\begin{document}
\begin{abstract}
We study the geometry of (generalized) twistor triangles $\triangle J_1J_2J_3$ in the period domain
of compact complex tori of  complex dimension $2n$ by the means of the representation theory 
of algebras (of real dimension 8) generated by the complex structure operators $J_1,J_2,J_3$. 
Considering the period domain as a homogeneous space for $G=GL_{4n}(\RR)$,
we introduce on it a $G$-invariant pseudometric and define
pseudometric invariants, helping us (generally) to distinguish triangles from a reasonably defined class up to $G$-equivalence.
\end{abstract}
\setcounter{tocdepth}{1}
\sloppy
\maketitle
\tableofcontents
\section{Introduction}

We call a manifold $M$ of a real dimension $4m$ a {\it hypercomplex} manifold, if  there exist 
(integrable) complex structures
$I,J,K$ on $M$ satisfying the quaternionic relations $$I^2=J^2=K^2=-Id, IJ=-JI=K.$$
The ordered triple $(I,J,K)$ is called a {\it hypercomplex structure} on $M$.


A Riemannian $4m$-manifold $M$ with a metric $g$ is called {\it hyperk\"ahler} with 
respect to $g$ (see \cite[p. 548]{Hitchin}), 
if there exist complex structures $I$, $J$ and $K$ on $M$, such that 
$I,J,K$ are covariantly constant and 
are isometries of the tangent bundle $TM$ with respect to $g$, satisfying
the above quaternionic relations.

We call the  ordered triple $(I,J,K)$ of such complex structures {\it a hyperk\"ahler
structure on $M$ compatible with $g$}. 

Every hyperk\"ahler manifold $M$ naturally carries the  underlying hypercomplex structure
and is thus hypercomplex.
A hypercomplex 

 structure $(I,J,K)$ gives rise to a sphere $S^2$ of complex structures on $M$:
$$S^2=\{aI+bJ+cK| a^2+b^2+c^2=1\}.$$ 


We call the family $\mathcal M=\{(M,\lambda)| \lambda \in S^2\}  \rightarrow S^2$ a {\it twistor family over the twistor sphere $S^2$}. The family $\mathcal M$ can be endowed with a complex structure,
so that it becomes a complex manifold and the fiber $\mathcal M_\lambda$ is biholomorphic to
the complex manifold $(M,\lambda)$, see \cite[p. 554]{Hitchin}.


The well known examples of compact hyperk\"ahler manifolds are even-dimensional complex tori and irreducible holomorphic symplectic manifolds ({\it IHS manifolds}).
We recall that an IHS manifold is a simply connected compact K\"ahler manifold $M$  with $H^0(M,{\Omega}_M^2)$ generated by
an everywhere non-degenerate holomorphic 2-form $\sigma$.


It is known that in the period domain of 
an IHS manifold any two periods can be connected
by a path of twistor lines arising from the corresponding hyperk\"ahler structures, see 
the work of Verbitsky, \cite{Verb-Torelli}, and its short exposition in  \cite{Bourbaki}. 


The twistor path connectivity of each of the two connected components of the period domain
of complex tori was proved in \cite{Twistor-lines}. 

Let us recall the construction of this period domain.
Let   $V_\RR$ be a real vector space of real dimension
$4n$.  The compact complex tori of complex dimension $2n$,
as real smooth manifolds, are quotients $V_\RR/\Gamma$ of $V_\RR$ by a lattice $\Gamma$
and the complex structure of such a torus is given by an endomorphism $I\colon V_\RR \rightarrow V_\RR, I^2=-Id$.
Following \cite{Twistor-lines}, we denote the period domain of compact complex tori of complex dimension $2n$ by $Compl$, as a
set of imaginary endomorphisms of $V_\RR$ it is diffeomorphic to the orbit $G\cdot I$, where 
$G=GL(V_\RR)=GL(4n,\RR)$ acts via the adjoint action, $g \cdot I =g(I)=gIg^{-1}$. We have $G\cdot I\cong G/G_I$,
where $G_I$ is the adjoint action stabilizer of $I$, $G_I=GL((V_\RR,I))\cong GL_{2n}(\CC)$. 
As the $Ad\, G$-action 
is the only action we will be dealing with, we will simply refer to it as the $G$-action. 
A twistor sphere $S=S(I,J)\subset Compl \subset End\,V_\RR$ determines an embedding 
of the algebra of quaternions $\HH \hookrightarrow End\,V_\RR$, we call the image of such embedding
{\it the algebra of quaternions associated with $S$}. We define $G_\HH\subset G$ to be the pointwise stabilizer of  $\HH$ in $End\,V_\RR$, or, what is the same, of the sphere $S$.
 We obviously have $G_\HH=G_I\cap G_J$. 

The $G$-action on $Compl$ naturally extends to the $G$-action on subsets of $Compl$,
in particular, on twistor lines and on configurations of those.


 The period domain $Compl$ consists
of two connected components, corresponding to two connected components of
$G$. We have the embedding of $Compl$ into the Grassmanian $Gr(2n,V_\CC)$ of $2n$-dimensional
complex subspaces in $V_\CC=V_\RR\otimes \CC$ given by $$Compl \ni I \mapsto (Id-iI)V_\RR \in Gr(2n,V_\CC),$$ which maps $Compl$ biholomorphically onto an open subset of $Gr(2n,V_\CC)$, whose
complement is the real-analytic locus $\mathcal L_\RR =\{U \in Gr(2n,V_\CC)|\, U\cap V_\RR \neq \{0\}\}$
of $2n$-dimensional complex subspaces in $V_\CC$ having nontrivial intersection with $V_\RR$.
This locus $\mathcal L_\RR$ is of real codimension 1 in $Gr(2n,V_\CC)$ and it cuts $Gr(2n,V_\CC)$ into two
pieces each of which is the corresponding component of $Compl$, the components correspond,
non-canonically, to the connected components $GL^+(V_\RR),GL^-(V_\RR)$ 
of $GL(V_\RR)$. 


For further discussion of twistor lines and the configurations of those we 
need the following lemma, which summarizes technical results proved in  \cite{Twistor-lines}, see also  \cite{Generalized-twistor-lines}.
\begin{lem}
\label{Two-point-lemma}
Let $S_1,S_2\subset Compl \subset End\,V_\RR$ be any two  twistor lines. 
If the intersection $S_1\cap S_2$ contains  points that are linearly independent as vectors in 
$End\,V_\RR$, 
then $S_1=S_2$. In particular, any two distinct twistor lines $S_1,S_2\subset Compl$
are either disjoint or $S_1\cap S_2$ consists of a pair of antipodal points $\pm I$. 
If $S\subset Compl$ is a twistor line and $I_1,I_2\in S$ are linearly independent,
then $G_{I_1}\cap G_{I_2}=G_\HH$, where $\HH\subset End\,V_\RR$ is the algebra of quaternions associated
with $S$. 
\end{lem}

This lemma tells us that every twistor line $S$ is uniquely determined by  any two non-proportional points
$I_1,I_2$ in $S$, allowing us to write $S=S(I_1,I_2)$ (here $I_1,I_2$ need not anticommute). Note,
that it is not true that any two points $I_1,I_2 \in Compl$ belong to a twistor sphere (this will actually be explained later).

Let $I_1,I_2,I_3$ be complex structure operators on  $V_\RR$, belonging to the same twistor sphere $S\subset Compl$  


  and linearly independent as vectors in $End\,V_\RR$. 
We are not assuming here that $I_1,I_2$ and $I_3$ satisfy quaternionic identities.


By Lemma \ref{Two-point-lemma} we have $G_\HH=G_{I_1}\cap G_{I_2}=G_{I_1}\cap G_{I_2}\cap G_{I_3}$. 

The main result of \cite{Twistor-lines} is that the triple intersection of submanifolds
$G_{I_1}/G_{\HH}, G_{I_2}/G_\HH$ and $G_{I_3}/G_\HH$ in $G/G_\HH$
at $eG_\HH$ is transversal (\cite[Prop. 3.5]{Twistor-lines}).


