\section{A categorical Heisenberg double}
\label{sec:CategoricalHeisenberg}
\label{sec:heisenbergaction}
\label{ssec:heisenbergcategorification}
\subsection{The statement}
Our starting point in this section is the attempt to construct a categorification of the Fock space representation of the Heisenberg double discussed in \autoref{sec:algebraFock}. There we described how this representation is constructed using the PSH algebra structure on $A$. Presently from a self-adjoint Hopf category structure on a category $\CCC$ we want to construct a categorical action on $\CCC$ which descends to the Fock space representation of $\Heis(K(\CCC))$.

Denote the left and right adjoints of the functor $m:\CS{\fset{2}}\rightarrow \CCC$ by $\Delta^l$ and $\Delta^r$. 
For the straightforward categorification of \autoref{prop:heisenbergaction} in the case of the Fock space action we need to construct the following for the SSH category $\CCC$:\\
\begin{enumerate}
\label{list:categoricalFock}
\item Endofunctors $m_F,\Delta^r_F$ of $\CCC$, for any $F\in\CCC$, and endofunctors $(\Delta^r)^2_\Phi$ of $\CS{[2]}$ for any $\Phi\in\CS{[2]}$.
\item Isomorphisms $m_Fm_G\cong m_{F\otimes G}$ and $\Delta^r_G\Delta^r_F\cong \Delta^r_{F\otimes G}$.
\item Isomorphisms $\Delta^r_F m\cong m {(\Delta^r)^2}_{\Delta^l(F)}$
\end{enumerate}

\begin{Remark}
\label{rem:leftright}
The construction of the isomorphism $3$ as outlined in the proof of Theorem \ref{th:deltam} below forces us to use both the left and right adjoint of the multiplication, although the roles may be reversed. Briefly, this happens because we use the unit of \emph{left} adjunction to construct a certain canonical square of functors which then has a commutative \emph{right} mate. This is evocative of the construction in \cite{savageyacobi} where the authors construct lifts of the Heisenberg relations in several examples where the left and right adjoints are not isomorphic.
\end{Remark}


In the view of the \autoref{prop:heisenbergaction} this would give us

\begin{Theorem}
\label{Th:HeisCat}
Let $\CCC$ be an SSH category and denote by $\EndAd(\CCC)$ the category of endofunctors of $\CCC$ admitting adjoints.\\
The functor $\CCC\otimes\CCC^{op}\rightarrow \EndAd(\CCC)$ given by $F\boxtimes G\mapsto m_F\circ \Delta^r_G$ can be naturally constructed using the SSH structure on $\CCC$ and descends to the Fock space representation of the Heisenberg double $\Heis(K(\CCC))$.
\end{Theorem}
In the following section we will use the SSH structure on $\CCC$ to construct the adjoint functors $i_F,j^r_F,j^l_F$ and define $m_{F}=m\circ i_F, \Delta^r_F=j^r_F\circ\Delta^r, \Delta^l_F=j^l_F\circ\Delta^l$, and similarly for the extensions to $\CS{[2]}$. We will then show that the $2$-morphisms in question are constructed naturally from the SSH structure. Their invertibility is a continuity statement (see \autoref{prop:bccart}), which follows from the continuity of the SSH functor (or, equivalently, from the fact that certain squares satisfy the Beck-Chevalley condition).

In \autoref{ssec:FockP} we will apply our general construction to the case of the SSH category $\PP$ of polynomial functors and explicitly describe the $1$- and $2$-morphisms given by it in this example. 


\subsection{Constructions}
\label{ssec:CatHeisenbergConstructions}
Consider the functor $\boxtimes:\CCC^{\times 2}\rightarrow\CS{\fset{2}}$ that is the image under $\Hop$ of the coCartesian arrow $(\{1\},\{2\}) \cartesianarrow \{1,2\}$. We define for any object $F\in\CCC$, the functor $i_F$ as
\begin{align*}
i_F:\CCC & \rightarrow\CS{\fset{2}}, & i_F(X)=F\boxtimes X
\end{align*}
Note that by our requirements $i_F$ has left and right adjoints which we denote by $j^l_F,j^r_F$
\[
j^l_F,j^r_F: \CS{\fset{2}}\rightarrow \CCC
\]
We define $m_F,\Delta^l_F,\Delta^r_F$ by
\begin{align*}
m_{F}&=m\circ i_F & \Delta^r_F&=j^r_F\circ\Delta^r & \Delta^l_F&=j^l_F\circ\Delta^l 
\end{align*} 
$\Delta^r_F, \Delta^l_F$ are respectively right and left adjoint to $m_F$ for any $F \in \CCC$. 

