0]d$ for $P=B$ explicitly.

We believe that there is no irreducible component of smaller dimension (see Remark~\ref{rem:dimofirr}), but we do not have a proof.


First consider the case $G = SL(2)$. By \lemref{lem:dimU} we have
$\dim U^d = 1$, and hence $\Uh[B,0]d$ has only one $(2d-1)$-dimensional
irreducible component. As we have
observed in the previous subsection, it is the closure of
$\Uh[B,0]d\cap \Bun{d}$. In \subsecref{sec:spaceVd}, it will be shown
that $\Uh[B,0]d\cap \Bun{d}$ consists of rank 2 vector bundles $E$
arising from a short exact sequence
\begin{equation}
    0 \to \shfO\to E \to \mathcal I\to 0
\end{equation}
compatible with framing, where $\mathcal I$ is an ideal sheaf of
colength $d$.

For a general $G$, consider the diagram \eqref{eq:CDpj} with $M = T$,
$L = L_i$ the Levi subgroup corresponding to a simple root $\alpha_i$.

Note that $[L_i,L_i] \cong SL(2)$, and hence $\Uh[{L_i}]d$ is
homeomorphic to $\Uh[SL(2)]d$. Therefore
$\Uh[B_{L_i},0]d\cap\Bun[L_i]d$ is irreducible of dimension $2d-1$ by
the above consideration.

\begin{Proposition}\label{prop:irred-comp}
    The irreducible components of $\Uh[B,0]d$ of dimension $dh^\vee-1$
    are the closures of $p^{-1}(\Uh[B_{L_i},0]d\cap \Bun[L_i]d)$ for
    $i\in I$.

Let us denote the closure by $Y_i$.\index{Yi@$Y_i$}
\end{Proposition}

\begin{proof}
    Consider the upper right part of \eqref{eq:CDpj}, which is
    \eqref{eq:1}.

    Its restriction to the open subset $\Bun[L_i]{d}$ has been
    described in \subsecref{sec:open}. As $p$ is a vector bundle whose
    rank is equal to the half of the codimension of $\Bun[L_i]{d}$ in
    $\Bun[G]{d}$, it follows that the inverse image
    $p^{-1}(\Uh[B_{L_i},0]d\cap \Bun[L_i]d)$ is irreducible and has
    dimension $dh^\vee - 1$.

    Therefore the closure of $p^{-1}(\Uh[B_{L_i},0]d\cap \Bun[L_i]d)$
    is an irreducible component of $\Uh[B,0]d$.

    Since $\dim U^d = \rank G$ by \lemref{lem:dimU}, it is enough to
    check that $p^{-1}(\Uh[B_{L_i},0]d\cap \Bun[L_i]d)\neq
    p^{-1}(\Uh[B_{L_j},0]d\cap \Bun[L_j]d)$ if $i\neq j$. When $G =
    SL(r)$, $\Uh[B,0]d\cap \Bun[G]{d}$ consists of vector bundles $E$
    having a filtration $0 = E_0\subset E_1\subset\cdots\subset E_r =
    E$ compatible with framing.  Moreover $p^{-1}(\Uh[B_{L_i},0]d\cap
    \Bun[L_i]d)$ consists of those with $c_2(E_i/E_{i-1}) = d$ and
    $c_2(E_j/E_{j-1}) = 0$ for $j\neq i$.

    Therefore $p^{-1}(\Uh[B_{L_i},0]d\cap \Bun[L_i]d)\neq
    p^{-1}(\Uh[B_{L_j},0]d\cap \Bun[L_j]d)$ for $i\neq j$.  (See
    \subsecref{sec:spaceVd} for detail.)

    For a general $G$, we embed $G$ into $SL(N)$. Then we need to
    replace $B$ by a parabolic $P$, but $p^{-1}(\Uh[B_{L_i},0]d\cap
    \Bun[L_i]d)$ is embedded into a corresponding space, and the same
    argument still works.
\end{proof}

\subsection{A pairing on \texorpdfstring{$U^d$}{Ud}}\label{sec:pairing}

Let us introduce a pairing between $U^{d,P}$ and $U^{d,P_-}$ in this
subsection.

We combine Braden's isomorphism \eqref{eq:Braden} with the natural
homomorphism $\xi_0^!\to \xi_0^*$ to get
\begin{equation}
  H^0(\xi_0^! i^* j^!\IC(\cUh{d})) \to
  H^0(\xi_0^* i_-^! j_-^*\IC(\cUh{d})).
\end{equation}
The right hand side is dual to
\begin{equation}
  U^{d,P_-} = H^0(\xi_0^! i_-^* j_-^!\IC(\cUh{d})).
\end{equation}
Thus we have a pairing between $U^{d,P}$ and $U^{d,P_-}$.

Following the convention in \cite[3.1.3]{MO}, we multiply the pairing
by the sign $(-1)^{\dim \cUh{d}/2} = (-1)^{dh^\vee-1}$.
\begin{NB}
  Let us explain why we take $\dim\cUh{d}/2 = dh^\vee-1$ instead of
  $\dim\Uh{d}/2 = dh^\vee$. Consider the Gieseker space with
  $h^\vee=1$, $d=2$. We have $\Gi{1}^1 = \AA^2\times T^*\proj^1$,
  $\cGi{1}^1 = T^*\proj^1$. The intersection pairing of $\proj^1$ in
  $T^*\proj^1$ is $-2$. But we want to identify it with $d (e_i,e_i) =
  d = 2$, where $e_i$ is the coordinate vector of $\Z^{h^\vee}$ with
  the standard inner product.
\end{NB}
Let us denote it by $\la\ ,\ \ra$.\index{< , >@$\la\ ,\ \ra$} When we want to
emphasize that it depends on the choice of the parabolic subgroup $P$,
we denote it by $\la\ ,\ \ra_P$.

Since $\xi_0^! \CC_{d\cdot 0} \to \xi_0^* \CC_{d\cdot 0}$ is obviously
an isomorphism, this pairing is nondegenerate.

The transpose of the homomorphism $U^{d,P}\to
(U^{d,P_-})^\vee$ is a linear map $U^{d,P_-}\to
(U^{d,P})^\vee$. It is
\begin{equation}
  H^0(\xi_0^! i_-^* j_-^!\IC(\cUh{d})) \to
  H^0(\xi_0^* i^! j^*\IC(\cUh{d})),
\end{equation}
given by the transpose of the composite of $\xi_0^!\to\xi_0^*$ and
Braden's isomorphism $i^*j^!\to i_-^! j_-^*$. They are the same as
original homomorphisms $\xi_0^!\to\xi_0^*$ and $i_-^* j_-^!\to i^!
j^*$ respectively.
\begin{NB}
  The transpose of $\xi_0^! \to \xi_0^*$ is $\xi_0^! = (\xi_0^*)^\vee
  \to (\xi_0^!)^\vee = \xi_0^*$. This is the same as the original
  $\xi_0^!\to \xi_0^*$.
\end{NB}
It means that
\begin{equation}\label{eq:41}
  \la u, v\ra_{P} = \la v, u\ra_{P_-} \quad
  \text{for $u\in U^{d,P}$, $v\in U^{d,P_-}$},
\end{equation}
where $\la\ ,\ \ra_{P_-}$ is the pairing defined with respect to the
opposite parabolic, i.e., one given after exchanging $i$, $j$ and $i_-$,
$j_-$ respectively.

\begin{NB}
  The following is the previous formulation, defining $\la\ ,\ \ra$ as a
  pairing on $U^d_{L,G}$. But it does not work as $P$ and $P_-$ are not
  conjugate under $G$ in general:

Recall that our hyperbolic restriction $\Phi_{L,G}$ depends on the
choice of a parabolic subgroup $P$, and also on $L$. However as in the
argument in \cite[Proof of Th.~3.6]{MV2}, we observe that $U^d_{L,G} =
H_{[0]}(\Uh[P,0]{d})$ forms a local system when we move $L\subset P$
in $G/L_0$ with a fixed $L_0$. It is a trivial local system as $G/L_0$
is simply connected.

\begin{NB2}
  I wonder that we need a stronger dimension estimate is necessary for
  this argument.
\end{NB2}
\begin{NB2}
  Let us also remark that the local system is embedded in the local
  system of cohomology over the affine Grassmannian, which is trivial
  in \cite{MV2}. But the latter does not make sense, so we need to use
  that $G/L_0$ is simply connected.
\end{NB2}
Therefore we have a canonical identification among different choices
of $P$. In particular, we have an isomorphism
\begin{equation}
  U^d_{L,G} =
  H^0(\xi_0^! i^* j^!\IC(\cUh{d}))
  \cong
  H^0(\xi_0^! i_-^* j_-^! \IC(\cUh{d})).
\end{equation}
Note that the constant sheaf $\CC_{d\cdot 0}$ at the point $d\cdot 0$
appears instead of $\cC_{S_{(d)}\AA^2}$ in \eqref{eq:Ud} in the
centered situation. Therefore $\xi_0^!$ does nothing.

The right hand side is isomorphic to
\begin{equation}\label{eq:opp}
  H^0(\xi_0^! i^! j^* \IC(\cUh{d}))
\end{equation}
by \cite[Th.~1]{Braden}. We compose the natural
homomorphism $\xi_0^!\to \xi_0^*$ and see that $H^0(\xi_0^* i^! j^*
(\IC(\cUh{d})))$ is the dual space to $H^0(\xi_0^! i^* j^!\IC(\cUh{d}))$.

Thus we have a pairing on $U^d_{L,G} = H^0(\xi_0^! i^* j^!\IC(\cUh{d}))$.

Following the convention in \cite[3.1.3]{MO}, we multiply the pairing
by the sign $(-1)^{\dim \cUh{d}/2} = (-1)^{dh^\vee-1}$.
\begin{NB2}
  Let us explain why we take $\dim\cUh{d}/2 = dh^\vee-1$ instead of
  $\dim\Uh{d}/2 = dh^\vee$. Consider the Gieseker space with
  $h^\vee=1$, $d=2$. We have $\Gi{1}^1 = \AA^2\times T^*\proj^1$,
  $\cGi{1}^1 = T^*\proj^1$. The intersection pairing of $\proj^1$ in
  $T^*\proj^1$ is $-2$. But we want to identify it with $d (e_i,e_i) =
  d = 2$, where $e_i$ is the coordinate vector of $\Z^{h^\vee}$ with
  the standard inner product.
\end{NB2}
Let us denote it by $\la\ ,\ \ra$.

Since $\xi_0^! \CC_{d\cdot 0} \to \xi_0^* \CC_{d\cdot 0}$ is obviously
an isomorphism, this pairing is nondegenerate.

\begin{Lemma}
  The pairing $\la\ ,\ \ra$ is symmetric.
\end{Lemma}

\begin{proof}
  If we denote by $\varphi$ the isomorphism
  $U^d_{L,G}\xrightarrow{\cong} (U^d_{L,G})^\vee$, the assertion means
  $\varphi^\vee = \varphi$. Going back to the definition of $\varphi$,
  we see that it is equivalent to the assertion that two homomorphisms
  from $U^d_{L,G}$ to \eqref{eq:opp}, one as above and another again
  by the composition of Braden's isomorphism and the identification by
  the triviality of the local system, but in the opposite order. This
  assertion is clear from the definition of Braden's isomorphism.
\end{proof}

\begin{NB2}
  If we do not like the centered Uhlenbeck space, we can work on
  $\Uh{d}$: We consider $\xi_0\colon \{d\cdot 0\}\to \Uh[L]{d}$, then
  $\xi_0^* \cC_{S_{(d)}\AA^2} = \xi_0^* \CC_{S_{(d)}\AA^2}[2] =
  \CC_{d\cdot 0}[2]$, $\xi_0^! \cC_{S_{(d)}\AA^2} =
  (\xi_0^*\cC_{S_{(d)}\AA^2})^\vee = (\CC_{d\cdot 0}[2])^\vee =
  \CC_{d\cdot 0}[-2]$. Therefore the pairing becomes just $0$. But we
  work on the equivariant setting and consider $H^{-2}(\xi_0^! i^!j^*
  \IC(\Uh{d}))\to H^2(\xi_0^*i^!j^* \IC(\Uh{d}))$, the natural homomorphism
  $\xi_0^!\to\xi_0^*$ multiplied by $-\ve_1\ve_2$.
\end{NB2}
\end{NB}

\subsection{Another base of \texorpdfstring{$U^d$}{Ud}}\label{sec:anotherbase}

We next construct another base of $U^d = U^d_{T,G}$ for $L=T$, which
is ($\rank G$)-dimensional by \lemref{lem:dimU}.

This new base is better behaved under hyperbolic restrictions than the
previous one given by irreducible components.

This subsection is preliminary, and the construction will be completed
in \subsecref{sec:anotherbase2}.

We study $U^d_{T,G}$, using the associativity of the hyperbolic
localization (\propref{prop:trans}) for $M = T$, $L = L_i$ the Levi
subgroup corresponding to a simple root $\alpha_i$.

Since various Levi subgroups appear, we use the notation $U^d_{T,G}$
indicating groups we are considering.


Note that $[L_i,L_i] \cong SL(2)$, and hence $\Uh[{L_i}]d$ is
homeomorphic to $\Uh[SL(2)]d$. We understand $\IC(\Uh[{L_i}]d)$ as
$\IC(\Uh[SL(2)]d)$ and apply \lemref{lem:dimU} to see that
\begin{equation}\label{eq:1_Li}
  U_{T,L_i}^d = \Hom_{\Perv(\Uh[T]d)}
  (\cC_{S_{(d)}\AA^2}, \Phi_{T,L_i}(\IC(\Uh[{L_i}]d)))
\end{equation}
is $1$-dimensional.
\begin{NB}
    I hope that there is a canonical element $1_{L_i}$ in
    $U_{T,L_i}^d$.

    Dec. 5, 2012: This could be given by the
    fundamental class of a certain variety, which turns out to be
    irreducible in this situation.

    Apr. 30, 2013: Unfortunately this is not true. We need to choose a
    basis by MO's stable envelop.
\end{NB}
In the next section, we shall introduce an element $1^d_{L_i}$ in
$U_{T,L_i}^d$ using the theory of the stable envelop in \cite{MO}.
\index{1Li@$1^d_{L_i}$|textit}

\begin{NB}
  Sep. 16 :
  A paragraph explaining $\Hom_{\Perv(\UhL{d})}(\IC(\UhL{d}),
  \Phi_{L,G}(\IC(\Uh{d})))\cong\CC$ is moved to \subsecref{sec:open}.
\end{NB}

Taking $L = L_i$ in the construction in \subsecref{sec:open}, we apply
the functor $\Phi_{T,L_i}$. By \propref{prop:trans} we get an element
\begin{equation}
    \label{eq:11}
\Phi_{T,L_i}(1^d_{L_i,G})\in
    \Hom_{\Perv(\Uh[T]d)}(\Phi_{T,L_i}(\IC(\Uh[{L_i}]d)),
    \Phi_{T,G}(\IC(\Uh{d}))).
\end{equation}
\begin{NB}
We have used $\Phi_{T,L_i}\Phi_{L_i,G}(\IC(\Uh{d}))
=\Phi_{T,G}(\IC(\Uh{d}))$.
\end{NB}
Composing with the element $1^d_{L_i}$ in $U_{T,L_i}^d$ mentioned just
above, we get
\begin{equation}
    \label{eq:12}
 \Phi_{T,L_i}(1^d_{L_i,G})\circ 1^d_{L_i}\in U^d_{T,G}.
\end{equation}

We have ($\rank G$)-choices of $i$. Then we will show that
\begin{equation}\label{eq:anotherbase}
  \{ \Tilde\alpha^d_i \defeq
  \Phi_{T,L_i}(\pol 1^d_{L_i,G})\circ 1^d_{L_i}\}_i
\index{alphatildai@$\Tilde\alpha^d_i$|textit}
\end{equation}
gives a basis of $U^d_{T,G}$ in the next subsection.

Here we will introduce an appropriate polarization $\pol = \pm 1$,
using a consideration of rank $2$ case.  See \eqref{eq:polarization}.

Moreover, this will gives us an identification $U^d_{T,G}$ with the
Cartan subalgebra $\h$ of $\g$ such that $\Tilde\alpha^d_i$ is sent to
the $i^{\mathrm{th}}$ simple coroot $\alpha_i^\vee$.
See a remark after \propref{prop:HeisRel}.
\begin{NB}
  Here is a copy of my message on Dec.\ 19, 2012:

  We still need to check that various Heisenberg algebras coming from
  have correct commutation relations, i.e., . But this can be checked
  by the reduction to the rank 2 case. Since we only consider ADE, we
  only need to consider the cases and , and these cases are already OK
  if we know that our construction is the same as [MO]. This is what I
  wrote `We may need a rank 2 consideration for the proof...' in my
  note, p.5. And this I learn what Sasha told us. Once we know the
  correct commutation relation, the linear independence of the base of
  (3.12) is clear, I believe.
\end{NB}

We normalize the inclusion $\IC(\Uh[L_i]d)\to\Phi_{L_i,G}(\IC(\Uh{d}))$
by $\pol 1^d_{L_i,G}$ as above. Then the projection
$\Phi_{L_i,G}(\IC(\Uh{d}))\to \IC(\Uh[L_i]d)$ is also determined, as
$\IC(\Uh[L_i]d)$ has multiplicity $1$ in $\Phi_{L_i,G}(\IC(\Uh{d}))$
(see \subsecref{sec:open}).

Therefore we have the canonical isomorphism
\begin{equation}\label{eq:decomp}
   \Phi_{L_i,G}(\IC(\Uh{d})) \cong
   \IC(\Uh[L_i]d) \oplus \IC(\Uh[L_i]d)^\perp,
\end{equation}
where $\IC(\Uh[L_i]d)^\perp$ is the sum of isotypical components for
simple factors not isomorphic to $\IC(\Uh[L_i]d)$.

Applying $\Phi_{T,L_i}$ and using $\Phi_{T,L_i}\Phi_{L_i,G} =
\Phi_{T,G}$, we get an induced decomposition
\begin{equation}\label{eq:decomp2}
  U^d_{T,G} = U^d_{T,L_i}\oplus (U^d_{T,L_i})^\perp.
\end{equation}

This decomposition is orthogonal with respect to the pairing in
\subsecref{sec:pairing} in the following sense.

We have the decomposition
\(
  U^{d,B_-}_{T,G} = U^{d,B_-\cap L_i}_{T,L_i}\oplus
  (U^{d,B_-\cap L_i}_{T,L_i})^\perp
\)
for the opposite Borel $B_-$, and
\begin{equation}\label{eq:110}
  \la U^d_{T,L_i}, (U^{d,B_-\cap L_i}_{T,L_i})^\perp\ra = 0
  = \la (U^{d}_{T,L_i})^\perp, U^{d,B_-\cap L_i}_{T,L_i}\ra.
\end{equation}

Moreover the restriction of the pairing to $U^{d,B\cap L_i}_{T,L_i}$,
$U^{d,B_-\cap L_i}_{T,L_i}$ coincides with one defined via $\Uh[L_i]d$.

\begin{NB}
    These assertions follow from as the diagram
    \begin{equation}
        \begin{CD}
        (p'\circ p'')_* (j \circ j'')^! @>>>
        (p_-'\circ p_-'')_! (j_-\circ j_-'')^!
\\
   @| @|
\\
       p'_* j^{\prime !} p_* j^! @>>>
       p'_{-!} j_-^{\prime *} p_{-!} j_-^*
        \end{CD}
    \end{equation}
    is commutative, where the vertical arrows are as in the proof of
    \propref{prop:trans}, and the horizontal arrows are Braden's
    isomorphisms. See 2013-11-20 Note.
\end{NB}

\subsection{Dual base}

Let $\Tilde\alpha^{d,-}_i$ denote the element defined as
$\Tilde\alpha^{d}_i$ for the opposite Borel. We shall prove
\begin{equation}
    \label{eq:111}
    \la [Y_j],\Tilde\alpha^{d,-}_i\ra = \pm \delta_{ij}(-1)^{d-1} d
\end{equation}
modulo the computation for $G=SL(2)$, corresponding to the case $i=j$
in this subsection. The computation for $G=SL(2)$ will be given in
\remref{rem:SL(2)}.

This formula means that $\Tilde\alpha^{d,-}_i$ is the dual base to the
base given by irreducible components $Y_j$ with respect to the pairing
$\frac{(-1)^{d-1}}d\la\ ,\ \ra$ up to sign.

Consider the diagram \eqref{eq:CDpj} for the centered version, where
we take $M=T$, $L=L_i$ as in \subsecref{sec:anotherbase}. Let us
consider the open embedding of $\cBun[L_i]d$ to $\cUh[L_i]d$. We
have the corresponding restriction homomorphism
\begin{equation}
    \label{eq:104}
    \begin{split}
    U^d_{T,G} &= H^0(\xi_0^! (p'\circ p'')_* (j\circ j'')^! \IC(\cUh[G]d))
\\
   &\cong H^0(\xi_0^! p'_* j^{\prime!} \Phi_{L_i,G}(\IC(\cUh[G]d)))
\\
   &\cong H^0(p^{\prime-1}(d\cdot 0),
   \Tilde j^{\prime !}\Phi_{L_i,G}(\IC(\cUh[G]d)))
\\
   & \quad \to
   H^0(p^{\prime-1}(d\cdot 0)\cap\cBun[L_i]d,
   \Tilde j^{\prime !}\Phi_{L_i,G}(\IC(\cUh[G]d))),
    \end{split}
\end{equation}
where $\Tilde{j'}$ is the restriction of $j'$ to $p^{\prime-1}(d\cdot 0)$.

When we restrict $\Phi_{L_i,G}(\IC(\cUh[G]d))$ to the open set
$\cBun[L_i]d$, the first summand $\IC(\Uh[L_i]d)$ in the
decomposition \eqref{eq:decomp} is replaced by the constant sheaf
$\cC_{\cBun[L_i]d}$, and the second summand is killed.

Therefore we have an isomorphism
\begin{multline*}
   H^0(p^{\prime-1}(d\cdot 0)\cap\cBun[L_i]d,
   \Tilde{j'}^!\Phi_{L_i,G}(\IC(\cUh[G]d)))
\\
   \cong H_{[0]}(p^{\prime-1}(d\cdot 0)\cap\cBun[L_i]d, \CC)
   \cong U^d_{T,L_i},
\end{multline*}
where the second isomorphism is nothing but \eqref{eq:107} for $G$
replaced by $L_i$.

Thus the projection $U^d_{T,G}\to U^d_{T,L_i}$ to the first summand in
\eqref{eq:decomp2} is nothing but the restriction homomorphism
we have just constructed.

Let us further consider the restriction of the upper right corner of
the diagram \eqref{eq:CDpj} to the open subset $\cBun[L_i]d$.

Then
\begin{equation*}
    p^{-1}(p^{\prime-1}(d\cdot 0)\cap\cBun[L_i]d) =
    p^{-1}(\Uh[B_{L_i},0]d\cap\cBun[L_i]d)
\end{equation*}
has been studied in \subsecref{sec:irred-comp}: Its closure is an
irreducible component of $\Uh[B,0]d$.

By the base change the restriction to $\cBun[L_i]d$ is replaced by one
to $p^{-1}(\Uh[B_{L_i},0]d\cap\cBun[L_i]d)$, and we can replace
relevant $\IC$ sheaves by constant sheaves. The Thom isomorphism gives
us $p_*j^! \cC_{\Bun[G]d} \cong \cC_{\Bun[L_i]d}$ as in \subsecref{sec:open}.

Note that the intersection of an irreducible component $Y_j$ of
\propref{prop:irred-comp} with the open subset
$p^{-1}(\Uh[B_{L_i},0]d\cap\cBun[L_i]d)$ is lower-dimensional if
$i\neq j$, as $p^{-1}(\Uh[B_{L_i},0]d\cap\cBun[L_i]d)$ is irreducible.

Therefore the fundamental class of $Y_j$ goes to $0$ under the
restriction.

Hence we have \eqref{eq:111} for $i\neq j$ by \eqref{eq:110}.

In fact, we will see that $Y_j\cap
p^{-1}(\Uh[B_{L_i},0]d\cap\cBun[L_i]d) = \emptyset$ for type A in
\subsecref{sec:spaceVd}, and the same is true for any $G$ thanks to an
embedding $G\to SL(N)$.

The Thom isomorphism sends $[Y_i]$ to $[\Uh[B_{L_i},0]d]$ from the
definition of $Y_i$.

The sign in \eqref{eq:111} appears as we multiply the Thom isomorphism
by a polarization $\delta$ (see \eqref{eq:polarization} below).

Therefore the computation of \eqref{eq:111} for
$i=j$ is reduced to the case $G=SL(2)$.

The relevant computation will be given in \remref{rem:SL(2)} as we
mentioned above.
\begin{NB}
    Since $\pm = o(i)^d$, we have $\pm (-1)^{d-1} =
    -(-o(i))^d$. Compare this with \eqref{eq:112} for $d=1$.
\end{NB}

\subsection{\texorpdfstring{$\Aut(G)$}{Aut(G)} invariance}\label{sec:autg-invariance}

Let $\Aut(G)$ be the group of automorphisms of $G$. Its natural action
on $\Bun{d}$ extends to $\Uh{d}$ (\cite[\S6.1]{BFG}).
\begin{NB}
    When a principal bundle is given by a transition function
    $\varphi_{\alpha\beta}\colon U_\alpha\cap U_\beta\to G$, we define
    a new principal bundle by the transition function
    $g(\varphi_{\alpha\beta})$. The cocycle condition
    $\varphi_{\alpha\beta}\varphi_{\beta\gamma}\varphi_{\gamma\alpha}
    = 1$ is preserved, as $g$ is a group automorphism. If $g(\bullet)
    = g_0\bullet g_0^{-1}$ for $g_0\in G$, the principal bundle is
    isomorphic to the original one, and the action is just given by a
    change of the framing.
\end{NB}

\begin{NB}
  We must be careful on the $\Aut(G)$-action in the following
  section. For type $A$, $\Aut(G)/G = \{ \pm 1\}$ is the Dynkin
  diagram automorphism given by the reflection at the center. In terms
  of $\Bun{d}$, it corresponds to associate a {\it dual\/} vector
  bundle. In particular, it does not extend to an action on the Gieseker
  space $\Gi{r}^d$.

  In the ADHM description, the diagram automorphism is given by
  \begin{equation}
      \label{eq:45}
      [(B_1,B_2,I,J)]\mapsto [(B_1^t, B_2^t, -J^t, I^t)].
  \end{equation}
  This does not preserve the stability condition.
\end{NB}

Let us fix a cocharacter $\lambda\colon \GG_m\to G$, and consider our
construction with respect to $\sigma\circ\lambda$ for $\sigma\in \Aut(G)$.

Here $L = G^{\lambda(\GG_m)}$ is considered as a fixed Levi subgroup.
Substituting $\sigma\circ\lambda$ into $\lambda$ in the formula
\eqref{eq:PLoneparam}, we define a pair $(P^\sigma,L^\sigma)$ of a parabolic
subgroup and its Levi part. The action $\varphi_\sigma\colon \Uh{d}\to
\Uh{d}$ induces $\varphi_\sigma\colon \Uh[P]d\to \Uh[P^{\sigma}]d$,
$\varphi_\sigma\colon \Uh[L]d\to \Uh[L^{\sigma}]d$, and we have a commutative
diagram
\begin{equation}    \label{eq:39}
  \begin{CD}
    \Uh[L]{d} @>i>> \Uh[P]d @>j>> \Uh{d}
\\
     @V{\varphi_\sigma}VV @V{\varphi_\sigma}VV @V{\varphi_\sigma}VV
\\
    \Uh[L^{\sigma}]{d} @>>{i_{\sigma}}> \Uh[P^{\sigma}]{d} @>>{j_{\sigma}}> \Uh{d},
  \end{CD}
\end{equation}


where the subscript $\sigma$ indicates morphisms between spaces for
$\sigma\in \Aut(G)$.

Since $\IC(\Uh{d})$ is an $\Aut(G)$-equivariant perverse sheaf, we have
an isomorphism $\varphi_\sigma^*\IC(\Uh{d})\cong \IC(\Uh{d})$. Therefore we
have an isomorphism
\begin{equation}\label{eq:44}
  i^* j^! \IC(\Uh{d}) \cong \varphi_\sigma^* i_{\sigma}^* j_{\sigma}^! \IC(\Uh{d}).
\end{equation}

The isomorphism \eqref{eq:44} is equivariant in the following sense:
The right hand side is a $\TT^\sigma=T^\sigma\times\CC^*\times\CC^*$-equivariant
perverse sheaf, while the left hand side is $\TT$-equivariant. The
isomorphism \eqref{eq:44} respects equivariant structures under the
group isomorphism $\sigma\colon \TT\xrightarrow{\cong} \TT^{\sigma}$. In
particular, we have an isomorphism
\begin{equation}
    \label{eq:43}
    \varphi_\sigma\colon
    H^*_{\TT}(\Uh[L]d, i^*j^! \IC(\Uh{d}))
    \xrightarrow{\cong} H^*_{\TT^{\sigma}}(\Uh[L^{\sigma}]d, i_{\sigma}^*j_{\sigma}^! \IC(\Uh{d})),
\end{equation}
which respects the $H^*_{\TT}(\mathrm{pt})$ and
$H^*_{\TT^{\sigma}}(\mathrm{pt})$ structures via $\TT\cong \TT^{\sigma}$.

In the same way, we obtain a canonical isomorphism
\begin{equation}
    \label{eq:40}
    U_{L,G}^{d,P} \xrightarrow{\cong} U_{L^{\sigma},G}^{d,P^{\sigma}},
\end{equation}
which is denoted also by $\varphi_\sigma$ for brevity.

The pairing $\la\ ,\ \ra$ in \subsecref{sec:pairing} is compatible
with $\varphi_\sigma$: Let us denote by $\la\ ,\ \ra_{P^\sigma}$ the
pairing between $U_{L^\sigma,G}^{d,P^\sigma}$ and $U_{L^\sigma, G}^{d,
  P^\sigma_-}$.