The transversality
at  $eG_\HH$ means that for  every triple $(g_1,g_2,g_3)$ with $g_{j} \in G_{I_j}, j=1,2,3$, close enough to $e \in G$ 
(and thus determining $g_jG_\HH \in G/G_\HH$ 
close enough, in the respective topology, to $eG_\HH$),
we have that $g_{1}g_{2}g_{3} \in G_\HH$ if and only if $g_{j} \in G_\HH$ for every $j=1,2,3$.
Speaking informally, the groups $G_{I_1},G_{I_2},G_{I_3}$  are independent (modulo $G_\HH$) near 
$G_{I_1}\cap G_{I_2}\cap G_{I_3}=G_\HH$. 


One may ask if there exist general (not necessarily close to $e$)

$g_{1} \in G_{I_1},g_{2}\in G_{I_2},g_{3}\in G_{I_3}$,

 such that
we have the relation $g_{1}g_{2}g_{3} \in G_\HH$, and, in general, 
one can ask
what is the whole fiber $m^{-1}(G_\HH)$ of the multiplication map
$$m\colon G_{I_1}\times G_{I_2}\times G_{I_3} \rightarrow G, (g_1,g_2,g_3)\mapsto g_1g_2g_3,$$
where the Cartesian product is merely  a product of sets.
\begin{comment}
Here we consider the fiber over $G_\HH$ rather than over a single element of it just because
further we will involve the twistor sphere $S$ into our quest for relations between $G_{I_j}$'s
and every element of $G_\HH$ acts on $S$ trivially, so it is convenient not to choose
a distinguished element of it. 
\end{comment}


Again, informally, this question is about how ``independent'' 
the subgroups $G_{I_j}\subset G$ are in global and what kind of relations of the specified type may arise. 
We answer this question in Theorem \ref{Theorem-relation}, where we give an explicit description of the fiber  $m^{-1}(G_\HH)$. On our way to the formulation of Theorem \ref{Theorem-relation} we need to develop
some geometry of twistor lines related to $m^{-1}(G_\HH)$.  
\begin{comment}
First of all, we trivially have that $G_\HH\times G_\HH \times G_\HH \subset m^{-1}(G_\HH)$.

If any of the $g_i$'s in the question satisfying $g_1g_2g_3 \in G_\HH$ belongs to $G_\HH$, then
actually all of them must be in $G_\HH$, so that
$m^{-1}(G_\HH) \cap G_\HH \times G_\HH \times G_\HH=G_\HH\times G_\HH \times G_\HH$. 
Indeed,
for example, from $g_1g_2g_3(S)=S$ we can easily see that if $g_2\in G_\HH$ then $I_3\in g_2(S)=(g_1g_2)^{-1}(S)\ni I_1$,
implying $g_1,g_3\in G_{I_1,S}\cap G_{I_3,S}=G_\HH$, other cases are analogous.

There is a less trivial subset in $m^{-1}(G_\HH)$ 
 if we require $g_j\in G_{I_j,S}\setminus G_\HH$, see part 2) of Theorem \ref{Theorem-relation},
which describes the part of the fiber $m^{-1}(G_\HH)\cap G_{I_1,S} \times G_{I_2,S} \times G_{I_3,S}$
lying in the subgroup  $G_{I_1,S}G_{I_2,S}G_{I_3, S}\subset G, G_{I_1,S}G_{I_2,S}G_{I_3, S}
\cong SO(3)\times G_\HH$, 
which, modulo $G_\HH$,
is just $SO(3)$ and there is a clear geometric picture behind three independent rotations in $SO(3)$
whose product is the identity element.


If any of the elements $g_1,g_3$ in the product $g_1g_2g_3$ such that $g_1g_2g_3\in G_\HH$
takes $S$ to itself then each of the three takes $S$ to itself, and so we are again dealing with part 2)
of Theorem \ref{Theorem-relation}. 

If we look for $g_1,g_2,g_3$ such that $g_1g_2g_3\in G_\HH$ and only $g_2(S)=S$, then,
as we must have $g_1g_2g_3(I_3)=g_1(g_2(I_3)) =I_3$ we clearly have a contradiction
with the fact that the only points of $S$ that are taken to $S$ by $g_1$ are $\pm I_1$.


Now we want to describe the most nontrivial part of the fiber, that is,
$m^{-1}(G_\HH)\cap (G_{I_1}\setminus G_{I_1,S}) \times (G_{I_2}\setminus G_{I_2,S}) \times (G_{I_3}\setminus G_{I_3,S})$. 

\end{comment}
\subsection{Triangles} 


Let us consider a more general relation $g_{1}g_{2}g_{3}(S)=S$, that is,
$(g_1,g_2,g_3) \in m^{-1}(G_S)$, where $G_S$ is the stabilizer in $G$ of $S$
as a set. 

Assume there is a triple $(g_{1},g_{2},g_{3}) \in m^{-1}(G_S)$ 

 


 and this triple is sufficiently nontrivial,
in the sense that the twistor lines $S, g_{2}(S)$ and $g_{2}g_{3}(S)=g_{1}^{-1}(S)$ 
are all distinct.

Then these twistor lines $S, g_{2}(S)$ and $g_{2}g_{3}(S)=g_{1}^{-1}(S)$
are consecutive, that is, their pairwise intersections are nonempty, and those are actually the pairs
of points (listed in the respective order) $\{\pm I_2\}=S \cap g_{2}(S), \{\pm g_2(I_3)=\pm g_{2}(g_{3}(I_3))\} =g_{2}(S \cap g_{3}(S))=g_{2}(S)\cap g_{1}^{-1}(S)$ and 
$\{\pm I_1\}=g_{1}^{-1}(S)\cap S$. 


Thus we obtain a triangle, formed by the three consecutive twistor lines.

\vspace*{0.5cm}
\input{triangle1.pic}
\vspace*{-12.5cm}
\begin{center}
Picture 1: Obtaining a twistor triangle from $g_{1}g_{2}g_{3}(S)=S$.
\end{center}


On the opposite, given three consecutive twistor lines, we can find three complex structures $I_1,I_2,I_3 \in S$,
where $S$ is one of these lines, and elements $g_{j}\in G_{I_j},j=1,2,3$, 
such that $(g_1,g_2,g_3)\in m^{-1}(G_S)$, that is, $S=g_{1}g_{2}g_{3}(S),g_{2}(S),g_{2}(g_{3}(S))$ constitute our triple of consecutive lines. Indeed, let $S_1,S_2,S_3$ be the consecutive twistor lines. 

Choose $I_1\in S_1 \cap S_3$ and $I_2\in S_1 \cap S_2$. As we know from \cite{Twistor-lines}
 or \cite{Generalized-twistor-lines}, 
the $G$-action stabilizer $G_{I}\subset G$ of $I\in Compl$ acts transitively on the set of twistor spheres containing $I$, so that we can find elements $g_2\in G_{I_2}$ such that $S_2=g_2(S_1)$
and $g_1 \in G_{I_1}$ such that $S_3=g_1^{-1}(S_1)$. 
 Next, choose $J_3\in S_2 \cap S_3$ and set $I_3=g_2^{-1}(J_3)\in S$, choose 
$f_3\in G_{J_3}$ such that $f_3(S_2)=S_3=g_1^{-1}(S_1)$. Then setting $g_3=g_2^{-1}f_3g_2$
we get that $g_3\in G_{I_3}$ and $g_1g_2g_3(S_1)=g_1g_2\cdot g_2^{-1}f_3g_2(S_1)=
g_1f_3(S_2)=g_1(g_1^{-1}(S_1))=S_1$, so that $(g_1,g_2,g_3)\in m^{-1}(G_{S_1})$, as required.
 
 


Further we give a rigorous definition of 
a (generalized) twistor triangle and relate to every twistor triangle

a certain real associative algebra $\mathcal H$ of dimension $8$. 


The  properties of the algebras $\mathcal H$ are formulated in  Theorem \ref{Algebraic-theorem}. 
The representation theory of such algebras, summarized in Theorem \ref{Main-theorem}, 
will allow us to prove Theorem \ref{Theorem-relation}. 
\begin{comment}
Now, given any element
The description of the fiber $m^{-1}(G_\HH)$ is given by Theorem \ref{Theorem-relation}, the proof of the
theorem  relies on classification of  the representations $\rho\colon \mathcal H \rightarrow End\,V_\RR$
of the algebra $\mathcal H$, arising from any triple $(g_1,g_2,g_3)\in m^{-1}(G_\HH)$, up to $G$-equivalence. Every $(g_1,g_2,g_3)\in m^{-1}(G_\HH)$ The classification of representations is given in Theorem \ref{Main-theorem}
and is based on the algebraic properties of $\mathcal H$, expressed in terms of its 
set of generators corresponding to the ``vertices'' of the twistor triangle,  proved in Theorem \ref{Algebraic-theorem}. 
\end{comment}


 


\subsection{Generalized triangles}It is natural to generalize
the notion of a twistor triangle, in order to proceed with the classification of
representations of the associated algebras $\mathcal H$. Let us explain this
generalization. 