The construction of first two isomorphisms in \eqref{list:categoricalFock} is fairly straightforward. To construct the isomorphism $m_Fm_G\cong m_{F\otimes G}$ we first note that the functor $m_{F\otimes G}$ is canonically isomorphic to the functor $m_{m_F G}$ - the isomorphism $F\otimes G \cong m_F G$ is a part of the self-adjoint Hopf structure on $\CCC$ as explained in \autoref{sec:SSHDef}. The functors $m_Fm_G$ and $m_{m_F G}$ are the images of the following morphisms in $\FinDisj$ under $\Hop$:

\begin{align*}
m_Fm_GH: & {} & \nnstik{1}{
(F) \fset{1} \ar{r} \& \fset{1} \ar{r} \& \fset{1} \ar{dr} \& \\
(G) \fset{1} \ar{dr} \& \& \& \fset{2}\ar{r} \& \fset{1} \\
\& \fset{2} \ar{r} \& \fset{1} \ar{ur} \\
(H) \fset{1} \ar{ur}
}\\
m_{m_F G}H: & {} & \nnstik{1}{
(F) \fset{1} \ar{dr} \\
\& \fset{2} \ar{r} \& \fset{1} \ar{dr} \\
(G) \fset{1} \ar{ur} \& \& \& \fset{2}\ar{r} \& \fset{1} \\
(H) \fset{1} \ar{r} \& \fset{1} \ar{r} \& \fset{1} \ar{ur}
}
\end{align*}
These morphisms form a commutative diagram in $\FinDisj$, hence we have an isomorphism $m_Fm_G \cong m_{m_F G}$ whenever we have a self-adjoint Hopf structure on $\CCC$.

Passing to adjoints we get an isomorphism
\[
\Delta^r_{F\otimes G}\cong \Delta^r_G\Delta^r_F
\]
It remains to give a categorical analog of the relation $\Delta_x m= m {(\Delta)^2}_{\Delta(x)}$. 
Let $\bar{c}$ be the arrow $(\fset{2},\fset{2})\rightarrow(\fset{4})$ in $\FinDisj$ given by the maps $\mapstack{1\mapsto 1}{2\mapsto 3},\mapstack{1\mapsto 2}{2\mapsto 4}$. For $\Phi\in\CS{\fset{2}}$ we define the functor $i_\Phi:\CS{\fset{2}}\rightarrow \CS{\fset{4}}$ by
\[i_{\Phi}(X)=\Hop(\bar{c})(\Phi,X)\]
Define adjoint functors $m^2_\Phi\dashv \Delta^2_\Phi$ as follows:
\begin{align*}
m^2_\Phi&=m^2\circ i_\Phi\\
(\Delta^r)^2_\Phi &=j^r_\Phi\circ(\Delta^r)^2
\end{align*}
then in this notation we have 

\begin{Theorem}
\label{th:deltam}
There is a canonical isomorphism 
\begin{equation}
\label{deltam}
\Delta^r_F m\cong m\circ (\Delta^r)^2_{\Delta^l(F)}
\end{equation}
coming from the SSH structure on $\CCC$.
\end{Theorem}

\begin{Corollary}
We have a canonical isomorphism \[
\Delta^r_Fm_G=\Delta^r_F m\circ i_G \cong m\circ (\Delta^r)^2_{\Delta(F)}\circ i_G
\]
\end{Corollary}

\begin{proof}
Note first that the isomorphism \eqref{deltam} can be represented as a square

\begin{equation}
\label{heisbcsquare}
\tik
\CS{2} \arrow[]{r}[above]{m}  \arrow[shorten >=0.5cm,shorten <=0.5cm,Rightarrow]{dr}[above,sloped]{\sim} & \CCC   \\
\CS{2} \arrow[]{u}{(\Delta^r)^2_{\Delta^l(F)}} \arrow[]{r}[below]{m} & \CCC \arrow[]{u}[right]{\Delta^r_F}
\tak
\end{equation}

The form of this square suggests that we should try to construct it as the right mate of a square of the form
\begin{equation}
\label{heisquare}
\tik
\CS{2} \arrow[]{d}[left]{m^2_{\Delta^l(F)}} \arrow[]{r}[above]{m} & \CCC \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{\alpha}  \arrow[]{d}[right]{m_F}\\
\CS{2} \arrow[]{r}[below]{m}& \CCC 
\tak
\end{equation}
satisfying the Beck-Chevalley condition. Our objective is to construct such square from the SSH structure on $\CCC$. We start by rewriting it as a composition of squares:
\[
\tik
\CS{2} \arrow[]{d}[left]{i_{\Delta^l(F)}} \arrow[]{r}{m} & \CCC \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{\beta}  \arrow[]{d}{i_F}\\
\CS{4} \arrow[]{r}{\overline m} \arrow[]{d}[left]{m^2}& \CS{2} \arrow[]{d}{m}\arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{\sim}\\
\CS{2} \arrow[]{r}{m} & \CCC
\tak
\]
where $\overline m$ corresponds to the maps of sets $\mapstack{1\mapsto 1}{3\mapsto 1},\mapstack{2\mapsto 2}{4\mapsto 2}$ and the lower square is the image of a Cartesian square under the SSH functor. The right mate of the lower square is therefore invertible, and hence it is enough to construct a square $\beta$ satisfying the Beck-Chevalley condition. 

The square $\beta$ can be obtained as a composition in $\CatAdTen$:
\begin{equation}
\label{bcdecomposition}
\stik{1}{
(\CCC^{\otimes 2})\ar{d}[left]{(\Delta^l(F),\Id)} \ar{r}{m} \& (\CCC) \ar{d}{(F,\Id)} \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{(\eta,\Id)}\\
(\CCC^{\otimes 2},\CCC^{\otimes 2}) \ar{d}[left]{\Hop(\bar{c})} \ar{r}{(m,m)} \& (\CCC,\CCC) \ar{d}{\Hop(c)} \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{\sim}\\
(\CCC^{\otimes 4}) \ar{r}{\overline{m}}  \& (\CCC^{\otimes 2})
}
\end{equation}

We will now prove that each of these squares satisfies the Beck-Chevalley condition, and hence also their composition does. In \autoref{heuristics} we explain how the decomposition of $\beta$ above can be obtained using the equivalence of Beck-Chevalley squares and \comma squares in $\CatAd$ from \autoref{sec:2Cartesian}.