We have $\varphi_\sigma \colon U_{L,G}^{d,P_-}\xrightarrow{\cong}
U_{L^\sigma, G}^{d, P^\sigma_-}$ as above, and the following holds
\begin{equation}
  \la\varphi_\sigma(u), \varphi_\sigma(v)\ra_{P^\sigma} = \la u,v\ra_P,
  \qquad u\in U_{L,G}^{d,P}, v\in U_{L,G}^{d,P_-}.
\end{equation}

The decomposition \eqref{eq:decomp} is transferred under $\varphi_\sigma$ to
\begin{equation}
    \label{eq:46}
    i_\sigma^* j_\sigma^! \IC(\Uh{d}) \cong
    \pm \IC(\Uh[L_i^\sigma]{d})\oplus \IC(\Uh[L_i^\sigma]{d})^\perp.
\end{equation}
Here the sign $\pm$ means that we multiply the projection to
$\IC(\Uh[L_i^\sigma]{d})$ by $\pm$, according to whether $\sigma$
respects the polarization $\pol$ for $\Uh[L_i]{d}$ and
$\Uh[L_i^\sigma]{d}$ or not.

Our polarization will be invariant under inner automorphisms, so the
sign depends on diagram automorphisms $\Aut(G)/\operatorname{Inn}(G)$.
\begin{NB}
  Remember that the polarization problem should be discussed later.
\end{NB}
The decomposition \eqref{eq:decomp2} is mapped to
\begin{equation}
    \label{eq:47}
    U^{d,B^\sigma}_{T^\sigma,G} = U^{d,B^\sigma\cap L_i^\sigma}_{T^\sigma, L_i^\sigma}\oplus
    (U^{d,B^\sigma\cap L_i^\sigma}_{T^\sigma, L_i^\sigma})^\perp.
\end{equation}

Suppose $\sigma\in L$. We have $L^\sigma = L$, $P^\sigma = P$, $i_\sigma = i$, $j_\sigma =
j$. Then $i^* j^!  \IC(\Uh{d})$ is an $L$-equivariant perverse sheaf,
and \eqref{eq:44} is the isomorphism induced by the equivariant
structure.

Let us further assume $L=T$. Then $T$ acts trivially on $\Uh[T]{d} =
S^d\AA^2$, and $\varphi_\sigma|_{\Uh[T]d} = \id$. The equivariant structure
of the $T$-equivariant perverse sheaf $i^* j^! \IC(\Uh{d})$ is {\it
  trivial}.

\begin{NB}
In the definition \cite[2.1.3]{BL}, the diagram is
\(
   X \xleftarrow{p} P = X\times E \xrightarrow{q} \overline{P} = X\times B,
\)
where $E\to B$ is an $n$-acyclic free $T$-space. Then a {\it
  trivial\/} equivariant object $F\in D^b_T(X)$ is a triple $(F_X,
\overline{F}, \beta)$ such that $\overline{F}\in D^b(\overline{P})$ is
a pull-back of $F_X$ and $\beta\colon p^*(F_X)\cong q^*(\overline{F})$
is the trivial isomorphism.
\end{NB}
In particular, the isomorphism \eqref{eq:44} is the identity.
\begin{NB}
    This argument does not work if we do not know that $i^* j^!
    \IC(\Uh{d})$ is perverse. So, for example, I do not know what we
    can say on $i^!j^!\IC(\Uh{d})$.
\end{NB}
Therefore \eqref{eq:44} is well-defined for $\sigma\in \Aut(G)/(T/Z(G))$,
where $Z(G)$ is the center of $G$.

Note that chambers of hyperbolic restrictions for $L=T$ are Weyl
chambers. They appear as a subfamily for $W = N_G(T)/T$ in
$\Aut(G)/(T/Z(G))$.

Let us take $\sigma = w_0$, the longest element of the Weyl group. Then
$B^{w_0} = B_-$. We come back to $B$ via \eqref{eq:41}, and hence we
get
\begin{equation}
  \la u,v\ra_B = \la\varphi_{w_0}(u), \varphi_{w_0}(v)\ra_{B_-}
  = \la\varphi_{w_0}(v), \varphi_{w_0}(u)\ra_B
\end{equation}
for $u\in U^{d,B}_{T,G}$,$v\in U^{d,B_-}_{T,G}$.
\begin{NB}
  Suppose that $\g$ is of type $B_2$ or $G_2$. Then
  $\varphi_{w_0}(\Tilde\alpha_i^d) = \Tilde\alpha_i^{d,-}$
  ($i=1,2$). Therefore
  \begin{equation}
    \la\Tilde\alpha_1^d, \Tilde\alpha_2^{d,-}\ra_B
    = \la\varphi_{w_0}(\Tilde\alpha_2^{d,-}),
    \varphi_{w_0}(\Tilde\alpha_1^d)\ra_B
    = \la\Tilde\alpha_2^d, \Tilde\alpha_1^{d,-}\ra_B.
  \end{equation}
  This argument does not work for $A_2$. So I need to use the outer
  automorphism.
\end{NB}

We can take $\sigma\in\Aut(G)$, which preserves $T$ and the set of positive
roots, and induces a Dynkin diagram automorphism. Then $B^\sigma =
B$. Hence $U^{d,B}_{T,G}$ is a representation of the group of Dynkin
diagram automorphisms. The inner product is preserved.

We have $L_i^\sigma = L_{\sigma(i)}$, where $\sigma(i)$ is the vertex of the Dynkin
diagram, the image of $i$ under the corresponding Dynkin diagram
automorphism. From \eqref{eq:46} $\varphi_\sigma (\tilde\alpha^d_i)$ is
equal to $\tilde\alpha^d_{\sigma(i)}$ up to scalar. We will prove the
following in \subsecref{sec:autG-invar}.
\begin{Lemma}\label{lem:autg}
  We have
  \begin{equation}
    \varphi_\sigma (\tilde\alpha^d_i) = \pm \tilde\alpha^d_{\sigma(i)},
  \end{equation}
  where $\pm$ is the ratio of the polarizations for $\Bun[L_i]d$ and
  $\Bun[L_{\sigma(i)}]d$, compared under $\varphi_\sigma$.
\end{Lemma}

\section{Hyperbolic restriction in type \texorpdfstring{$A$}{A}}\label{sec:typeA}

We shall study the case $G = SL(r)$ in detail in this section.

We have the moduli space $\Gi{r}^d$\index{UhTilder@$\Gi{r}^d$} of
framed torsion free sheaves $(E,\varphi)$ of rank $r$, second Chern
class $d$ over $\proj^2$. It is called the {\it Gieseker space}.

We have a projective morphism $\pi$ (the {\it Gieseker-Uhlenbeck
  morphism\/}) from $\Gi{r}^d$ to the corresponding Uhlenbeck space
$\Uh{d}$.

It is known that $\Gi{r}^d$ is smooth and $\pi$ is a semi-small
resolution of singularities. Therefore we can study $\IC(\Uh{d})$ via
the constant sheaf $\cC_{\Gi{r}^d}$ over $\Gi{r}^d$.

(See \cite[Ch.~3,5,6]{Lecture}, where $\Gi{r}^d$, $\Uh{d}$ are denoted by
$\mathcal M(n,r)$, $\mathcal M_0(n,r)$ respectively. See also
\cite[Ch.3]{MR2095899} for further detail.)

If $r=1$, we understand $\Gi{1}^d$ as the Hilbert scheme
$\operatorname{Hilb}^{d}(\AA^2)$ of $d$ points on $\AA^2$, while
$\Uh[{SL(1)}]d$ is the symmetric power $S^d\AA^2$.

\subsection{Gieseker-Uhlenbeck}

Let us first explain the relation between $\IC(\Uh{d})$ and
$\cC_{\Gi{r}^d}$ in more detail.

We have the Gieseker-Uhlenbeck morphism $\pi\colon \Gi{r}^d\to
\Uh{d}$\index{pai@$\pi$ (Gieseker-Uhlenbeck morphism)}. It is
semi-small with respect to the standard
stratification~\eqref{eq:strat}. All strata are relevant and fibers
are irreducible.

Therefore
\begin{equation}\label{eq:GU}
  \pi_! \cC_{\Gi{r}^d} \cong \bigoplus_{d_1+|\lambda|=d}
  H_{\topdeg}(\pi^{-1}(x^{d_1}_\lambda))\otimes
  \IC(\BunGl{d_1}),
\end{equation}
where $x^{d_1}_\lambda$ is a point in the stratum
$\BunGl{d_1}$. See \cite[\S3]{Baranovsky}.

We identify $H_{\topdeg}(\pi^{-1}(x^{d_1}_{\lambda}))\cong \CC$ by the
fundamental class $[\pi^{-1}(x^{d_1}_{\lambda})]$. And
$\IC(\UhGl{d_1})$ is isomorphic to the pushforward of
$\IC(\Uh{d_1})\boxtimes \cC_{\overline{S_\lambda\AA^2}}$ under the
finite morphism \eqref{eq:finite}.

Then we have
\begin{equation}\label{eq:IH}
  H_{\TT}^{[*]}(\Gi{r}^d) \cong
  \bigoplus \IH_{\TT}^{[*]}(\Uh{d_1})\otimes
  H_{\topdeg}(\pi^{-1}(x^{d_1}_\lambda))\otimes
  H_{\TT}^{[*]}(\overline{S_\lambda\AA^2}).
\end{equation}
We also have the corresponding isomorphism for the cohomology with
compact support.

\subsection{Heisenberg operators}\label{sec:BaraOp}

For $r = 1$, the third author and Grojnowski independently constructed
operators acting on the direct sum of homology groups of $\Gi{1}^d$
satisfying the Heisenberg relation (see \cite[Ch.~8]{Lecture}).

It was extended by Baranovsky to higher rank case \cite{Baranovsky}.

Let us review his construction in this subsection.

We consider here both $H_\TT^{[*]}(\Gi{r}^d)$ and
$H_{\TT,c}^{[*]}(\Gi{r}^d)$, the equivariant cohomology with arbitrary
and compact support, which is Poincar\'e dual to Borel-Moore and the
ordinary equivariant homology
groups. \index{HTc@$H_{\TT,c}^{[*]}(\Gi{r}^d)$}

To save the notation, we use the notation
$H_{\TT(,c)}^{[*]}(\Gi{r}^d)$ meaning either of cohomology groups.

For $n > 0$ we consider subvariety
\begin{equation}
  P_n \subset \bigsqcup_d \Gi{r}^d \times \Gi{r}^{d+n} \times \AA^2,
\end{equation}
consisting of triples $(E_1,E_2,x)$ such that $E_1\supset E_2$ and
$E_1/E_2$ is supported at $x$.

We have
\begin{Proposition}
  $P_n$ is half-dimensional in $\Gi{r}^d \times \Gi{r}^{d+n} \times
  \AA^2$ for each $d$.
\end{Proposition}

Let us denote the projection to the third factor by $\Pi$. For a
cohomology class $\alpha\in H^{[*]}_{\TT(,c)}(\AA^2)$, we consider
$P_{-n}^\Delta(\alpha) = [P_n]\cap
\Pi^*(\alpha)$\index{P@$P_n^\Delta(\alpha)$ (diagonal Heisenberg
  generator)} as a correspondence in $\Gi{r}^d\times\Gi{r}^{d+k}$.

Then we have the convolution product
\begin{equation}
    P_{-n}^\Delta(\alpha) \colon H^{[*]}_{\TT(,c)}(\Gi{r}^d) \to
    H^{[* + \deg\alpha]}_{\TT(,c)}(\Gi{r}^{d+n}).
\end{equation}
Thanks to the previous proposition, the shift of the degree is simple
in our perverse degree convention.
\begin{NB}
    I am still wondering for the convention for cohomology
    degree. Here I take the notation from \cite[\S8.9]{CG}.

    However, it is a little annoying as the Heisenberg algebra we
    consider is not of degree $0$. In particular, its representation,
    the homology group of the fiber, is not of degree $0$.
\end{NB}
The reason why we put $\Delta$ in the notation will be clear later.

We define $P_n^\Delta(\alpha)$ as the adjoint operator
\begin{equation}
  P_n^\Delta(\alpha) \colon H^{[*]}_{\TT(,c)}(\Gi{r}^{d+n}) \to
  H^{[*+\deg\alpha]}_{\TT(,c)}(\Gi{r}^{d}).
\end{equation}
Here we have two remarks. First we follow the sign convention in
\cite[3.1.3]{MO} for the intersection pairing
\begin{equation}\label{eq:73}
  \la\bullet,\bullet\ra = (-1)^{\dim X/2} \int_X \bullet\cup \bullet.
\end{equation}

Second, we take $\alpha\in H^{[*]}_{\TT,c}(\AA^2)$ for
$H^{[*]}_\TT(\Gi{r}^d)$ and $\alpha\in H^{[*]}_{\TT}(\AA^2)$ for
$H^{[*]}_{\TT,c}(\Gi{r}^d)$. Then the operators are well-defined,
though various projections are not proper. (See
\cite[\S8.3]{Lecture}.)

We have the commutator relation
\begin{equation}
  [P_m^\Delta(\alpha), P_n^\Delta(\beta)]
  = \la\alpha,\beta\ra m\delta_{m+n,0}\, r.
\end{equation}
If $m+n=0$, one of $\alpha$ or $\beta$ is in $H^{[*]}_{\TT}(\AA^2)$
and another is in $H^{[*]}_{\TT,c}(\AA^2)$. Hence $\la\alpha,\beta\ra$
is well-defined.

Since the construction is linear over $H^*_\TT(\mathrm{pt})$, and
$H^{[*]}_{\TT,c}(\AA^2)$, $H^{[*]}_\TT(\AA^2)$ are free of rank $1$,
we can choose $\alpha$ to be their generators, i.e., the Poincar\'e
dual of $[0]$ for $H^{[*]}_{\TT,c}(\AA^2)$, and $1$ (dual of $[\AA^2]$)
for $H^{[*]}_{\TT}(\AA^2)$.
\begin{NB}
    $\deg [0] = 2$, $\deg 1 = -2$.
\end{NB}
We assume these choices hereafter until \secref{sec:w-algebra-repr}.
\begin{NB}
    For the integral form of the Heisenberg algebra, we choose a
    different convention.
\end{NB}
Note also that $\la[0],1\ra = -1$ in our sign convention.


We take the direct sum over $d$ in \eqref{eq:IH}:
\begin{equation}\label{eq:IH2}
  \bigoplus_d H_\TT^{[*]}(\Gi{r}^d)
  \cong \bigoplus_{d} \IH_\TT^{[*]}(\Uh{d})
  \otimes \bigoplus_{\lambda}
  H_{\topdeg}(\pi^{-1}(x^{d}_\lambda))\otimes
  H_\TT^{[*]}({\overline{S_\lambda\AA^2}}).
\end{equation}
Note that $H_\TT^{[*]}({\overline{S_\lambda\AA^2}}) \cong
H_{\TT}^*(\mathrm{pt}) \cdot 1$, as $\overline{S_\lambda\AA^2}$ is
equivariantly contractible. Here $1\in
H^{0}_\TT({\overline{S_\lambda\AA^2}})
= H^{[-2l(\lambda)]}_\TT({\overline{S_\lambda\AA^2}})$.

From the definition of the Heisenberg operators, it acts only on the
second factor of \eqref{eq:IH2}: $\lambda=\emptyset$ are killed by
$P_k^\Delta([0])$ ($k>0$) and the summand for $\lambda =
(1^{n_1}2^{n_2}\cdots)$ is spanned by the monomial in
$P_{-1}(1)^{n_1}/n_1!  \cdot P_{-2}(1)^{n_2}/n_2! \cdots$. The second
factor is isomorphic to the Fock space.

Let us give another representation of the Heisenberg algebra. Let $0$
denote the point $d\cdot 0\in S_{(d)}\AA^2$, and consider the inverse
image $\pi^{-1}(0)\subset\Gi{r}^{d}$, and denote it by
$\Gi{r,0}^{d}$.\index{UhTilder0@$\Gi{r,0}^d$} It is the Quot scheme
parametrizing quotients of $\shfO_{\proj^2}^{\oplus r}$ of length $d$
whose support is $0$.

The following is known, and was already used in \eqref{eq:GU}.
\begin{Proposition}\label{prop:quot}
  $\Gi{r,0}^{d}$ is an irreducible $(dr - 1)$-dimensional subvariety
  in $\Gi{r}^d$, unless $d=0$.
\end{Proposition}

It is needless to say that we have $\Gi{r,0}^0 = \Gi{r}^0 =
\mathrm{pt}$.

The convolution product by $P_{\pm k}^\Delta(\alpha)$ sends
$H_{[*]}^\TT(\Gi{r,0}^d)$ to $H_{[*-\deg\alpha]}^\TT(\Gi{r,0}^{d\pm
  k})$, where $\alpha\in H^*_{\TT,c}(\AA^2)$ for $k < 0$, $\alpha\in
H^*_{\TT}(\AA^2)$ for $k > 0$. Therefore
\begin{equation}
    \label{eq:19}
    \bigoplus_d H^\TT_{[*]}(\Gi{r,0}^d)
\end{equation}
is a representation of the Heisenberg algebra.

It is known that $\Gi{r,0}^d$ is homotopy equivalent to $\Gi{r}^d$,
hence $H^\TT_{[*]}(\Gi{r,0}^d)$ is isomorphic to the ordinary homology
group of $\Gi{r}^d$, and hence to $H^{[-*]}_{\TT,c}(\Gi{r}^d)$ by the
Poincar\'e duality.

\begin{NB}
  It seems that the following paragraphs should be put in
  Introduction, and is not relevant here.

Maulik-Okounkov \cite{MO} and Schiffmann-Vasserot \cite{SV} extended
the representation to $W(\gl_r)$, the $\scW$-algebra of $\gl_r$, which is
the tensor product of $W(\algsl_r)$, the $\scW$-algebra of $\algsl_r$ and
the Heisenberg algebra.

Since $\bigoplus_{d} \IH_\TT^{[*]}(\Uh{d})$ consists of highest weight
vectors of the Heisenberg algebra, it is a module of
$W(\algsl_r)$. This is the special case of our main result for $G =
SL_r$.

In order to generalize it to arbitrary $G$, we will reformulate the
argument of \cite{MO} in terms of the hyperbolic restriction functor
on the Uhlenbeck space after \subsecref{sec:sheaf}.
\end{NB}

\subsection{Fixed points and polarization}\label{sec:Gifixed}

Let us take a decomposition $r = r_1 + r_2 + \cdots + r_N$. We have
the corresponding $(N-1)$-dimensional torus, which is the connected
center $A = Z(L)^0$ of the Levi subgroup $L = S(GL(r_1)\times \cdots\times
GL(r_N))\subset SL(r)$. We have the corresponding parabolic subgroup
$P$ consisting of block upper triangular elements.

Let us consider the fixed point set $\Gi{L}^d =
(\Gi{r}^d)^{A}$.\index{UhTildeL@$\Gi{L}^d$} It consists of framed
sheaves, which is a direct sum of sheaves of rank $r_1$, $r_2$, \dots,
$r_N$. Thus we have
\begin{equation}\label{eq:product}
  \Gi{L}^d =
  \bigsqcup_{d=d_1+\dots+d_N} \Gi{r_1}^{d_1}\times\cdots\times
  \Gi{r_N}^{d_N}.
\end{equation}

We omit the superscript $d$, when there is no fear of confusion.

\begin{NB}
  Sep. 16 : Following Sasha's suggestion, I explain the polarization
  in more detail.
\end{NB}

Following \cite[Ex.3.3.3]{MO}, we choose a polarization $\pol$ for
each component of $\Gi{L}$, as a quiver variety associated with the
Jordan quiver.

Let us review the construction quickly. See the original paper for
more detail:

We represent $\Gi{L}$ as the space of quadruples $(B_1,B_2,I,J)$
satisfying certain conditions.

We decide to choose pairs, say $(B_1,I)$, from quadruples. The choice
gives us a decomposition of the tangent bundle of $\Gi{r}$ as
\begin{equation}
  T\Gi{r} = T^{1/2} + (T^{1/2})^\vee
\end{equation}
in the equivariant $K$-theory with respect to the $A$-action on
$\Gi{r}^d$. We also have the decomposition of $T\Gi{L}$, and hence
also of the normal bundle.

Then we choose a polarization $\pol$ of $\Gi{L}$ in $\Gi{r}$ as product
of weights in the normal bundle part of $(T^{1/2})^\vee$.

Let us also explain another description of the polarization
$\pol$\index{delta@$\pol$ (polarization)} given in
\cite[\S12.1.5]{MO}. We consider the following Quot scheme
\begin{equation}
  Q_r = \{ (E,\varphi) \mid x_2 \shfO^{\oplus r}\subset E\subset
  \shfO^{\oplus r} \} \subset \Gi{r},
\end{equation}
where $x_2$ is one of coordinates of $\AA^2$.

This is a fixed point component of a certain $\CC^*$-action, and is a
smooth lagrangian subvariety in $\Gi{r}$. In the ADHM description, it
is given by the equation $B_2 = 0 = J$.

Now $(T^{1/2})^\vee$ is the normal direction to $Q_r$ at a point in
$Q_r$. Since any component of $\Gi{L}$ intersects with $Q_r$, and the
intersection is again a smooth lagrangian subvariety, $Q_r$ gives us
the polarization.

Note that the polarization is invariant under the action of $G =
SL(r)$ on $\Gi{G}$, as we promised in \subsecref{sec:autg-invariance}.

We calculate the sign $\pm$ of the ratio of this polarization $\pol$
and the repellent one $\pol_{\text{rep}}$, of $\Gi{2}^d\times\Gi{1}^0$
and $\Gi{1}^0\times\Gi{2}^d$ in $\Gi{3}^d$ for a later purpose. Here
$L = S(GL(2)\times GL(1))$ in the first case and $L= S(GL(1)\times
GL(2))$ for the latter case.

\begin{Lemma}\label{lem:polSL3}
  We have $\pol_{\mathrm{rep}}/\pol = 1$ for $\Gi{2}^d\times\Gi{1}^0$,
  $\pol_{\mathrm{rep}}/\pol = (-1)^d$ for $\Gi{1}^0\times\Gi{2}^d$.
\end{Lemma}

\begin{NB}
  Consider the attracting sets $\Gi{P,0}^d$ corresponding to two Levi
  subgroups. By \propref{prop:Un} the first one is collapsed under the
  Hilbert-Chow morphism $\pi$. It seems that the only ratio of two
  polarizations is relevant for our later purpose, but still need a
  clarification.

  The ratio is much simpler to compute. We consider $g\in SL(3)$
  corresponding to the longest element of the Weyl group. Then the
  first fixed point component is mapped to the second by $g$. The
  repellent direction is mapped to the attracting one. And $\pol$ is
  invariant under $g$. Therefore the ratio is
  $\pol_{\mathrm{rel}}\pol_{\mathrm{att}} = (-1)^{\codim/2} \pol_{\mathrm{rel}}^2
  = (-1)^{\codim/2} = (-1)^d$.
\end{NB}

\begin{proof}
  Both components $\Gi{2}^d\times\Gi{1}^0$, $\Gi{1}^0\times\Gi{2}^d$
  intersect with the open set $\pi^{-1}_{a,G}(S_{(1^d)}\AA^1)$, the
  inverse image of the open stratum under the factorization morphism.

  Since the normal bundle decomposes according to the factorization,
  the polarization is of the form $(\pm 1)^d$. Hence it is enough to
  determine the case $d=1$.

  We factor out $\AA^2$ in $\Gi{r}^1$ and
  consider the centered Gieseker
  spaces. \index{UhTildercG@$\cGi{r}^d$|textit} We have
\begin{gather}
\cGi{3}^1 \cong T^* \proj^{2},
\\
\cGi{2}^1\times\cGi{1}^0 \cong T^*(z_2=0), \quad
\cGi{1}^0\times\cGi{1}^2 \cong T^*(z_0=0),
\end{gather}
where $[z_0:z_1:z_2]$ is the homogeneous coordinate system of
$\proj^2$. The polarization $\pol$ above is given by the base
direction of the cotangent bundle.

On the other hand, the repellent directions are base in the first
case and fibers in the second case. Therefore we have
$\pol_{\text{rep}}/\pol = 1$ in the first case and $-1$ in the second
case.
\end{proof}

\subsection{Stable envelop}\label{sec:envelop}

Recall we considered the attracting set $\UhP{}$ in the Uhlenbeck
space $\Uh{}$.  Let us denote its inverse image $\pi^{-1}(\UhP{})$ in
$\Gi{r}$ by $\Gi{P}$.\index{UhTildeP@$\Gi{P}^d$} This is the {\it tensor
  product variety}, denoted by $\fT$ in \cite{tensor2}, where $\UhP{}$
is denoted by $\fT_0$.

(In \cite{Na-Tensor} $\fT$ was denoted by $\mathfrak Z$.)

We have the following moduli theoretic description:
\begin{equation}
  \Gi{P} = \left\{ (E,\varphi)\in \Gi{r}\, \middle|\,
  \begin{minipage}{0.6\linewidth}
    $E$ admits a filtration
$0 = E_0 \subset E_1 \subset \cdots \subset E_N = E$ with
$\rank E_i/E_{i-1} = r_i$, compatible with $\varphi$.
  \end{minipage}
  \right\}.
\end{equation}
See \subsecref{sec:hyperb-restr}.

We consider the fiber product $Z_P$ of $\Gi{P}$ and $\Gi{L}$
over $\UhL{}$:
\begin{equation}
  Z_P = \Gi{P}\times_{\UhL{}} \Gi{L}, \index{ZP@$Z_P$}
\end{equation}
where the map from $\Gi{L}$ to $\UhL{}$ is the restriction of $\pi$,
and the map from $\Gi{P}$ to $\UhL{}$ is the composition of the
restriction
\(
  \Gi{P}\to \UhP{}
\)
of $\pi$ and the map $p$ in \subsecref{sec:hyperb-restr}.  In the
above description of $\Gi{P}$, it is just given as the direct sum
$\bigoplus (E_i/E_{i-1})^{\vee\vee}$ plus the sum of singularities of
$E_i/E_{i-1}$.

One can show that $Z_P$ is a lagrangian subvariety in $\Gi{r}\times
\Gi{L}$. See \cite[Prop.~1]{tensor2}. (There are no lower dimensional
irreducible components, as all strata are relevant for the semismall
morphism $\pi\colon \Gi{r}\to \Uh{}$.)

Maulik-Okounkov stable envelop is a `canonical' lagrangian cycle class
$\cL$\index{L@$\cL$ (stable envelop)} in $Z_P$:
\begin{equation}
  \cL\in H_{[0]}(Z_P).
\end{equation}
See \cite[\S3.5]{MO}. Note that $\cL$ depends on the choice of the
parabolic subgroup $P$ as well as the polarization $\pol$. Since they
are canonically chosen, we suppress them in the notation $\cL$.

The convolution by $\cL$ defines a homomorphism
\begin{equation}\label{eq:conv}
  \cL\ast - = p_{1*}(p_2^*(-)\cap \cL)
  \colon H_{[*]}(\Gi{L}) \to H_{[*]}(\Gi{P}).
\end{equation}
It is known that $\cL\ast-$ is an isomorphism (see
\cite[\S4.2]{tensor2}), and it {\it does\/} also make sense for
equivariant homology groups, as $H_{[0]}(Z_P) \cong H_{[0]}^\TT(Z_P)$

We have $H_{[*]}(\Gi{L}) \cong H^{[*]}(\Gi{L})$ by the Poincar\'e duality.

Then we have
\begin{equation}\label{eq:stable}
  H^{[*]}_\TT(\Gi{L}) \to H^{[*]}_\TT(\Gi{r})
\end{equation}
as the composite of $\cL\ast -$ and the pushforward with respect to
the inclusion $\Gi{P}\subset \Gi{r}$.

This is the original formulation of {\it stable envelop\/} in
\cite[Ch.~3]{MO}, and properties of $\cL$ are often stated in terms of
this homomorphism there.

Let $x\in\UhL{}$. Let $\Gi{L,x}$ denote the inverse image of $x$ under
the Gieseker-Uhlenbeck morphism $\Gi{L}\to\UhL{}$. Similarly let
$\Gi{P,x}$ denote the inverse image of $x$ under the composition
$\Gi{P}\to \UhP{}\to \UhL{}$.\index{UhTildePx@$\Gi{P,x}^d$} Then the
convolution $\cL\ast -$ also defines
\begin{equation}\label{eq:convx}
  \cL\ast-\colon H^{\TT_x}_{[*]}(\Gi{L,x})\to H^{\TT_x}_{[*]}(\Gi{P,x}),
\end{equation}
where $\TT_x$ is the stabilizer of $x$.

\begin{NB}
  The degree is the perverse degree as before. Therefore
  $H_{[*]}(\Gi{P,x}) = H_{\dim\Gi{r}+*}(\Gi{P,x})$. If $x=0$ as in the
  next subsection, $\dim \Gi{P,0} = \dim\Gi{r}/2 - 1$, so $*=-2$ is
  the top degree, unless $d=0$.
\end{NB}

\begin{NB}
  We can also consider $\Gi{r,x}$ the inverse image of $x$ under
  $\pi\colon \Gi{r}\to \Uh{}$, where $x\in\UhL{}$ is considered as a
  point in $\Uh{}$ by $j\circ i\colon \UhL{}\to\Uh{}$. Then we have
  $\Gi{r,x}\subset \Gi{P,x}$ by definition. But $\Gi{P,x}$ is larger than
  $\Gi{r,x}$ in general. We have the pushforward homomorphism
  $H^\TT_*(\Gi{r,x})\to H^\TT_*(\Gi{P,x})$.
\end{NB}

\subsection{Tensor product module}\label{sec:tensor}

Let $0 = d\cdot 0$ as before and consider the inverse image $\Gi{P,0}^d$
of $0$ under $\Gi{P}^d\to \UhL{d}$ as in the previous subsection.

We consider the direct sum
\begin{equation}
    \label{eq:21}
 \bigoplus_d H^\TT_{[*]}(\Gi{P,0}^d).
\end{equation}
The Heisenberg algebra acts on the sum: This follows from a general
theory of the convolution algebra: it is enough to check that
$\Gi{P,0}^d \circ (P_n\cap\Pi^{-1}(0))\subset \Gi{P,0}^{d+n}$ (for
$k>0$). If $(E_1,E_2,x)\in P_n\cap\Pi^{-1}(0)$, then $\pi(E_2) =
\pi(E_1) + n\cdot 0$. Therefore the assertion follows.

The stable envelop $\cL\ast-$ gives an isomorphism $\bigoplus_d
H^\TT_{[*]}(\Gi{L,0}^d)\cong \bigoplus_d H^\TT_{[*]}(\Gi{P,0}^d)$,
where the left hand side is the tensor product
\begin{equation}
    \label{eq:20}
    \bigoplus_{d_1, \cdots, d_N} H^\TT_{[*]}(\Gi{r_1,0}^{d_1})
    \otimes\cdots\otimes
    H^\TT_{[*]}(\Gi{r_N,0}^{d_N})
\end{equation}
by \eqref{eq:product}.