It is easy to see, and this is explained in \cite{Generalized-twistor-lines} that two non-proportional complex structures
$J_1,J_2$ belong to the same, uniquely defined, twistor sphere $S$ if and only if $J_1J_2+J_2J_1=2\alpha Id$
for some $\alpha \in \RR$ such that $|\alpha|<1$ (such $J_1$ and $J_2$ generate the subalgebra  
$\HH\subset End\, V_\RR$ associated with $S$). This fact provides a natural generalization
of the notion of a twistor sphere, namely,  if $J_1J_2+J_2J_1=2\alpha Id$ for some general 
$\alpha \in \RR$ and $J_1\neq \pm J_2$, then there is a canonically defined 
complex-analytic curve $S(J_1,J_2)$
in $Compl$ containing $\pm J_1,\pm J_2$, it is the intersection of the subalgebra in $End\,V_\RR$, generated by $J_1,J_2$ with $Compl\subset End\,V_\RR$.  


In case of $|\alpha|\geqslant 1$ this curve is a non-compact 
curve that we will call {\it a non-compact twistor line}, as opposed to the earlier considered compact twistor lines. If we do not specify whether a twistor line is compact or not, we can talk about it as 
a {\it generalized twistor line}.
The geometry of such curves is studied in \cite{Generalized-twistor-lines}, 
where it is shown, in particular, that the (analytic or Zariski topology) closures of non-compact twistor lines 
in $Gr(2n,V_\CC)\supset Compl$ are $\mathbb{P}^1$'s. 


We generalize accordingly  the notion of a twistor triangle, namely we call 
an ordered triple  of complex structures $(J_1,J_2,J_3)$
{\it a (generalized) twistor triangle $\triangle J_1 J_2 J_3$}, if $J_1J_2+J_2J_1=2\alpha Id$,
$J_2J_3+J_3J_2=2\beta Id$ and 
$J_1J_3+J_3J_1=2\gamma Id$ for some $\alpha, \beta,\gamma \in \RR$ (with no restrictions
on their absolute values now). It is natural  not to require that the sides $S(J_1,J_2), S(J_2,J_3), S(J_3,J_1)$
are all distinct.  Two triangles $\triangle J_1 J_2 J_3$ and $\triangle K_1 K_2 K_3$ are said to be
$G$-{\it equivalent}, if there is $g \in G$ such that $K_l=g(J_l),l=1,2,3$, we emphasize the
importance of the order of vertices. 


Note, that an ordered triple of distinct, pairwise intersecting twistor lines $S_1,S_2,S_3$
does not determine uniquely a twistor triangle, as the intersection
of any two twistor lines $S_i\cap S_j$ consists of two distinct points, so that we indeed need to specify
an ordered triple of points, not only a triple of sides.
\begin{comment}
This generalization is indeed geometric, in the sense that  the pairs $J_k,J_l$
determine certain complex analytic curves in $Compl$ containing the respective complex structures.
These curves need not be compact for general $\alpha,\beta,\gamma$, the geometry of the curves
and their compactifications in $Gr(2n,4n)\supset Compl$ is analyzed in Section                             
 \ref{Generalized-twistor-lines}. Just as in the case of compact twistor curves (spheres)
we have that every generalized twistor line is uniquely determined by any pair of its
nonproportional complex structures $J_1\neq\pm J_2$, that is, we will also use the notation $S(J_1,J_2)$
for them.
\end{comment}

\begin{comment}
In this text we are going to describe some basic geometry of generalized twistor triangles in $Compl$.
The ideal goal would be to classify all generalized twistor triangles up to $G$-action, but this goal has
not been achieved.  What we do here is a {\it rough} classification up to $G$-action of
generalized triangles from a certain ($G$-invariant) class, which properly contains the class of compact twistor triangles. ``Rough'' here means that
\end{comment}

\begin{comment}
There is a naturally defined pseudo-riemannian metric on $Compl$, invariant under 
the $G$-action.
We are going to construct invariants distinguishing the triangles
up to $G$-action, except a certain special case. There will be shown to exist three major types
of triangles: the triangles of  the first one are uniquely characterized by lengths of their sides. While the triples of lengths 
are not sufficient for the rest two types, they still work well enough in the sense that
for each triple of lengths there is only a finite (depending on $n$ only) number of $G$-inequivalent triangles
with these lengths. One of the rest types requires adding one more numerical
invariant to distinguish $G$-inequivalent triangles. And the remaining one is the above mentioned 
special one, when the additional invariant does not help (we actually think it is possible to show
that there are no  natural metric invariants arising from our pseudometric distinguishing
the cylindrical triangles). 
\end{comment}

Now we introduce the algebra $\mathcal H$, associated to $\triangle J_1 J_2 J_3$,
\begin{multline}
\label{H-definition}
\mathcal H=\mathcal H_{\alpha,\beta,\gamma}=\mathcal H(e_1,e_2,e_3)
=\langle e_1,e_2,e_3\,|\,e_1^2=e_2^2=e_2^2=-1, \\
e_1e_2+e_2e_1=2\alpha, 
e_2e_3+e_3e_2=2\beta,e_3e_1+e_1e_3=2\gamma\rangle.
\end{multline}

This algebra has real dimension $8$. 
By $\mathcal H(J_1,J_2,J_3)$ we denote the homomorphic image of $\mathcal H(e_1,e_2,e_3)$ in $End\, V_\RR$ under the homomorphism $e_i\mapsto J_i, i=1,2,3$. 

The problem of classification of the twistor triangles up to $G$-action is equivalent to the problem of 
 classification of all representations $\rho\colon \mathcal H \rightarrow End\, V_\RR$
up to $G$-isomorphism ($G$-equivalence). 


It is easy to study the irreducible representations of $\mathcal H$ (and thus arbitrary representations) 
when  the 8-dimensional algebra $\mathcal H$ contains the algebra of quaternions $\HH$
as a subalgebra, as then $\dim_\HH \mathcal H=2$ and it is really easy to write down 
the (left or right) regular representations for such $\mathcal H$. 

This is the case, as we have seen, for example, when one of $|\alpha|,|\beta|,|\gamma|$ is strictly less than 1.
In fact, as we will see later, $\mathcal H$ may contain $\HH$ even when none of these strict inequalities holds. 

The above mentioned restricted class of triangles is defined to be the set of those triangles $\triangle J_1 J_2 J_3$ for which the respective algebra $\mathcal H_{\alpha,\beta,\gamma}$ 
(and hence $\mathcal H(J_1,J_2,J_3)$)
contains $\HH$ as a subalgebra, we call such algebra $\mathcal H=\mathcal H_{\alpha,\beta,\gamma}$ {\it quaternionic}. 


\begin{comment}
It is easy to study the decomposition of the regular (left or right) representation of the algebra $\mathcal H$
into irreducible representations in the considered case when 
one of $|\alpha|,|\beta|,|\gamma|$ is strictly less than 1,
because in that case $\mathcal H$ contains the algebra of quaternions $\HH$, 

so that the only proper $\mathcal H$-submodules of $\mathcal H$ (giving rise
to the subrepresentations of $\rho_{reg}$) are the left (or right) ideals in $\mathcal H$,
which are cyclic modules over $\HH$.
\end{comment}

The classification of representations of quaternionic $\mathcal H$, and, thus,
of the triangles from the restricted class, 

 is the content of  Theorem \ref{Main-theorem}.  
Theorem \ref{Main-theorem} relies heavily on Theorem \ref{Algebraic-theorem}, which 
specifies the necessary and sufficient conditions on $\alpha,\beta,\gamma$ in order for  $\mathcal H_{\alpha,\beta,\gamma}$
to contain $\HH$, and proves, in particular,
that, up to isomorphism, there are just three quaternionic algebras $\mathcal H$. 


After all we return to the original question of describing the 

fiber $m^{-1}(G_\HH)$ which is done, as we said  earlier, 
in Theorem \ref{Theorem-relation}.


Let us now get to introducing a machinery, which allows to formulate the ``quaternionic restrictions''
on $(\alpha,\beta,\gamma)$ in a convenient, compact, form.