The top square is given by two squares (as described in  \autoref{def:CatAdTen}):
\begin{align*}
\stik{1}{
\Vect\ar{d}[left]{\Delta^l(F)} \ar[equal]{r} \& \Vect \ar{d}{F} \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{\eta}\\
\CCC^{\otimes 2} \ar{r}{m} \& \CCC}
&\stik{1}{}
&
\stik{1}{\CCC^{\otimes 2}\ar[equal]{d} \ar{r}{m} \& \CCC \ar[equal]{d} \arrow[shorten >=0.4cm,shorten <=0.4cm,equal]{dl}\\
\CCC^{\otimes 2} \ar{r}{m} \& \CCC }
\end{align*}

where $\eta:F\rightarrow m\Delta^l(F)$ comes from the unit of the adjunction.

Both of these squares satisfy the left Beck-Chevalley condition. For the left square this follows from the definition of adjunction of functors, and the right square is degenerate.

The bottom square in \eqref{bcdecomposition} is the image of the Cartesian square in $\FinDisj$
\[
\stik{1}{
(\fset{2},\fset{2}) \ar{d}[left]{(\bar{c})} \ar{r}{(m,m)} \& (\fset{1},\fset{1}) \ar{d}{(c)}\\
(\fset{4}) \ar{r}{\overline{m}}  \& (\fset{2})
}
\]

under $\Hop$, where $c$ is the pair of maps $1\mapsto 1,1\mapsto 2$ and $\bar{c}$ is computed accordingly as the pullback.

We therefore have left to prove:

\begin{Lemma}
\label{lem:delignetensorbc}
The square
\[
\stik{1}{
(\fset{2},\fset{2}) \ar{d}[left]{(\bar{c})} \ar{r}{(m,m)} \& (\fset{1},\fset{1}) \ar{d}{(c)}\\
(\fset{4}) \ar{r}{\overline{m}}  \& (\fset{2})
}
\]
in $\FinDisj$ maps to a square in $\CatAdTen$ that satisfies the left Beck-Chevalley condition.
\end{Lemma}

\begin{proof}
Note that the verticals are maps presenting a set as a disjoint union of two sets, and hence are coCartesian arrows in $\FinDisj$. By assumption they must go to maps presenting the Deligne tensor, so it is enough for us to prove the following more general assertion:

Let 
\[
\stik{1}{
(A,B) \ar{d} \ar{r} \& (C,D) \ar{d} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\alpha} \\
(A\otimes B) \ar{r}  \& (C\otimes D)
}
\]
be a square in $\CatAdTen$ where the verticals are presentations of the respective Deligne tensors (and so we denote the target category as the tensor), and $\alpha$ is invertible, then it satisfies the left Beck-Chevalley condition.

The idea of the proof is that the square with $\alpha$ should be a coCartesian arrow in the category of arrows fibered over the category of arrows of $\FinSet_*$, and so its mate should be a coCartesian arrow in the opposite category. This is because taking the mate is in some sense an invertible operation. Then the invertibility of $\alpha$ implies the invertibility of the mate.