This is a representation of $N$ copies of Heisenberg algebras. Under
the stable envelop, $P_{-k}^\Delta([0])$ on \eqref{eq:21} is mapped to
\begin{equation}
    \label{eq:22}
    \sum_{i=1}^N 1\otimes\cdots\otimes
    \underbrace{P_{-k}^\Delta([0])}_{\text{$i^{\mathrm{th}}$ factor}}
    \otimes\cdots\otimes 1.
\end{equation}
This is \cite[Th.~12.2.1]{MO}. Our Heisenberg generators are diagonal
in this sense, and hence we put $\Delta$ in the notation.

This result is compatible with the decomposition $W(\mathfrak{gl}_r) =
W(\algsl_r)\otimes\Heis$, where $W(\algsl_r)$ is contained in the
tensor product of the remaining $(N-1)$ copies of Heisenberg algebras,
orthogonal to the diagonal one.

\subsection{Sheaf theoretic analysis}\label{sec:sheaf}

By \cite[\S4, Lem.~4]{tensor2} we have a natural isomorphism
\begin{equation}\label{eq:isom}
  H_{[0]}(Z_P) \cong \Hom_{\Perv(\UhL{})}
  \left(p_! j^* \pi_! \cC_{\Gi{r}}, \pi_! \cC_{\Gi{L}}\right)
\end{equation}
where $j$, $p$ are as in \subsecref{sec:hyperb-restr} and we use the
same symbol $\pi$ for Gieseker-Uhlenbeck morphisms for $\Gi{r}$ and
$\Gi{L}$.
\begin{NB}
  The proof will be given in \subsecref{subsec:stHom}.
\end{NB}

The Verdier duality gives us an isomorphism
\begin{equation}
  \Hom(p_!j^* \pi_!\cC_{\Gi{r}}, \pi_! \cC_{\Gi{L}})
  \cong
  \Hom(\pi_* \cC_{\Gi{L}}, p_*j^! \pi_*\cC_{\Gi{r}}).
\end{equation}
Therefore the stable envelop gives us the canonical isomorphism
\begin{equation}\label{eq:stac}
  \pi_! \cC_{\Gi{L}} \xrightarrow[\cong]{\cL}
  \Phi_{L,G}(\pi_! \cC_{\Gi{r}}) =
  p_* j^! \pi_! \cC_{\Gi{r}},
\end{equation}
as $\pi_! = \pi_*$.

This is nothing but \thmref{nakajima-symplectic}(2) in Introduction.

Let $x\in \UhL{}$ and $i_x$ denote the inclusion of $x$ in
$\UhL{}$. Then
\(
   \cL\in \Hom(\pi_! \cC_{\Gi{L}}, p_*j^! \pi_!\cC_{\Gi{r}})
\)
defines an operator
\begin{equation}\label{eq:stalk}
  \begin{CD}
  H^*(i_x^! \pi_! \cC_{\Gi{L}}) @>>>
  H^*(i_x^! p_*j^! \pi_!\cC_{\Gi{r}})
\\
  @| @|
\\
  H_{[-*]}(\Gi{L,x}) @.
  H_{[-*]}(\Gi{P,x}).
  \end{CD}
\end{equation}
This is equal to $\cL\ast -$ in \eqref{eq:convx} under the isomorphism


\eqref{eq:isom}. See \cite[\S4.4]{tensor2}.

\begin{NB}
\subsection{Proof of \texorpdfstring{\eqref{eq:isom}}{}}\label{subsec:stHom}

We give a proof of \eqref{eq:isom}.

We have the diagram
\begin{equation}
  \begin{CD}
    Z_P = \Gi{P} \times_{\UhL{}} \Gi{L}
    @>{\tilde \delta}>> \Gi{P} \times \Gi{L} @>{\tilde j\times\id}>>
    \Gi{r} \times \Gi{L}
\\
  @V{\alpha}VV @VV{\tilde p\times\pi}V @.
\\
  \UhL{} @>\delta>> \UhL{}\times \UhL{},
  \end{CD}
\end{equation}
where $\tilde p$ is the composition of $\pi\colon \Gi{P}\to \UhP{}$ and
$p\colon\UhP{}\to \UhL{}$. Therefore we have
\begin{equation}
  \begin{split}
    H_{[-*]}(Z_P) &= H^*(Z_P,((\tilde j\times\id)\circ\tilde \delta)^!
    \cC_{\Gi{r}\times \Gi{L}})
    \\
    &= H^*(\UhL{}, \alpha_* \tilde \delta^! (\tilde j\times\id)^!
    (\cC_{\Gi{r}}\boxtimes \cC_{\Gi{L}}))
\\
    &= H^*(\UhL{}, \delta^! (\tilde p\times\pi)_*
    (\tilde j^! \cC_{\Gi{r}}\boxtimes \cC_{\Gi{L}}))
\\
    &= H^*(\UhL{}, \delta^! (\tilde p_* \tilde j^!
    \cC_{\Gi{r}}\boxtimes \pi_* \cC_{\Gi{L}}))
\\
    &= H^*(\UhL{}, (\tilde p_! \tilde j^* \cC_{\Gi{r}})^\vee \overset{!}{\otimes}
    \pi_* \cC_{\Gi{L}})
\\
    &= H^*(\UhL{}, {\mathcal Hom}(\tilde p_! \tilde j^* \cC_{\Gi{r}},
    \pi_* \cC_{\Gi{L}}))
\\
    &= \Ext^*(\tilde p_!\tilde j^* \cC_{\Gi{r}}, \pi_* \cC_{\Gi{L}}).
  \end{split}
\end{equation}
We have $\pi_* \cC_{\Gi{L}} = \pi_! \cC_{\Gi{L}}$ as $\pi$ is proper.

We also have a Cartesian diagram
\begin{equation}
  \begin{CD}
    \Gi{P} @>{\tilde j}>> \Gi{r}
\\
    @V{\pi}VV @VV{\pi}V
\\
    \UhP{} @>j>> \Uh{}.
  \end{CD}
\end{equation}
Therefore
\begin{equation}
  \tilde p_! \tilde j^* \cC_{\Gi{r}}
  = p_! \pi_! \tilde j^* \cC_{\Gi{r}}
  = p_! j^* \pi_! \cC_{\Gi{r}}.
\end{equation}
This finishes the proof of \eqref{eq:isom}.
\end{NB}

\subsection{The associativity of stable envelops}

Let us take parabolic subgroups $Q\subset P \subset G$ and the
corresponding Levi subgroup $M\subset L$ as in
\subsecref{sec:ass}. ($G$ is still $SL(r)$.) Let $Q_L$ be the image of
$Q$ in $L$.

Let us denote by $\cL_{L,G}$ the isomorphism given by the stable
envelop in \eqref{eq:stac}:
\begin{equation}\label{eq:st}
    \pi_! \cC_{\Gi{L}}\xrightarrow[\cong]{\cL_{L,G}}
    \Phi_{L,G}(\pi_! \cC_{\Gi{r}}).
\end{equation}
We similarly have isomorphisms
\begin{equation}
  \pi_! \cC_{\Gi{M}}\xrightarrow[\cong]{\cL_{M,G}}
    \Phi_{M,G}(\pi_!\cC_{\Gi{r}}),\quad
  \pi_! \cC_{\Gi{M}} \xrightarrow[\cong]{\cL_{M,L}}
    \Phi_{M,L}(\pi_!\cC_{\Gi{L}}).
\end{equation}

Then stable envelops are compatible with the associativity
\eqref{eq:5} of the hyperbolic restriction:
\begin{Proposition}
We have a commutative diagram
\begin{equation}\label{eq:ass2}
  \begin{CD}
  \pi_!\cC_{\Gi{M}}@>\cong>{\cL_{M,G}}>  \Phi_{M,G}(\pi_! \cC_{\Gi{r}})
\\
  @V{\cong}V{\cL_{M,L}}V \eqref{eq:5}@|
\\
  \Phi_{M,L}(\pi_!\cC_{\Gi{L}})
 @>{\Phi_{M,L}(\cL_{L,G})}>\cong>
  \Phi_{M,L} \Phi_{L,G}(\pi_! \cC_{\Gi{r}}).
  \end{CD}
\end{equation}
\end{Proposition}

Let us check that this follows from the proof of
\cite[Lemma~3.6.1]{MO}.

(To compare the following with the original paper, the reader should
note that the tori $A\supset A'$ were used in \cite{MO}, which
correspond to $Z(M)^0\supset Z(L)^0$ respectively in our situation.)

We consider
\begin{equation}
  Z_{P} = \Gi{P}\times_{\Uh[{L}]{}} \Gi{L},
\quad
  Z_{Q} = \Gi{Q}\times_{\Uh[{M}]{}} \Gi{M},
\quad
  Z_{Q_L} = \Gi{Q_L}\times_{\Uh[M]{}} \Gi{M}.
\end{equation}

The stable envelops $\cL_{L,G}$, $\cL_{M,G}$, $\cL_{M,L}$ are classes in
$H_{[0]}(Z_{P})$, $H_{[0]}(Z_{Q})$, $H_{[0]}(Z_{Q_L})$
respectively.

We consider the convolution product
\begin{equation}
  \cL_{L,G}\ast \cL_{M,L}\in
  H_{[0]}(Z_{P}\circ Z_{Q_L}).
\end{equation}
Note that $Z_{P}\circ Z_{Q_L}$ consists of $(x_1,x_3)\in
\Gi{P}\times\Gi{M}$ such that there exists $x_2\in \Gi{Q_L}\subset
\Gi{L}$ with $(x_1,x_2)\in Z_{P}$, $(x_2,x_3)\in Z_{Q_L}$ by
definition. This is nothing but $Z_{Q}$. Therefore
$\cL^P\ast\cL^{Q_L}$ is a class in $H_{[0]}(Z_{Q})$.

The proof in \cite[Lemma~3.6.1]{MO} actually gives $\cL_{L,G}\ast
\cL_{M,L} = \cL_{M,G}$.

Therefore the commutativity of \eqref{eq:ass2} follows, once we check
that the convolution product corresponds to the composition of
homomorphisms (Yoneda product) under the isomorphism \eqref{eq:isom}.

This is not covered by \cite[Prop.~8.6.35]{CG}, as the base spaces of
fiber products are different: $\UhL{}$ and $\Uh[M]{}$. But we can
easily modify its proof to our situation.
\begin{NB}
  Really ?
\end{NB}

\subsection{Space \texorpdfstring{$V^d$}{Vd} and its base given by irreducible components}\label{sec:spaceVd}

Let us write $d$ for the instanton number again. Similarly to
\eqref{eq:Ud} we define
\begin{equation}
    \label{eq:13}
    \begin{split}
    V^d_{L,G} \equiv V^d \index{VLG@$V^d_{L,G}$}
    &\defeq \Hom(\cC_{S_{(d)}\AA^2}, \Phi_{L,G}(\pi_!\cC_{\Gi{r}^d})) \\
    & = H^{-2}(S_{(d)}\AA^2, \xi^!\Phi_{L,G}(\pi_! \cC_{\Gi{r}^d})),
    \end{split}
\end{equation}
where $\xi\colon S_{(d)}\AA^2\to \UhL{d}$ is as before. We denote by
$V^{d,P}_{L,G}$ or $V^{d,P}$ when we want to emphasize
$P$. \index{VLGdP@$V^{d,P}_{L,G}$}

As in \lemref{lem:Ud} we have
\begin{equation}\label{eq:tv}
  V^d \cong H_{[0]}(\Gi{P,0}^d),
\end{equation}
and $V^d$ has a base given by $(dh^\vee-1)$-dimensional irreducible
components of $\Gi{P,0}^d$.

\begin{NB}
We have
\begin{equation}
    \label{eq:14}
    \begin{split}
    V^d &= H^{-2}(S_{(d)}\AA^2, \xi^!p_*j^{!} \pi_! \cC_{\Gi{r}^d})
\\
    &\cong H_{[2]}(\tilde p^{-1}(S_{(d)}\AA^2)) \cong
    H_{[-2]}(\Gi{P,0}^d),
    \end{split}
\end{equation}
where $\tilde p = p \circ\pi\colon \Gi{P_-}\to \UhL{d}$, $0$ is
the point $d\cdot 0$ in $S_{(d)}\AA^2$, and
$\Gi{P,0}^d$ is the inverse image $\tilde p^{-1}(d\cdot 0)$.

The last isomorphism comes from $\tilde p^{-1}(S_{(d)}\AA^2)\cong
\Gi{P,0}\times\AA^2$. In fact, consider the fiber square
\begin{equation}
    \begin{CD}
        p^{-1}(S_{(d)}\AA^2) @>{\tilde \xi}>> \UhP{d}
\\
       @V{p}VV @VV{p}V
\\
       S_{(d)}\AA^2 @>{\xi}>> \UhL{d}.
    \end{CD}
\end{equation}
We have $\xi^! p_* = p_* \tilde \xi^!$. Hence
\begin{equation}
    H^{-2}(S_{(d)}\AA^2, \xi^! p_* j^{!} \pi_!\cC_{\Gi{r}^d})
    = H^{-2}(p^{-1}(S_{(d)}\AA^2),
    (j\circ\tilde \xi)^{!} \pi_!\cC_{\Gi{r}^d}).
\end{equation}

Next we consider
\begin{equation}
    \label{eq:16}
    \begin{CD}
        \tilde p^{-1}(S_{(d)}\AA^2) @>\alpha>> \Gi{r}^d
\\
     @V{\pi}VV      @V{\pi}VV
\\
    p^{-1}(S_{(d)}\AA^2) @>>{j\circ \tilde \xi}> \Uh{d}.
    \end{CD}
\end{equation}
Therefore the above is equal to
\begin{equation}
    \label{eq:18}
    H^{-2-2dh^\vee}(\tilde p^{-1}(S_{(d)}\AA^2), \alpha^! \mathbb D_{\Gi{r}^d}).
\end{equation}
This is the Borel-Moore homology of $\tilde p^{-1}(S_{(d)}\AA^2)$.
\end{NB}

On the other hand, $H_{[0]}(\Gi{P,0}^d)$ is isomorphic to
$H_{[0]}(\Gi{L,0}^d)$ by the stable envelop. In the description
\eqref{eq:product}, note that the fiber $\Gi{r_i,0}^{d_i}$ has $\dim =
\dim \Gi{r_i}^{d_i}/2 - 1$ by \propref{prop:quot} unless $d_i =
0$. Therefore we can achieve the degree $[0]=\dim\Gi{L}^d-2 = \sum
\dim\Gi{r_i}^{d_i}-2$ only when all $d_i = 0$ except one. There are $N$
choices $i=1$, \dots, $N$. Therefore $\dim V^d = N$.

Let us study $V^d = H_{[0]}(\Gi{P,0}^d)$ in more detail. This will
give the detail left over from \subsecref{sec:irred-comp}. By
\cite[\S3]{Na-Tensor} we have a decomposition
\begin{equation}
  \Gi{P,0}^d = \bigsqcup_{d_1+\cdots+d_N = n}
  \fT(d_1,\dots,d_N)_0,
\end{equation}
where
\begin{equation}\label{eq:tenstr}
  \fT(d_1,\dots,d_N)_0 =
  \left\{ (E,\varphi)\, \middle|\,
  \begin{minipage}{0.55\linewidth}
    $E$ admits a filtration $0 = E_0 \subset E_1 \subset \cdots
    \subset E_N = E$ with $E_i/E_{i-1} \in \Gi{r_i,0}^{d_i}$
    compatible with $\varphi$.
  \end{minipage}
  \right\}.
\end{equation}
\index{Td1@$\fT(d_1,\dots,d_N)_0$}
We have a projection
\begin{equation}
  \fT(d_1,\dots,d_N)_0 \to \Gi{r_1,0}^{d_1}\times\cdots\times
  \Gi{r_N,0}^{d_N},
\end{equation}
which is a vector bundle of rank $dr - \sum d_i r_i$.

Note that
\begin{equation}
  \dim \Gi{r_i,0}^{d_i} =
  \begin{cases}
    0 & \text{if $d_i = 0$},
\\
    d_i r_i - 1 & \text{if $d_i\neq 0$}.
  \end{cases}
\end{equation}
(See \propref{prop:quot}.)

Therefore
\begin{equation}
  \dim \fT(d_1,\dots,d_N)_0
  = dr - \# \{ i \mid d_i \neq 0\} \le dr - 1.
\end{equation}
The equality holds if and only if there is only one $i$ with $d_i \neq
0$. Therefore $H_{[0]}(\Gi{P_-,0}^d)$ is spanned by fundamental
cycles
\begin{equation}\label{eq:fT_0}
  [\overline{\fT(d,0,\dots,0)_0}], \quad \cdots \quad,
  [\overline{\fT(0,\dots,0,d)_0}].
\end{equation}
Thus it is $N$-dimensional, as expected.

In the remainder of this subsection, we study the corresponding space
$U^d = H_{[0]}(\Uh[P,0]d)$ for the Uhlenbeck space. Note that we have
projective morphism $\pi\colon \Gi{P,0}^d\to \Uh[P,0]d$, and the
Quot scheme $\pi^{-1}(d\cdot 0) = \Gi{r,0}^d$ is contained in
$\Gi{P,0}^d$.

The class of fiber $\Gi{r,0}^d$ is given by $P_{-d}([0])[\Uh{0}]$, and
$H_{[0]}(\Uh[P,0]d)$ is killed by Baranovsky's Heisenberg operators
by the construction.

\begin{Proposition}\label{prop:Un}
  Among $N$ cycles in \eqref{eq:fT_0}, the first one
  $[\overline{\fT(d,0,\dots,0)_0}]$ is $[\Gi{r,0}^d]$. The remaining
  cycles give a base of $U^d = H_{[0]}(\Uh[P,0]d)$ under $\pi_*$.
\end{Proposition}

From the definition, this description of irreducible components
\begin{NB}
    $\pi(\overline{\fT(0,d,0,\dots,0)_0})$, \dots,
    $\pi(\overline{\fT(0,\dots,0,d)_0})$
\end{NB}
of $\Uh[P,0]d$ is the same as one in \propref{prop:irred-comp} when
$G = SL(r)$, $P = B$.

\begin{NB}
I first thought that we need the following. But it seems not.....

\begin{Claim}
  Suppose that $F$ is a torsion free sheaf on a nonsingular surface
  and $E$ is its subsheaf. Then the natural homomorphism
  $E^{\vee\vee}\to F^{\vee\vee}$ is injective.
\end{Claim}

\begin{proof}
  Since the statements are local, we may assume $E$, $F$ are modules
  over the local ring $A$ at a point. Let $K$ be the quotient field of
  $A$. Then $E\otimes_A K\to F\otimes_A K$ is injective as $K$ is flat
  over $A$. Since $E$, $F$ are torsion free, $E\to E\otimes_A K$,
  $F\to F\otimes_A K$ are injective.

  We have $E^\vee = \{ \varphi\in (E\otimes_A K)^\vee \mid
  \varphi(E)\subset A\}$, and similarly for $E^{\vee\vee}$. Thus we
  have $E\subset E^{\vee\vee}\subset E\otimes_A K$ and the same for
  $F$. The assertion is clear.
\end{proof}
\end{NB}

\begin{proof}
Suppose that $E\in\fT(d,0,\dots,0)_0$. Then we have a short exact
sequence
\begin{equation}
  0 \to E_1 \to E \to \shfO^{\oplus r_2 + \cdots + r_N} \to 0
\end{equation}
with $E_1\in \Gi{r_1,0}^d$.
\begin{NB}
  We have
  \begin{equation*}
    \begin{split}
    & 0 \to E_2/E_1 \to E_3/E_1 \to E_3/E_2 \to 0,
\\
    & 0 \to E_3/E_1 \to E_4/E_1 \to E_4/E_3 \to 0,
\\
    & \qquad\qquad\qquad \cdots
\\
    & 0 \to E_{N-1}/E_1 \to E/E_1 \to E_N/E_{N-1} \to 0.
    \end{split}
  \end{equation*}
  From our assumption we have $E_i/E_{i-1} \cong \shfO^{\oplus r_i}$
  for $i\neq 1$. By the induction, we see that $E_i/E_1$ is
  trivial. In particular, $E/E_1$ is trivial.
\end{NB}
Consider
\begin{equation}
  0 \to \shfO^{\oplus r_2 + \cdots + r_N} \to E^\vee \to E_1^\vee \to
  \mathcal Ext^1(\shfO^{\oplus r_2 + \cdots + r_N}, \shfO)
\end{equation}
Since $\mathcal Ext^1(\shfO^{\oplus r_2 + \cdots + r_N}, \shfO) = 0$,
the last homomorphism $E^\vee\to E_1^\vee$ is surjective. Therefore
this is a short exact sequence. Dualizing again, we get
\begin{equation}
  0 \to E_1^{\vee\vee} \to E^{\vee\vee} \to \shfO^{\oplus r_2 + \cdots + r_N} \to 0.
\end{equation}
The last homomorphism $E^{\vee\vee} \to \shfO^{\oplus r_2 + \cdots +
  r_N}$ is surjective as $E\to \shfO^{\oplus r_2 + \cdots + r_N}$ is
so. (Or, we observe $E_1^{\vee\vee} \cong \shfO^{\oplus r_1}$ as
$E_1\in \Gi{r_1,0}^d$, and $\mathcal Ext^1(\shfO^{\oplus r_1},
\shfO) = 0$.)

Therefore this is also exact. We have $E_1^{\vee\vee} = \shfO^{\oplus
  r_1}$ as $E_1\in \Gi{r_1,0}^{d_1}$. Since the extension between the
trivial sheaves is zero on $\proj^2$, we have $E^{\vee\vee} =
\shfO^{\oplus r}$. Therefore $E\in \Gi{r,0}^d$.

Thus we have $\overline{\fT(d,0,\dots,0)_0}\subset \Gi{r,0}^d$. Since
both are $(dr-1)$-dimensional, and $\Gi{r,0}^d$ is irreducible, they
must coincide. This shows the first claim.

For the second claim, we need to show that any of
$\fT(0,d,0,\dots,0)_0$, \dots,
$\fT(0,\dots,0,d)_0$ contains a locally free sheaf.

Then it it is enough to consider the case $N=2$ and check that
$\fT(0,d)_0$ contains a locally free sheaf, as
an extension of a locally free sheaf by a locally sheaf is again
locally free.
\begin{NB}
  We have
  \begin{equation*}
    \begin{split}
    & 0 \to E_{N-1} \to E \to E_N/E_{N-1} \to 0,
\\
    & 0 \to E_{N-2} \to E_{N-1} \to E_{N-1}/E_{N-2} \to 0,
\\
    & \qquad\qquad\qquad \cdots
\\
    & 0 \to E_1 \to E_2 \to E_2/E_1 \to 0.
    \end{split}
  \end{equation*}
\end{NB}
Furthermore we may assume $r=2$ and $r_1 = r_2 = 1$.
\begin{NB}
For $n=1$, we
consider the Koszul resolution
\begin{equation*}
  0 \to \Omega^2_{\AA^2} \to \Omega^1_{\AA^2} \to \Omega^0_{\AA^2}
  \to \CC_0 \to 0,
\end{equation*}
where $\CC_0$ is the skyscraper sheaf at the origin $0$. We replace the
last part by the maximal ideal $\mathfrak m_0$, we get
\begin{equation*}
  0 \to \Omega^2_{\AA^2} \to \Omega^1_{\AA^2} \to \mathfrak m_0 \to 0.
\end{equation*}
This exact sequence does not give us an example, as the first Chern
classes does not vanish. But it gives a local example.
\end{NB}

We use the ADHM description. Let
\begin{equation*}
  \begin{split}
  & B_1 =  \begin{pmatrix}
   0      & 1        & 0       & \hdots & 0          \\
          & 0        & 1       & \hdots  & 0   \\
          &          & \ddots  & \ddots & \vdots     \\
          &          &         & 0      & 1          \\
   \bigzerol &       &         &         & 0
  \end{pmatrix}, \;
  B_2 = 0, \;
\\
  & I =
\begin{pmatrix}
  0 & 0 \\ \vdots & \vdots \\ 0 & 0 \\ 1 & 0
\end{pmatrix},
\;
  J =
  \begin{pmatrix}
    0 & 0 & \hdots & 0 \\
    1 & 0 & \hdots & 0
  \end{pmatrix}.
  \end{split}
\end{equation*}
We have $[B_1,B_2]+IJ = 0$. We see $(B_1,B_2,I,J)$ is stable, i.e., a
subspace $S\subset\CC^d$ containing the image of $I$ and invariant
under $B_1$, $B_2$ must be $S = \CC^d$. We also see that
$(B_1,B_2,I,J)$ is costable, i.e., a subspace $S\subset\CC^d$
contained in the kernel of $J$ and invariant under $B_1$, $B_2$ must
be $S = 0$. Therefore $(B_1,B_2,I,J)$ defines a framed locally free
sheaf $(E,\varphi)$, i.e., an element in $\Bun[SL(2)]d$. We consider
a subspace $\{
\begin{pmatrix}
  0 \\ *
\end{pmatrix}\}\subset \CC^2$, which is the kernel of $a$. Taking $0$
as a subspace in $\CC^d$, we have a subrepresentation of a
quiver. Therefore $E$ contains the trivial rank $1$ sheaf
$\shfO_{\proj^2}$ correspondingly. The quotient $E/\shfO_{\proj^2}$ is
given by the data
\begin{equation*}
  I =
\begin{pmatrix}
  0 \\ \vdots  \\ 0 \\ 1
\end{pmatrix}
\end{equation*}
and $J = 0$, $B_1$, $B_2$ the same as above. This is the ideal sheaf
$(x^d,y)$, and hence in $\Gi{1,0}^d$. Thus $E$ is a point in $\fT(0,d)_0$.
\end{proof}

\begin{NB}
  The following is a copy of an old computation. It is kept as a
  record. Note that some notation is different.

\subsection{Study of the space \texorpdfstring{$U_n$}{Un}}

We give a similar description for $U_n^{L,G} =
\Hom(\Phi_{L,G}(\IC(\Uh{n})), \cC_{S_{(n)}\AA^2})$.

Consider the diagram
\begin{equation}
  \begin{CD}
    \UhP{n}
    @>{\id\times p}>>
    \UhP{n} \times \UhL{n}
    @>{i\times \id}>> \Uh{n}\times \UhL{n}
\\
  @V{p}VV @VV{p\times\id}V @.
\\
  \UhL{n} @>j>> \UhL{n}\times \UhL{n}.
  \end{CD}
\end{equation}
Then
\begin{equation}
  \begin{split}
    & \Ext^*(p_! i^* \IC(\Uh{n}), \cC_{S_{(n)}\AA^2})
\\
    =\; &
    H^*(\UhL{n}, {\mathcal Hom}(p_! i^* \IC(\Uh{n}),
    \cC_{S_{(n)}\AA^2}))
\\
    =\; & H^*(\UhL{n}, (p_! i^* \IC(\Uh{n}))^\vee
    \overset{!}{\otimes}
    \cC_{S_{(n)}\AA^2})
\\
    =\; & H^*(\UhL{n}, j^! (p_* i^! \IC(\Uh{n})
    \boxtimes \cC_{S_{(n)}\AA^2}))
\\
    =\; & H^*(\UhL{n}, j^! (p\times\id)_*
    (i^! \IC(\Uh{n}) \boxtimes \cC_{S_{(n)}\AA^2}))
\\
    =\; & H^*(\UhL{n}, p_* (\id\times p)^!  (i\times\id)^!
    \IC(\Uh{n}) \boxtimes \cC_{S_{(n)}\AA^2}
    )
\\
    =\; & H^*(\UhP{n},(i\times p)^!
    \IC(\Uh{n}) \boxtimes \cC_{S_{(n)}\AA^2}
    ).
  \end{split}
\end{equation}

We next consider the diagram
\begin{equation}
  \begin{CD}
   (\UhP{n})_0 \times\AA^2 @>\cong>> p^{-1}(S_{(n)}\AA^2)
   @>{i\times p}>> \Uh{n}\times S_{(n)}\AA^2
\\
  @. @V{\beta}VV @VV{\id\times\gamma}V
\\
  @. \UhP{n} @>{i\times p}>> \Uh{n}\times \UhL{n},
  \end{CD}
\end{equation}
where $\beta$, $\gamma$ are inclusions. We have
\begin{equation}
  \begin{split}
  & H^*(\UhP{n},(i\times p)^!
    \IC(\Uh{n}) \boxtimes \cC_{S_{(n)}\AA^2}
    )
\\
  = \; &
  H^*(\UhP{n},(i\times p)^! (\id\times\gamma)_*
    \IC(\Uh{n}) \boxtimes \cC_{S_{(n)}\AA^2}
    )
\\
  = \; &
  H^*(\UhP{n},\beta_* (i\times p)^!
    \IC(\Uh{n}) \boxtimes \cC_{S_{(n)}\AA^2}
    )
\\
  = \; &
  H^*((\UhP{n})_0\times\AA^2, i^!
    \IC(\Uh{n}) \boxtimes \cC_{\AA^2})
\\
  = \; &
  H^*((\UhP{n})_0, i^!\IC(\Uh{n})).
\end{split}
\end{equation}
By considering the degree, we have
\begin{equation}
  \Hom(p_!i^* \IC(\Uh{n}), \cC_{S_{(n)}\AA^2})
  \cong H^{\topdeg}((\UhP{n})_0, i^! \IC(\Uh{n})).
\end{equation}
\end{NB}

\subsection{A pairing on \texorpdfstring{$V^d$}{Vd}}\label{sec:pairV}

In the same way as \subsecref{sec:pairing}, we can define a
nondegenerate pairing between $V^{d,P}$ and $V^{d,P_-}$.

\begin{NB}
For a later purpose, we do not use the trivialization of the local
system over $G/L_0$ and keep $i_-$, $j_-$. Then
\end{NB}
We have an isomorphism
\begin{equation}
  H^0(\xi_0^! i^* j^! \pi_* \cC_{\cGi{r}^d})
  \xrightarrow{\cong} H^0(\xi_0^* i_-^! j_-^* \pi_! \cC_{\cGi{r}^d}),
\end{equation}
where we also used $\pi_* = \pi_!$ as $\pi$ is proper. By the base
change and the replacement $i^*$, $i_-^!$ to $p_*$, $(p_-)_{!}$, we
can identify this with
\begin{equation}
  H_{[0]}( \Gi{P,0}^d) \to H_c^{[0]}(\Gi{P_-,0}^d),
\end{equation}
and we have a pairing
\begin{equation}
  \la\ ,\ \ra\colon
  H_{[0]}( \Gi{P,0}^d) \otimes H_{[0]}(\Gi{P_-,0}^d) \to \CC.
\end{equation}

Note that we also have the intersection pairing in the centered
Gieseker space $\cGi{r}^d$.\index{UhTildercG@$\cGi{r}^d$} As the
intersection $\Gi{P,0}^d\cap \Gi{P_-,0}^d$ consists of a {\it
  compact\/} space $\Gi{L,0}^d$, the pairing is well-defined, and
takes values in $\CC$. We multiply the sign $(-1)^{\dim \cGi{r}^d/2} =
(-1)^{dr-1}$ as before.

\begin{Lemma}\label{lem:inter=pair}
  The pairing is equal to the intersection pairing.
\end{Lemma}

\begin{proof}
  \begin{NB}
    Dec. 22 : The following idea does not seem to work, as the square
    \begin{equation}
      \begin{CD}
        \Gi{L}^d @>>> \Gi{P}^d
\\
         @VVV @VVV
\\
        \Uh[L]{d} @>>> \Uh[P]{d}
      \end{CD}
    \end{equation}
    is {\it not\/} a fiber square. Therefore we cannot move to
    $\Gi{L}^d$. Moreover, we do not have a natural homomorphism
    $\cC_{M^T}\to i^* j^! \cC_M$. In fact, the stable envelop gives us
    a homomorphism.