\subsection{The pseudometric}
\label{subsec-pseudometric}
In this subsection we introduce a pseudometric on $Compl$, which will later be used for defining
the pseudometric invariants of our twistor triangles, that will help us in distinguishing them up to 
$G$-action.  
We define a symmetric bilinear form on $End\, V_\RR$ 
by  $$(A,B)=-\frac{1}{4n}tr(AB).$$
This form is clearly positive on  the vectors corresponding to complex structure operators, that is, vectors
 in $Compl$. 
If we choose  an 
 inner product on $V_\RR$, then there 
is the decomposition $End\,V_\RR =A\oplus S$, where $A$ and $S$ are subspaces of antisymmetric and, respectively, symmetric
operators. The decomposition is orthogonal with respect to the form $(\cdot,\cdot)$, the form $(\cdot,\cdot)$ is positive on $A$, negative on $S$, 

 so that it  has the signature $(8n^2-2n,8n^2+2n)$ (we write the signature of a non-degenerate form as
a pair $(n_+,n_-)$). 
This form is clearly $G$-invariant. 


Let us choose a complex structure operator $I\in Compl$, orthogonal with respect to the inner form on $V_\RR$. 
 Identifying $T_ICompl$ with the subspace of operators, anticommuting with $I$, and further decomposing
$T_ICompl\cong A_I\oplus S_I$ into the respective subspaces of antisymmetric and symmetric operators,
 we can see that $(\cdot,\cdot)|_{T_ICompl}$
has signature $(4n^2-2n, 4n^2+2n)$. As $G$ acts transitively on $Compl$ we see that the restriction of $(\cdot,\cdot)$
to $T_{I_1}Compl$ for every $I_1 \in Compl$ has the same signature, thus $(\cdot,\cdot)|_{TCompl}$ determines a pseudo-riemannian metric
on $Compl$. Note that for a tangent vector $J \in T_ICompl$, $JI=-IJ$ and $J^2=-Id$ we have that 
$(J,J)=-\frac{1}{4n}tr(J^2)>0$,
thus the tangent 2-plane $T_IS$ to an arbitrary compact twistor spheres $S=S(I,J)$, which is explicitly written as 
$T_IS=\langle J,K\rangle$ for $K=IJ$,  is positive  with respect to this pseudo-riemannian metric.

For the case of a generalized twistor line determined by a pair $J_1\neq \pm J_2$ of complex structures,
$J_1J_2+J_2J_1=2\alpha Id$,
we have that the restriction of our indefinite metric to the plane $\langle J_1,J_2\rangle_\RR$
is positive definite if and only if $|\alpha| < 1$, thus, in the latter case
we can define  $\cos \, \sphericalangle J_1J_2 = \frac {(J_1,J_2)}{\sqrt{(J_1,J_1)}\sqrt{(J_2,J_2)}}=-\alpha$.
\begin{comment}
\begin{rem}
\label{Validity-cosine-remark}
The latter equality justifies that the restriction of our pseudo-riemannian metric to the plane $\langle J_1,J_2\rangle_\RR$
is positive definite, thus the definition of  $\cos \, \sphericalangle J_1J_2$ as $\frac {(J_1,J_2)}{\sqrt{(J_1,J_1)}\sqrt{(J_2,J_2)}}$ is indeed valid. 
\end{rem}
\end{comment}


 For $J_1,J_2$ determining a compact twistor sphere 
 the angle $\sphericalangle J_1J_2$ is the length of one of two arcs of the great circle in $S(J_1,J_2)$
through $J_1$ and $J_2$. This is easy to see using the parametrization $t \mapsto e^{tJ}J_1e^{-tJ}$ of the great circle 
in $S$ containing $J_1, J_2$, where
$J \in S$ is a complex structure

 anticommuting with both $J_1,J_2$). 

If $|\alpha|\geqslant 1$ the twistor line determined by $J_1,J_2$ is non-compact, in this case the restriction $(\cdot,\cdot)|_{\langle J_1,J_2\rangle}$
is indefinite, being degenerate precisely when $|\alpha|=1$.

\subsection{The invariants and the formulations of the results}


\begin{comment}
\begin{nota}
We denote the (generalized) twistor triangle with vertices $\pm J_1,\pm J_2, \pm J_3$  by $\triangle J_1J_2J_3$.
\end{nota}

\begin{defi}
For every triangle $\triangle J_1J_2J_3$ the triple of traces

$$T(\triangle J_1J_2J_3)=(-\cos\sphericalangle (J_2,J_3), -\cos\sphericalangle (J_3,J_1))=
(\frac{1}{4n}Tr\, J_1J_2, \frac{1}{4n}Tr\, J_2J_3, \frac{1}{4n}Tr\, J_3J_1),$$
 is called {\it the triple invariant}  or {\it the $T$-invariant of $\triangle J_1J_2J_3$}.

The quadruple 
$$Q(\triangle J_1J_2J_3)=(\frac{1}{4n}Tr\, J_1J_2, \frac{1}{4n}Tr\, J_2J_3, \frac{1}{4n}Tr\, J_3J_1, 
\frac{1}{4n}Tr\, J_1J_2J_3)$$
is called {\it the quadruple invariant}, or {\it the $Q$-invariant of $\triangle J_1J_2J_3$}.
\end{defi}


 


The triple invariant and the quadruple invariant are obviously invariant under the
$Ad \,G$-action $g\colon J\mapsto gJg^{-1}$ on the twistor triangles. 
\end{comment}


For a generalized twistor triangle $\triangle J_1J_2J_3$ introduce $$T(\triangle J_1J_2J_3):=

\left(\frac{1}{4n}Tr\, J_1J_2, \frac{1}{4n}Tr\, J_2J_3, \frac{1}{4n}Tr\, J_3J_1\right)=(\alpha,\beta,\gamma).$$
If the triangle $\triangle J_1 J_2 J_3$ is compact, then, as follows from the above discussion, the triple
$T(\triangle J_1 J_2 J_3)$ has a clear geometric meaning, namely 
$T(\triangle J_1 J_2 J_3)=(-\cos\sphericalangle J_1 J_2, -\cos\sphericalangle J_2 J_3, -\cos\sphericalangle J_3 J_1)$. 
\begin{comment}
Introduce  the following $G$-invariant of the triangle $\triangle J_1J_2J_3$, $$|\triangle J_1J_2J_3|=\alpha^2+\beta^2+\gamma^2+2\alpha\beta\gamma-1,$$
the $G$-invariance will be shown later. 
We will also denote the corresponding function on triples by $f$, 
$f(\alpha,\beta,\gamma)=\alpha^2+\beta^2+\gamma^2+2\alpha\beta\gamma-1$.
\end{comment}


\begin{comment}
Let $\triangle J_1J_2J_3$ be a twistor triangle with $T(\triangle J_1J_2J_3)=(\alpha,\beta,\gamma)$.
An important algebraic structure associated to $\triangle J_1J_2J_3$ is the subalgebra
$\mathcal H(J_1,J_2,J_3) \subset End\, V_\RR$ generated by $J_1,J_2,J_3$.

 The vertices $J_1,J_2,J_3$ (up to choosing among $\pm J_k$)
satisfy equalities $$J_1J_2+J_2J_1=2\alpha Id, J_2J_3+J_3J_2=2\beta Id, J_3J_1+J_1J_3=2\gamma Id.$$
\end{comment}
Formula (\ref{H-definition}) introduces a real associative algebra 
$\mathcal H$ 

of dimension 8 on three letters $e_1,e_2,e_3$.  


In general, a set of generators $f_1,f_2,f_3$ of the algebra $\mathcal H$ that are imaginary units, that is, $f_i^2=-1$, $i=1,2,3$, satisfying the relations $f_1f_2+f_2f_1=2\alpha^\prime,f_2f_3+f_3f_2=2\beta^\prime,f_3f_1+f_1f_3=2\gamma^\prime$

 is called a {\it standard set of generators corresponding to $(\alpha^\prime,\beta^\prime,\gamma^\prime)$} and we say that the triple $(\alpha^\prime,\beta^\prime,\gamma^\prime)$ {\it represents}
$\mathcal H$.  
Note that algebra $\mathcal H$ may be represented by sufficiently different triples, 

so that we may have an isomorphism $\mathcal H=\mathcal H_{\alpha,\beta,\gamma}\cong 
\mathcal H_{\alpha^{\prime},\beta^{\prime},\gamma^{\prime}}$  
for the triple $({\alpha^{\prime},\beta^{\prime},\gamma^{\prime}})$ not reducing to permutations of
the original triple $(\alpha,\beta,\gamma)$ and scalings of the kind $\alpha \mapsto -\alpha$. 