More concretely, consider an incomplete cube
\[
\stik{1}{
{} \& (A,B) \ar{rr}[name=topback,below]{} \ar{dd}[name=backleft,sloped,xshift=-1.5pc]{} \& {} \& (C,D) \ar{dd}\\
(A,B)\ar[crossing over]{rr}[name=topfront,above]{}\ar{dd}[name=frontleft]{} \ar[equals]{ur}[name=topleft,below]{} \& {} \& (C,D) \ar{dd}[name=frontright,sloped,xshift=1pc]{} \ar[equals]{ur}[name=topright,below]{} \& {}\\
{} \& (E) \ar{rr} \& {} \& (F)\\
(A\otimes B) \ar{rr}[name=bottomfront,below]{} \& {} \& (C\otimes D) 
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)}, "\alpha"]{}{}
\arrow[equal,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(topback) to[bend left] (backleft)}, "\beta"]{}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-3}{4-3}{}
\latearrow{/tikz/commutative diagrams/crossing over,/tikz/commutative diagrams/equal}{2-3}{1-4}{}
}
\]
where the top is degenerate. We note that the property of the Deligne tensor allows us, first of all, to fill in the sides with an isomorphism, and given that, to fill in the bottom in a unique way. So we get a cube
\[
\stik{1}{
{} \& (A,B) \ar{rr}[name=topback,below]{} \ar{dd}[name=backleft,sloped,xshift=-1.5pc]{} \& {} \& (C,D) \ar{dd}\\
(A,B)\ar[crossing over]{rr}[name=topfront,above]{}\ar{dd}[name=frontleft]{} \ar[equals]{ur}[name=topleft,below]{} \& {} \& (C,D) \ar{dd}[name=frontright,sloped,xshift=1pc]{} \ar[equals]{ur}[name=topright,below]{} \& {}\\
{} \& (E) \ar{rr} \& {} \& (F)\\
(A\otimes B) \ar{ur}[name=bottomleft,sloped,yshift=-0.2pc]{} \ar{rr}[name=bottomfront,below]{} \& {} \& (C\otimes D) \ar{ur}
\arrow[equal,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)}, "\alpha"]{}{}
\arrow[Rightarrow,crossing over,to path={(frontright) to[bend right] (topright)}]{}{}
\arrow[Rightarrow,to path={(bottomfront) to[bend right] (bottomleft)},"f"]{}{}
\arrow[Rightarrow,to path={(bottomleft) to[bend left] (backleft)}]{}{}
\arrow[Rightarrow,to path={(topback) to[bend left] (backleft)}, "\beta"]{}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-3}{4-3}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-1}{2-3}{}
\latearrow{/tikz/commutative diagrams/crossing over,/tikz/commutative diagrams/equal}{2-3}{1-4}{}
}
\]
Passing to the left mate, as described in \autoref{sec:BC}, we get a cube
\[
\stik{1}{
{} \& (A,B) \ar[equals]{rr}[name=topback,below]{} \ar{dd}[name=backleft,sloped,xshift=-1.5pc]{} \& {} \& (A,B) \ar{dd}\\
(C,D)\ar[equals,crossing over]{rr}[name=topfront,above]{}\ar{dd}[name=frontleft]{} \ar[dashed,red,thick]{ur}[name=topleft,below]{} \& {} \& (C,D) \ar{dd}[name=frontright,sloped,xshift=1pc]{} \ar[dashed,red,thick]{ur}[name=topright,below]{} \& {}\\
{} \& (A\otimes B) \ar{rr} \& {} \& (E)\\
(C\otimes D) \ar[dashed,red,thick]{ur}[name=bottomleft,sloped,yshift=-0.2pc]{} \ar{rr}[name=bottomfront,below]{} \& {} \& (F) \ar[dashed,red,thick]{ur}
\arrow[equal,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)}]{}{}
\arrow[Rightarrow,red,crossing over,to path={(frontright) to[bend right] (topright)}, "\beta_L"]{}{}
\arrow[Rightarrow,red,to path={(bottomfront) to[bend right] (bottomleft)},"f_L"]{}{}
\arrow[Rightarrow,red,to path={(bottomleft) to[bend left] (backleft)}, "\alpha_L"]{}{}
\arrow[Rightarrow,to path={(topback) to[bend left] (backleft)}]{}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-3}{4-3}{}
\latearrow{/tikz/commutative diagrams/crossing over,/tikz/commutative diagrams/equal}{2-1}{2-3}{}
\latearrow{/tikz/commutative diagrams/crossing over,dashed,red,thick}{2-3}{1-4}{}
}
\]
Now, if the unit/counit data is fixed, the operations of left mate and right mate are inverse to each other, so we get that $\alpha_L$ has a property that (up to choosing isomorphisms coming from the Deligne tensor property for the front and back - noting also that these faces are not touched by the mate operations) any square
\[
\stik{1}{
(C,D) \ar{d} \ar{r} \& (A,B) \ar{d} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\gamma} \\
(F) \ar{r}  \& (E)
}
\]
factors uniquely through $\alpha_L$. It is now straitforward to see that this implies $\alpha_L$ is an isomorphism.
\end{proof}

We have proved that both squares and therefore their composition satisfies the left Beck-Chevalley condition. It follows (see \autoref{app:BeckChevalley}) the composition square also satisfies the right Beck-Chevalley condition since for this square both left and right mates are defined. The right Beck-Chevalley condition is what we need to obtain the isomorphism \eqref{deltam}.
\end{proof}


\subsection{The Heisenberg double category and a conjectural generalization of the Heisenberg relation}
\label{heuristics}
\begin{Note}
Throughout this section we will write $\Delta$ for $\Delta^l$ - the left adjoint of the multiplication.
\end{Note}

In this section we outline a more conceptual  approach for constructing and working with the Beck-Chevalley squares of the form

\begin{equation}
\stik{1}{
\label{originalheissquare}
\CS{2} \arrow[]{d}[left]{i_{\Delta(F)}} \arrow[]{r}[above]{m} \& \CCC \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{\beta}  \arrow[]{d}[right]{i_F}\\
\CS{4} \arrow[]{r}[below]{\overline{m}} \& \CS{2} 
}
\end{equation}
in an SSH category $\CCC$. The key idea is to use the equivalence between Beck-Chevalley and \comma squares from \autoref{sec:2Cartesian} and think about the square $\beta$ as a pullback of its bottom right corner. This section describes the method for computing this pullback which in particular provides us with the decomposition \eqref{bcdecomposition} into Beck-Chevalley squares used in the previous section to construct a canonical 
lift of the Heisenberg double relation. Our hope is that the construction described in this section can be generalized and made precise in various contexts to produce new canonical categorifications and therefore is of independent interest. 