    Oct. 24 : Misha says that it is a consequence of $\cC_{M^T} \cong
    i^* j^! \cC_M \to i_-^! j_-^* \cC_M = \cC_{M^T}$ is an identity
    for a smooth space $M$. I need to check it.

  \end{NB}
The pairing is the restriction of that on equivariant cohomology groups:
\begin{equation}
    \la\ ,\ \ra\colon
    H^{*}_\TT(\xi_0^!i^*j^!\pi_*\cC_{\cGi{r}^d})
    \otimes
    H^{*}_\TT(\xi_0^!i_-^*j_-^!\pi_*\cC_{\cGi{r}^d})
   \to H^*_\TT(\mathrm{pt}).
\end{equation}
By the localization theorem, natural homomorphisms
\begin{gather}
     H^{*}_\TT(\xi_0^!i^!j^!\pi_*\cC_{\cGi{r}^d})\to
     H^{*}_\TT(\xi_0^!i^*j^!\pi_*\cC_{\cGi{r}^d}),
\\
    H^{*}_\TT(\xi_0^!i_-^*j_-^!\pi_*\cC_{\cGi{r}^d})\to
    H^{*}_\TT(\xi_0^*i_-^*j_-^*\pi_*\cC_{\cGi{r}^d})
\end{gather}
become isomorphisms over the fractional field of
$H^*_\TT(\mathrm{pt})$. Then the pairing between
$H^{*}_\TT(\xi_0^!i^!j^!\pi_*\cC_{\cGi{r}^d})$ and
$H^{*}_\TT(\xi_0^*i_-^*j_-^*\pi_*\cC_{\cGi{r}^d})$ is equal to the
intersection pairing by \cite[\S8.5]{CG}. Therefore we only need to
show that the composition
\begin{equation}
    i^! j^! \to i^* j^! \to i_-^! j_-^*
\end{equation}
is equal to $i^! j^! = i_-^! j_-^! \to i_-^! j_-^*$. This is a
consequence of the following general statement :
\begin{NB}
    See Sasha's message on May 13, 2014.
\end{NB}
Let $T$ be a torus
action on $X$ and $Y = X^T$ (more generally, it can be a closed
invariant subset containing $X^T$). Let $a\colon Y\to X$ be the
embedding. Let $F$ be a functor from $D_T(X)$ to $D_T(Y)$. Assume that
we have two morphisms of functors $\alpha$, $\beta\colon a^!\to
F$. Then $\alpha=\beta$ if and only if it is so on the image of
$a_!\colon D_T(Y)\to D_T(X)$. We apply this claim to $a=ji$, $F=i_-^!
j_-^*$. In our case, $\alpha = \beta$ on $a_!D_T(Y)$ is evident, as all
the involved morphisms are identities on the fixed point set $Y$.

Let us give the proof of the claim. We consider a natural map $a_!a^!
\mathcal F \to\mathcal F$ for $\mathcal F\in D_T(X)$. It becomes an
isomorphism if we apply $a^!$ by the base change. We set $\mathcal G =
a_! a^!\mathcal F$. We have $\alpha_{\mathcal G} = \beta_{\mathcal
  G}$, as homomorphisms $a^!\mathcal G\to F(\mathcal G)$, from
the assumption. Then we have $\alpha_{\mathcal F} = \beta_{\mathcal
  F}$ as the composition of $\alpha_{\mathcal G} = \beta_{\mathcal G}$
and $F(\mathcal F)\to F(\mathcal G)$.
\begin{NB}
Record of the original text:

This directly follows from the definition of Braden's isomorphism $i^*
j^! \to i_-^!  j_-^*$.
\begin{NB2}
  For the global case, $i^!j^!\to i^* j^!$, $i_-^!j_-^!\to i_-^!
  j_-^*$ are identified with
  \begin{equation}
    H^*(X,X\setminus X^A,\scF) \to H^*(X, X\setminus
    \mathcal A_X,\scF),
    \quad
    H^*(X,X\setminus X^A,\scF) \to
    H^*(\mathcal R_X, \mathcal R_X\setminus X^A,\scF)
  \end{equation}
respectively. Braden's isomorphism is the restriction
\begin{equation}
  H^*(X, X\setminus \mathcal A_X,\scF)\to
  H^*(\mathcal R_X, \mathcal R_X\setminus X^A,\scF).
\end{equation}
Therefore the assertion is clear.

Recall the definition of Braden's morphism:
$$i^!j^!\to i^*j^!\to i^*j^!j_{-*}j_-^*\to i^*i_*i_-^!j_-^*\to i_-^!j_-^*$$
where the third arrow comes from the base change, and the fourth arrow
comes from the adjointness of $i^*$ and $i_*$.

The desired equality follows from the fact that the composition
$$i^!j^!=i^*j^!j_{-*}j_-^!\to i^*i_*i_-^!j_-^!\to i_-^!j_-^!$$ is the identity
automorphism of $(j\circ i)^!=(j_-\circ i_-)^!$, where the first arrow
comes from the base change, while the second arrow comes from the adjointness
of $i^*$ and $i_*$.
\end{NB2}
\end{NB}
\end{proof}

\subsection{Another base of \texorpdfstring{$V^d$}{Vd}}\label{sec:comp}

Recall we have the canonical isomorphism
\(
 \pi_! \cC_{\Gi{L}^d}
  \xrightarrow[\cong]{\cL}\Phi_{L,G}(\pi_! \cC_{\Gi{r}^d})
\)
in \eqref{eq:stac}.

Thanks to the decomposition \eqref{eq:product},
the morphism $\pi\colon \Gi{L}^d\to \UhL{d}$ is the composite of
\begin{equation}\label{eq:pipi}
\pi\times\cdots\times\pi \colon \Gi{r_1}^{d_1}\times\cdots\times
  \Gi{r_N}^{d_N}
\to
\Uh[{SL(r_1)}]{d_1}\times\cdots\times \Uh[{SL(r_N)}]{d_N}
\end{equation}
with the sum map
\begin{equation}
  \kappa\colon
  \Uh[{SL(r_1)}]{d_1}\times\cdots\times \Uh[{SL(r_N)}]{d_N}
  \to \UhL{d}.
\end{equation}
The latter is a finite birational morphism.

Then $\cC_{\Gi{L}^d}$ decomposes under \eqref{eq:pipi} as in \eqref{eq:GU}:
\begin{equation}\label{eq:piA}
  \pi_! \cC_{\Gi{L}^d}
  = \bigoplus \kappa_! \IC(\Bun[{SL(r_1),\lambda_1}]{d_1}\times
  \cdots\times
  \Bun[{SL(r_N),\lambda_N}]{d_N}),
\end{equation}
where $\lambda_1$, \dots, $\lambda_N$ are partitions with
$d = d_1 + |\lambda_1| + \cdots +d_N + |\lambda_N|$. (These $d_1$, \dots, $d_N$ are different from above.)

The image of the closure of $\Bun[{SL(r_1),\lambda_1}]{d_1}\times
  \cdots\times
  \Bun[{SL(r_N),\lambda_N}]{d_N}$ under $\kappa$ is the closure of
\begin{equation}
   \Bun[{SL(r_1)}]{d_1}\times \cdots\times \Bun[{SL(r_N)}]{d_N}\times S_\mu\AA^2,
\end{equation}
where $\mu = \lambda_1\sqcup\cdots\sqcup\lambda_N$. Let us denote this
stratum by $\Bun[{L,\mu}]{d_1,\dots,d_N}$. Then as $\kappa$ is a
finite morphism, we have
\begin{equation}\label{eq:kappa}
    \kappa_! \IC(\Bun[{SL(r_1),\lambda_1}]{d_1}\times
  \cdots\times
  \Bun[{SL(r_N),\lambda_N}]{d_N})
  \cong
  \IC(\Bun[{L,\mu}]{d_1,\dots,d_N},\rho),
\end{equation}
where $\rho$ is the local system corresponding to the covering
\begin{equation}\label{eq:cover}
  S_{\lambda_1}\AA^2\times
  \cdots\times S_{\lambda_N}\AA^2\setminus\text{diagonal}
  \to
  S_{\mu}\AA^2
\end{equation}
and $\mu = \lambda_1\sqcup\cdots\sqcup\lambda_N$. Taking sum over
$\lambda_1$, $\lambda_2$, \dots which give the same $\mu$, we get
\begin{multline}
  \bigoplus_{\lambda_1\sqcup\cdots\sqcup\lambda_N = \mu}
  \kappa_! \IC(\Bun[{SL(r_1),\lambda_1}]{d_1}\times
  \cdots\times
  \Bun[{SL(r_N),\lambda_N}]{d_N})
\\
  \cong \IC(\Bun[{L,\mu}]{d_1,\dots,d_N},\rho),
\end{multline}
where $\rho$ is now given by the permutation representation
\begin{equation}\label{eq:rho}
  (V^1)^{\otimes n_1}\otimes (V^2)^{\otimes n_2}\otimes \cdots
\end{equation}
of $S_{n_1}\times S_{n_2}\times \cdots$ if $\mu = (1^{n_1}
2^{n_2}\cdots)$ with $\dim V^d = N$. Let us understand that this $V^d$
is temporarily different from the previous one \eqref{eq:13}.

Here we define $V^d$ as the cohomology of the union of the fibers of
\eqref{eq:cover} for the special case when $\mu$ is the partition
$(d)$ with the single entry $d$, where the union runs over $\lambda_1$, \dots,
$\lambda_N$:
\begin{equation}\label{eq:sigma}
  \sigma \colon
  \bigsqcup_{\lambda_1\sqcup\cdots\sqcup\lambda_N = (d)} S_{\lambda_1}\AA^2
  \times \cdots\times S_{\lambda_N}\AA^2 \setminus\text{diagonal}


  \to S_{(d)}\AA^2,
\end{equation}
and
\begin{equation}
  V^d = H_0( \sigma^{-1}(d\cdot 0)).
\end{equation}
If $\mu = (d)$, one of $\lambda_1$, \dots, $\lambda_N$ is $(d)$ and
others are the empty partition $\emptyset$. Therefore we have $\dim
V^d = N$.

Moreover
\(
  \Hom_{\Perv(\UhL{d})}(\cC_{S_{(d)}\AA^2},
  \pi_! \cC_{\Gi{L}^d})
\)
is given by the component $\IC(\Bun[{L,(d)}]{0,\dots,0},\rho)$, where
$\rho$ is the trivial representation of $S_1$ on $V^d$. Therefore
we have a canonical isomorphism
\begin{equation}\label{eq:Vd}
  \begin{split}
  V^d &\cong \Hom_{\Perv(\UhL{d})}(\cC_{S_{(d)}\AA^2},
  \pi_! \cC_{\Gi{L}^d})\\
  &\cong \Hom_{\Perv(\UhL{d})}(\cC_{S_{(d)}\AA^2},
  \Phi_{L,G}(\pi_! \cC_{\Gi{r}^d})),
  \end{split}
\end{equation}
where the second isomorphism is given by the stable envelop $\cL$.
Thus our $V^d$ is canonically isomorphic to the previous one
\eqref{eq:13}.

We have just shown
\begin{equation}
  \pi_!\cC_{\Gi{L}^d} \cong \bigoplus \IC(\BunLl{d_1,\dots,d_N},\rho).
\end{equation}
This is similar to \propref{prop:hyperbolic}, where we used the
factorization argument to construct an isomorphism.

Our argument looks slightly different, as we have not used the
projection $a\colon\AA^2\to \AA^1$. But the isomorphism is the same as
one given by the factorization argument from the above construction,
together with the observation that $a$, $i$, $j$ commute with the
projection $\pi^d_{a,?}$ ($? = G, P, L$).

Note that we have $e_i\in V^d = H_0(\sigma^{-1}(d\cdot 0))$ corresponding to
the component of $\sigma^{-1}(d\cdot 0)$ in
\begin{equation}
    S_{\emptyset}\AA^2\times\cdots\times
    \underbrace{S_{(d)}\AA^2}_{\text{$i^{\mathrm{th}}$ factor}}
    \times\cdots\times S_{\emptyset}\AA^2.
\end{equation}
Then $e_1,\dots, e_N$ gives a base of $V^d$.
\index{ei@$e_i$ (base element of $V^d$)}

If we view $V^d$ as $\Hom_{\Perv(\UhL{d})}(\cC_{S_{(d)}\AA^2}, \pi_!
\cC_{\Gi{L}^d})$, $e_i$ is the composite of homomorphisms
\begin{equation}
    \label{eq:25}
    \cC_{S_{(d)}\AA^2}\to \pi_! \cC_{\Gi{r_i}^d} \to
    \pi_! \cC_{\Gi{L}^d},
\end{equation}
where the left homomorphism is given by the fundamental class
$[\pi^{-1}(d\cdot 0)]$, and the right one is given by the inclusion
of the component $d_i = d$, $d_j = 0$ ($j\neq i$) in the decomposition
\eqref{eq:product}.

\begin{Example}
For $d=1$, $\Gi{r}^1$ (resp.\ $\Uh{1}$) is isomorphic to the product
of $\AA^2$ and the cotangent bundle of $\proj^{r-1}$ (resp.\ the
closure of the minimal nilpotent orbit of $\algsl_r$). Further suppose
$N=r$ and $r_1=\cdots=r_N = 1$.

Then \cite[Remark~3.5.3]{MO} gives us the relation:
\begin{equation}
    \label{eq:23}
    [\overline{\fT(0,\cdots,0,
    \underbrace{1}_{\text{$k^{\mathrm{th}}$ factor}},
    0,\cdots, 0)}]
  = (-1)^{k-1}(e_k + e_{k+1} + \cdots + e_r).
\end{equation}
Here the sign $(-1)^{k-1}$ comes from the polarization, mentioned
in~\subsecref{sec:Gifixed}.
\end{Example}

\begin{NB}
Let us give the computation.

This is taken from my message on Dec.~17, 2012.

Let us consider the special case $n=1$. Then it is well-known that
$\Mreg_0(1,r)$ is the product of $\AA^2$ and the minimal nilpotent orbit
of $\algsl_r$. And $M(1,r)$ is the product of $\AA^2$ and the
cotangent bundle of $\proj^{r-1}$.

The torus fixed points are numbered $1$, \dots, $r$, and given a
positive chamber $\mathfrak C$, we have the corresponding
Bialynicki-Birula stratification
$\proj^0\subset\proj^1\subset\cdots\subset\proj^{r-1}$.

The fundamental classes of the conormal bundles
$T^*_{\proj^{k-1}}\proj^{r-1}$ ($k=1,\dots,r$) gives a basis of $V_1 =
H_{\topdeg}(\fT(1)_0)$. We have relations
\begin{equation*}
  \begin{gathered}
  T^*_{\proj^{r-1}}\proj^{r-1} = \overline{\fT(1,0,\dots,0)}, \quad
  T^*_{\proj^{r-2}}\proj^{r-1} = \overline{\fT(0,1,0, \dots,0)},
  \quad\cdots\quad
\\
  T^*_{\proj^0}\proj^{r-1} = \overline{\fT(0,\dots,0,1)}
  \end{gathered}
\end{equation*}
in the notation in the previous subsection.

According to \cite[Remark~3.5.3]{MO}, the stable envelop component
corresponding to the torus-fixed point is the union of two conormal
bundles
\begin{equation*}
  T^*_{\proj^k}\proj^{r-1}\cup T^*_{\proj^{k-1}}\proj^r.
\end{equation*}
Let us give a little more detail:
\begin{NB2}
  Misha's message on Nov.~29, 2012.
\end{NB2}
Since all the strata closures are smooth in our situation, it is easy
to compute $j_{k!}(\CC)$: it is an extension $0\to \CC_{\proj^{k-1}}\to
j_{k!}(\CC)\to \CC_{\proj^{k}}\to 0$. Now the characteristic cycles are
additive in short exact sequences, and the characteristic cycles of
constant sub- and quotient- sheaves are just the conormal bundles to
their supports. This gives the above.

There is a sign issue here, called a {\it polarization\/} in
\cite[3.3.2]{MO}. For quiver varieties, in particular, Gieseker
spaces, they choose a canonical choice of a polarization, which is
explained in \cite[2.2.7]{MO}.

In our case, we have
\begin{equation}\label{eq:ek}
  \begin{split}
  e_k &= (-1)^{k-1} ([T^*_{\proj^{r-k}}\proj^{r-1}]
  + [T^*_{\proj^{r-k-1}}\proj^{r-1}])
\\
  &= (-1)^{k-1}
  \begin{aligned}[t]
  & \big([\overline{\fT(0,\cdots,0,
    \underbrace{1}_{\text{$k^{\mathrm{th}}$ factor}},
    0,\cdots, 0)}]
\\
  & \quad +
  [\overline{\fT(0,\cdots,0,
    \underbrace{1}_{\text{$(k+1)^{\mathrm{th}}$ factor}},
    0,\cdots, 0)}]\big)
  \end{aligned}
  \end{split}
\end{equation}
for $k=1,\dots,r$. Then gives a basis of for the vector space $V_1$
from the hyperbolic restriction from $G = SL_r$ to $T$, applied to the
pushforward of the constant sheaf for the Gieseker space. We have
\begin{equation}
  \sum_{k=1}^{r} e_k = [T^*_{\proj^{r-1}}\proj^{r-1}].
\end{equation}
The right hand side is $\beta_{-1}[M(0,r)]$, where $\beta_{-1}$ is the
Baranovsky (creation) operator.

This means that the diagonal gives the Baranovsky's Heisenberg.
It is compatible with \cite[Th.~12.2.1]{MO}.

More generally, we have
\begin{equation}
  [T^*_{\proj^{r-k}}\proj^{r-1}]
  = (-1)^{k-1}(e_k + e_{k+1} + \cdots + e_r).
\end{equation}
\end{NB}

\begin{Example}\label{ex:base}
  We know that $[\overline{\fT(d,0,\dots,0)_0}] = [\Gi{r,0}^d]$
  (\propref{prop:Un}), and hence
  \begin{equation}
    [\overline{\fT(d,0,\dots,0)_0}] = e_1+\cdots+e_N
  \end{equation}
  by \eqref{eq:22}.

  On the other hand, the opposite extreme
  $[\overline{\fT(0,0,\dots,d)_0}]$ is equal to $e_N$ up to sign by
  the support property of the stable envelop \cite[Th.~3.3.4
  (i)]{MO}. The polarization is opposite, therefore the sign is the
  half of the codimension of the corresponding fixed point
  component. We get
  \begin{equation}
    [\overline{\fT(0,0,\dots,d)_0}] = (-1)^{d(r-r_N)} e_N.
  \end{equation}

  If $N=2$, two elements exhaust the base.
\end{Example}

The transition matrix between two bases for $d > 1$, $N > 2$ can be
calculated from \eqref{eq:111} together with \eqref{eq:109}
below. Though \eqref{eq:111} determines $[\overline{\fT(0,\cdots,0,
  {d}, 0,\cdots, 0)}]$ ($d$ is in the {$k^{\mathrm{th}}$ entry) up to
  $\CC[\overline{\fT(d,0,\dots,0)_0}]$, it is a linear span of $e_k$,
  \dots, $e_N$ thanks to the support property of the stable envelop.
  Therefore we can fix the ambiguity.

\begin{NB}
    Let me describe $\Uh[B,0]1$ even more explicitly. It consists of
    rank $1$ upper triangular matrices $X$. Let
    \begin{equation}
        \label{eq:49}
        X = \left(
          \begin{array}{c:c}
              0 & X' \\
              \hdashline
              0 & 0
          \end{array}
        \right),
    \end{equation}
    where $X'$ is a $(k-1)\times(r-k+1)$ matrix of rank $\le 1$. Such
    matrices $X$ is the irreducible component in \eqref{eq:23}.
\end{NB}


\subsection{Computation of the pairing}

Let us relate the pairing in \subsecref{sec:pairV} to the pairing
defined on $\Gi{L}^d$ using the stable envelop
\begin{equation}
  \cL \colon H_{[0]}(\Gi{L,0}^d) \xrightarrow{\cong} H_{[0]}(\Gi{P,0}^d).
\end{equation}

Let us temporarily denote the stable envelop with respect to the
opposite parabolic by $\cL^-$. Then we want to compute
\begin{equation}
  \la\cL(\alpha),\cL^-(\beta)\ra,
\end{equation}
which is equal to the intersection pairing times
$(-1)^{\dim\cGi{r}^d}$ by \lemref{lem:inter=pair}.

Suppose that $\alpha$, $\beta$ are classes on a component $Z$ of
$\Gi{L,0}^d$.

Let us take equivariant lifts of $\alpha$, $\beta$ to
$Z(L)^0$-equivariant cohomology.

Since the supports of $\cL(\alpha)$ and $\cL^-(\beta)$ intersects
along $Z$ by one of characterizing properties of the stable envelop
\cite[Th.~3.3.4(i)]{MO}, we need to compute the restriction of the
(Poincar\'e dual of) $\cL(\alpha)$, $\cL^-(\beta)$ to the fixed point
$Z$. Again by a property of the stable envelop
\cite[Th.~3.3.4(ii)]{MO}, we have $\cL(\alpha)|_{Z} =
(\pol_{\text{rep}}/\pol) e(N^-)\cup \alpha$ and $\cL^-(\beta)|_{Z} =
(\pol_{\text{att}}/\pol) e(N^+)\cup \beta$, where $\pol_{\text{rep}}$,
$\pol_{\text{att}}$ are the polarizations given by attracting and
repellent directions. Then we have
\begin{equation}
  \int_{\cGi{r}^d} \cL(\alpha)\cup \cL^-(\beta)
  = \frac{\pol_{\text{rep}} \pol_{\text{att}}}{\pol^2}
  \int_{\cGi{r}^d} e(N) \cup \alpha\cup\beta
  = (-1)^{\codim Z/2} \int_Z \alpha\cup\beta
\end{equation}
by the fixed point formula.
\begin{NB}
  Note that $\pol_{\text{rep}} \pol_{\text{att}} = (-1)^{\codim Z/2}
  \pol_{\text{rep}}^2 = (-1)^{\codim Z/2}\pol^2$.
\end{NB}
Therefore if we multiply $(-1)^{\dim \cGi{r}^d/2}$, we get
$(-1)^{\dim Z/2}\int_X \alpha\cup\beta = \la\alpha,\beta\ra$.

If $\alpha$, $\beta$ are supported on different components $Z$, $Z'$
of $\Gi{L,0}^d$ respectively, we use a property \cite[Th.~3.7.5]{MO},
which says the restrictions of $\cL(\alpha)$, $\cL(\beta)$ to
components other than $Z$, $Z'$ are zero. Then it is clear that
$\la\cL(\alpha),\cL^-(\beta)\ra = 0$.

\begin{NB}
  I probably need to explain the identification by the trivialization
  over $G/P_0$ at some point...... I think that it does not affect the
  sign.
\end{NB}

As an application of this formula, let us we compute $\la
e_i,e^-_j\ra$, where $e_i\in V^d$ as in the previous subsection, and
$e_j^-\in V^{d,P_-}$ is defined in the same way using the opposite
hyperbolic restriction $\cL^-$.\index{eiminus@$e_i^-$}

This is reduced to the computation of the self-intersection number of
the punctual Quot scheme $\Gi{r_i,0}^d$ in the centered Gieseker
space $\cGi{r_i}^d$. This is given by $(-1)^{r_id-1}d r_i = (-1)^{\dim
  \cGi{r_i}^d/2} d r_i$ (\cite[\S4]{Baranovsky}).

Therefore we get
\begin{Proposition}\label{prop:pairinge}
  We have
  \begin{equation}\label{eq:109}
    \la e_i,e_j^-\ra = d r_i \delta_{ij}.
  \end{equation}
\end{Proposition}

\begin{NB}
  This is a comment for the previous formulation.

  I am no longer sure about this formula. We need to use the
  identification via the trivialization of the local system. I am not
  sure that the following is commutative.
  \begin{equation}
    \begin{CD}
      H_{[0]}(\Gi{L,0}^d) @>\cL>> H_{[0]}(\Gi{P,0}^d)
\\
      @| @VV{\cong}V
\\
      H_{[0]}(\Gi{L,0}^d) @>\cL^->> H_{[0]}(\Gi{P_-,0}^d)
    \end{CD}
  \end{equation}
\end{NB}

\subsection{Relation between \texorpdfstring{$V^d$}{Vd} and \texorpdfstring{$U^d$}{Ud}}\label{subsec:VU}
Let us apply the decomposition \eqref{eq:GU} to \eqref{eq:Vd}. We have
\begin{equation}
  \begin{split}
  V^d &= \Hom(\cC_{S_{(d)}\AA^2}, \Phi_{L,G}(\pi_!\cC_{\Gi{r}^d}))
\\
  &= \bigoplus_{d_1+|\lambda|=d}
  H_{\topdeg}(\pi^{-1}(x^{d_1}_\lambda)) \otimes
  \Hom(\cC_{S_{(d)}\AA^2}, \Phi_{L,G}(\IC(\BunGl{d_1}))).
  \end{split}
\end{equation}
Then
\(
  \Hom(\cC_{S_{(d)}\AA^2}, \Phi_{L,G}(\IC(\BunGl{d_1})))
\)
is nonzero only in either of the following cases:
\begin{enumerate}
\item $d_1=d$ and $\lambda=\emptyset$,
\item $d_1=0$ and $\lambda=(d)$.
\end{enumerate}
In the first case, it is $U^{d}$ by definition. And in the second
case, it is
\begin{equation}
  \Hom(\cC_{S_{(d)}\AA^2}, \Phi_{L,G}(\cC_{S_{(d)}\AA^2}))
  =
  \Hom(\cC_{S_{(d)}\AA^2},
  \cC_{S_{(d)}\AA^2}) \cong \CC\id.
\end{equation}
Thus
\begin{equation}\label{eq:VU}
  V^d \cong \left(H_{\topdeg}(\pi^{-1}(x^d_{\emptyset})) \otimes U^d\right)
  \oplus H_{\topdeg}(\pi^{-1}(x^0_{(d)})).
\end{equation}
Note that $\pi^{-1}(x^d_{\emptyset})$ is a single point. Therefore we
have the canonical isomorphism
$H_{\topdeg}(\pi^{-1}(x^d_{\emptyset}))\cong \CC$.

Now the homomorphism $\pi_*\colon V^d \cong H_{[0]}(\Gi{P,0}^d) \to
U^d \cong H_{[0]}(\Uh[P,0]d)$ is identified with the projection to
the first component in \eqref{eq:VU}. In particular, bases of $U^d$
and $V^d$ given by irreducible components (see \lemref{lem:Ud} and
\propref{prop:Un}) are related by the projection.
\begin{NB}
The following is not quite true. See \eqref{eq:51}.

Now it is clear that the isomorphism \eqref{eq:VU} is compatible with
bases of $U^d$ and $V^d$ given by irreducible components (see
\lemref{lem:Ud} and \propref{prop:Un}).
\end{NB}

The subspace $H_{\topdeg}(\pi^{-1}(x^0_{(d)}))$ is $1$-dimensional
space spanned by the fundamental class $[\pi^{-1}(x^0_{(d)})]$, or
equivalently $P_{-d}([0])\cdot [\Uh{0}]$ where $P_{-d}([0])$ is the
Heisenberg operator, and $[\Uh{0}] = 1\in H^0_{\TT}(\Uh{0})$.

Recall that the Baranovsky's Heisenberg operator is mapped to the
diagonal operator under the stable envelop, see \subsecref{sec:tensor}.
\begin{NB}
  I do not check that this is true even for general $L\neq T$. It is
  stated only for $L = T$. Probably the compatibility implies the
  general case.
\end{NB}
It means that $[\pi^{-1}(x^0_{(d)})]$ is equal to
\begin{equation}
  e_1 + \cdots + e_N,
\end{equation}
where $\{ e_i\}$ is the base of $V^d$ in the previous subsection.

And $U^d$ is the subspace killed by the Heisenberg operator
$P_d(1)$. Therefore
\begin{equation}\label{eq:baseofU}
  U^d
  \cong
  \left\{ \lambda_1 e_1 + \cdots + \lambda_N e_N\, \middle|\,
  \lambda_1 + \cdots + \lambda_N = 0 \right\}.
\end{equation}

We have a base $\{ e_i - e_{i+1} \}_{i=1,\dots, N-1}$ of $U^d$.

It is also clear that the decomposition \eqref{eq:VU} is orthogonal
with respect to the pairing in \subsecref{sec:pairV}. And the
restriction of the pairing to $U^d$ is equal to one in
\subsecref{sec:pairing}. Therefore we can calculate the pairing
between $U^{d,P}$ and $U^{d,P_-}$. Let us consider the case $P = B$
for brevity. We have
\begin{equation}\label{eq:Cartan}
  \la e_i - e_{i+1}, e_j^- - e_{j+1}^-\ra =
  \begin{cases}
    2d & \text{if $i=j$}, \\
    -d & \text{if $|i-j|=1$}, \\
    0 & \text{otherwise}
  \end{cases}
\end{equation}
by \propref{prop:pairinge}.