Introduce  a bilinear form on $\mathcal H$, $$q(u,v)=\frac{1}{\dim_\RR\, \mathcal H}Tr(\rho_{reg}(uv)),{\dim_\RR\, \mathcal H}=8,$$
where $\rho_{reg}\colon \mathcal H \to End\,\RR^8$ is the (left or right) regular representation of $\mathcal H$, 
and set $Q_{\alpha,\beta,\gamma}=q|_{\langle e_1,e_2,e_3\rangle}$. The relations of $\mathcal H$ 
easily imply  that the matrix
of $Q_{\alpha,\beta,\gamma}$ in the  basis $e_1,e_2,e_3$ is
\[
Q_{\alpha,\beta,\gamma}=\left(\begin{array}{ccc}
-1 & \alpha & \gamma \\
\alpha & -1 & \beta \\
\gamma & \beta & -1 \\
\end{array}\right).
\]
 
We will also denote such $Q_{\alpha,\beta,\gamma}$  by $Q$. 
We have 
$\det\, Q_{\alpha,\beta,\gamma}=\alpha^2+\beta^2+\gamma^2+2\alpha\beta\gamma-1$. 


The triangle $\triangle \rho_{reg}(e_1)\rho_{reg}(e_2)\rho_{reg}(e_3) \subset  End\, \RR^8,i=1,2,3$  
will be denoted for short by $\triangle e_1 e_2 e_3$.


If the algebra homomorphism $\mathcal H(e_1,e_2,e_3) \rightarrow \mathcal H(J_1,J_2,J_3),e_i\mapsto J_i,$ is 
not an isomorphism, we say that the triangle $\triangle J_1J_2J_3$ is {\it degenerate}.


Introduce the following algebras $\mathcal H(\varepsilon)$ for $\varepsilon=-1,0,1$, 
$$\mathcal H(\varepsilon)=\langle i,j,c|\, i^2=j^2=-1,ij+ji=0, ic=ci,jc=cj,c^2=\varepsilon\rangle\cong \HH\oplus \HH\cdot c$$ with the center $Z(\mathcal H(\varepsilon))=\langle 1,c \rangle$.
The algebra $\mathcal H(-1)$ is classically known as the {\it algebra of biquaternions},
$\mathcal H(1)$ is known as the {\it algebra of split-biquaternions}, and
$\mathcal H(0)$ is  known as the {\it algebra of dual quaternions}.

As we said above, our main result, Theorem \ref{Main-theorem} relies on the
following result, where the signature of a non-degenerate form is written as a pair $(n_+,n_-)$
and the signature of a degenerate form is written as a triple $(n_+,n_-,n_0)$. 
 
Fix an algebra $\mathcal H$ defined by Formula (\ref{H-definition}). 

\begin{thm}
\label{Algebraic-theorem} Let $(\alpha,\beta,\gamma)$ be any triple of real numbers 
representing $\mathcal H$, $\mathcal H \cong \mathcal H_{\alpha,\beta,\gamma}$.   
The signature of the form $Q=Q_{\alpha,\beta,\gamma}$ 

does not depend on the choice of such $(\alpha,\beta,\gamma)$ and 

is thus an isomorphism invariant of the algebra  $\mathcal H$.
All possible signatures of such forms $Q$ are the nondegenerate cases $(0,3)$, $(2,1)$, $(1,2)$ and the degenerate cases
$(0,2,1)$, $(1,1,1)$, $(0,1,2)$. 


The algebra $\mathcal H$ contains a subalgebra of quaternions
$\HH$ precisely when the signature of $Q$ is $(0,3)$, $(0,2,1)$ or $(1,2)$.


The center $Z(\mathcal H)$ has dimension 3 in the case, when $rank\,Q = 1$, that is, 
$Q$ has signature $(0,1,2)$, and this condition determines $\mathcal H$ uniquely, up to isomorphism.
 In all other cases the center has dimension 2.


\medskip
{\bf The case of signature $(0,3)$ (necessarily $\det\, Q< 0$):}
in this case $\mathcal H\cong \mathcal H(1)$. 


 The algebra $\mathcal H$ contains exactly two nontrivial two-sided 
ideals $\HH(1+c), \HH(1-c)$. 

The regular representation $\rho_{reg}$ decomposes as $\rho_1\oplus\rho_2$, where the non-faithful representations $\rho_1,\rho_2$ 
are the only, up to isomorphism, irreducible (non-faithful) 4-representations of $\mathcal H$, $\rho_1,\rho_2$ are given by restrictions
of the regular representations
$\rho_{1}(h)=\rho_{reg}(h)|_{\HH(1+c)},\rho_2(h)=\rho_{reg}(h)|_{\HH(1-c)}$ for all $h \in \mathcal H$, and are isomorphic to the regular representation $\rho_{\HH}$ precomposed, respectively,  with   
 the quotient maps  $\mathcal H\rightarrow \mathcal H/\HH(1-c)\cong\HH$ and $ 
\mathcal H\rightarrow \mathcal H/\HH(1+c)\cong \HH$. 


\medskip

{\bf The case of signature $(0,2,1)$
(necessarily $\det\, Q=0$):} in this case $\mathcal H\cong \mathcal H(0)$. 


The algebra $\mathcal H$ contains exactly one nontrivial two-sided 
ideal $\HH c$.  
The 8-representation $\rho_8=\rho_{reg}$ is reducible but not completely reducible. There
are no faithful 4-representations of $\mathcal H$ and there exists exactly one, up to isomorphism, non-faithful
4-representation $\rho_4$, $\rho_4(h)=\rho_{reg}(h)|_{\HH c}, h \in \mathcal H$, isomorphic to the regular
representation $\rho_{\HH}$ precomposed with the quotient map 
$\mathcal H\rightarrow \mathcal H/\HH c \cong \HH$.

\medskip

{\bf The case of signature $(1,2)$ (necessarily $\det\, Q> 0$):}
 in this case $\mathcal H\cong \mathcal H(-1)$. 


 The algebra $\mathcal H$ 
 has no nontrivial two-sided ideals.
All proper left ideals are subspaces of dimension 4 in $\mathcal H$, that are of the form $\HH(w+c), w\in S^2\subset\HH, w^2=-1$.


The regular representation $\rho_{reg}$ decomposes as $\rho_0\oplus \rho_0$, where 
$\rho_0$ is the unique, up to isomorphism, irreducible (faithful) 4-representation of $\mathcal H$,
arising from any proper left ideal $\HH(w+c)$, $\rho_0(h)=\rho_{reg}(h)|_{\HH(w+c)}, h\in \mathcal H$. 
\end{thm}

\begin{rem}
\label{Noncanonical-remark}
 If $e_1,e_2,e_3$ is any standard set of generators corresponding to 
$(\alpha,\beta,\gamma)$, then
the central element $c$ in the formulation of Theorem \ref{Algebraic-theorem}
is proportional to the element $\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3$, 
see Proposition \ref{Prop-properties-of-H}. The normalized such $c$, that is,
satisfying $c^2=-1,0$ or $1$ is determined up to a scalar multiple, so the 
definitions of the two-sided ideals
of $\mathcal H\cong \mathcal H(1)$ in Theorem \ref{Algebraic-theorem} and hence the representations $\rho_1,\rho_2$ depend on the (non-canonical!) choice of $c$. 
\begin{comment}
In fact, the subalgebra structure of  $Z(\mathcal H)$, described by Proposition \ref{Prop-properties-of-H}, implies, at least in the case when $\dim\, Z(\mathcal H)=2$,  
that the line $\RR(\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3)\subset \mathcal H$ does not depend
on the choice of the standard set of generators. 
\end{comment}
\end{rem}

\begin{comment}
\begin{rem}
In the cases when all $|\alpha|,|\beta|, |\gamma| \geqslant 1$ we obviously have that
none of the sides of the triangle $\triangle e_1e_2e_3$ is a compact twistor line. 
But even in such case it is still possible that $\mathcal H(e_1,e_2,e_3)$ contains a (negative) 3-subspace corresponding to a compact twistor line in $End\, \RR^8$.
\end{rem}
\end{comment}
\begin{comment}
\begin{rem}
We will see in the process of proving Theorem \ref{Algebraic-theorem}
that the case $(0,2,1)$ can be considered as a degeneration of each of the
two subcases,  $(0,3)$ ($\det\, Q<0$) and $(1,2)$ ($\det\, Q>0$). 