We start from noting that  \autoref{lem:delignetensorbc} makes it reasonable to expect that images under $\Hop$ of a certain class of Cartesian squares in $\FinDisj$ (not necessarily lying in the fiber over $\pset{1}$) are Beck-Chevalley squares. We would like to state this in the language of preserving \comma squares. We introduce a notion of \emph{active} square in $\FinDisj$: \begin{Definition}
A map $f:\pset{n}\rightarrow \pset{m}$ in $\FinSet_*$ is called \emph{active} if $f^{-1}\{*\}=\{*\}$.
\end{Definition}

Consider a square in $\FinDisj$:
\[
\stik{1}{
(A_i)_{i\in I} \ar{r}{\varphi} \ar{d}{\psi} \& (B_j)_{j\in J} \ar{d}{\tau} \\
(C_k)_{k\in K} \ar{r}{\nu} \& (D_l)_{l\in L}
}
\]
and suppose that it lies over a square of active maps in $\FinSet_*$. We say in this case that it is an \emph{active} square. The subcategory of active maps in $\FinSet_*$ is closed on pullbacks in $\FinSet_*$, so a Cartesian square of active maps is just a usual Cartesian square in the subcategory $(\FinSet_*)_{ac}\cong\FinSet$. The following gives an easy way of computing pullbacks of active maps.
\begin{Lemma} \label{lem:commafinset} 
An active square is Cartesian in $\FinDisj$, iff the square it lies over, \stik{0.7}{\pset{I}\ar{r}\ar{d} \& \pset{J}\ar{d}\\
\pset{K}\ar{r} \& \pset{L}}, is Cartesian in $\FinSet_*$, and all the squares of sets involved in it are Cartesian in $\FinSet$.
\end{Lemma}

\begin{proof}
We may assume that $L$ has one point, and then we have an explicit construction for a Cartesian square with given bottom right corner. Namely, we consider sets $T_{k,j},k\in K, j\in J$ given by $T_{k,j}=B_j\times_D C_k$ with the obvious projections to $B_\bullet,C_\bullet$. The lemma then follows immediately.
\end{proof}

\begin{Conjecture} \label{prop:generalcomma} \label{conj:activecomma}
An SSH functor $\Hop$ takes active \comma squares in $\FinDisj$ to (necessarily active) \comma squares in $\CatAdTen$.
\end{Conjecture}

We believe that the an ideologically correct proof of this general statement should follow from factoring $\Hop$ through a functor to a category of algebra objects in $\CatAd$. This kind of approach necessitates a construction and careful consideration of such a framework which is outside the scope of this article. 

Instead, we restrict ourselves to proving this for the specific square needed for the construction of Heisenberg relations in \autoref{lem:delignetensorbc}. However, writing the statement in this general form allows us to generalize our approach as follows:

Restrict $\Hop$ over the subcategory of active maps in $\FinSet_*$ to get a functor 
\[
\Hop_{ac}:(\FinDisj)_{ac}\rightarrow(\CatAdTen)_{ac}
\]
and consider a bicategory $\Heis(\Hop)$  defined by the following comma square of bicategories

\begin{equation}\label{heisenbergsquare}
\stik{1}{\Heis(\Hop) \ar{r} \ar{d}[left]{p} \& \CatAd \ar{d}{i} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{} \\
(\FinDisj)_{ac} \ar{r}{\Hop_{ac}} \& (\CatAdTen)_{ac}
}
\end{equation}
where $\CatAd\xrightarrow{i}(\CatAdTen)_{ac}$ is the obvious imbedding. We call the bicategory $\Heis(\Hop)$ the \emph{categorical Heisenberg double} of an SSH functor $\Hop$.

The objects and 1-morphisms in $\Heis(\Hop)$ can be described as follows
\begin{itemize}
\item Objects are pairs $(C,S=(S_k)_{k\in K})$ with $C\in\CatAd$ and $S\in\FinDisj$, along with a morphism $i(C)\xrightarrow{a}\Hop(S)$.
\item A morphism $(C,S)\rightarrow(D,T)$ is a pair $(C\xrightarrow{\varphi}D,S\xrightarrow{f}T)$, and a square in $\CatAdTen$\[
\stik{1}{
i(C) \ar{d} \ar{r}{i(\varphi)} \& i(D) \ar{d} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\alpha}\\
\Hop(S) \ar{r}[below]{\Hop(f)} \& \Hop(T)
}
\]
We call $\varphi$ the \emph{functor part} of the morphism, and $\alpha$ the \emph{transformation part} of the morphism.
\item 2-morphisms are defined in the obvious way.
\end{itemize}

In particular, let $(\CCC,(S_i),a)$ be an object in $\Heis(\Hop)$ where the map $a$ sits over an injection $\pset{1}\rightarrow\pset{n}$ which sends $1$ to $k$. Let $\Hop(S)=(\Hop(S_1),\ldots,\Hop(S_n))$, then $a$ amounts to a choice of elements in all but the $k^{\text{th}}$ category, and a map $\CCC\rightarrow \Hop(S_k)$.

Both functors forming the bottom right corner of the pullback square \eqref{heisenbergsquare} preserve \comma squares (whose definition closely resembles that of a weighted limit) hence it is reasonable to expect

\begin{Conjecture}
\label{conj:heiscomma}
The functors out of $\Heis(\Hop)$ preserve \comma squares.
\end{Conjecture}

Let us use the above to provide a canonical construction of the square $\beta$ from the beginning of this section. We will compute the pullback of a preimage of its lower right corner in $\Heis(\Hop)$. If we believe the conjectures in this section, the resulting $\comma$ square in $\Heis(\Hop)$ - which is a canonical object - should then project to an \comma square in $\CatAd$, which is the same as a Beck-Chevalley square.