Thus the pairing between $U^{d,B}$ and $U^{d,B_-}$ is identified with
the natural pairing on the Cartan subalgebra $\h$ of $\algsl_r$
multiplied by $d$, under the identification $e_i - e_{i+1}$ and
$e_i^--e_{i+1}^-$ with the simple coroot $\alpha_i^\vee$

\begin{NB}
  Let us denote \eqref{eq:23} by $t_{k-1}$ ($k=1,2,\dots,r$). Its
  projection to $U^{d=1}$ is given by
    \begin{equation}\label{eq:51}
        \begin{split}
        & \frac{(-1)^k}r\left(
          (r-k+1) (e_1 + \cdots + e_{k-1})
          - (k-1) (e_k + \cdots + e_r) \right)
\\
   =\; & \frac{(-1)^k}r\left(
     \begin{aligned}[c]
          & (r-k+1) (e_1 - e_2) + 2(r-k+1)(e_2 - e_3) + \cdots
          + (k-1)(r-k+1) (e_{k-1} - e_k)
\\
          & \quad
          + (k-1)(r-k)(e_k - e_{k+1}) + \cdots + (k-1) (e_{r-1} - e_r)
     \end{aligned}\right).
        \end{split}
    \end{equation}
For $k=r$, we have
\begin{equation}
  \frac{(-1)^r}r (e_1 + \cdots + e_{r-1} - (r-1) e_r).
\end{equation}
We have
\begin{equation}\label{eq:112}
  (-1)^{k}\la pr_{U^1} t_{k-1}, e^-_i - e^-_{i+1}\ra =
  \begin{cases}
    1 & \text{if $i=k-1$,} \\ 0 & \text{otherwise}.
  \end{cases}
\end{equation}
\end{NB}

\subsection{Compatibility}\label{subsec:comp}

Let us take $L=T$. We shall show that the base $\{ e_i - e_{i+1} \}_i$
of $U^d$ is compatible with the construction in
\subsecref{sec:anotherbase} in this subsection.

We fix the Borel subgroup $B$ consisting of upper triangular matrices,
and let $P_i$ be the parabolic subgroup corresponding to a simple root
$\alpha_i$ and $L_i$ be the Levi subgroup ($i=1,\dots r-1$).

Recall that we have taken
\begin{equation}
  1^d_{L_i,G}\in
  \Hom_{\Perv(\Uh[{L_i}]{d})}(\IC(\Uh[{L_i}]{d}), \Phi_{L_i,G}(\IC(\Uh{d}))).
\end{equation}
(See \eqref{eq:1LG}.)

Let us consider the corresponding fixed point set $\Gi{L_i}^d =
(\Gi{r}^d)^{Z(L_i)}$ in the Gieseker space. The decomposition
\eqref{eq:product} in our case is
\begin{equation}\label{eq:121}
\bigsqcup_{d_1+\dots+\widehat{d_{i+1}}
    + \dots + d_r = d} \Gi{1}^{d_1}\times\cdots \times \Gi{1}^{d_{i-1}}
  \times \Gi{2}^{d_i}\times \Gi{1}^{d_{i+2}}\times\cdots \times\Gi{1}^{d_r}.
\end{equation}

There is a distinguished connected component, isomorphic to $\Gi{2}^d$
with $d_i = d$, $d_j = 0$ for $j\neq i$. Let us denote it by $Z$.

Recall that $\Uh[{L_i}]d$ is equal to $\Uh[SL(2)]d$ as a topological
space and the open subvariety $\Bun[{L_i}]d$ is equal to
$\Bun[SL(2)]d$. The connected component $Z$ is characterized among all
components of $\Gi{L_i}^d$, as it contains
$\Bun[{L_i}]d$.

We denote by $\pol$ the polarization of $Z$ in $\Gi{r}^d$ in
\subsecref{sec:Gifixed}. We understand it is $\pm 1$, according to
whether it is equal to the polarization given by attracting directions
or not, as in \subsecref{sec:polarization}.

We correct $1^d_{L_i,G}$ by $\pol 1^d_{L_i,G}$ so that it will be
compatible with the stable envelop.

Let us consider the diagram
\begin{equation}\label{eq:CD}
  \begin{CD}
  \IC(\Uh[L_i]{d})@>{\pol 1^d_{L_i,G}}>> \Phi_{L_i,G}(\IC(\Uh{d}))
\\
@AAA @AAA
\\
  \pi_! \cC_{\Gi{L_i}^d} @>\cong>{\cL_{L_i,G}}>
  \Phi_{L_i,G}(\pi_!\cC_{\Gi{r}^d}).
  \end{CD}
\end{equation}
The upper arrow is given just above, and the bottom arrow is the
stable envelop.

The right vertical arrow comes from the natural projection to the
direct summand
\(
   \pi_!(\cC_{\Gi{r}^d})\to \IC(\Uh{d})
\)
in \eqref{eq:GU},

which is the identity homomorphism on the open subset $\Bun{d}$ of
$\Uh{d}$.

The left vertical arrow is defined as follows. We have the
distinguished component $Z$ of $\Gi{L_i}^d$ isomorphic to
$\Gi{2}^d$. We have $\IC(\Uh[{L_i}]d) = \IC(\Uh[SL(2)]d)$, and hence
have a natural projection $\pi_! \cC_{\Gi{2}^d}\to \IC(\Uh[L_i]{d})$,
as for the right vertical arrow. Composing with the restriction to the
distinguished component $\pi_!\cC_{\Gi{L_i}^d}\to \pi_!  \cC_{Z}$ , we
define the left vertical arrow.

\begin{Proposition}
  The diagram \eqref{eq:CD} is commutative.
\end{Proposition}

\begin{proof}
  From the construction of the diagram, it is clear that we need to
  check the commutativity on the open subset $\Bun[{L_i}]{d}$. Then
  the commutativity is clear, as two constructions $\pol 1^d_{L_i,G}$
  and $\mathcal L_{L_i,G}$ are the same: Both are given by the Thom
  isomorphism corrected by polarization. See \cite[Th.~3.3.4(ii)]{MO}
  for the stable envelop.
\end{proof}

Recall also that we have proposed that there exists a canonical
element
\begin{equation}
  \begin{split}
  1^d_{L_i}\in & \Hom(\cC_{S_{(d)}\AA^2}, \Phi_{T,L_i}(\IC(\Uh[L_i]{d})))\\
  & \cong \Hom(\cC_{S_{(d)}\AA^2},\Phi_{\CC^*,SL_2}(\IC(\Uh[{SL_2}]{d})))
  \end{split}
\end{equation}
in \subsecref{sec:anotherbase}. \index{1Li@$1^d_{L_i}$}

We define it so that the following diagram is commutative:
\begin{equation}
    \label{eq:24}
    \begin{CD}
     \cC_{S_{(d)}\AA^2} @>1^d_{L_i}>> \Phi_{\CC^*,SL(2)}(\IC(\Uh[SL(2)]{d}))
\\
  @VVV @AAA
\\
    \pi_! \cC_{\Gi{\CC^*}^d}
    @>\cong>\cL_{\CC^*,SL(2)}> \Phi_{\CC^*,SL(2)}(\pi_!\cC_{\Gi{2}^d}),
    \end{CD}
\end{equation}
where we choose the parabolic subgroup in $SL(2)\cong [L_i,L_i]$
corresponding to the chosen Borel subgroup $B$ to define the
hyperbolic restriction $\cL_{\CC^*,SL(2)}$. The right vertical arrow
is the projection to the direct summand as before.

The left vertical arrow is $e_i - e_{i+1}$, where $\{e_i, e_{i+1}\}$
is the base of $V^d_{\CC^*,SL(2)} \cong \Hom(\cC_{S_{(d)}\AA^2},
\pi_!\cC_{\Gi{\CC^*}^d})$, i.e., $e_i$ corresponds to
$\Gi{1}^d\times\Gi{1}^0\subset \Gi{\CC^*}^d = \bigsqcup
\Gi{1}^{d_1}\times \Gi{1}^{d_2}$, and $e_{i+1}$ corresponds
to $\Gi{1}^0\times\Gi{1}^d$.

We enlarge the bottom row as
\begin{equation}\label{eq:CD2}
  \begin{CD}
  \cC_{S_{(d)}\AA^2} @>1^d_{L_i}>>   \Phi_{T,L_i}(\IC(\Uh[L_i]{d}))
\\
  @VV{e_i-e_{i+1}}V @AAA
\\
     \pi_! \cC_{\Gi{T}^d}
    @>\cong>\cL_{T,L_i}>\Phi_{T,L_i}(\pi_!\cC_{\Gi{L_i}^d}).
  \end{CD}
\end{equation}
Here we identify $\Gi{2}^d$ with the distinguished component $Z$.

We similarly consider $\Gi{\CC^*}^d$ as a union of components
of $\Gi{T}^d$, putting it in $i$ and $(i+1)^{\mathrm{th}}$ components.

The left vertical arrow is $e_i - e_{i+1}$, where $\{ e_1,\dots,
e_r\}$ is a base of $V^d_{T,G}$.

Two bases are obviously compatible, so it is safe to use the same
notation.

\begin{NB}
  Remark that this shows that our guess that $1_{L_i}$ is given by an
  irreducible component of the relevant variety is WRONG. See Misha's
  message on Nov.~29, 2012. Recall (see \eqref{eq:ek})
  \begin{equation}
    e_1 - e_2 = [T^*_{\proj^1}\proj^1] + 2[T^*_{\proj^0}\proj^1].
  \end{equation}

Under the projection $\pi\colon \Gi{P}^1\to \UhP{1}$, this is mapped to
$2[T^*_{\proj^0}\proj^1]$. If we divide this by $2$, we need to change
$1_{L_i}$ to
  \begin{equation}
    \frac12 (e_1 - e_2) = \frac12 [T^*_{\proj^1}\proj^1] + [T^*_{\proj^0}\proj^1].
  \end{equation}
  This seems unnatural to me.
\end{NB}


We apply $\Phi_{T,L_i}$ to the commutative diagram \eqref{eq:CD} and
combine it with \eqref{eq:CD2}:
\begin{equation}\label{eq:CD3}
  \begin{CD}
  \cC_{S_{(d)}\AA^2} @>1^d_{L_i}>>
  \Phi_{T,L_i}(\IC(\Uh[L_i]{d}))
  @>{\Phi_{T,L_i}(\pol 1_{L_i,G})}>> \Phi_{T,G}(\IC(\Uh{d}))
\\
 @VV{e_i-e_{i+1}}V @AAA @AAA
\\
  \pi_! \cC_{\Gi{T}^n}@>\cong>\cL_{T,L_i}>
  \Phi_{T,L_i}(\pi_! \cC_{\Gi{L_i}^d})
  @>\cong>{\Phi_{T,L_i}(\cL_{L_i,G})}>
  \Phi_{T,G}(\pi_!\cC_{\Gi{r}^d}).
  \end{CD}
\end{equation}
The composite of lower horizontal arrows is $\cL_{T,G}$ by the
commutativity \eqref{eq:ass2}.

Recall we made an identification of $V^d$ by $\cL_{T,G}$ (see
\eqref{eq:Vd}). Therefore $e_i - e_{i+1}\in V^d$ considered as a
homomorphism $\Hom(\cC_{S_{(d)}\AA^2},
\Phi_{T,G}(\pi_!\cC_{\Gi{r}^d}))$ is the composition of arrows from
the upper left corner to the lower right corner.

It is also clear that the homomorphism $V^d\to U^d$ given by the
composition of the rightmost upper arrow coincides with the projection
in \eqref{eq:VU}.

We thus see that $\{ \Tilde\alpha^d_i = \Phi_{T,L_i}(\pol
1^d_{L_i,G})\circ 1^d_{L_i}\}_i$ coincides with the base $\{ e_i -
e_{i+1}\}$ of $U^d$. This gives the construction promised in
\subsecref{sec:anotherbase} when $G$ is of type $A$.

\begin{Remark}\label{rem:SL(2)}
    Suppose $G=SL(2)$. Thanks to Example~\ref{ex:base}, we have
    $[\overline{\fT(0,d)_0}] = (-1)^{d} e_2^d$. (Here $r_1=r_2=1$.)
    Therefore we have
    \begin{equation*}
        \la [\overline{\fT(0,d)_0}], \Tilde\alpha^{d,-}_1\ra
        = (-1)^d \la e_2^d, e_1^{d,-} - e_2^{d,-}\ra
        = (-1)^{d+1}d
    \end{equation*}
    by \propref{prop:pairinge}. This completes the proof of
    \eqref{eq:111}.
\end{Remark}

\subsection{\texorpdfstring{$\Aut(G)$}{Aut(G)} invariance}
\label{sec:autG-invar}

Recall that we have studied $\Aut(G)$ invariance of various
constructions for $\Uh{d}$ in \subsecref{sec:autg-invariance}. The
same applies also to the Gieseker space $\Gi{r}^d$, if we restrict to
the inner automorphism $\operatorname{Inn}(G)$.

This is because $\operatorname{Inn}(G)$ acts on $\Gi{r}^d$, and hence
the same applies.

Let us consider $\Aut(G)/\operatorname{Inn}(G)$. It is $\{ \pm 1\}$
for type $A$, and is the Dynkin diagram automorphism given by the
reflection at the center. It is represented modulo inner automorphisms
by a group automorphism $g\mapsto {}^tg^{-1}$.

In terms of $\Bun{d}$, it corresponds to associate a {\it dual\/}
vector bundle. In particular, it does not extend to an action on the
Gieseker space $\Gi{r}^d$, as the second Chern class may drop when we
take a dual of a sheaf.

In the ADHM description, the diagram automorphism is given by
\begin{equation}
  [(B_1,B_2,I,J)]\mapsto [(B_1^t, B_2^t, -J^t, I^t)].
\end{equation}
This does not preserve the stability condition. Therefore we must be
careful when we study what happens under this automorphism.

Nevertheless we give

\begin{proof}[Proof of Lemma~{\rm\ref{lem:autg}}]
  Recall $\sigma\in\Aut(G)$ preserves $T$, $B$, and corresponds to a Dynkin
  diagram automorphism. Recall also $\tilde\alpha_i^d =
  \Phi_{T,L_i}(\delta 1^d_{L_i,G})\circ 1^d_{L_i}$.

  It is clear that $\Phi_{T,L_i}(1^d_{L_i,G})$ is sent to
  $\Phi_{T,L_{\sigma(i)}}(1^d_{L_{\sigma(i)},G})$ under $\varphi_\sigma$ from its
  definition.

  Next consider $1^d_{L_i} \in U^d_{T,L_i}$. In view of
  \lemref{lem:Ud}, $U^d_{T,L_i}$ is $H_{[0]}(\Uh[B\cap L_i,0]d)$,
  which is $1$-dimensional space spanned by the irreducible component
  $[\Uh[B\cap L_i,0]d]$. The class $[\Uh[B\cap L_i,0]d]$ is sent to
  $[\Uh[B\cap L_{\sigma(i)},0]d]$ under $\varphi_\sigma$, as it is induced from
  the isomorphism $\Uh[B\cap L_i,0]d\to \Uh[B\cap L_{\sigma(i)},0]d$.

  On the other hand, $[\Uh[B\cap L_i,0]d]$ is the image of
  $[\overline{\fT(0,d)_0}]$ under $\pi_*\colon H_{[0]}(\Gi{B\cap
    L_i,0}^d)\to H_{[0]}(\Uh[B\cap L_i,0]d)$. We have
  $[\overline{\fT(0,d)_0}] = (-1)^d e_2$ by Example~\ref{ex:base}.
  \begin{NB}
    Its projection is $\frac{(-1)^{d+1}}2 (e_1 - e_2)$.
  \end{NB}
  Hence $[\Uh[B\cap L_i,0]d] = {(-1)^{d+1}} 1^d_{L_i}/2$. Combining
  with the above observation, we deduce the assertion.
\end{proof}

\begin{NB}
    Let us consider the case $d=1$. Recall that $\cUh[SL(r)]1$ is the
    space of nilpotent matrices of rank $\le 1$. Let
    \begin{equation}
\def\hsymbl#1{\smash{\hbox{\huge$#1$}}}
\def\hsymbu#1{\smash{\lower1.7ex\hbox{\huge$#1$}}}
        g_0 =
        (-1)^{r(r-1)/2}
        \begin{pmatrix}
            \hsymbu{0} & & & 1
\\
           & & \reflectbox{$\ddots$} &
\\
           & \reflectbox{$\ddots$} & &
\\
          1 & & & \hsymbl{0}
         \end{pmatrix}.
    \end{equation}
    Then the map
    \begin{equation}
        \label{eq:54}
        X \longmapsto - g_0 {}^t X g_0
    \end{equation}
    is an involution on $\cUh[SL(r)]1$. It is the composite of the
    morphism induced by $g\mapsto {}^t g^{-1}$ and the action of
    $g_0\in SL(r)$.

    From the description \eqref{eq:49}, we see that the
    $k^{\mathrm{th}}$ irreducible component $t_k$ is mapped to
    $t_{r-k+2}$, as $(k-1)$ and $(r-k-1)$ are swapped.

    So consider the involution $t_k\mapsto t_{r-k+2}$
    ($k=2,3,\dots,r$) in \eqref{eq:51}. It is given by
    \begin{equation}
        \label{eq:52}
         
         
         
         
         
         
         
         
         
        e_k - e_{k+1} \mapsto (-1)^r (e_{r-k} - e_{r-k+1})
        \qquad (k=1,\dots,r-1).
    \end{equation}
\end{NB}


\section{\texorpdfstring{$\scW$}{W}-algebra representation on
  localized equivariant cohomology}\label{sec:w-algebra-repr}

The goal of this section is to define a representation of the
$\scW$-algebra $\scW_k(\g)$\index{Wkg@$\scW_k(\g)$|textit} on the direct sum of equivariant intersection
cohomology groups $\IH^*_\TT(\Uh{d})$ over $d$, isomorphic to the Verma
module with the level and highest weight, given by the equivariant
variables by
\begin{equation}\label{eq:level}
  k + h^\vee = -\frac{\ve_2}{\ve_1}, \quad
  \lambda = \frac{\ba}{\ve_1} - \rho, \quad
  \text{where $\ba = (a^1,\dots,a^\ell)$}
    \index{a_vec@$\ba= (a^1,\dots,a^\ell)$}
\end{equation}
respectively. Here $\ba$ is a collection of variables, but will be
regarded also as a variable in the Cartan subalgebra $\h$ so that $a^i
= \alpha_i(\ba)$ for a simple root $\alpha_i$.
\begin{NB}
  So $\ba = \sum_i a^i \varpi_i^\vee$.
\end{NB}

\begin{NB}
  I need to check the normalization of the highest weight. Sometimes,
  it is simply given by invariant polynomials in $a^i$. Sometimes one
  shift them by some functions in level.
\end{NB}

Since the level is a rational function in $\ve_1$, $\ve_2$, we
must be careful over which ring the representation is defined.

In geometric terms, it corresponds to that we need to consider {\it
  localized\/} equivariant cohomology groups.

The equivariant cohomology group $H^*_\TT(\ )$ is a module over
$H^*_\TT(\mathrm{pt}) = \CC[\operatorname{Lie}\TT] =
\CC[\ve_1,\ve_2,\ba]$. Let us denote this polynomial ring by $\bA_T$
and its quotient field by $\bF_T$.
\index{AAT@$\bA_T = \CC[\ve_1,\ve_2,\ba]$}
\index{FFT@$\bF_T = \CC(\ve_1,\ve_2,\ba)$}

In algebraic terms, it means that our $\scW$-algebra is defined over
$\CC(\ve_1,\ve_2)$. Then the level $k$ is a {\it generic point\/} in
$\mathbb A^1$. Moreover we consider a Verma module whose highest
weight is in $\h^*\otimes\bF_T$. This means that the highest weight is
also generic.

More precisely, we regard $\ba$ as a canonical element in
$\h^*\otimes\bF_T = \h^*\otimes
\operatorname{Frac}(S(\h^*)[\ve_1,\ve_2])$ given by the inner product
on $\h$. Here we have used the Langlands duality implicitly : we first
consider $\ba$ as an element in $\h\otimes\bF_T$ via the identity
$\h\otimes\h^*$. Then we regard the first $\h$ as the dual of the
Cartan subalgebra of the Langlands dual of $\g$. But the Langlands dual
is $\g$ itself as we are considering ADE cases.
\begin{NB}
    More detail was given in my message on Apr.3,~2014.
\end{NB}


We will construct a representation on
\begin{equation}\label{eq:locIC}
  \bigoplus_d \IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T =
  \bigoplus_d H^*_\TT(\Uh{d}, \IC(\Uh{d}))\otimes_{\bA_T}\bF_T.
\end{equation}
By the localization theorem and \lemref{lem:centralfiber}, natural
homomorphisms
\begin{equation}\label{eq:4sp}
  \begin{split}
  & \IH^*_{\TT,c}(\Uh{d})
  \cong H^*_{\TT,c}(\Uh[T]{d}, i^!j^! (\IC(\Uh{d})))
\\
  & \to H^*_{\TT,c}(\Uh[T]{d}, \Phi_{T,G}(\IC(\Uh{d})))
\\
  & \to H^*_{\TT}(\Uh[T]{d}, \Phi_{T,G}(\IC(\Uh{d})))
\\
  & \to H^*_{\TT}(\Uh[T]{d}, i^*j^* (\IC(\Uh{d})))
  \cong \IH^*_\TT(\Uh{d})
  \end{split}
\end{equation}
all become isomorphisms over $\bF_T$. Thus over $\bF_T$, we could use
any of these four spaces. Let us denote its direct sum by $M_\bF(\ba)$:
\begin{equation}\label{eq:localhyper}
  M_\bF(\ba) = \bigoplus_d \IH^*_{\TT,c}(\Uh{d})\otimes_{\bA_T}\bF_T.
\end{equation}
\index{MF@$M_\bF(\ba)$}
\begin{NB}
  I denote it by $M$, as it will be the Verma module. On the other
  hand, other three spaces (nonequivariant ones) will be identified
  with Wakimoto modules, dual Wakimoto modules, and dual Verma module
  respectively.
\end{NB}

In fact, we will construct representations of {\it integral forms\/}
(i.e., $\bA_T$-forms) of Heisenberg and Virasoro algebras on {\it
  non-localized\/} equivariant cohomology groups
\(
   \bigoplus_d H^*_{\TT,c}(\Uh[T]{d},\Phi_{T,G}(\IC(\Uh{d})))
\)
of hyperbolic restrictions in this section. This construction will be
the first step towards a construction of the $\scW$-algebra
representation on {\it non-localized\/} equivariant cohomology groups.

To follow the remaining argument, the reader needs to read our
definition of an integral form of the $\scW$-algebra given in
\secref{sec:intW}. Therefore the whole construction will be postponed
to \subsecref{sec:univ-verma-modul}.

Let us denote the fundamental class $1\in \IH^0_\TT(\Uh{0}) =
\IH^0_{\TT,c}(\Uh{0}) = H^0_\TT(\mathrm{pt})$ by $|\ba\rangle$.
\index{a_vecr@$\vert\ba\rangle$}
It will be identified with the highest weight vector (or the vacuum
vector) of the Verma module. See \propref{prop:lambda} below.

We also use the following notation:
\begin{equation*}
  \bA = \CC[\ve_1,\ve_2], \quad \bF = \CC(\ve_1,\ve_2).
\end{equation*}
\index{AA@$\bA = \CC[\ve_1,\ve_2]$}\index{FF@$\bF = \CC(\ve_1,\ve_2)$}

\subsection{Freeness}

\begin{Lemma}\label{lem:free}
  Four modules appearing in \eqref{eq:4sp} are free over $\bA_T$.
\end{Lemma}

\begin{proof}
  By \lemref{lem:centralfiber} all four modules are pure, as
  $(\Uh{d})^{\TT}$ is a single point, and they are stalks at the
  point. Now freeness follows as in \cite[Th.~14.1(8)]{GKM}.

  Or we have odd cohomology vanishing by \cite[Th.~7.10]{BFG}. So
  it also follows from \cite[Th.~14.1(1)]{GKM}.
\end{proof}

\begin{NB}
   Since $\Uh{d}$ is not projective,
  it is not clear whether we can apply \cite[Th.~14.1(7)]{GKM}. We
  must explain this somewhere.

  \begin{NB2}
    If you know the purity of $\IH^*(\Uh{d})$, it is also enough.
  \end{NB2}
\end{NB}

In particular, homomorphisms in \eqref{eq:4sp} are all injective.

\subsection{Another base of \texorpdfstring{$U^d$}{Ud}, continued}\label{sec:anotherbase2}

Let $U^d = U^d_{T,G}$ be as in \subsecref{sec:Ud}. Let $L_i$ be the
Levi subgroup corresponding to a simple root
$\alpha_i$\index{alphai@$\alpha_i$} and consider $U^d_{T,L_i}$ as in
\subsecref{sec:anotherbase}.

We identify $\IC(\Uh[L_i]{d})$ with $\IC(\Uh[SL(2)]{d})$ by the
bijective morphism $\Uh[SL(2)]d\to \Uh[L_i]{d}$ (see
\propref{prop:LL}).

We have a maximal torus and a Borel subgroup induced from those of
$G$.

Then $U^d_{T,L_i}$ has the base in \eqref{eq:baseofU}, where it
consists of a single element as $N=2$. Let us denote the element by
$1^d_{L_i}$, as we promised in \subsecref{sec:anotherbase}.

Next consider $1^d_{L_i,G}$ given by the Thom isomorphism as in
\subsecref{sec:open}. We have the repellent polarization
$\pol_{\text{rep}}$ of $\Bun[L_i]{d}$ in $\Bun[G]{d}$. We modify it to
$\pol$\index{delta@$\pol$ (polarization)} according to
\lemref{lem:polSL3}.

We choose and fix a bipartite coloring of the vertices of the Dynkin
diagram, i.e., $o\colon I\to \{ \pm 1\}$ such that $o(i) = -o(j)$ if
$i$ and $j$ are connected in the diagram. Then we set
\begin{equation}\label{eq:polarization}
  \pol = o(i)^d \pol_{\text{rep}}.
\end{equation}
This is our polarization, which was promised in \eqref{eq:anotherbase}.

Let us write
\begin{equation}
  \tilde\alpha^d_i \defeq \Phi_{T,L_i}(\pol 1^d_{L_i,G})\circ 1^d_{L_i}.
\end{equation}
\index{alphatildai@$\Tilde\alpha^d_i$}
This gives us a collection $\{\tilde\alpha^d_i\}_i$ of elements in
$U^d$ labeled by $I$.

Thanks to \eqref{eq:111}, it is a base of $U^d$.

This will follow also from \propref{prop:HeisRel}.
\begin{NB}
We will prove that it is a base of $U^d$ in \corref{cor:Ubase}.
\end{NB}

\subsection{Heisenberg algebra associated with the Cartan subalgebra}\label{sec:heis-algebra-assoc}

We construct a representation of the Heisenberg algebra associated
with the Cartan subalgebra $\mathfrak h$ of $\g$ on the direct sum of
\eqref{eq:locIC} in this subsection. It will be the first step towards
the $\scW$-algebra representation.

Let us first review the construction of the Heisenberg algebra
representation in \subsecref{sec:tensor} for the case $r=2$ and $L =
S(GL(1)\times GL(1)) = \CC^*$. We consider Heisenberg operators
$P_n^\Delta\equiv P_n^\Delta(1)$ associated with the cohomology class
$1\in H^{[*]}_\TT(\AA^2)$. We omit $(1)$ hereafter.

They are not well-defined on $\bigoplus_d H^\TT_{[*]}(\Gi{P}^d)$ if
$d > 0$, but are well-defined on the localized equivariant homology
group
\(
   \bigoplus_d H^\TT_{[*]}(\Gi{P}^d)\otimes_{\bA_T}\bF_T,
\)
and satisfy the commutation relations
\begin{equation}
  [P_m^\Delta, P_n^\Delta] = - 2m\delta_{m,-n}\frac1{\ve_1\ve_2}.
\end{equation}
\begin{NB}
  In fact, we must invert $\ve_1$, $\ve_2$, but not $\ba$. Therefore
  we can work on $\otimes_{\bA}\bF$. Does this make any difference ?
\end{NB}

Via the stable envelop, we have the isomorphism
\begin{equation}
  \bigoplus_d H^\TT_{[*]}(\Gi{P}^d) \cong \bigoplus_d H^\TT_{[*]}(\Gi{L}^d)
  = \bigoplus_{d_1,d_2}H^\TT_{[*]}(\Gi{1}^{d_1})\otimes
  H^\TT_{[*]}(\Gi{1}^{d_1}),
\end{equation}
and we have the representation of the tensor product of two copies of
Heisenberg algebras, given by $P_n^{(1)} = P_n\otimes 1$ and
$P_n^{(2)} = 1\otimes P_n$ on the localized equivariant
homology group, where $P_n$ is the Heisenberg generator for $r=1$.
The above Heisenberg generator $P_n^{\Delta}$ is the
diagonal $P_n^{(1)} + P_n^{(2)}$. See \subsecref{sec:tensor}.
\index{P@$P_n^{(i)}$}

We have
\begin{equation}
  H^\TT_{[-*]}(\Gi{P}^d)
  \cong H^{*}_{\TT}(\Uh[\CC^*]d, p_* j^! \pi_!\cC_{\Gi{2}^d})
  = H^{*}_{\TT}(\Uh[\CC^*]d, \Phi_{\CC^*,SL(2)}(\pi_!\cC_{\Gi{2}^d}))
\end{equation}
by \subsecref{sec:sheaf} and $\pi_! = \pi_*$. This homology group
contains
\begin{equation}
  H^{*}_\TT(\Uh[\CC^*]d, \Phi_{\CC^*,SL(2)} (\IC(\Uh[SL(2)]d)))
\end{equation}
as a direct summand, and the {\it anti-diagonal\/} Heisenberg algebra
generated by $P_n^{(1)} - P_n^{(2)}$ acts on its direct sum over
$d$. (See \subsecref{subsec:VU}.)

Let us return back to general $G$.