\end{rem}
\end{comment}


Let $$\mathcal D=\{(\alpha,\beta,\gamma) \in \RR^3\,|\, Q_{\alpha,\beta,\gamma}\mbox{ has signature } (0,3),(1,2)
\mbox{ or }(0,2,1)\}.$$

From now we restrict ourselves to  triangles $\triangle J_1J_2J_3$ with
$T(\triangle J_1J_2J_3) \in \mathcal D$, this, by Theorem \ref{Algebraic-theorem},  is the (above mentioned) proper class of twistor triangles,
whose corresponding algebras $\mathcal H_{T(\triangle J_1 J_2 J_3)}$ contain the algebra of quaternions $\HH$. By Theorem \ref{Algebraic-theorem}, up to isomorphism, there are just three such algebras.

The triples of $\alpha,\beta,\gamma$ with $|\alpha|,|\beta|,|\gamma|<1$ corresponding to 
 compact twistor triangles, that is, with compact sides, form a proper subset in $\mathcal D$. 


\begin{defi}
\label{geometric-definition}
The triangle $\triangle J_1J_2J_3$ is called {\it hyperbolic}, if $\det Q_{T(\triangle J_1J_2J_3)}>0$, {\it spherical}, if $\det Q_{T(\triangle J_1J_2J_3)}<0$,
and {\it cylindrical}, if  $\det Q_{T(\triangle J_1J_2J_3)}=0$. The algebra 
$\mathcal H=\mathcal H_{T(\triangle J_1 J_2 J_3)}$ is called {\it hyperbolic} 
($\mathcal H\cong \mathcal H (-1)$), {\it spherical} ($\mathcal H\cong \mathcal H (1)$), or {\it cylindrical} 
($\mathcal H\cong \mathcal H (0)$), 
if the triangle $\triangle J_1J_2J_3$ is such. 
\end{defi}

The part of the latter definition involving the algebra $\mathcal H_{\alpha,\beta,\gamma}$ is correct
because Theorem \ref{Algebraic-theorem} provides that the signature of
the form $Q_{\alpha,\beta,\gamma}$ does not depend on the choice of the 
representing triple $(\alpha,\beta,\gamma)$ 

and the sign of $\det\, Q_{\alpha,\beta,\gamma}$ 
uniquely identifies the signature of $Q_{\alpha,\beta,\gamma}$ when
$(\alpha,\beta,\gamma) \in \mathcal D$. 

\begin{comment}
  Moreover, Theorem \ref{Algebraic-theorem}
provides a trichotomy with respect to the number of two-sided ideals in 
$\mathcal H_{\alpha,\beta,\gamma}$ 
(no such ideals in case $\det\, Q_{\alpha,\beta,\gamma}>0$, one ideal if $\det\, Q_{\alpha,\beta,\gamma}=0$ and two ideals if $\det \,Q_{\alpha,\beta,\gamma}<0$), 
that agrees with our above signature trichotomy.
\end{comment}
\begin{rem}
Let $\triangle J_1 J_2 J_3$ be a nondegenerate generalized  twistor triangle,  with $T(\triangle J_1 J_2 J_3)=(\alpha,\beta,\gamma) \in \mathcal D$. 
For the  algebra $\mathcal H_{\alpha,\beta,\gamma}\cong \mathcal H(J_1,J_2,J_3) \subset End\,V_\RR$ the subset
$\langle J_1,J_2,J_3\rangle \cap Compl=\{aJ_1+bJ_2+cJ_3|(aJ_1+bJ_2+cJ_3)^2=-Id, a,b,c\in \RR\}$ in the cases $\det\,Q_{\alpha,\beta,\gamma}>0, \det\,Q_{\alpha,\beta,\gamma}<0,\det\,Q_{\alpha,\beta,\gamma}=0$, is, respectively, a one-sheeted hyperboloid, a sphere, a cylinder, that contains the 
``geodesic'' segments $\langle J_k,J_l\rangle \cap S(J_k,J_l), 1\leqslant k<l\leqslant 3$, of the 
respective twistor lines, 

joining the vertices $J_1,J_2,J_3$ of our twistor triangle.
 This explains the geometric terminology introduced in  Definition \ref{geometric-definition}.
\end{rem}

\begin{rem}
For a triangle $\triangle J_1J_2J_3$ to be of spherical type means that 
$(\alpha,\beta,\gamma)=T(\triangle J_1 J_2 J_3)$ 
is a triple of minus cosines of lengths of sides of a (geodesic) triangle on a unit 2-sphere.

Note that here we compare triangles using only sides lengths, not saying anything about
comparing their angles. In fact, due to our form $(\cdot,\cdot)$ being indefinite,
it is not always possible to define (in a geometrically meaningful way) the angle 
between the sides of $\triangle J_1J_2J_3$. The exceptional situation, when 
the angles of a spherical twistor triangle are defined and equal, up to taking complements to $\pi$, 
 to the corresponding angles of the 
respective geodesic triangle on a sphere, 
is discussed in Theorem \ref{Theorem-relation}. 
\end{rem}


Let $e_1,e_2,e_3$ be any standard set of generators of $\mathcal H_{\alpha,\beta,\gamma}$
 corresponding to $(\alpha,\beta,\gamma)$ and $c\in \RR(\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3) \subset Z(\mathcal H_{\alpha,\beta,\gamma})$ (where, as we know, $c^2\in \RR$) be normalized as in Theorem \ref{Algebraic-theorem}, so that $c^2=-1,0$ or $1$.  We are further using the notations for the irreducible
representations of $\mathcal H_{\alpha,\beta,\gamma}$ introduced in Theorem \ref{Algebraic-theorem}. 


 


Let us now formulate our result about twistor triangles, in terms of the representation theory of the 
respective algebras $\mathcal H$.

\begin{thm}
\label{Main-theorem}

{\bf Existence.} For every triple $ (\alpha,\beta,\gamma) \in \mathcal{D}$
there exists a (possibly non-faithful) 
representation $\rho\colon \mathcal H=\mathcal H_{\alpha,\beta,\gamma} \rightarrow End\, V_\RR$, $\dim_\RR V_\RR=4n$. 
Moreover, 
if one of the two additional
conditions holds$\colon $ 
 
a) $\mathcal H$ 
is hyperbolic;


or

b) $\mathcal H$ 
 is either spherical or cylindrical 

 and $n>1$;


then there exists a faithful such  $\rho$.
 If $\mathcal H$ 
 is either spherical or cylindrical,

and $n=1$, 
then only a non-faithful $\rho$ exists, 
whose image is a subalgebra of quaternions $\HH \subset End\, V_\RR$. 

 {\bf The number of non-$G$-equivalent representations.}
1) For a hyperbolic algebra $\mathcal H$ there is a unique, up to $G$-equivalence, representation 
$\rho \colon \mathcal H \rightarrow End\, V_\RR$, $\rho=n\rho_0$, which is faithful;

2) For a spherical algebra $\mathcal H$ there are total of $n+1$ classes of $G$-equivalent representations  
$\rho \colon \mathcal H \rightarrow End\,V_\RR, \rho=k\rho_1\oplus (n-k)\rho_2$
(among which there are precisely two non-faithful ones, they correspond to $k=0$ and $k=n$, mapping $\mathcal H$ to $\HH\subset End\,V_\RR$). 
Here $k$ is uniquely identified as $k=\frac{1}{8}(Tr(\rho(c))+4n)$.


3) For a cylindrical algebra $\mathcal H$ there are total of $\lfloor \frac{n}{2}\rfloor +1$ 

non-$G$-equivalent representations $\rho=(n-2k)\rho_4\oplus k\rho_8$ (including the only 
non-faithful one, 
corresponding to $k=0$). Here $k$ is uniquely identified as $k=\frac{1}{4}rk\,\rho(c)$. 