We have the obvious preimages for the maps $i_F,\overline{m}$ (abbreviating $(\fset{k})$ for $\Hop(\fset{k})$):
\begin{align*}
\stik{1}{
(\fset{1}) \ar{d}[left]{(F,\Id)} \ar{r}{i_F} \& (\fset{2}) \ar[equal]{d} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}\\
(\fset{1},\fset{1}) \ar{r}[below]{\Hop(c)} \& (\fset{2})
}
& \stik{1}{}
& \stik{1}{
(\fset{4}) \ar[equal]{d} \ar{r}{\overline{m}} \& (\fset{2}) \ar[equal]{d} \ar[equal,shorten <=0.7em,shorten >=0.7em]{dl}\\
(\fset{4}) \ar{r}{\overline{m}} \& (\fset{2})
}
\end{align*}
where $c=(1\mapsto 1,1\mapsto 2)$ and in the left square the 2-morphism is the unitor morphism $\Id\circ(f\circ g)\rightarrow  f\circ g$.

So we consider the following bottom right corner in $\Heis(\Hop)$:
\[
\stik{1}{
{} \& {} \& {} \& (\fset{1},\fset{1}) \ar{dd}\\
{}  \& {} \& (\fset{1}) \ar{dd}[right,yshift=1pc]{i_F} \ar{ur}{(F,\Id)} \& {}\\
{} \& (\fset{4}) \ar{rr}[xshift=-2pc]{\overline{m}} \& {} \& (\fset{2})\\
(\fset{4}) \ar{rr}{\overline{m}} \ar[equal]{ur} \& {} \& (\fset{2}) \ar[equal]{ur}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\]
And we want to complete it to an \comma square. By \autoref{conj:heiscomma} the back face should come from an \comma square in $\FinDisj$, so first we compute the product in the back (as in \autoref{lem:commafinset}) to get
\[
\stik{1}{
{} \& (\fset{2},\fset{2}) \ar{dd}[left]{(\bar{c})} \ar{rr}{(m,m)}\& {} \& (\fset{1},\fset{1}) \ar{dd}{(c)}\\
{}  \& {} \& (\fset{1}) \ar{dd}[right,yshift=1pc]{i_F} \ar{ur}{(F,\Id)} \& {}\\
{} \& (\fset{4}) \ar{rr}[xshift=-2pc]{\overline{m}} \& {} \& (\fset{2})\\
(\fset{4}) \ar{rr}{\overline{m}} \ar[equal]{ur} \& {} \& (\fset{2}) \ar[equal]{ur}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\]
where $\bar{c}=(\mapstack{1\mapsto 1}{1\mapsto 3},\mapstack{1\mapsto 2}{1\mapsto 4})$. Note that the back face is the image of an active \comma square under $\Hop$, so is an \comma square in $\CatAd$ by \autoref{conj:activecomma}.

To proceed we need to make an ansatz regarding \comma squares:

Suppose that we have completed what we have to a cube representing an \comma square in $\Heis(\Hop)$. 

\[
\stik{1}{
{} \& (\fset{2},\fset{2}) \ar{dd}[left,yshift=-1pc]{(\bar{c})} \ar{rr}{(m,m)}\& {} \& (\fset{1},\fset{1}) \ar{dd}{(c)}\\
(?) \ar{dd}[left]{?} \ar[crossing over]{rr}[xshift=1pc]{?} \ar{ur}{?}  \& {} \& (\fset{1}) \ar{dd}[right,yshift=1pc]{i_F} \ar{ur}{(F,\Id)} \& {}\\
{} \& (\fset{4}) \ar{rr}[xshift=-2pc]{\overline{m}} \& {} \& (\fset{2})\\
(\fset{4}) \ar{rr}{\overline{m}} \ar[equal]{ur} \& {} \& (\fset{2}) \ar[equal]{ur}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\]

The right side of the cube represents a morphism in $\Heis(\Hop)$ which has a fully faithful functor part and an invertible transformation part. This is reminiscient of a monomorphism, and so it seems reasonable to conjecture that the left face, being a "pullback" of the right along the bottom, should also be of this type. Given this, and that by \autoref{conj:heiscomma} the front face is an \comma square in $\CatAd$, an extension of the pasting lemma for pullbacks to the \comma case would give us that the top square must also be \comma.

Therefore, to proceed we compute the product on the top. Consider the corner 
\[
\stik{1}{
{} \& (\fset{1}) \ar{d}{(F,\Id)} \\
(\fset{2},\fset{2}) \ar{r}{(m,m)} \& (\fset{1},\fset{1})
}
\]
Since the element in the bottom right is over $\pset{2}$, a square with this corner is given by two squares, with corners:
\begin{align*}
\stik{1}{
{} \& (\fset{1}) \ar[equal]{d} \\
(\fset{2}) \ar{r}{(m)} \& (\fset{1})
}
&\stik{1}{}
&
\stik{1}{
{} \& () \ar{d}{(F)} \\
(\fset{2}) \ar{r}{(m)} \& (\fset{1})
}
\end{align*}
and we should compute the product separately for each one.
The first corner fits in a degenerate square, and any degenerate square is \comma. 
Let us compute the pullback for the second corner:

Suppose we have a square with this bottom right corner, i.e.
\[
\stik{1}{
(\CCC_i) \ar{r} \ar{d}{\varphi} \& () \ar{d} \\
(\fset{2}) \ar{r}{m} \& (\fset{1})
}
\]