Let $L_i$ be the Levi subgroup as in the previous subsection. We
identify $\IC(\Uh[SL(2)]d)$ with $\IC(\Uh[L_i]d)$ as before, and we have
a(n anti-diagonal) Heisenberg algebra representation on
\begin{equation}
  \bigoplus_d H^*_\TT(\Uh[T]d, \Phi_{T,L_i}(\IC(\Uh[L_i]d)))
  \otimes_{\bA_T}\bF_T.
\end{equation}

Using the decomposition \eqref{eq:decomp} and $\Phi_{T,L_i}\Phi_{L_i,G}
= \Phi_{T,G}$, we have an induced Heisenberg algebra representation on
$M_\bF(\ba)$ in \eqref{eq:localhyper}.

Let us denote the Heisenberg generator by $P^i_n$.
\index{P@$P_n^i$ (Heisenberg generator)}

By \lemref{lem:Fock}, the space $M_\bF(\ba)$ is isomorphic to
\begin{equation}\label{eq:symprod}
  
  
  
  \Sym((U^1\oplus U^2\oplus\cdots)\otimes_\CC\bF_T),
\end{equation}
where $\Sym$ denotes the symmetric power.  ($U^d = U^d_{T,G}$ as
before.)

Let us describe $P^i_n$ in this space. Recall that we have the
orthogonal decomposition $U^d = U^d_{T,L_i}\oplus
(U^d_{T,L_i})^\perp$ in \eqref{eq:decomp2}.

Then we have the factorization
\begin{multline}\label{eq:53}
  \operatorname{Sym}((U^1\oplus U^2\oplus\cdots)\otimes_\CC\bF_T)
\\
  \cong \operatorname{Sym}((U^1_{T,L_i}\oplus U^2_{T,L_i}\oplus\cdots)
  \otimes_\CC\bF_T)  \otimes_{\bF_T}
\\
  \operatorname{Sym}(((U^1_{T,L_i})^\perp\oplus (U^2_{T,L_i})^\perp\oplus\cdots)
  \otimes_\CC\bF_T)
\end{multline}
The first factor of the right hand side is the usual Fock space
associated with the Cartan subalgebra $\h_{\algsl_2}$ of
$\algsl_2$. In fact, using $U^d_{T,L_i} \cong \CC 1^d_{L_i}$, we
identify $U^d_{T,L_i}$ with $\h_{\algsl_2}$.  The pairing is
multiplied by $-1/\ve_1\ve_2$ from the natural one. Then the factor is
$\operatorname{Sym}(z^{-1}\h_{\algsl_2}[z^{-1}])$ and the Heisenberg
algebra acts in the standard way.

From its definition, our Heisenberg operator $P^i_n$ is given by the
tensor product of the Heisenberg operator for
$\operatorname{Sym}(z^{-1}\h_{\algsl_2}[z^{-1}])$, and the identity.

The following means that the operators $P^i_n$ define the Heisenberg
algebra $\Heis(\mathfrak h)$\index{Heis@$\Heis(\mathfrak h)$}
associated with the Cartan subalgebra $\mathfrak h$ of $\g$.

\begin{Proposition}\label{prop:HeisRel}
  Heisenberg generators satisfy commutation relations
  \begin{equation}
    [P^i_m, P^j_n] = - m\delta_{m,-n}(\alpha_i,\alpha_j)\frac1{\ve_1\ve_2}.
  \end{equation}
\end{Proposition}

If we normalize the generator by $\widehat h^i_n = \ve_2 P^i_n$, the
relations match with a standard convention with level $-\ve_2/\ve_1 =
k+h^\vee$. See \eqref{eq:hhatcomm}.

From the construction, $P^i_{-d}$ applied to the vacuum vector
$|\ba\ra\in H^0_{\TT}(\Uh[T]{0}, \Phi_{T,G}(\IC(\Uh{0})))$ is equal to
$\Phi_{T,L_i}(\pol 1^d_{L_i,G})\circ 1^d_{L_i}\in U^d$ divided by
$\ve_1\ve_2$, considered as an element in \eqref{eq:symprod}.

\begin{NB}
    The following is commented out as we already knew
    $\{ \Tilde\alpha^d_i\}$ is a base.

Hence we deduce the following, as promised in
\subsecref{sec:anotherbase} and \subsecref{sec:anotherbase2}.

\begin{Corollary}\label{cor:Ubase}
\(
  \{ \tilde\alpha^d_i = \Phi_{T,L_i}(\pol 1^d_{L_i,G})\circ 1^d_{L_i}
  \}_i
\)
is a base of $U^d$.
\end{Corollary}
\end{NB}

From the construction, \eqref{eq:symprod} is the Fock space of the
Heisenberg algebra associated with the Cartan subalgebra $\mathfrak
h$. It is $\operatorname{Sym}(z^{-1}\h[z^{-1}])$, where the base field
is $\bF_T$. The element $\tilde\alpha^d_i

$ is a linear function, living on $z^{-d}\mathfrak h$.

\begin{proof}[Proof of Proposition~{\rm\ref{prop:HeisRel}}]
  The case $(\alpha_i,\alpha_j) = 2$, i.e., $i=j$ is obvious from the
  construction.

  Next consider the case $(\alpha_i,\alpha_j) = -1$. Then $i$ and $j$
  are connected by an edge in the Dynkin diagram. Let us take the
  parabolic subgroup $P$ corresponding to the subset consisting of two
  vertices $i$ and $j$, and the corresponding Levi subgroup $L$. We
  have $[L,L] \cong SL(3)$. Then from our construction and the
  compatibility of the stable envelop with the hyperbolic restriction
  functor in \subsecref{subsec:comp}, the assertion follows from the
  $SL(3)$-case, which is clear as Heisenberg algebra generators are
  given by
  \begin{equation}
    P^i_n = P_n\otimes 1\otimes 1 - 1\otimes P_n\otimes 1,
\quad
    P^j_n = 1\otimes P_n\otimes 1 - 1\otimes 1\otimes P_n.
  \end{equation}
  Note also that our polarization $\pol$ in \eqref{eq:polarization} was
  chosen so that it is the same as the polarization for $\Gi{SL(3)}^d$
  via \lemref{lem:polSL3} up to overall sign independent of $d$.
  \begin{NB}
    Here the only ambiguity comes from the choice of a bijection $\{
    i,j\} \leftrightarrow \{ 1,2\}$.
  \end{NB}

  Finally consider the case $(\alpha_i,\alpha_j) = 0$. We argue as
  above by taking the corresponding Levi subgroup $L$ with $[L,L]\cong
  SL(2)\times SL(2)$. Then it is clear that Heisenberg generators
  commute.

  If a reader would wonder that $SL(2)\times SL(2)$ is not considered
  in \secref{sec:typeA}, we instead take a type $A_k$ subdiagram
  containing $i$, $j$ and take the corresponding Levi subgroup $L$
  with $[L,L]\cong SL(k+1)$. Then it is clear that the Heisenberg
  generators $P^i_m$, $P^j_n$ commute for $SL(k+1)$. Therefore they
  commute also for $G$.
\end{proof}

Let us consider Heisenberg operators
\(
  P^i_n([0]) = \ve_1\ve_2 P^i_n,
\)
coupled with the Poincar\'e dual of $[0]\in H^\TT_0(\AA^2)$, and
denote them by $\widetilde{P}^i_n$.
\index{P@$\widetilde P_n^i$}
Then they are well-defined on non-localized equivariant cohomology
groups
\begin{equation}
  \bigoplus_d H^*_\TT(\Uh[T]d, \Phi_{T,G}(\IC(\Uh{d}))),
\end{equation}
and satisfy the commutation relations
\begin{equation}\label{eq:mPrel}
  [\widetilde{P}^i_m, \widetilde{P}^j_n]
  = - m\delta_{m,-n}(\alpha_i,\alpha_j){\ve_1\ve_2}.
\end{equation}
The same is true for the non-localized equivariant cohomology with
compact supports.

We define the $\bA$-form $\Heis_\bA(\mathfrak h)$
\index{HeisA@$\Heis_\bA(\mathfrak h)$} of the Heisenberg vertex
algebra as the vertex $\bA$-subalgebra of $\Heis(\mathfrak h)$
generated by $\widetilde{P}^i_m$.

\subsection{Virasoro algebra}\label{sec:Vir}

Let us introduce $0$-mode operators $P^i_0$. In \subsecref{sec:BaraOp}
we did not introduce them. Since they commute with all other
operators, we can set them any scalars. We follow the convention in
\cite[\S13.1.5, \S14.3.1]{MO}, that is
\begin{equation}
  P^i_0 = \frac{a^i}{\ve_1\ve_2}.
\end{equation}
Here $a^i$ is the $i^{\mathrm{th}}$ simple root, and should be
identified with $a_i - a_{i+1}$ in \cite{MO} in the Fock space
$\mathbf F(a_1)\otimes \cdots\otimes \mathbf F(a_r)$ corresponding to
the equivariant cohomology of Gieseker spaces for rank $r$ sheaves.
\begin{NB}
Note also that the convention in \cite[\S13.1.5]{MO} is
  \begin{equation}
    \alpha_0(\gamma)|\eta\rangle
    = - (\gamma,\eta)|\eta\rangle,
  \end{equation}
and $-(1,\eta) = \int_{\AA^2}\eta = \eta/\ve_1\ve_2$.
\end{NB}

We also set $\widetilde{P}_0^i = \ve_1\ve_2 P_0^i = a^i$.

We then introduce Virasoro generators by
\begin{equation}\label{eq:Vira}
  {L}^i_n = - \frac14 \ve_1\ve_2 \sum_m \normal{P^i_m P^i_{n-m}}
  - \frac{n}2 (\ve_1+\ve_2) P^i_n
  + \frac{(\ve_1+\ve_2)^2}{4\ve_1\ve_2}\delta_{n,0}.
\end{equation}
See \cite[(13.10),(14.10)]{MO}.\index{Lin@$L^i_n$ (Virasoro generator)}

Let us briefly explain how to derive the above expression from \cite{MO}:

The Virasoro field $T(\gamma,\kappa) = \sum L_n(\gamma,\kappa) z^{-n}$
in \cite[(13.10)]{MO} is given by
\begin{equation}
    T(\gamma,\kappa) = \frac12 \normal{\boldsymbol\alpha^2}(\gamma)
    + \partial\boldsymbol\alpha(\gamma\kappa) - \frac12 \tau(\gamma\kappa^2),
\end{equation}
where $\boldsymbol\alpha(\gamma) = \sum \alpha_n(\gamma) z^{-n}$ is
the free field. Note that $T$ and $\boldsymbol\alpha$ are different
from the usual convention, as the exponents are not $-n-1$, $-n-2$
respectively. Also $\partial = z\partial_z$.

We take $\gamma = 1$, the fundamental class of $H^0_\TT(\AA^2)$.

Next note that $\boldsymbol\alpha = {\boldsymbol\alpha^-}/{\sqrt{2}}$
\cite[(14.8)]{MO}, and our $P^i$ is identified with
$\boldsymbol\alpha^-$. This is the reason we have $1/4$ instead of
$1/2$.

  The remaining factor $-\ve_1\ve_2$ comes from
\(
    1^\Delta = - 1\otimes\mathrm{pt}
\)
in \cite[\S13.3.2]{MO}.

For the second term, note $\kappa = \hbar/\sqrt{2}$ (see
\cite[(14.8)]{MO}), $\hbar = -t_1-t_2$ (see
\cite[\S17.1.1,(18.10)]{MO} for example). We denote their $t_1$, $t_2$
by $\ve_1$, $\ve_2$ instead.

For the last constant term, we have
\(
  -\gamma\kappa^2 = - (\ve_1+\ve_2)^2/2
\)
and
\(
   \tau(1) = -\int_{\AA^2} 1 = - 1/\ve_1\ve_2.
\)

\begin{NB}
  Lehn's formula \cite[3.10]{Lehn} says
  \begin{equation}
    \frac1n [c_1(\shfO^{[n]}_X), q_n(\gamma)]
    = L_n(\gamma) + \frac{|n|-1}2 q_n(K\gamma).
  \end{equation}
  I should compare it with \cite[(14.1), \S15.1]{MO}.
\end{NB}

The Virasoro algebra commutation relations are
\begin{equation}
  [L^i_m, L^i_n]
    = (m-n) L^i_{m+n} +
    \left(1 + \frac{6(\ve_1+\ve_2)^2}{\ve_1\ve_2}\right)
    \delta_{m,-n}\frac{m^3 - m}{12}.
\end{equation}
See \cite[\S13.3.2]{MO}. And the highest weight is given by
\begin{equation}\label{eq:92}
  L_0^i |\ba\rangle = -\frac14 \left( \frac{(a^i)^2}{\ve_1\ve_2} -
    \frac{(\ve_1+\ve_2)^2}{\ve_1\ve_2}
    \right)|\ba\rangle.
\end{equation}
See \cite[\S13.3.5]{MO}.
\begin{NB}
  Here the highest weight is shifted by a function in level. I must be
  careful to check that this is a correct formulation.

  In \cite[(13.15)]{MO}, the highest weight is
  \begin{equation}
      \frac12\tau(\eta^2 - \kappa^2)
      = -\frac1{4\ve_1\ve_2}\left( (a_1 - a_2)^2 - (\ve_1+\ve_2)^2\right).
  \end{equation}

  \cite[(285)]{Arakawa2007} says
  \begin{equation}
      \label{eq:60}
      L(0) |\gamma_{\overline{\lambda}}\rangle
      = \Delta_{\overline{\lambda}} |\gamma_{\overline{\lambda}}\rangle
  \end{equation}
with
\begin{equation}
    \label{eq:61}
    \Delta_{\overline{\lambda}} =
    \frac{|\overline{\lambda}+\overline{\rho}|^2}{2(k+h^\vee)}
    - \frac{\rank\overline\g}{24} + \frac{c(k)}{24},
\end{equation}
where
\begin{equation}
      \label{eq:59}
      \begin{split}
      c(k) &= \rank\overline\g - 12\left(
        (k+h^\vee)|\overline\rho^\vee|^2
        - 2\langle\overline\rho,\overline\rho^\vee\rangle
        + \frac1{k+h^\vee}|\overline\rho|^2\right),
\\
     &= \rank\overline\g - 12|\overline\rho^\vee|^2\left(
        (k+h^\vee)
        - 2
        + \frac1{k+h^\vee}\right)\quad
     \text{for $ADE$}.
      \end{split}
\end{equation}
Hence
\begin{equation}
    \label{eq:62}
    \Delta_{\overline\lambda}
    = \frac{|\overline{\lambda}+\overline{\rho}|^2}{2(k+h^\vee)}
    - \frac{|\overline\rho^\vee|^2}2
    \left(
        (k+h^\vee)
        - 2
        + \frac1{k+h^\vee}\right).
\end{equation}


Let us also note that highest weight modules are invariant under
the shifted action \cite[(57)]{Arakawa2007}:
\begin{equation}
    \begin{split}
    & \gamma_{\overline{\lambda}} = \gamma_{\overline{w}\circ\overline{\lambda}},
\\
    & \overline{w}\circ\overline\lambda
    = \overline{w}(\overline{\lambda}+\overline\rho)
    - \overline\rho.
    \end{split}
\end{equation}
In particular, $\Delta_{\overline\lambda} =
\Delta_{\overline{w}\circ\overline\lambda}$ as
\begin{equation}
    |\overline{w}\circ\overline\lambda + \overline\rho|^2
    = |\overline{w}(\overline\lambda+\overline\rho)|^2
    = |\overline\lambda+\overline\rho|^2.
\end{equation}
\end{NB}

In order to apply the result of Feigin-Frenkel to our situation later,
we shift $P^i_n$ in \eqref{eq:Vira} as $P^i_n -
(\ve_1+\ve_2)/\ve_1\ve_2 \delta_{n,0}$ (see \cite[\S19.2.5]{MO}) so
that
\begin{equation}
    \label{eq:57}
      {L}^i_n = - \frac14 \ve_1\ve_2 \sum_m \normal{P^i_m P^i_{n-m}}
  - \frac{n+1}2 (\ve_1+\ve_2) P^i_n.
\end{equation}
This is a standard embedding of the Virasoro algebra in the Heisenberg
algebra, given as the kernel of the screening operator (see
\cite[\S15.4.14]{F-BZ}). We have
\begin{equation}
    \label{eq:58}
  P^i_0 = \frac1{\ve_1\ve_2}\left( a^i - (\ve_1+\ve_2)\right)
\end{equation}
in this convention.

We modify \eqref{eq:Vira} as
\begin{equation}
    \begin{split}
  {L}^i_n &=
  \begin{aligned}[t]
    & - \frac14 \ve_1\ve_2 \sum_m
  \normal{(P^i_m - \frac{\ve_1+\ve_2}{\ve_1\ve_2} \delta_{m,0})
    (P^i_{n-m}- \frac{\ve_1+\ve_2}{\ve_1\ve_2} \delta_{m,n})
  }
  \\
  & \qquad - \frac12 (\ve_1+\ve_2) P^i_n
  + \frac{(\ve_1+\ve_2)^2}{4\ve_1\ve_2} \delta_{n,0}
  \\
  & \quad - \frac{n}2 (\ve_1+\ve_2) P^i_n
  + \frac{(\ve_1+\ve_2)^2}{4\ve_1\ve_2}\delta_{n,0}
  \end{aligned}
\\
  &=
  \begin{aligned}[t]
    & - \frac14 \ve_1\ve_2 \sum_m
  \normal{(P^i_m - \frac{\ve_1+\ve_2}{\ve_1\ve_2} \delta_{m,0})
    (P^i_{n-m}- \frac{\ve_1+\ve_2}{\ve_1\ve_2} \delta_{m,n})
  }
  \\
  & \quad - \frac{n+1}2 (\ve_1+\ve_2)
  (P^i_n - \frac{\ve_1+\ve_2}{\ve_1\ve_2} \delta_{n,0}).
  \end{aligned}
    \end{split}
\end{equation}
Therefore if we replace $P^i_n$ by $P^i_n - (\ve_1+\ve_2)/\ve_1\ve_2
\delta_{n,0}$, we get the above expression.

\begin{NB}
  We have
  \begin{equation}
    \ve_2 P^i_0 = \frac{a^i}{\ve_1} - 1 - \frac{\ve_2}{\ve_1}
    = (\alpha_i,\frac{\ba}{\ve_1} - \bar\rho) + k + h^\vee.
  \end{equation}
  Therefore we should make an identification of the highest weight
  vector as
  \begin{equation}
    \bar\lambda = \frac{\ba}{\ve_1} - \bar\rho
  \end{equation}
  in \cite[\S5.1]{Arakawa2007}.
\end{NB}

\begin{NB}
  Let me check the central charge formula again. We have
  (\cite[\S2.5.9]{F-BZ})
  \begin{equation}
    \begin{split}
    & Y(\omega_\lambda,z) = \frac12 \normal{b(z)^2} + \lambda\partial_z b(z)
\\
  =\; & \sum_n \left(\frac12 \sum_{m} \normal{b_m b_{n-m}}
    - \lambda(n+1) b_n\right) z^{-n-2}.
    \end{split}
  \end{equation}
  From the Heisenberg commutator relation we should make an
  identification
  \begin{equation}
     b_n = \frac{\sqrt{-\ve_1\ve_2}P_n}{\sqrt{2}}.
  \end{equation}

  Therefore the above is
  \begin{equation}
    \sum_n\left(-\frac{\ve_1\ve_2}4 \sum_{m} \normal{P_m P_{n-m}}
    - \lambda(n+1) \frac{\sqrt{-\ve_1\ve_2}P_n}{\sqrt{2}} \right) z^{-n-2}.
  \end{equation}
Comparing with \eqref{eq:57}, we find
\begin{equation}
  \lambda = \frac{\ve_1+\ve_2}{\sqrt{-2\ve_1\ve_2}}.
\end{equation}
Therefore the central charge is
\begin{equation}
  c_\lambda = 1 - 12 \lambda^2 = 1 + \frac{6(\ve_1+\ve_2)^2}{\ve_1\ve_2}.
\end{equation}
\end{NB}


We denote by $\Vir_i$ the Virasoro vertex subalgebra of $\Heis(\h)$
generated by $L^i_n$. \index{Vir@$\Vir_i$}

Let us introduce a modified Virasoro generator $\mL^i_n = \ve_1\ve_2
L^i_n$. \index{Linm@$\mL^i_n$}
We have
\begin{equation}\label{eq:mVir}
    \mL^i_n = - \frac14 \sum_m \normal{\widetilde{P}^i_m \widetilde{P}^i_{n-m}}
  
  
  - \frac{n+1}2 (\ve_1+\ve_2) \widetilde{P}^i_n.
\end{equation}
Hence $\mL^i_n$ is an element in $\Heis_\bA(\mathfrak h)$. We denote
the corresponding vertex $\bA$-subalgebra by $\Vir_{i,\bA}$.\index{ViriA@$\Vir_{i,\bA}$}

Note that the central charge $1 + {6(\ve_1+\ve_2)^2}/{\ve_1\ve_2}$ is
equal to that of Virasoro algebras, appearing in the construction of the
$\scW$-algebra $\scW_k(\g)$\index{Wkg@$\scW_k(\g)$|textit} as the BRST
reduction of the affine vertex algebra at level $k$, if we have the
relation
\begin{equation}
    -\frac{(\ve_1+\ve_2)^2}{\ve_1\ve_2}
  = k + h^\vee - 2 + \frac1{k+h^\vee},
\end{equation}
see \cite[\S15.4.14]{F-BZ} and \corref{cor:duality} below.

In other words,
\(
  k + h^\vee = -{\ve_2}/{\ve_1}
\)
or
\(
   -{\ve_1}/{\ve_2}.
\)
It is known that the $\scW$-algebra for type $ADE$ has a symmetry under
$k+h^\vee\leftrightarrow (k+h^\vee)^{-1}$
\cite[Prop.~15.4.16]{F-BZ}. Therefore either choice gives the same
result. We here take
\(
  k + h^\vee = -{\ve_2}/{\ve_1},
\)
see \eqref{eq:level}.

It is remarkable that the symmetry $k+h^\vee\leftrightarrow
(k+h^\vee)^{-1}$ corresponds to a trivial symmetry
$\ve_1\leftrightarrow\ve_2$ in geometry.
\begin{NB}
  Let us give more detail. The central charge is given by $c(\beta) =
  1 - 6(\beta/\sqrt{2}-\sqrt{2}/\beta)^2$ for the vertex operator
  $V_\beta(z)$. This follows from the computation in
  \cite[\S15.4.14]{F-BZ}, where we substitute $\beta =
  -\sqrt{2/(k+2)}$ to $c(k) = 1- 6(k+1)^2/(k+2)$. Or the formula
  $c(\lambda) = 1 - 12\lambda^2$ in \cite[(5.3.10)]{F-BZ} and $\lambda
  = \beta/2 - 1/\beta$ in \cite[\S15.4.14]{F-BZ}.

  Since the $i^{\mathrm{th}}$ screening operator for the $\scW$-algebra
  is given with $\beta = -\alpha_i/\sqrt{k+h^\vee}$, we get
  \begin{equation}
    c(\beta) = 1 - 6\left(\frac{2(k+h^\vee)}{(\alpha_i,\alpha_i)}
    - 2 + \frac{(\alpha_i,\alpha_i)}{2(k+h^\vee)}\right).
  \end{equation}
  This is the central charge for the $i^{\mathrm{th}}$ Virasoro
  algebra, given as the kernel of the $i^{\mathrm{th}}$ screening
  operator. Since we assume $\g$ is of type $ADE$, we get the above
  relation.
\end{NB}

\subsection{The first Chern class of the tautological bundle}

Let us explain a geometric meaning of the Virasoro generators in the
previous subsection. It was obtained in \cite[Th.~14.2.3]{MO},
based on an earlier work by Lehn \cite{Lehn} for the rank $1$ case.

Let us first consider the rank $2$ case.

Consider the Gieseker space $\Gi{2}^d$ of rank $2$ framed sheaves on
$\proj^2$ with $c_2 = d$. For $(E,\varphi)\in \Gi{2}^d$, consider
$H^1(\proj^2, E(-\linf))$. Other cohomology groups vanish, and hence
it has dimension equal to $d$ by the Riemann-Roch formula. In the ADHM
description, it is identified with the vector space $V$. When we vary
$E$, it forms a vector bundle over $\Gi{2}^d$, which we denote by
$\mathcal V$. Its first Chern class $c_1(\mathcal
V)$\index{c1V@$c_1(\mathcal V)$} can be considered as an operator on
$H^*_\TT(\Gi{2}^d)$ acting by the cup product. Then its commutator
with the diagonal Heisenberg generator, restricted to
$\IH^*_{\TT}(\Uh[SL_2]{d})$, is the Virasoro generator up to constant:
\begin{equation}\label{eq:32}
  \left.[c_1(\mathcal V), P_n^\Delta]\right|_{\IH^*_{\TT}(\Uh[SL_2]{d})}
  = n L_n,
\end{equation}
where we denote $L_n^i$ in the previous subsection by $L_n$ since
$G=SL_2$. (See \cite[Th.~14.2.3]{MO}.)
\begin{NB}
    Corrected from $2nL_n$. Feb.~18, See Misha's messages on Feb.~17.
\end{NB}

\begin{NB}
  The following was not correct. If I do not restrict to
  $\IH^*_{\TT}(\Uh[SL_2]{d})$, there are additional terms written by
$P_n^{\Delta}$. See Note 2013-12-03:

Then its commutator with the diagonal Heisenberg generator is the
Virasoro generator up to constant:
\begin{equation}
  [c_1(\mathcal V), P_n^\Delta] = 2n L_n.
\end{equation}
See \cite[Th.~14.2.3]{MO}.
\begin{NB2}
  I have calculated the commutator of \cite[(14.6)]{MO} with
  $P_n^\Delta$. It is better to check it again.
\end{NB2}
\end{NB}

Let us remark that $c_1(\mathcal V)$ is defined on {\it
  non-localized\/} equivariant cohomology groups
$\IH^*_{\TT,c}(\Gi{2}^d)$. Therefore $\mL_n = \ve_1\ve_2 L_n$
is also well-defined on non-localized equivariant cohomology groups
for $n\neq 0$.

The operator $\mL_0 = \ve_1\ve_2 L_0$ is also well-defined as it is
the grading operator (\cite[Lem.~13.1.1]{MO}).

Returning back to general $G$, we see that $\mL^i_n$ is well-defined on
\begin{equation}\label{eq:71}
  \bigoplus_d H^*_{\TT,c}(\Uh[L_i]d, \Phi_{L_i,G}(\IC(\Uh{d})))
\end{equation}
thanks to the decomposition \eqref{eq:decomp}. Namely this space is a
module over $\Vir_{i,\bA}$.

It lies in between the first two spaces in \eqref{eq:4sp}:
\begin{equation}
  \IH^*_{\TT,c}(\Uh{d}) \to
  H^*_{\TT,c}(\Uh[L_i]d, \Phi_{L_i,G}(\IC(\Uh{d})))
  \to
  H^*_{\TT,c}(\Uh[T]d, \Phi_{T,G}(\IC(\Uh{d}))).
\end{equation}
The formula \eqref{eq:mVir} relates operators $\mL^i_n$ and
$\widetilde{P}^i_n$ acting on the middle and right spaces respectively
via the second homomorphism.

\subsection{\texorpdfstring{$\scW$}{W}-algebra representation}

Let us consider the vertex algebra associated with the Heisenberg
algebra, and denote it by the same notation $\Heis(\mathfrak h)$ for
brevity.

It is regarded as a vertex algebra over $\bF$.

We have the Virasoro vertex subalgebra $\Vir_i$ corresponding to each
simple root $\alpha_i$ as in \subsecref{sec:Vir}.
\begin{NB}
  I must consider whether the zero mode must be corrected so that we
  have correct highest weight or not.

  Zero mode is corrected, and the notation $\Vir_i$ is already
  introduced above.
\end{NB}
Consider the orthogonal complement $\alpha_i^\perp$ of $\CC\alpha_i$
in $\mathfrak h$, and the corresponding Heisenberg vertex algebra
$\Heis(\alpha_i^\perp)$. It commutes with $\Vir_i$, and the tensor
product $\Vir_i\otimes \Heis(\alpha_i^\perp)$ is a vertex subalgebra
of $\Heis(\mathfrak h)$.

By a result of Feigin-Frenkel (see \cite[Th.~15.4.12]{F-BZ}), the
$\scW$-algebra $\scW_k(\g)$\index{Wkg@$\scW_k(\g)$} is identified with the intersection
\begin{equation}\label{eq:W=Vir}
  \bigcap_i \Vir_i\otimes \Heis(\alpha_i^\perp)
\end{equation}
in $\Heis(\mathfrak h)$ when the level $k$ is generic. More precisely,
$\Vir_i\otimes \Heis(\alpha_i^\perp)$ is given by the kernel of a
screening operator on $\Heis(\mathfrak h)$, and $\scW_k(\g)$ is the
intersection of the kernel of screening operators.

Now $\scW_k(\g)$ has a representation on the direct sum of localized
equivariant cohomology groups $M_\bF(\ba)$ (see \eqref{eq:localhyper}),
as a vertex subalgebra of $\Heis(\mathfrak h)$.


\begin{NB}
  At this stage, we cannot state much on the $\scW_k(\g)$-module
  structure on $M_\bF(\ba)$. It is known that the Fock space
  $M_\bF(\ba)$ is irreducible as a $\scW_k(\g)$-module, when the central
  charge and the highest weight is generic.

  I would like to say that Fourier modes of vertex operators in
  $\scW_k(\g)$ are in something like the direct product
  \begin{equation}
    \prod_{d_1,d_2} \Ext(\IC(\Uh{d_1}), \IC(\Uh{d_2})),
  \end{equation}
  probably localized one at this stage. I do not consider seriously,
  but probably we embed $\Uh{d_1}$ and $\Uh{d_2}$ into a larger space
  $\Uh{d}$ adding singularities concentrated at $0$, then consider
  $\Ext_{D^b(\Uh{d})}(\IC(\Uh{d_1}), \IC(\Uh{d_2}))$. This space should
  be independent of $d$.

  If we want to go further, we need to have a characterization of this
  kind of Ext algebra in the Ext algebra for the Heisenberg algebra. I
  do not know how to do at this moment. Therefore I only have a very
  weak statement at this point.
\end{NB}

\subsection{Highest weight}\label{sec:highest-weight}

In this subsection we explain that we can identify $\ba$ with the
highest weight of the $\scW_k(\g)$-module $M_\bF(\ba)$, where the highest
weight vector is $|\ba\rangle$.