\end{thm}

\begin{comment}
\begin{thm}
\label{Main-theorem}

{\bf Realizability.} For every triple $ (\alpha,\beta,\gamma) \in \mathcal{D}$
there exist a (possibly degenerate) twistor triangle $\triangle J_1J_2J_3$ with $T(\triangle J_1J_2J_3)=(\alpha,\beta,\gamma)$.
Moreover, 
if one of the two additional
condition holds$\colon $  a) $\det\,Q_{\alpha,\beta,\gamma}=f(\alpha,\beta,\gamma)> 0$
(that is, $Q_{\alpha,\beta,\gamma}$ has signature $(1,2)$); 

b) $\det\,Q_{\alpha,\beta\gamma}=f(\alpha,\beta,\gamma)\leqslant 0$ (that is, the signature
of $Q_{\alpha,\beta,\gamma}$ is $(0,3)$ or $(0,2,1)$)
and $n>1$;


then the triangle
$\triangle J_1J_2J_3$ can be chosen non-degenerate. If $f(\alpha,\beta,\gamma)\leqslant 0$ and $n=1$
then only a degenerate triangle $\triangle J_1J_2J_3$ exists (all $J_1,J_2,J_3$ belong to the same
compact twistor sphere).

 {\bf Uniqueness.}
1) The group $G=GL(V_\RR)$ acts transitively on the set of all hyperbolic twistor triangles $\triangle J_1J_2J_3$
having the same (up to permutation) fixed triple of invariants $T$.

2) There are total of $n+1$ $G$-nonequivalent spherical triangles with the same invariant $T$
(among which there are two degenerate triangles).
They are all distinguished by the invariant $Q$. 


3) There are total of $\lfloor \frac{n}{2}\rfloor +1$ 

 $G$-nonquivalent special triangles (including the degenerate one) with the same triple invariant $T$.
\end{thm}
\end{comment}


\begin{comment}
\begin{rem}
In the above theorem formulation it does not matter if we compare the traces $Tr(\rho_i(e_1e_2e_3))$ 
in the 
case of a cylindrical $\mathcal H$, actually  $Tr(\rho_i(e_1e_2e_3))$ is always zero;
In the case of cylindrical triangles we do not know if there exists a set of $G$-invariants distinguishing
these $\lfloor \frac{n}{2}\rfloor +1$ triangles.


 

\end{rem}
\end{comment}
 The abstract representation theory of algebras $\mathcal H(\varepsilon)$ is elementary
and must be a folklore, nevertheless the nontrivial point of Theorem \ref{Main-theorem} is that the theorem explains this representation theory
with respect to a standard set of generators, that is, with respect to a triangle, from which
our $\mathcal H$ originates.

\begin{defi}
We say that a representation $\rho\colon \mathcal H \rightarrow End\,V_\RR$
of an algebra $\mathcal H$ of spherical type is {\it balanced} if $\rho=k\rho_1\oplus k\rho_2$,
that is, the multiplicities of both $\rho_1$ and $\rho_2$ are equal.
\end{defi}
Theorem \ref{Main-theorem} tells us that a balanced representation of a spherical algebra  
$\mathcal H$ 
exists if and only if $n=\frac{1}{4}dim\,V_\RR$ is an even number.

Let us introduce additional subgroups of the group $G$. Let $S$ be a compact twistor line 
(that is, a 2-sphere), $\HH \subset End\,V_\RR$ be the algebra of quaternions associated to $S$, 
and $I\in S$ be a period.  
 We set $G_{I,S}\subset G_I$ to be the $G_I$-adjoint action stabilizer of $S$ as a set. We note that,
as $G_{I,S}$ is the subgroup of elements of $G_I$, acting as rotations of $S$ about
the ``axis" $\{I,-I\}\subset S$, we have 

 $G_{I,S}\cong 
\langle \exp(tI)h, t\in \RR,h\in G_\HH \rangle \cong SO(2)\times G_\HH
\subset G_S$. 
Then if $I_1,I_2,I_3$ are linearly independent complex structures in $S$, we have
that  $G_S$ is generated by its subgroups $G_{I_j,S},j=1,2,3$, 
and so we have an isomorphism $G_S\cong SO(3)\times G_\HH$.
Fix such $I_1,I_2,I_3\in S$. 
Set $(\alpha,\beta,\gamma)=T(\triangle I_1 I_2 I_3)$. 

 Theorem \ref{Theorem-relation} below states that $\mathcal H_{\alpha,\beta,\gamma}$
is spherical. 

For such $\mathcal H_{\alpha,\beta,\gamma}$ 

Theorem \ref{Main-theorem} allows us to
 choose  representatives $\rho_k\colon \mathcal H_{\alpha,\beta,\gamma} \rightarrow
End\,V_\RR,k=0,\dots,n$, of $n+1$ $G$-equivalence classes of representations of $\mathcal H_{\alpha,\beta,\gamma} \rightarrow
End\,V_\RR$ (not to be confused with the above introduced {\it irreducible} representations!), such that $\rho_k(e_1)=I_1,\rho_k(e_2)=I_2, Tr(\rho_k(c))=4(2k-n), k=0,\dots,n$ 
(again, $\rho_k$ are defined non-canonically, as follows from Remark \ref{Noncanonical-remark}). 
The values $k=0$ and $n$ correspond to the two non-faithful representations, 
$\rho_0(\mathcal H_{\alpha,\beta,\gamma})=\rho_n(\mathcal H_{\alpha,\beta,\gamma})=\HH\subset End\,V_\RR$.  For a representation $\rho\colon \mathcal H_{\alpha,\beta,\gamma} \rightarrow
End\,V_\RR, \rho(e_1)=I_1,\rho(e_2)=I_2,$  we denote by $G_{\HH,\rho}$ the $G_\HH$-action stabilizer of  $\rho$. We set $G_\HH^l$ to be the $l$-fold Cartesian product of $G_\HH$. 
We introduce the following action of $G_\HH^4$ on $m^{-1}(G_\HH)$,
$$(h_1,h_2,h_3,h_4)\cdot (f_1,f_2,f_3)= (h_1f_1h_2^{-1},h_2f_2h_3^{-1},h_3f_3h_4^{-1}).$$ 

Finally we formulate the following answer to our original question about 
the fiber $ m^{-1}(G_\HH)$. 

\begin{thm}
\label{Theorem-relation}
Given three linearly independent complex structures $I_1,I_2,I_3$
in a compact twistor line $S$  with $(\alpha,\beta,\gamma)=T(\triangle I_1 I_2 I_3)$,
the algebra $\mathcal H_{\alpha,\beta,\gamma}$ is spherical. 
The fiber $m^{-1}(G_\HH)$ consists of $n+1$ connected components that are in 
one-to-one correspondence with $G$-equivalence classes of representations $\rho_k\colon
\mathcal H_{\alpha,\beta,\gamma} \rightarrow End\,V_\RR, Tr(\rho_k(c))=4(2k-n),  0\leqslant k \leqslant n$. 

{\bf General components.} 
Each component is a subset of the form $$\{(f_1,f_2,f_3)\in m^{-1}(G_\HH)\,|\, f_2(I_3)\in G_\HH\cdot \rho_k(e_3)\},$$
each such set is an orbit under the action of $G_\HH^4$, 

the orbit is diffeomorphic 
to $G_\HH^4/Stab_{G_\HH^4}(f_1,f_2,f_3)$, where the stabilizer $Stab_{G_\HH^4}(f_1,f_2,f_3)\cong 
G_{\HH,\rho_k}$,
$\dim\, G_\HH^4/Stab_{G_\HH^4}(f_1,f_2,f_3)=12n^2+8nk-8k^2$. 