First, the map $\varphi$ must sit over a map which sends everything to the base point, in order for the whole square to sit over a commutative square. So we may assume that $(C_i)=()$ and that $\varphi$ is the selection of an element of $\fset{2}$. In all, we can assume that the square is of the form 
\[
\stik{1}{
() \ar{d}[left]{X} \ar[equal]{r} \& () \ar{d}{F} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\gamma}\\
(\fset{2}) \ar{r}{m} \& (\fset{1})
}
\]

i.e. a map $F\xrightarrow{\gamma}mX$. If this square is to be \comma, the map $\gamma$ must satisfy the property that for any map $F\xrightarrow{\psi}mY$ we have a unique map $X\xrightarrow{\tau}Y$ such that $\psi=m(\tau)\circ\gamma$, i.e. that $X$ together with $\gamma$ is a local left adjoint for $m$ at $F$. Since $m$ has a global left adjoint, this implies that $X$ and $\gamma$ are canonically isomorphic to $\Delta(F)$ and $F\xrightarrow{\eta}m\Delta(F)$.


So now we have
\[
\stik{1}{
{} \& (\fset{2},\fset{2}) \ar{dd}[left,yshift=-1pc]{(\bar{c})} \ar{rr}{(m,m)}\& {} \& (\fset{1},\fset{1}) \ar{dd}{(c)}\\
(\fset{2}) \ar[crossing over]{rr}[xshift=1pc]{m} \ar{ur}{(\Delta(F),\Id)}  \& {} \& (\fset{1}) \ar{dd}[right,yshift=1pc]{i_F} \ar{ur}{(F,\Id)} \& {}\\
{} \& (\fset{4}) \ar{rr}[xshift=-2pc]{\overline{m}} \& {} \& (\fset{2})\\
(\fset{4}) \ar{rr}{\overline{m}} \ar[equal]{ur} \& {} \& (\fset{2}) \ar[equal]{ur}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\]
and we complete to a cube by setting the missing 1-morphism to be the composition $(\bar{c})\circ(\Delta(F),\Id)=i_{\Delta(F)}$; the left face to be the unitor isomorphism to the composition (to be compatible with our ansatz); and the front face to be the unique 2-morphism that makes the cube commute as in \autoref{app:cubecommutes}.

In all we have the cube which should represent an \comma square in $\Heis(\Hop)$
\[
\stik{1}{
{} \& (\fset{2},\fset{2}) \ar{dd}[left,yshift=-1pc]{(\bar{c})} \ar{rr}{(m,m)}\& {} \& (\fset{1},\fset{1}) \ar{dd}{(c)}\\
(\fset{2}) \ar{dd}[left]{i_{\Delta(F)}} \ar[crossing over]{rr}[xshift=1pc]{m} \ar{ur}{(\Delta(F),\Id)}  \& {} \& (\fset{1}) \ar{dd}[right,yshift=1pc]{i_F} \ar{ur}{(F,\Id)} \& {}\\
{} \& (\fset{4}) \ar{rr}[xshift=-2pc]{\overline{m}} \& {} \& (\fset{2})\\
(\fset{4}) \ar{rr}{\overline{m}} \ar[equal]{ur} \& {} \& (\fset{2}) \ar[equal]{ur}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\]
Now, the front face is exactly the square $\beta$ and assuming \autoref{conj:heiscomma}, it is an \comma square in $\CatAd$. 

Independently of the conjectures, from the commutativity of the above cube we get the decomposition \eqref{bcdecomposition}, used to lift the Heisenberg relation in the previous section.In this sense one can consider the material in this chapter as providing a heuristic method for constructing such canonical squares.


\subsection{A categorification of the Fock space action of the infinite dimensional Heisenberg algebra}
\label{ssec:FockP}
In this section we consider the example of the SSH category $\PP$ of polynomial functors and apply our general construction of a categorical Heisenberg double to this example. This gives us a canonically constructed categorification of the Fock space representation of infinite dimensional Heisenberg algebra, which is the Heisenberg double associated to the Grothendieck group $K(\PP)$.

Consider the coCartesian arrow $c=(c_1,c_2):(\{1\},\{2\}) \cartesianarrow \{1,2\}$ in $\FinDisj$. Then, from the definition of the functor $\Hop$ giving the SSH structure on $\PP$, for any $F\in\PP$ the functor $i_F: \PP \rightarrow\PPS{\fset{2}}$ is given by the formula:
\begin{align*}
i_FX(V)=F(c_0^*V)\otimes X(c_1^*V)
\end{align*}
where $X \in \PP, V\in \Sh(\{1,2\})$.Its left and right adjoints are both given by the functor $j_F$:
\begin{align*}
j_F:\PPS{[2]} & \rightarrow\PP, &  j_F\Phi(V)=\Hom_\PP(F,\Phi(\boxtimes_c(-,V))
\end{align*}
where $\boxtimes_c(W, V)$ is the sheaf defined by $c_0^*\boxtimes_c(W, V)=W,c_1^*\boxtimes_c(W, V)=V$. 
The functors $m_F,\Delta_F$ are defined by composing
\begin{align*}
m_{F}&=m\circ i_F & \Delta_F&=j_F\circ\Delta
\end{align*} 

Note that these functors are biadjoint, so we will not distinguish between $\Delta^r_F$ and $\Delta^l_F$ and just write $\Delta_F$. 