Let us first briefly review the definition of Verma modules of the
$\scW$-algebra to set up the notation. See \cite[\S5]{Arakawa2007} for
detail.

Let $\fU(\scW_k(\g))$ be the current algebra of the $\scW$-algebra as in
\cite[\S4]{Arakawa2007}. (The finite dimensional Lie algebra is
denoted by $\bar\g$, while $\g$ is the corresponding untwisted affine
Lie algebra in \cite{Arakawa2007}.) We denote the current algebra of
the Heisenberg algebra by $\fU(\Heis(\h))$. It is a completion of the
universal enveloping algebra of the Heisenberg Lie algebra. The
embedding $\scW_k(\g)\subset\Heis(\h)$ induces an embedding
$\fU(\scW_k(\g))\to\fU(\Heis(\h))$.
\index{UWkg@$\fU(\scW_k(\g))$}
\index{UHh@$\fU(\Heis(\h))$}

We have decompositions
\(
   \fU(\scW_k(\g)) = \bigoplus_d \fU(\scW_k(\g))_d,
\)
\(
   \fU(\Heis(\h)) = \bigoplus_d \fU(\Heis(\h))_d
\)
by degree. Two decompositions are compatible under the embedding.

Let
\begin{equation}
   \fU(\scW_k(\g))_{\ge 0} \defeq \bigoplus_{d \ge 0} \fU(\scW_k(\g))_d,
   \quad
   \fU(\scW_k(\g))_{>0} \defeq \bigoplus_{d > 0} \fU(\scW_k(\g))_d.
\end{equation}
The {\it Zhu algebra\/} of $\fU(\scW_k(\g))$ is given by
\begin{equation}
  \mathfrak{Zh}(\scW_k(\g)) \defeq \fU(\scW_k(\g))_0/
  \overline{\sum_{r>0} \fU(\scW_k(\g))_{-r}\fU(\scW_k(\g))_r}.
\end{equation}
Then it is isomorphic to the center $Z(\g)$ of the universal
enveloping algebra $U(\g)$ of $\g$ (\cite[Th.~4.16.3]{Arakawa2007}).
\index{ZhuWkg@$\mathfrak{Zh}(\scW_k(\g))$}

We further identify it with the Weyl group invariant part of the
symmetric algebra of $\h$ (\cite[(55)]{Arakawa2007}):
\begin{equation}\label{eq:76}
  \mathfrak{Zh}(\scW_k(\g))\cong Z(\g) \cong S(\h)^W.
\end{equation}

We have an induced embedding $\mathfrak{Zh}(\scW_k(\g))\to
\mathfrak{Zh}(\fU(\Heis(\h)))$, where the latter is the subalgebra
generated by zero modes. We have
\begin{equation}
  \label{eq:75}
  \mathfrak{Zh}(\fU(\Heis(\h)))\cong S(\h).
\end{equation}

\begin{Lemma}\label{lem:shift}
  Under the identifications \textup{(\ref{eq:76}, \ref{eq:75})}, the
  embedding $\mathfrak{Zh}(\scW_k(\g))\to \mathfrak{Zh}(\fU(\Heis(\h)))$
  is induced by
  \begin{equation}
    h^i \mapsto h^i + (k+h^\vee),
  \end{equation}
  where $h^i$ is a simple coroot of $\h$.
\end{Lemma}

\begin{proof}
  The assertion follows from \cite[Th.~4.16.4]{Arakawa2007}, together
  with an isomorphism $\widehat t_{-\bar\rho^\vee}$ which sends the
  old Zhu algebra, denoted by
  $H^0(\mathfrak{Zh}(C_k(\bar\g)''_{\mathrm{old}}))$ there, to a new
  one $H^0(\mathfrak{Zh}(C_k(\bar\g)''_{\mathrm{new}}))$. The zero
  mode is written as $\widehat{J_i}(0)$ there.

  We can calculate $\widehat t_{-\bar\rho^\vee}(\widehat{J_i}(0)) =
  \widehat{J_i}(0) + k+h^\vee$ by formulas in \cite[bottom of
  p.276]{Arakawa2007}.
  \begin{NB}
    See note 2013-11-18.
  \end{NB}
\end{proof}

We regard $\lambda\in\h^*$ as a homomorphism $S(\h)^W\to \CC$ by the
evaluation at $\lambda+\rho$, where $\rho$ is the half sum of positive
roots of $\g$.

(It is denoted by $\gamma_{\bar{\lambda}}$ in
\cite[\S5]{Arakawa2007}.)

We further regard $\CC$ as a $\mathfrak{Zh}(\scW_k(\g))$-module by the
above isomorphism, and denote it by $\CC_\lambda$. We extend it to a
$\fU(\scW_k(\g))_{\ge 0}$-module on which $\fU(\scW_k(\g))_{>0}$ acts
trivially. Then we define
\begin{equation}
  M(\lambda) \defeq \fU(\scW_k(\g))\otimes_{\fU(\scW_k(\g))_{\ge 0}}\CC_\lambda.
\end{equation}
This is called the {\it Verma module with highest weight\/} $\lambda$.
\index{Mlambda@$M(\lambda)$}

Now we turn to our $\scW_k(\g)$-module $M_\bF(\ba)$. We identify $\h =
\operatorname{Lie}T$ with $\h^*$ by the invariant bilinear form $(\ ,\
)$. Then we have an identification
\begin{equation}
  S(\h)^W \cong
  \CC[\operatorname{Lie}T]^W = H^*_T(\mathrm{pt})^W
  \begin{NB}
    = H^*_G(\mathrm{pt})
  \end{NB}
  .
\end{equation}
We regard the collection $\ba = (a^1,\dots,a^\ell)$ as a variable in
$\operatorname{Lie}T$ by considering $a^i$ its coordinate. Hence $\ba$
has value in $\h^*$ by the above identification.

Recall that $|\ba\ra$ is the fundamental class $1\in
\IH^0_\TT(\Uh{0})$. Since the degree $d$ corresponds to an instanton
number, $\fU(\scW_k(\g))_{\ge 0}$ acts via a homomorphism
$\fU(\scW_k(\g))_{\ge 0}\to\bF_T$ induced from $\mathfrak{Zh}(\scW_k(\g))\to
\bF_T$ on $\bF_T|\ba\ra = \IH^*_\TT(\Uh{0})\otimes_{\bA_T}\bF_T$.

Hence we have a $\scW_k(\g)$-homomorphism $M(\lambda)\to M_\bF(\ba)$, sending
$1\in\CC_\lambda\subset M(\lambda)$ to $|\ba\ra\in M_\bF(\ba)$. Here we
generalize the above definition to $\lambda\colon
\mathfrak{Zh}(\scW_k(\g))\to\bF_T$.

\begin{Proposition}\label{prop:lambda}
  \textup{(1)} The highest weight $\lambda$ is given by
  \begin{equation}
    \lambda = \frac{\ba}{\ve_1} - \rho.
  \end{equation}

  \textup{(2)} $M_\bF(\ba)$ is irreducible as a $\scW_k(\g)$-module, and
  isomorphic to $M(\lambda)$.
\end{Proposition}

Note that the Weyl group action on $\ba$ corresponds to the
dot action on $\lambda$,
\(
    w\circ\lambda = w(\lambda+\rho) - \rho.
\)

\begin{proof}
  (1) Recall that our Heisenberg generators and standard generators
  are related by $\widehat h^i_n = \ve_2 P^i_n$. Then the zero mode
  acts by
  \begin{equation}
    \frac{a^i}{\ve_1} - 1 - \frac{\ve_2}{\ve_1}
    = (\alpha_i,\frac{\ba}{\ve_1} - \rho) + k + h^\vee
  \end{equation}
  thanks to \eqref{eq:58}.

  We compare this formula with a realization of $M(\lambda)$ in
  \cite[\S5.2]{Arakawa2007}. Our $\widehat h^i_n$ is $\widehat
  t_{-\bar\rho^\vee}(\widehat{J_i}(n))\in
  \fU(C_k(\bar\g)''_{\mathrm{old}})$, and $\widehat
  t_{-\bar\rho^\vee}(\widehat{J_i}(0)) = \widehat{J_i}(0) + k+h^\vee$
  as in \lemref{lem:shift}.

  Since $\widehat{J_i}(0)$ acts by $\lambda(J_i)$ on $M(\lambda)$, we
  obtain $\lambda = \ba/\ve_1 - \rho$.

  (2) It is well-known that $M(\lambda)$ is irreducible when $\lambda$
  is generic. It follows, for example, from the fact that the determinant of the
  Kac-Shapovalov form is a nonzero rational function, hence the form
  is nondegenerate if $\lambda$ is neither a zero nor  a pole. (See below
  for the Kac-Shapovalov form.)  It also means that the form is
  nondegenerate when one views $\lambda$ as a rational function like
  us. Therefore $M(\lambda)\to M_\bF(\ba)$ is injective.

  Now we compare the graded characters. The character of $M(\lambda)$
  is the same as the character of $S(t{\mathfrak h}[t])$ where $\deg(t)=1$.
  We have $M_\bF(\ba)=\bigoplus_{d\in{\mathbb N}}\IH^*(\Uh{d})\otimes\bF_T$.
  According to~\cite[Theorem~7.10]{BFG}, the character of $M_\bF(\ba)$
  (with grading by the instanton number) is the same as the character of
  $S(t{\mathfrak g}^f[t])$ where $f$ is a principal nilpotent. Since
  $\dim{\mathfrak g}^f=\dim{\mathfrak h}$, the graded characters of
  $M(\lambda)$ and $M_\bF(\ba)$ coincide.
\end{proof}


\begin{NB}
  The argument should be used again, probably in more refined manner,
  in \propref{prop:span}.
\end{NB}


\subsection{Kac-Shapovalov form}\label{sec:kac-shapovalov-form}

We shall identify the Kac-Shapovalov form on $M(\lambda)$ with a
natural pairing on $M_\bF(\ba)$ given by the Verdier duality in this
subsection.

\begin{NB}
The following paragraph added on Sep.14, 2014.
\end{NB}

Let $\sigma$ be the Dynkin diagram automorphism given by
$-\alpha_{\sigma(i)} = w_0(\alpha_i)$. We denote the corresponding
element in $\Aut(G)$ also by $\sigma$. We have an induced isomorphism
\begin{equation*}
    \varphi_\sigma\varphi_{w_0}\colon 
    \IH^*_{\TT,c}(\Uh{d})\to \IH^*_{\TT,c}(\Uh{d}),
\end{equation*}
which is $\bA_T = H^*_\TT(\mathrm{pt})$-linear if we twist the
$\bA_T$-structure on the second $\IH^*_{\TT,c}(\Uh{d})$ by composing
the automorphism $\ba\mapsto -\ba$ of $\bA_T$. This is explained in
the paragraph after \eqref{eq:44}.

Let us denote the natural perfect pairing by
\begin{equation}
  \la\ ,\ \ra\colon \IH^*_{\TT,c}(\Uh{d})\otimes_{\bA_T}
  \IH^*_{\TT}(\Uh{d})\to\bA_T,
\end{equation}
where we compose the above $\varphi_\sigma\varphi_{w_0}$ for the first
factor.
\begin{NB}
    Added, on Sep. 14, 2014.
\end{NB}
We also multiply it by $(-1)^{dh^\vee}$ as in \eqref{eq:73}. The
notation conflicts with the pairing between $U^{d,P}$ and $U^{d,P_-}$
in \subsecref{sec:pairing}. But the two pairings are closely related,
so the same notation does not give us any confusion. (See
\subsecref{sec:univ-verma-modul} for a more precise relation.)
\index{< , >@$\la\ ,\ \ra$}

By the localization theorem and \lemref{lem:centralfiber} we extend it
to a perfect pairing
\begin{equation}\label{eq:pairing}
    \la\ ,\ \ra\colon M_\bF(-\ba)\otimes M_\bF(\ba)\to\bF_T.
\end{equation}
(cf.\ \cite[\S2.6]{BraInstantonCountingI}.)
Here the highest weight of the first factor is $-\ba$ since we compose
the automorphism $\ba\mapsto -\ba$.
\begin{NB}
    The first factor is changed from $M_\bF(\ba)$ to $M_\bF(-\ba)$ on
    Sep.\ 14, 2014.
\end{NB}
\begin{NB}
We multiply it by
$(-1)^{\dim \Uh{d}/2}$ as in \eqref{eq:73}.
\end{NB}

When we localize the equivariant cohomology groups, there is no
distinction between compact support and arbitrary support. We then see
that \eqref{eq:pairing} is symmetric in the sense as in \eqref{eq:41}.
\begin{NB}
    Strictly speaking, we have two pairings one on $M_\bF(\ba)\otimes
    M_\bF(-\ba)$ and the other on $M_\bF(-\ba)\otimes
    M_\bF(\ba)$. Analog of \eqref{eq:41} is a more precise statement.
\end{NB}

We also have the pairing
\begin{equation}
  \la\ ,\ \ra\colon H^*_{\TT,c}(\Uh[T]{d}, \Phi_{T,G}(\IC(\Uh{d})))
  \otimes_{\bA_T}
  H^*_{\TT}(\Uh[T]{d}, \Phi_{T,G}(\IC(\Uh{d})))
  \to\bA_T,
\end{equation}
where we compose $\varphi_\sigma\varphi_{w_0}$ on the
first factor as above.

Since $\sigma w_0$ sends $B$ to the opposite Borel $B_-$, the above is
coming from the pairing between $H^*_{\TT,c}(\Uh[T]{d},
\Phi^{B_-}_{T,G}(\IC(\Uh{d})))$ and $H^*_{\TT}(\Uh[T]{d},
\Phi_{T,G}(\IC(\Uh{d})))$. Therefore it is a perfect pairing
thanks to Braden's isomorphism \eqref{eq:Braden}.
\begin{NB}
    Editted on Sep.~14, 2014.
\end{NB}
This pairing also extends to a pairing \eqref{eq:pairing}, which is
the same as defined above thanks to the compatibility between Braden's
isomorphism and $i^!j^!\to i^* j^!$ as in the proof of
\lemref{lem:inter=pair}.

The Heisenberg generator $P^i_n$ satisfies
\begin{equation}\label{eq:83}
  \la u, P^i_n v\ra = \la \theta(P^i_n) u, v\ra,
\end{equation}
where $\theta$ is an anti-involution on the Heisenberg algebra given
by
\begin{equation}\label{eq:81}
  \theta(P^i_n) = - P^i_{-n} - \frac{2(\ve_1+\ve_2)}{\ve_1\ve_2} \delta_{n0}.
\end{equation}
\begin{NB}
We can rewrite this as
  \begin{equation}
    \theta(\ve_2 P^i_n) = - \ve_2 P^i_{-n} -
    \frac{2(\ve_1+\ve_2)}{\ve_1} \delta_{n0}
  \end{equation}
in more standard convention. We have
\begin{equation}
  \frac{(\ve_1+\ve_2)}{\ve_1} = 1 + \frac{\ve_2}{\ve_1} =
  1 - (k+h^\vee).
\end{equation}
\end{NB}
Let us explain the reason for this formula of $\theta$. Thanks to a
standard property of convolution algebras, the diagonal Heisenberg
generator $P^\Delta_n$ in \subsecref{sec:BaraOp} was defined so that
$P^\Delta_n$ is adjoint to $P^\Delta_{-n}$. Since the intersection
pairing \eqref{eq:73} is compatible with the above one, we change $n$
to $-n$. Moreover, since $P^i_n$ is defined via the stable envelop and
we must use the opposite Borel as in \subsecref{sec:pairing}, we need
to swap $P^{(1)}_n$ and $P^{(2)}_n$ in
\subsecref{sec:heis-algebra-assoc}. Therefore we need to change the
sign of $P^i_{-n}$.

The zero mode $P^i_0$ was defined by hand as \eqref{eq:58}. We must
also change the sign of $a^i$, as the $\bA_T$-module structure is
twisted by $\ba\mapsto-\ba$ on the first factor.  
\begin{NB}
    The reason is corrected. Sep.~14, 2014
\end{NB}
Then we must correct $-P^i_0$ by $-2(\ve_1+\ve_2)/\ve_1\ve_2$.

\begin{NB}
    We need to change the sign as $a^i\to -a^i$, since the stable
    envelop is naturally replaced by its opposite. Also the sign of
    $P^i_n$ must be changed, as it is the anti-diagonal.
\end{NB}

The Virasoro generator $L^i_n$ is mapped to $L^i_{-n}$ by
$\theta$. This is clear from \eqref{eq:32}: $c_1(\mathcal V)$ is self
adjoint and $\theta(P^\Delta_n) = P^\Delta_{-n}$ as we have just
explained. It can be also checked by the formula \eqref{eq:57}.
\begin{NB}
  See note on 2013-11-27.
\end{NB}

Therefore $\theta$ preserves $\scW_k(\g)$, more precisely the associated
Lie algebra $\mathfrak L(\scW_k(\g))$ and the current algebra
$\fU(\scW_k(\g))$, thanks to \eqref{eq:W=Vir}. We have
\begin{equation}\label{eq:98}
  \la u, x v\ra = \la \theta(x)u, v\ra
\end{equation}
for $x\in\mathfrak L(\scW_k(\g))$, $u,v\in M_\bF(\ba)$.

On the other hand, $\mathfrak L(\scW_k(\g))$ has an anti-involution as in
\cite[\S5.5]{Arakawa2007}, denoted also by $\theta$.

\begin{Proposition}
  Our $\theta$ coincides with one in \cite[\S5.5]{Arakawa2007}.
\end{Proposition}

\begin{proof}
\begin{NB}
  Let us first check that the assertion is true for Verma modules.

  By \cite[Lem.~5.5.2]{Arakawa2007}, the highest weight is changed as
  $\lambda\mapsto -w_0(\lambda)$, where the action of $w_0$ is the
  ordinary one, instead of the shifted one. Then under the correspondence
  in \propref{prop:lambda}(1), we have
  \begin{equation}
    -w_0(\lambda) = -w_0(\frac{\ba}{\ve_1}) - \rho,
  \end{equation}
  as $w_0(\rho) = -\rho$. This means that the equivariant variable
  $\ba$ is replaced by $-w_0(\ba)$. Since the highest weight module is
  invariant under the Weyl group action, we can omit $w_0$, so we get
  $-\ba$. This is the same as ours. This checks that the assertion is
  true for $\mathfrak{Zh}(\scW_k(\g))$.
\end{NB}
We use the formula \cite[Prop.~3.9.1]{Arakawa2007} for the Heisenberg
vertex algebra. We follow various notation in \cite{Arakawa2007}.

Since $\widehat J_i(n)$ is a Fourier mode of the vertex operator
$Y(v,z) = \sum \widehat J_i(n) z^{-n-1}$ with $v= \widehat J_i({-1})|0
\ra$, we have
\begin{equation}
  \theta(\widehat J_i(n)) = - (e^{T^*} v)_{-n}.
\end{equation}
Here $T^*$ must be substituted by $T^*_{\mathrm{new}}$ in
\cite[(173)]{Arakawa2007}. Using
\begin{equation}
  v = \widehat J_i(-1)|0\ra = J_i({-1})|0\ra -
  \sum_{\alpha\in\Delta} \alpha(h^i) \psi_{-\alpha}(0) \psi_\alpha(-1) |0\ra
\end{equation}
(see \cite[the beginning of \S4.8]{Arakawa2007}), we can check
\begin{equation}
  e^{T^*} v = \widehat J_i(-1)|0\ra + 2(1 - (k+h^\vee))|0\ra.
\end{equation}
\begin{NB}
  See Note 2013-12-02.
\end{NB}
Therefore we get the same formula as \eqref{eq:81} under the
identification $\widehat J_i(-1) = \ve_2 P^i_n$. (This $\widehat
J_i(-1)$ is in $\fU(C_k(\bar\g)''_{\mathrm{new}})$ and we do not need
to apply $\widehat t_{-\bar\rho^\vee}$ in the proof of
\propref{prop:lambda}, as it is in $T^*_{\mathrm{new}}$.)
\end{proof}

\begin{Remark}\label{rem:Dual}
  We can identify the graded dual $D(M_\bF(\ba))$ of $M_\bF(\ba)$ with
  $M_\bF(-\ba)$
  \begin{NB}
      The second factor is changed from $M_\bF(\ba)$ to $M_\bF(-\ba)$
      on Sep.\ 14, 2014.
  \end{NB}
  via $\la\ ,\ \ra$. The graded dual has a $\scW_k(\g)$-module
  structure via $\theta$ and the formula \eqref{eq:98}. This is the
  duality functor $D$ in \cite[\S5.5]{Arakawa2007}. The isomorphism
  $D(M_\bF(\ba))\cong M_\bF(-\ba)$ respects $\scW_k(\g)$-module
  structures.

  When $\lambda$ is generic and $M(\lambda)$ is irreducible, the dual
  module $D(M(\lambda))$ is isomorphic to $M(-w_0(\lambda))$, where
  $w_0$ is the longest element in the Weyl group by
  \cite[Th.~5.5.4]{Arakawa2007}. Under the correspondence
  in \propref{prop:lambda}(1), we have
  \begin{equation}
    -w_0(\lambda) = -w_0(\frac{\ba}{\ve_1}) - \rho,
  \end{equation}
  as $w_0(\rho) = -\rho$. This means that the equivariant variable
  $\ba$ is replaced by $-w_0(\ba)$. Since the highest weight module is
  invariant under the Weyl group action, we can omit $w_0$. So the
  equivariant variable is $-\ba$ for $D(M_\bF(\ba))$. Therefore we
  have $D(M_\bF(\ba))\cong M_\bF(-\ba)$. This is what we already
  observed in a geometric way above.

  The pairing $\la \cdot,\cdot \ra$ is uniquely determined from \eqref{eq:98}
  and the normalization $\la -\ba|\ba\ra = 1$ for generic $\ba$.
  It is called the {\it Kac-Shapovalov form\/}. We thus see that the
  Poincar\'e pairing twisted by $\varphi_\sigma\varphi_{w_0}$ on
  $M_\bF(\ba)$ coincides with the Kac-Shapovalov form.
  \begin{NB}
      Corrected, Sep.~14, 2014.
  \end{NB}
\end{Remark}

\section{\texorpdfstring{$R$}{R}-matrix}\label{sec:R-matrix}

Recall that our hyperbolic restriction $\Phi_{L,G}$ depends on the
choice of a parabolic subgroup $P$. Following \cite[Ch.~4]{MO} (see
also \cite[\S1.3]{dynamicalW}), we introduce $R$-matrices giving
isomorphisms between various hyperbolic restrictions, and study their
properties. They are defined as rational functions in equivariant
variables, and their existence is an immediate corollary to
localization theorem in the previous section.

As for the usual $R$-matrices for Yangians, they satisfy the
Yang-Baxter equation and are ultimately related to the $\scW$-algebra.

As an application, we give a different proof of the Heisenberg
commutation relation (\propref{prop:HeisRel}) up to sign, which does
not depend on Gieseker spaces for $SL(3)$. We hope that this proof
could be generalized to other rank $2$ cases $B_2$, $G_2$.

Since the dependence on a parabolic subgroup is important, we denote
the hyperbolic restriction by $\Phi_{L,G}^P$ in this section.

\subsection{Definition}

Let us consider the diagram \eqref{eq:1} with respect to a parabolic
subgroup $P$. Let us consider the homomorphism in \eqref{eq:48}
\begin{equation}\label{eq:30}
  \hR_P\colon
  H^*_{\TT}(\Uh[L]d, \Phi^P_{L,G}(\scF))  \to
  H^*_{\TT}(\Uh[L]d, i^* j^*\scF)
  \cong H^*_{\TT}(\Uh{d}, \scF)
\end{equation}
for $\scF\in D^b_\TT(\Uh{d})$. This is an isomorphism over the
quotient field $\bF_T$ of $\bA_T = \CC[\operatorname{Lie}(\TT)]$.

When we want to emphasize $\scF$, we write $\hR^\scF_P$.

\begin{Definition}
  Let $P_1$, $P_2$ be two parabolic subgroups compatible with
  $(G,L)$. Let us introduce the {\it $R$-matrix\/}
  \begin{multline}\label{eq:R}
    \index{RPA@$R_{P_1,P_2}^\scF$}
    R_{P_1,P_2} = (\hR_{P_1})^{-1} \hR_{P_2}  \colon
    H^*_{\TT}(\Uh[L]d, \Phi_{L,G}^{P_2} (\scF))\otimes_{\bA_T}\bF_T
\\
    \to
    H^*_{\TT}(\Uh[L]d, \Phi_{L,G}^{P_1} (\scF))\otimes_{\bA_T}\bF_T
  \end{multline}
\end{Definition}

When we want to view $R_{P_1,P_2}$ as a rational function in
equivariant variables, we denote it by $R_{P_1P_2}(\ba)$. Dependence
on $\ve_1$, $\ve_2$ are not important, so they are omitted.

When we want to emphasize $\scF$, we write $R^\scF_{P_1,P_2}$.

From the definition, we have
\begin{equation}\label{eq:27}
  R_{P_1,P_2} R_{P_2,P_3} = R_{P_1,P_3}.
\end{equation}

\begin{NB}
The following subsection seems unnecessary.

\subsection{Adjoint operators}(cf.\ \protect\cite[\S4.4]{MO})

We have a nondegenerate pairing between
\(
    H^*_{\TT}(\Uh{d}, \scF)
\)
and
\(
    H^*_{\TT,c}(\Uh{d}, \scF^\vee).
\)

Similarly we have a natural nondegenerate pairing between
\(
  H^*_{\TT}(\Uh[L]d, \Phi^P_{L,G}(\scF))
\)
and
\(
  H^*_{\TT,c}(\Uh[L]d, \Phi^{P_-}_{L,G}(\scF^\vee)),
\)
where $P_-$ is the opposite parabolic to $P$.

Therefore the adjoint of $\hR^\scF_P$ is an operator
\begin{equation}
    \label{eq:35}
    (\hR^\scF_P)^t\colon
    H^*_{\TT,c}(\Uh{d}, \scF^\vee)
    \to
    H^*_{\TT,c}(\Uh[L]d, \Phi^{P_-}_{L,G}(\scF^\vee)).
\end{equation}
We consider $\otimes_{\bA_T}\bF_T$, then we do not need to distinguish
$H^*_{\TT,c}$ and $H^*_\TT$. Therefore it goes in the opposite
direction to \eqref{eq:30}.

\begin{Proposition}
    We have $\hR^\scF_P \circ (\hR^{\scF^\vee}_{P_-})^t = \operatorname{id}$.
\end{Proposition}

\begin{proof}
Note that \eqref{eq:35} is given by
\begin{equation}
    H^*_{\TT,c}(\Uh{d}, \scF^\vee)
    \cong H^*_{\TT,c}(\Uh[L]d, i^! j^! \scF^\vee)
    \to
    H^*_{\TT,c}(\Uh[L]{d}, i^! j^* \scF^\vee)
    \cong
    H^*_{\TT,c}(\Uh[L]{d}, i_-^* j_-^! \scF^\vee).
\end{equation}
We replace $P$ by $P_-$, $\scF$ by $\scF^\vee$ to get
\begin{equation}
    H^*_{\TT,c}(\Uh{d}, \scF)
    \cong H^*_{\TT,c}(\Uh[L]d, i_-^! j_-^! \scF)
    \to
    H^*_{\TT,c}(\Uh[L]{d}, i_-^! j_-^* \scF)
    \cong
    H^*_{\TT,c}(\Uh[L]{d}, i^* j^! \scF).
\end{equation}
Therefore the composite is given by
\begin{equation}
  (j_-\circ i_-)^! \to i_-^! j_-^* \to i^* j^! \to (j\circ i)^*.
\end{equation}
From the definition of $i_-^! j_-^* \to i^* j^!$ in \cite{Braden},
this is equal to the natural homomorphism
$(j_-\circ i_-)^! = (j\circ i)^!\to (j\circ i)^*$.

Hence we have the assertion.
\end{proof}

\begin{Corollary}
$(R^\scF_{P_1,P_2})^t = R^{\scF^\vee}_{(P_2)_-, (P_1)_-}$.
\end{Corollary}
\end{NB}

\subsection{Factorization}\label{sec:factorization-1}

Suppose that $Q_1\subset P$ be a pair of parabolic subgroups as in
\subsecref{sec:ass}. Let $M\subset L$ be the corresponding Levi
subgroups. We have $\Phi^{Q_{1,L}}_{M,L}\circ \Phi^P_{L,G} =
\Phi^{Q_1}_{M,G}$ by \propref{prop:trans}.

We further suppose that there is another parabolic subgroup $Q_2$
contained in $P$ such that the corresponding Levi subgroup is also
$M$:
\begin{equation}
    M \subset Q_1, Q_2 \subset P.
\end{equation}
Then we also have the factorization $\Phi^{Q_{2,L}}_{M,L}\circ
\Phi^P_{L,G} = \Phi^{Q_2}_{M,G}$. It is clear from the definition that
we have
\begin{equation}\label{eq:28}
  R_{Q_1,Q_2}^\scF = R_{Q_{1,L}, Q_{2,L}}^{\Phi^P_{L,G}(\scF)}.
\end{equation}

\begin{NB}
Consider the diagram
\begin{equation}
  \begin{CD}
    \Uh[Q_a]{} @>{j_a''}>> \UhP{} @>{j}>> \Uh{}
    \\
    @A{i''_a}AA @AA{i}A @.
    \\
    \Uh[Q_{a,L}]{} @>{j_a'}>> \UhL{} @.
    \\
    @A{i'_a}AA @. @.
    \\
    \Uh[M]{} @. @.
  \end{CD}
\end{equation}
Let $\Phi'_{M,G}$ denote the naive restriction $(j\circ j_a'' \circ
i_a'' \circ i_a')^*$, which is independent of the choice $a=1,2$. We
similarly write $\Phi'_{L,G}$, $\Phi'_{M,L}$.

We have
\begin{equation}
  \begin{CD}
    H^*_{\TT}(\Uh[M]d, \Phi^{Q_a}_{M,G}(\scF)) @>>>
    H^*_{\TT}(\Uh[G]d,\scF)\\
    @| @|
    \\
    H^*_{\TT}(\Uh[M]d, \Phi^{Q_{a,L}}_{M,L} \Phi^{P}_{L,G}(\scF))
    @>>> H^*_{\TT}(\Uh[M]d, \Phi'_{M,L}\Phi'_{L,G}(\scF))
    \\
    @| @AAA
    \\
    H^*_{\TT}(\Uh[M]d, \Phi^{Q_{a,L}}_{M,L} \Phi^{P}_{L,G}(\scF))
    @>>> H^*_{\TT}(\Uh[M]d, \Phi'_{M,L}\Phi^P_{L,G}(\scF)).
  \end{CD}
\end{equation}
Now the assertion is clear.
\end{NB}

Consider the case $L = T$.