The isomorphisms  
\begin{align*}
m_Fm_G\cong m_{F\otimes G} \\
\Delta_G\Delta_F\cong \Delta_{F\otimes G}
\end{align*}
come from the associator of the monoidal structure on the category $\Vect$. As we saw in section \autoref{ssec:CatHeisenbergConstructions}, the non-trivial part of the isomorphism $\Delta_F m\cong m \Delta^2_{\Delta(F)}$ is given by the mate of the top square in \eqref{bcdecomposition}. We want to check that the mate is invertible in the case of $\PP$. Since we are now working with a concrete example, we will just compute the mate using the formula from \autoref{sec:BC} and check it is an invertible morphism.

Consider the arrow $\bar{c}: (\fset{2},\fset{2})\rightarrow(\fset{4})$ in $\FinDisj$ given by the maps $\bar{c}_0:\mapstack{1\mapsto 1}{2\mapsto 3},\bar{c}_1:\mapstack{1\mapsto 2}{2\mapsto 4}$ and let
\begin{align*}
i_{\Phi}:\PPS{\fset{2}} & \rightarrow\PPS{\fset{4}} &i_{\Delta \Phi}\Psi(V)&=\Phi(\bar{c_0}^*V)\otimes \Psi(\bar{c_1}^*V)\\
j_{\Phi}:\PPS{\fset{4}} & \rightarrow\PPS{\fset{2}}, &  j_\Phi\Omega(V)&=\Hom(\Phi,\Omega(\boxtimes_{\bar{c}}(-,V))))
\end{align*}
The functors $i_\Phi, j_\Phi$ are biadjoint. Using the adjunction morphisms $\eta:\Id\rightarrow j_Fi_F$ and $\epsilon:i_{\Delta(F)}j_{\Delta(F)}\rightarrow\Id$, the mate of the top square in \eqref{bcdecomposition} is given by:
\begin{equation}
\label{rightmateofdoubleimage}
mj_{\Delta(F)}\xrightarrow{\eta}j_Fi_Fmj_{\Delta(F)}\xrightarrow{\alpha}j_F\overline{m} i_{\Delta(F)}j_{\Delta(F)}\xrightarrow{\epsilon}j_F\overline{m}
\end{equation}

The maps $\eta$ and $\epsilon$ are given by the following formulas:
\begin{equation}\label{i1j1runit}
\begin{gathered}
\begin{aligned}
G(V) \xrightarrow{\eta} j_Fi_F(G)(V) &=\Hom(F,F(-)\otimes G(V)) \\ &=\End(F) \otimes G(V)\end{aligned} \\  
x \mapsto \Id \otimes x\
\end{gathered}
\end{equation}
\begin{equation}\label{i1j1rcounit}
\begin{gathered}
i_{\Delta F}j_{\Delta F}(\Omega)(V)=\Delta F(\bar{c}_0^*V)\otimes \Hom (\Delta F,\Omega(\boxtimes_{\bar{c}}(-,\bar{c}_1^*V))) \xrightarrow{\epsilon} \Omega(V) \\ 
x\otimes\tau \mapsto \tau_{\bar{c}_0^*V}(x)
\end{gathered}
\end{equation}
(note that for any sheaf $V\in\Sh(\fset{4})$ the sheaf $\boxtimes_{\bar{c}}(\bar{c}_0^*V, \bar{c}_1^*V)$ is canonically isomorphic to $V$)

Take $\Omega\in\PPS{4},U\in\Sh(\fset{1})$ and let $a:\fset{2}\rightarrow \fset{1}$. Then 
\begin{equation}
mj_{\Delta(F)}\Omega(U)=\Hom(\Delta F,\Omega(\boxtimes_{\bar{c}}(-,\bar{c}_1^*U)))
\end{equation}

We want to take $x\in mj_{\Delta(F)}\Omega(U)$ and follow it through the map. Denote by $W$ some vector space and $v$ an element in $F(W)$, then
\begin{gather*}
\eta(x)_W=(v\mapsto v\otimes x)\\
\alpha(\eta(x))_W=(v\mapsto \eta_L(v)\otimes m(x)) \\
\epsilon(\alpha(\eta(x)))_W=(v\mapsto m(x)(\eta_L(v)))
\end{gather*}

where $\eta_L$ is the map $F\rightarrow m\Delta F$ from the adjunction $(m,\Delta)$ (see \autoref{app:pullpushadjsets} for details) and $m(x)$ is the application of $m$ to $x$, which is a map $m\Delta F$ to $m\Omega(\boxtimes_{\bar{c}}(-,\bar{c}_1^*U))$. Altogether, we have computed that the mate takes the map $x$ to $m(x)$ precomposed with the unit $\eta_L$, but this is exactly the construction of the isomorphism \[
\Hom(\Delta F,\Omega(\boxtimes_{\bar{c}}(-,\bar{c}_1^*U)))\xrightarrow{\sim}\Hom(A,m\Omega(\boxtimes_{\bar{c}}(-,\bar{c}_1^*U)))
\]
for the adjoint pair $m,\Delta$, so it is invertible.
{"config": "arxiv", "file": "1406.3973/categoricalheisenberg.tex"}