Note that Borel subgroups containing a fixed torus $T$ are
parametrized by the Weyl group $W$. Let us denote by $B^w$ the Borel
subgroup corresponding to $w\in W$, where $B^e = B$ is one which we
have fixed at the beginning.

From \eqref{eq:27} $R^\scF_{B^w,B^y}$ factors to a composition of
$R$-matrices for two Borel subgroups related by a simple reflection,
i.e.,
\(
  y = w s_i.
\)

Then we choose $P = P_i^w\supset B^w, B^{w s_i}$ for the parabolic
subgroup to use \eqref{eq:28}. We have
\begin{equation}
  R^\scF_{B^w, B^{ws_i}} = R^{\Phi^{P}_{L,G}(\scF)}_{B_{1,L}, B_{2,L}},
\end{equation}
where $L$ is the Levi subgroup of $P$ and $B_{1,L}$, $B_{2,L}$ are
images of $B^w$, $B^{ws_i}$ in $L$ respectively. As $[L,L] \cong
SL(2)$, we are reduced to study the $SL(2)$ case. The $R$-matrix for
$SL(2)$ was computed in \cite[Th.~14.3.1]{MO} and will be explained in
\subsecref{sec:sl2-case}.

\subsection{Intertwiner property}\label{sec:intertwiner-property}

Let $\scF\in D^b_\TT(\Uh{d})$. We have representations of the Ext algebra
$\Ext_{D^b_\TT(\Uh{d})}(\scF,\scF)$ on two cohomology groups in
\eqref{eq:30}. This is thanks to (\ref{eq:31}, \ref{eq:38}).

Since $\hR^\scF_P$ is defined by a natural transformation of
functors, it is a homomorphism of the Ext algebra. Therefore
\begin{Proposition}\label{prop:intertwiner}
    The $R$-matrix $R^\scF_{P_1,P_2}$ is a homomorphism of modules
    over the Ext algebra $\Ext_{D^b_\TT(\Uh{d})}(\scF,\scF)$.
\end{Proposition}

\subsection{Yang-Baxter equation}

Take $L=T$ and $\scF = \IC(\Uh{d})$ in this subsection.

By \eqref{eq:43} we can map all cohomology groups in \eqref{eq:R} to
the fixed one
\(
   H^*_{\TT}(\Uh[T]d, \Phi_{T,G}^{B}(\IC(\Uh{d})))\otimes_{\bA_T}\bF_T
   \cong M_\bF(\ba)
\)
by $\varphi_w$. We conjugate the $R$-matrix as
\begin{equation}\label{eq:50}
    \varphi_{w_1}^{-1} R_{B^{w_1}, B^{w_2}} \varphi_{w_2}
    \in \End(
    H^*_{\TT}(\Uh[T]d, \Phi_{T,G}^{B}(\IC(\Uh{d})))\otimes_{\bA_T}\bF_T
    ).
\end{equation}
Remark that $H^*_\TT(\mathrm{pt})$-structures are twisted by
isomorphisms $w_1, w_2\colon \TT\to \TT$, as mentioned after
\eqref{eq:44}. In practice, we change the equivariant variable $\ba$
according to $w_1$ ,$w_2$.

Since $\hR_P$ is $\varphi_w$-equivariant, \eqref{eq:50} depends only
on $w_1 w_2^{-1}$. Moreover by \eqref{eq:27} it is enough to consider
the case $w_1 w_2^{-1}$ is a simple reflection $s_i$. Therefore we
define
\begin{equation}
  \Check R_i \defeq \varphi_{s_i}^{-1} R_{B^{s_i}, B} \varphi_{e}.
  \index{Rcheck@$\Check R_i$}
\end{equation}

By the factorization (\subsecref{sec:factorization-1}), this is the
$R$-matrix for $SL(2)$. Since we only have two chambers, \eqref{eq:27}
implies
\begin{equation}
  \Check R_i(s_i\ba) \Check R_i(\ba) = 1.
\end{equation}
We change the equivariant variable to $s_i\ba$, as it is the
$R$-matrix from the opposite Borel to the original Borel.

In the conventional notation for the $R$-matrix, we write $u = \langle
\alpha_i,\ba\rangle$ for the variable. Then $\langle \alpha_i,
s_i\ba\rangle = -u$, so this equation means the unitarity of the
$R$-matrix.

Consider $R$-matrices $\Check R_i$, $\Check R_j$. By the factorization
(\subsecref{sec:factorization-1}), we consider them as the
$R$-matrices for the rank $2$ Levi subgroup $L$ containing $SL(2)$ for
$i$ and $j$. We compute the $R$-matrix from a Borel subgroup of $L$ to
the opposite Borel by \eqref{eq:27} in two ways to get

\begin{Theorem}
  \begin{gather}
    \Check R_i(s_j\ba) \Check R_j(\ba) = \Check R_j(s_i\ba)\Check R_i(\ba)
    \quad\text{if $(\alpha_i,\alpha_j) = 0$},
\\
\label{eq:42}
    \Check R_j(s_i s_j\ba) \Check R_i(s_j\ba) \Check R_j(\ba)
    = \Check R_i(s_j s_i \ba) \Check R_j(s_i\ba)\Check R_i(\ba)
    \quad\text{if $(\alpha_i,\alpha_j) = -1$}.
  \end{gather}
\end{Theorem}

\subsection{\texorpdfstring{$SL(2)$}{SL(2)}-case}\label{sec:sl2-case}

As we mentioned earlier, it is enough to compute the $R$-matrix for
$SL(2)$, which was given in \cite[Th.~14.3.1]{MO}. We briefly recall
the result, and point out a slight difference for the formulation.

By \propref{prop:intertwiner} and the observation that the left hand
side of the formula \eqref{eq:32} is contained in the Ext algebra, we
deduce that the $R$-matrix is an intertwiner of the Virasoro
algebra. This is a fundamental observation due to Maulik-Okounkov
\cite{MO}.

The highest weight is generic, since we work over $\bF_T$. Therefore
the intertwiner is unique up to scalar, and we normalize it so that it
preserves the highest weight vector $|\ba\rangle$.

In \cite{MO} the $R$-matrix is given as an endomorphism on the
localized equivariant cohomology group of the fixed point set via the
stable envelop. On the other hand, our $R$ is an endomorphism on
$H^*_\TT(\Uh[T]d, \Phi^B_{T,G}(\IC(\Uh{d})))$. Concretely
\begin{equation}
  \Check R = \left.P_{12} R^{\text{MO}} \right|_{\text{anti-diagonal part}},
\end{equation}
where $P_{12}$ is the exchange of factors of the Fock space $F\otimes
F$, as ${s_i} = P_{12}$.

By \cite[Prop.~4.1.3]{MO} we have
\begin{equation}\label{eq:55}
  \Check R = -1 + O(\ba^{-1}),
  \qquad \ba\to \infty.
\end{equation}

\subsection{\texorpdfstring{$\GG$}{G}-equivariant cohomology}
\label{subsec:G-equiv-coh}

Recall that a larger group $\GG = G\times\CC^*\times\CC^*$ acts on
$\Uh{d}$ so that $\IC(\Uh{d})$ is a $\GG$-equivariant perverse
sheaf. Therefore we can consider $\IH^*_\GG(\Uh{d}) =
H^*_\GG(\Uh{d},\IC(\Uh{d}))$. It is related to the $\TT$-equivariant
cohomology $\IH^*_\TT(\Uh{d})$ as follows.

Let $N(\TT)$ (resp.\ $N(T)$) be the normalizer of $\TT$ (resp.\ $T$)
in $\GG$ (resp.\ $G$). Then we have forgetful homomorphisms
$\IH^*_\GG(\Uh{d})\to \IH^*_{N(\TT)}(\Uh{d})\to \IH^*_\TT(\Uh{d})$. It is
well-known that the first homomorphism is an isomorphism, as the
cohomology of $\GG/N(\TT) = G/N(T)$ is $1$-dimensional (see e.g.,
\cite{MR0423384}).

The Weyl group $W = N(T)/T$ acts naturally on $\IH^*_\TT(\Uh{d})$,
induced from the $N(T)$-action on $\Uh{d}$.

Moreover we have
\begin{equation}\label{eq:105}
    \IH^*_\GG(\Uh{d})\xrightarrow{\cong} \IH^*_{N(\TT)}(\Uh{d})
    \xrightarrow{\cong} \IH^*_\TT(\Uh{d})^W.
\end{equation}

Let us consider the following diagram
\begin{equation}\label{eq:108}
    \begin{CD}
        H^*_\TT(\Uh[T]d, \Phi^B_{T,G}(\IC(\Uh{d})))\otimes_{\bA_T}\bF_T
        @>{\hR_B}>{\cong}> \IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T
\\
    @V{\Check R_i}VV @VV{s_i}V
\\
        H^*_\TT(\Uh[T]d, \Phi^B_{T,G}(\IC(\Uh{d})))\otimes_{\bA_T}\bF_T
        @>{\hR_B}>{\cong}> \IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T,
    \end{CD}
\end{equation}
where $s_i$ is a simple reflection of the above $W$-action.

\begin{Lemma}
    The diagram \eqref{eq:108} is commutative.
    \begin{NB}
    The $R$-matrix $\Check R_i$, viewed as an endomorphism of
    $\IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T$, coincides with the simple
    reflection $s_i$ of the above $W$-action.
    \end{NB}
\end{Lemma}

\begin{proof}
    We have $\Check R_i = \varphi_{s_i}^{-1} \hR_{B_i}^{-1}\hR_{B}
    \varphi_e$. As an endomorphism of
    $\IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T$, it is replaced by
    $\hR_{B}\varphi_{s_i}^{-1} \hR_{B_i}^{-1}$, as
    $\varphi_e=\operatorname{id}$.
    \begin{NB}
        Viewing an endomorphism of
        $\IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T$ means that we take the
        conjugation by
        \begin{equation}
            \hR_B\colon H^*_\TT(\Uh[T]{d},\Phi^{B}_{T,G}(\IC(\Uh{d})))
            \otimes_{\bA_T}\bF_T
            \xrightarrow{\cong} \IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T.
        \end{equation}
    \end{NB}

    From the definition of $\hR_{B_i}$ and the commutativity of the
    diagram \eqref{eq:39}, we have $\hR_{B_i} \varphi_{s_i} =
    \varphi_{s_i} \hR_{B}$, where $\varphi_{s_i}$ in the right hand
    side is the action on $\Uh{d}$, the rightmost arrow in
    \eqref{eq:39}. Since the $W$-action is induced from
    $\varphi_{\sigma}$, the assertion follows.
\end{proof}

\begin{Proposition}\label{prop:commute}
    The Weyl group action on $M_\bF(\ba) = \bigoplus_d
    \IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T$ commutes with the $\scW_k(\g)$
    action.

    Hence $\scW_k(\g)$ acts on the $W$-invariant part
\(
    M_\bF(\ba)^W = \bigoplus_d
    \IH^*_\GG(\Uh{d})\otimes_{\bA_G}\bF_G.
    \index{FFG@$\bF_G = \CC(\ve_1,\ve_2,\ba)^W$}
\)
\end{Proposition}

\begin{proof}
    Since $\scW_k(\g)$ is the intersection of $\Vir_i\otimes
    \Heis(\alpha_i^\perp)$ (see \eqref{eq:W=Vir}), it is enough to
    show that $\Vir_i\otimes \Heis(\alpha_i^\perp)$ commutes with
    $s_i$. By the previous lemma, $s_i$ is given by the
    $R$-matrix.

    Let us first factorize the hyperbolic restriction functors
    $\Phi_{T,G}^B$, $\Phi_{T,G}^{B^{s_i}}$ as
    \begin{equation*}
        \Phi_{T,G}^B = \Phi_{T,L_i}^{B_{L_i}} \Phi_{L_i,G}^{P_i},\qquad
        \Phi_{T,G}^{B^{s_i}} = \Phi_{T,L_i}^{B^{s_i}_{L_i}} \Phi_{L_i,G}^{P_i}
    \end{equation*}
    by \propref{prop:trans}. Then the same argument as in
    \propref{prop:intertwiner} shows that $\Check R_i$ commutes with
    the action of the Ext algebra of
    $\Phi_{L_i,G}^{P_i}(\IC(\Uh{d}))$.  Since the Virasoro generators
    $\mL^i_n$ are in this Ext algebra, the first assertion follows.

    For the second assertion, we only need to check
    \begin{equation*}
        (\IH^*_\TT(\Uh{d})\otimes_{\bA_T}\bF_T)^W \cong
        \IH^*_\GG(\Uh{d})\otimes_{\bA_G}\bF_G.
    \end{equation*}
    By \eqref{eq:105} we have a natural injective homomorphism form
    the right hand side to the left. On the other hand, if $m/f$
    ($f\in\bA_T$, $m\in \IH^*_\TT(\Uh{d})$) is fixed by $W$, we have
    \begin{equation*}
        \frac{m}f = \frac1{|W|}\sum_{\sigma\in W} \frac{\sigma m}{\sigma f}
        = \frac1{|W|} \left(\prod_{\sigma\in W} \sigma f\right)^{-1}
      \sum_{\sigma\in W} \sigma m \prod_{\tau\neq\sigma} \tau f.
    \end{equation*}
This is contained in the right hand side. Therefore the above follows.
\end{proof}

\subsection{A different proof of the Heisenberg commutation relation}

Let us go back the point before \propref{prop:HeisRel}.

Let $\Tilde\alpha^{d,-}_i$ be the element defined as in
$\Tilde\alpha^{d}_i$ for the opposite Borel. Since the pairing can be
computed from the $SL(2)\cong [L_i,L_i]$ case, we already know that
\begin{equation}
  \la\Tilde\alpha^d_i, \Tilde\alpha^{d,-}_i\ra = 2d.
\end{equation}

We generalize this to
\begin{Proposition}\label{prop:diff_proof}
  \begin{equation}
    \la\Tilde\alpha^d_i, \Tilde\alpha^{d,-}_j\ra = \pm d(\alpha_i,\alpha_j).
  \end{equation}
\end{Proposition}

The following proof does not determine $\pm$, though we know that it
is $+$ by the reduction to the $SL(3)$ case and the formula
\eqref{eq:Cartan}, which has been proved via Gieseker spaces.

\begin{NB}
    Fortunately, the linear independence now follows from
    \eqref{eq:111}, and hence the proof is completed.

\textcolor{red}{
  The following proof contains a gap, as I implicitly assume that
  $\Tilde\alpha_i^d$, $\Tilde\alpha_j^d$ are linearly independent.
}
\end{NB}

\begin{proof}
  We consider the case $(\alpha_i,\alpha_j) = -1$. The proof for the
  case $(\alpha_i,\alpha_j) = 0$ is similar (and simpler).

  Let us study the leading part of Yang-Baxter equation
  \eqref{eq:42}. We consider $R$-matrices as endomorphisms of the
  space \eqref{eq:symprod}.

  By the factorization \eqref{eq:28}, we can use the expansion
  \eqref{eq:55} for $SL(2)$. Then `$-1$' in \eqref{eq:55} is replaced
  by the direct sum of $(-1)$ on $U^d_{T,L_i}\cong \h_{\algsl_2}$
  and the identity on $(U^d_{T,L_i})^\perp$ in \eqref{eq:53}. Let us
  denote it by $\Tilde s_i$.

  Since $(U^d_{T,L_i})^\perp$ is the orthogonal complement of
  $\CC\Tilde\alpha_i^{d,-}$, we have
  \begin{equation}
    \Tilde s_i(x)
    \begin{NB}
      =
      x - \frac{2\la x,\Tilde\alpha_i^{d,-}\ra}
      {\la\Tilde\alpha_i^d,\Tilde\alpha_i^{d,-}\ra} \Tilde\alpha_i^{d}
    \end{NB}
    =
    x - \la x,\Tilde\alpha_i^{d,-}\ra
    \frac{\Tilde\alpha_i^d}d, \quad
    \text{for $x\in U^d$}.
  \end{equation}
  From the Yang-Baxter equation, we have the braid relation
  \begin{equation}
    \Tilde s_i \Tilde s_j \Tilde s_i = \Tilde s_j \Tilde s_i \Tilde s_j.
  \end{equation}

  Since we are considering the $SL(3)$-case, there is the diagram
  automorphism $\sigma$ exchanging $i$ and $j$. By \lemref{lem:autg}, we
  have $\varphi_\sigma(\Tilde\alpha^d_i) = (-1)^d \Tilde\alpha^d_j$. Since
  $\varphi_\sigma$ preserves the inner product, we get
  \begin{equation}
    \la\Tilde\alpha^d_i, \Tilde\alpha^{d,-}_j\ra =
    \la\Tilde\alpha^d_j, \Tilde\alpha^{d,-}_i\ra.
  \end{equation}

  Now $\Tilde s_i$ is the usual reflection with respect to the
  hyperplane $\Tilde\alpha^{d,-}_i = 0$. Hence we conclude
  $\la\Tilde\alpha^d_i,\Tilde\alpha^{d,-}_j\ra = \pm d$.

  Note that $\Tilde\alpha_i^d = \pm \Tilde\alpha_j^d$ are excluded
  thanks to \eqref{eq:111}, which has been proved without using
  Gieseker spaces for $SL(3)$.
\end{proof}

\begin{NB}
  Let us give a little more detail. We may assume $d=1$. We know
  \begin{equation}
    \la\tilde\alpha^d_1,\tilde\alpha^{d,-}_1\ra = 2 =
    \la\tilde\alpha^d_2, \tilde\alpha^{d,-}_2\ra.
  \end{equation}
  Set
  \begin{equation}
    a \defeq \la\tilde\alpha^d_1, \tilde\alpha^{d,-}_2\ra, \qquad
    b\defeq \la\tilde\alpha^d_2, \tilde\alpha^{d,-}_1\ra.
  \end{equation}
Then reflections $\Tilde s_1$, $\Tilde s_2$ are
\begin{equation}
  \Tilde s_1(\tilde\alpha^d_1) = -\tilde\alpha^d_1, \quad
  \Tilde s_1(\tilde\alpha^d_2) = \tilde\alpha^d_2 - a\tilde\alpha^d_1, \quad
  \Tilde s_2(\tilde\alpha^d_1) = \tilde\alpha^d_1 - b \tilde\alpha^d_2, \quad
  \Tilde s_2(\tilde\alpha^d_2) = -\tilde\alpha^d_2.
\end{equation}
In the matrix form, we have
\begin{equation}
  \Tilde s_1 =
  \begin{pmatrix}
    -1 & -a \\ 0 & 1
  \end{pmatrix},\quad
  \Tilde s_2 =
  \begin{pmatrix}
    1 & 0 \\
    -b & -1
  \end{pmatrix}.
\end{equation}
Then
\begin{equation}
  \Tilde s_1 \Tilde s_2 \Tilde s_1 =
  \begin{pmatrix}
    1 - ab & 2a - a^2 b \\ b & ab-1
  \end{pmatrix},\quad
  \Tilde s_2 \Tilde s_1 \Tilde s_2 =
  \begin{pmatrix}
    ab - 1 & a \\ 2b - ab^2 & 1 - ab
  \end{pmatrix}.
\end{equation}
We deduce $ab = 1$ from $\Tilde s_1 \Tilde s_2 \Tilde s_1 = \Tilde s_2
\Tilde s_1 \Tilde s_2$.

Since we know $a = b$, we get $a = \pm 1$.

Consider the case $(\alpha_i,\alpha_j) = 0$. We have
\begin{equation}
  \Tilde s_1 \Tilde s_2 =
  \begin{pmatrix}
    ab - 1 & a \\ -b & -1
  \end{pmatrix},
\quad
  \Tilde s_2 \Tilde s_1 =
  \begin{pmatrix}
    - 1 & -a \\ b & ab - 1
  \end{pmatrix}.
\end{equation}
Therefore $a=b=0$.
\end{NB}

Once we compute the inner product, the Heisenberg relation is a
consequence of the factorization \eqref{eq:53}.

The generator $P^i_n$ is the tensor product of the Heisenberg
generator for the first factor and the identity in \eqref{eq:53}.

\begin{NB}
  The following subsection is my personal note.

\subsection{}

Let us consider the case $G=SL(3)$ and $L = T$ the maximal torus.

Then $\h = \{ (a_1,a_2,a_3) \in\CC^3 \mid a_1+a_2+a_3=0\}$ and root
hyperplanes are $a_1 = a_2$, $a_2 = a_3$, $a_1 = a_3$. We have 6
chambers. The $R$-matrix between the fundamental chamber $a_1 > a_2 >
a_3$ and the opposite one factorizes into $R$-matrices of $SL(2)$-type
in two ways, and gives the Yang-Baxter equation
\begin{multline}
    \label{eq:33}
  R_{12}(a_1-a_2) R_{13}(a_1-a_3) R_{23}(a_2-a_3)
\\
  = R_{23}(a_2-a_3) R_{13}(a_1-a_3) R_{12}(a_1-a_2).
\end{multline}

*****************************************

In \cite{MO}, the Yang-Baxter equation is given by the following form:
\begin{equation}
  R_{12}(u) R_{13}(u+v) R_{23}(v) = R_{23}(v) R_{13}(u+v) R_{12}(u),
\end{equation}
where $R_{\alpha\beta}$ is the $R$-matrix acting of the
$(\alpha,\beta)$-factors in the triple product $F_1\otimes F_2\otimes
F_3$ of the Fock spaces, by acting the identity on the remaining
factor. And it is normalized so that
\begin{equation}\label{eq:classicalr}
  R_{12}(u) = 1 + \frac{\hbar}u \br_{12} + \cdots,
\end{equation}
where $\br_{12}$ is the classical $R$-matrix. Similar for $13$,
$23$. Note that it starts with the identity $1$.

This formulation does not have an immediate generalization to other
root systems,
\begin{NB2}
  We have three terms $R_{12}$, $R_{13}$, $R_{23}$, but the braid
  relation should involve only two terms.
\end{NB2}
so let us put the exchange matrix $P_{\alpha\beta}\colon
F_\alpha\otimes F_\beta\to F_\beta\otimes F_\alpha$ and introduce
$\Check R_{\alpha\beta} = P_{\alpha\beta}R_{\alpha\beta}$.

Note first that
\begin{equation}
    \label{eq:34}
    P_{12} R_{12}(u) P_{12} = R_{12}(-u)^{-1}.
\end{equation}
See \cite[(4.11)]{MO}. Therefore
\begin{equation}
    \label{eq:36}
    \Check R_{12}(s_{12} u) \Check R_{12}(u) = 1,
\end{equation}
as $s_{12}u = -u$.

The Yang-Baxter equation is written as
\begin{equation}
  \Check R_{12}(v) \Check R_{23}(u+v) \Check R_{12}(u)
  = \Check R_{23}(u) \Check R_{12}(u+v) \Check R_{23}(v).
\end{equation}
Two equations are equivalent: The left hand side is
\begin{equation}
  \begin{split}
  & (P_{13} P_{23} \Check R_{23}(v) P_{13} P_{12})
  (P_{12}\Check R_{13}(u+v) P_{12})\Check R_{12}(u)
\\
  = \; &P_{13} R_{23}(v) R_{13}(u+v) R_{12}(u),
  \end{split}
\end{equation}
and similarly the right hand side is equal to
\begin{equation}
  P_{13} R_{12}(u) R_{13}(u+v) R_{23}(v).
\end{equation}

Now we set $\Check R_{1} = \Check R_{12}$, $\Check R_{2} = \Check
R_{23}$ as usual, and we write the Yang-Baxter equation as
\begin{equation}
  \Check R_1 (u) \Check R_2(u+v) \Check R_1(v)
  =   \Check R_2 (v) \Check R_1(u+v) \Check R_2(u).
\end{equation}
Taking the leading term of this equation in the expansion
\eqref{eq:classicalr}, we see the braid relation
\begin{equation}
  s_1 s_2 s_1 = s_2 s_1 s_2.
\end{equation}
Also note that if we replace $v=a_1 - a_2$, $u=a_2 - a_3$ as above, we
can further write
\begin{multline}
    \label{eq:37}
  \Check R_1 (\langle \alpha_1, s_2 s_1(\ba)\rangle)
  \Check R_2( \langle \alpha_2, s_1(\ba)\rangle)
  \Check R_1( \langle \alpha_1, \ba\rangle)
\\
  =   \Check R_2(\langle \alpha_2, s_1 s_2(\ba)\rangle)
      \Check R_1(\langle \alpha_1, s_2(\ba)\rangle)
      \Check R_2(\langle \alpha_2, \ba\rangle).
\end{multline}
This looks more compatible with the root system.

In our setting $s_1$, $s_2$ appear as reflections with respect to the
orthogonal decomposition \eqref{eq:decomp2}. Therefore this equation
means that two lines $U^d_{T,L_i}$ ($i=1,2$) intersect at angle
$2\pi/3$. This implies that
\begin{equation}
  (\Tilde\alpha_1^d,\Tilde\alpha_2^d) = \pm d.
\end{equation}
In order to see that this is really equal to $-d$, we need to be a
little more careful, as it involves the sign for $\Tilde\alpha_i^d$.

Let us proceed a little further. The classical $R$-matrix is given in
\cite[Th.~12.4.4]{MO}
\begin{equation}
  \begin{split}
  & \br_{12} = - \ve_1\ve_2 \sum_{n > 0}
  (P_{-n}^{(1)} - P_{-n}^{(2)})(P_n^{(1)} - P_n^{(2)}), \\
&  \br_{23} = - \ve_1\ve_2 \sum_{n > 0}
  (P_{-n}^{(2)} - P_{-n}^{(3)})(P_n^{(2)} - P_n^{(3)}).
  \end{split}
\end{equation}
Let us switch to the root system notation and write
\begin{equation}
  \br_i = -\ve_1\ve_2 \sum_{n > 0} P^i_{-n} P_n^i
  =
  \begin{cases}
      \br_{12} & \text{if $i=1$}
\\
      \br_{23} & \text{if $i=2$}
  \end{cases}
  \quad
  
  \Check\br_i = s_i \br_i.
\end{equation}
We also have
\begin{equation}
  \br_{13} = -\ve_1\ve_2 \sum_{n>0} (P^1_{-n}+P^2_{-n})(P^1_{n}+P^2_{n}).
\end{equation}

Then the classical $R$-matrix satisfies
\begin{equation}
  \begin{split}
    & s_1 \Check\br_2 \Check\br_1 + \Check\br_1 s_2\Check\br_1
    = \Check\br_2 s_1\Check\br_2 + \Check\br_2\Check\br_1 s_2,
\\
    & \Check\br_1 s_2 \Check\br_1 + \Check\br_1 \Check\br_2 s_1
    = s_2 \Check\br_1 \Check\br_2 + \Check\br_2 s_1 \Check\br_2.
  \end{split}
\end{equation}
\begin{NB2}
    We also have $\Check\br_1 s_2 s_1 = s_2 s_1\Check\br_2$,
    $\Check\br_2 s_1 s_2 = s_1 s_2\Check\br_1$, $s_1\Check\br_2 s_1 =
    s_2\Check\br_1 s_2$. And also $\Check\br_1\Check\br_2\Check\br_1 =
    \Check\br_2\Check\br_1\Check\br_2$.
\end{NB2}

We already know that how $s_i$ acts on $\Check\br_j$ since we already
compute the inner product. We move all $s_i$ to the right and get
\begin{equation}
  \begin{split}
    & (\br_{13} \br_{23} + \br_{12} \br_{23})s_1s_2s_1
  = (\br_{23}\br_{12} + \br_{23}\br_{13}) s_2 s_1 s_2,
\\
    &
    (\br_{12} \br_{23} + \br_{12} \br_{13})s_1s_2s_1
  = (\br_{13}\br_{12} + \br_{23}\br_{12}) s_2 s_1 s_2.
  \end{split}
\end{equation}
Removing $s_1 s_2 s_1 = s_2 s_1 s_2$ and get
\begin{equation}
  [\br_{13} + \br_{12}, \br_{23}] = 0 =
  [\br_{12}, \br_{23} + \br_{13}].
\end{equation}
This is the equation in \cite[(5.15)]{MO}.
\begin{NB2}
  It seems that their classical Yang-Baxter equation \cite[(4.22)]{MO}
  is not quite correct, and \cite[(5.15)]{MO} can be directly deduced
  from the Yang-Baxter equation.
\end{NB2}

We have
\begin{equation}
  \begin{split}
    & \frac1{(\ve_1\ve_2)^4}[\br_{12}, \br_{23}] =
    \sum_{n>0} (P^1_{-n} [P^1_n, P^2_{-n}] P^2_n
    - P^2_{-n} [P^2_n, P^1_{-n}] P^1_n) ,
\\
    & \frac1{(\ve_1\ve_2)^4}[\br_{13}, \br_{23}]
    \begin{aligned}[t]
    &=
    \sum_{n>0} ((P^1_{-n} + P^2_{-n})[P^1_n + P^2_n, P^2_{-n}] P^2_n
    - P^2_{-n} [P^2_n, P^1_{-n} + P^2_{-n} ](P^1_n + P^2_n))
\\
    &=
    \sum_{n>0} (P^1_{-n} [P^1_n + P^2_n, P^2_{-n}] P^2_n
    - P^2_{-n} [P^2_n, P^1_{-n} + P^2_{-n} ]P^1_n ),
    \end{aligned}
  \end{split}
\end{equation}
as $[P^1_n+P^2_n, P^2_{-n}] = [P^2_n, P^1_{-n}+P^2_{-n}]$.
Therefore the sum is
\begin{equation}
  \sum_{n > 0}
   (P^1_{-n} [2P^1_n + P^2_n, P^2_{-n}] P^2_n
    - P^2_{-n} [P^2_n, 2P^1_{-n} + P^2_n]P^1_n).
\end{equation}
If $P^i_n, P^j_n$ have the correct commutation relation, this
vanishes.
\end{NB}


\input{Whittaker}

\appendix
\input{perverse}
\input{intW}

\bibliographystyle{myamsplain}
\bibliography{nakajima,mybib,tensor2,MO}

\def\indexname{List of notations}
\printindex
\end{document}
{"config": "arxiv", "file": "1406.2381/BFN.tex"}