{\bf The trivial and $SO(3)$-type components.}
For $k=0,n$ we have $G_{\HH,\rho_k}=G_\HH$ 
and the respective orbits are 
 diffeomorphic to $G_\HH^3$. The two orbits are, non-canonically in $k$, the subsets

 $$G_\HH \times G_\HH \times G_\HH \subset 
G_{I_1}\times G_{I_2}\times G_{I_3}$$
(the trivial component)
and 
$$g_1G_\HH \times g_2G_\HH \times g_3G_\HH \subset G_{I_1}\times G_{I_2}\times G_{I_3},$$
where  $g_{j}\in G_{I_j,S}\setminus G_\HH,j=1,2,3$ are unique, up to $G_\HH$, elements satisfying $g_{1}g_{2}g_{3}\in G_\HH$ (the  $SO(3)$-type component).


\begin{comment}
{\bf The three types of components of the fiber.} Given three linearly independent complex structures $I_1,I_2,I_3$
in a compact twistor line $S$, 
the fiber $m^{-1}(G_\HH)\subset G_{I_1}\times G_{I_2}\times G_{I_3}$ consists of $n+1$ disjoint components, each of which is diffeomorphic to
$G_\HH \times G_\HH \times G_\HH$, among which we have three types of components, enumerated
according to the increase of nontriviality:

1) the component $G_\HH \times G_\HH \times G_\HH \subset G_{I_1}\times G_{I_2}\times G_{I_3}$ itself, providing the ``trivial solution'';

2)  the component $g_1G_\HH \times g_2G_\HH \times g_3G_\HH$,
where  $g_{j}\in G_{I_j,S}\setminus G_\HH,j=1,2,3$ are unique, up to $G_\HH$, elements satisfying $g_{1}g_{2}g_{3}\in G_\HH$ (the  $SO(3)$-type component);


3) $n-1$ components that are 
 contained in  
$(G_{I_1}\setminus G_{I_1,S}) \times (G_{I_2}\setminus G_{I_2,S}) \times (G_{I_3}\setminus G_{I_3,S})$ and 
 are in natural bijective correspondence with all non-equivalent faithful representations 
$\rho\colon \mathcal H_{T(\triangle I_1 I_2 I_3)}
\rightarrow End\,V_\RR$ of the spherical algebra 
$\mathcal H_{T(\triangle I_1 I_2 I_3)}$. 


{\bf The $G_\HH$-structure of components.} Each of the components is a set of the form $\{(h_1 f_1, f_2h_2, h_2^{-1}f_3 h_3)|h_i \in G_\HH\}$,
where $f_j \in G_{I_j}\setminus G_{I_j, S}$ are some elements satisfying  $f_1f_2f_3\in G_\HH$.


\end{comment}

{\bf The geometry of triangles.} For every $(g_1,g_2,g_3) \in m^{-1}(G_\HH) \cap (G_{I_1}\setminus G_{I_1,S}) \times (G_{I_2}\setminus G_{I_2,S}) \times (G_{I_3}\setminus G_{I_3,S})$  the (compact) twistor triangle $\triangle I_1I_2J_3, J_3=g_{2}(I_3)$, in $Compl$
is nondegenerate and spherical, with $T(\triangle I_1 I_2 J_3)=T(\triangle I_1 I_2 I_3)$, 

so that $\mathcal H_{T(\triangle I_1 I_2 I_3)}=\mathcal H_{T(\triangle I_1 I_2 J_3)} \cong \mathcal H(I_1,I_2,J_3)$. 
If, in addition, the natural representation $\rho\colon \mathcal H(I_1,I_2,J_3) \rightarrow End\,V_\RR$
is balanced, the angles of $\triangle I_1I_2J_3$ are well defined, as 
the angles between the tangent subspaces to the twistor spheres at the vertices,
and they are equal, up to taking complements to $\pi$, to the respective angles of $\triangle I_1I_2I_3$.


\begin{comment}
3) if in the triangle $\triangle I_1I_2I_3 \subset S$ sides $I_2I_1$ and $I_2I_3$ have equal length,
there exist $g_{I_1}\in G_{I_1}\setminus G_{I_1,S},g_{I_3}\in G_{I_3}\setminus G_{I_3,S}$
and $g_{I_2}\in G_{I_2,S}$, where $g_{I_2}(I_3)=I_1$, such that $g_{I_1}g_{I_2}g_{I_3}\in G_\HH$.


\end{comment}

\end{thm}

\begin{comment}
\begin{rem}
The part 3 of Theorem \ref{Theorem-relation} makes a statement about how
the ``nontrivial'' relations arise from faithful representations. Actually, the $SO(3)$-type
relation of part 2 can also be considered as arising from any of the two non-faithful non-equivalent representations
$\rho_1,\rho_2 \colon \mathcal H \rightarrow End\,V_\RR$.
So together with the most trivial part of the fiber $m^{-1}(G_\HH)\cap G_\HH\times G_\HH \times G_\HH
=G_\HH\times G_\HH \times G_\HH$ we  get $n+1$ components diffeomorphic to $G_\HH\times G_\HH \times G_\HH$,
forming the whole fiber $m^{-1}(G_\HH)$.
\end{rem}
\end{comment}


So Theorem \ref{Theorem-relation} tells us that the independence 
of the subgroups $G_{I_1}, G_{I_2}, G_{I_3}$ in terms of the multiplication mapping 
$m\colon G_{I_1}\times G_{I_2}\times G_{I_3}\to G$ in general fails not only at the expected locus
$G_\HH^3$ and at the easy-to-guess locus ``of $SO(3)$-type'' (diffeomorphic to $G_\HH^3$), 
both of which correspond to degenerate triangles, 

but also at (a finite number of) higher dimensional loci in $G_{I_1}\times G_{I_2}\times G_{I_3}$, corresponding
to nondegenerate spherical triangles in $Compl$. 


We note here that the problem of the description of the specific fiber
$m^{-1}(G_\HH)$ is  extremely approachable, while it may be difficult, if not impossible at all,
to apply the same methods for describing fibers of the more general type $m^{-1}(g_1g_2g_3G_\HH)$
with $g_j\in G_{I_j}$. 

\begin{comment}
One may try to study what happens in the case when $S$ is a non-compact twistor line and $I_1,I_2,I_3\in S$ any linearly independent
complex structures. For non-compact curves there are ``curves of type $\HH(1)$''
and ``curves of type $\HH(0)$'', see \cite{Generalized-twistor-lines} for the terminology. In that case we introduce $G_{\HH(\varepsilon)}=G_{I_1}\cap G_{I_2}\cap G_{I_3},
\varepsilon=0,1$, 
and, if one specifies $\varepsilon=1$, we similarly have the transversality, analogous to that in  the compact case,
see \cite{Generalized-twistor-lines}. 
Then one may try to describe the  fiber  $m^{-1}(G_{\HH(1)})$ which would correspond to 
the relations of ``hyperbolic type'', compared to the established by Theorem \ref{Theorem-relation}
relations of ``spherical type''.
\end{comment}

Now let us sketch the plan of the paper.


In Section \ref{When-section} we prove the part of Theorem \ref{Algebraic-theorem} stating the isomorphism invariance of
the signature  of $Q$, classifying the
possible signatures and specifying the ones that correspond to $\mathcal H$ containing $\HH$.


 Besides that this section contains a summary of algebraic properties of the algebra 
$\mathcal H$, which has a lot of symmetry with respect to a standard set of generators.


In Section \ref{Regular-section} we write down the left regular representation of $\mathcal H$ 
and its irreducible representations, which completes the proof of Theorem \ref{Algebraic-theorem}. 
 Understanding the irreducible representations of $\mathcal H$ allows us to prove
Theorem \ref{Main-theorem}, see Subsection \ref{Main-theorem-subsection}.


Section \ref{Proof of Theorem-relation} 
proves Theorem \ref{Theorem-relation}.

Section \ref{Appendix} contains proofs of some technical statements regarding
the structure of the algebra $\mathcal H$, in particular it describes the center $Z(\mathcal H)$.

\begin{comment}
The following subsection is intended to give a bit broader, geometric, perspective
of the introduced invariants, the discussion there is not necessary for understanding the above formulated results. 


\subsection{A geometric digression} 
\label{Geometric-digression-subsection}
We define a bilinear form on $End\, V_\RR \cong Mat_{4n\times 4n}(\RR)$ (we fix a basis in $V_\RR$),
by  $$(A,B)=-tr(AB).$$
This inner product is clearly positive on  the vectors corresponding to matrices in $Compl$. 
The decomposition $Mat_{4n\times 4n}(\RR) =S\oplus A$, where $S$ and $A$ are subspaces of symmetric and, respectively, antisymmetric
matrices, is orthogonal with respect to $(\cdot,\cdot)$. The form $(\cdot,\cdot)$ is positive on $A$, negative on $S$, so it has 
signature $(8n^2-2n,8n^2+2n)$. 
This form is definitely $ad\,G$-invariant. Let $I$ be the complex structure with block-diagonal matrix in the basis that we fixed above, where 
the blocks on the diagonal are of the form 
$\left(
\begin{array}{cc}
0 & -1\\ 
1 & 0 
\end{array}\right)$.
Then identifying $T_ICompl$ with the subspace of matrices anticommuting with $I$ and further decomposing
$T_ICompl\cong A_I\oplus S_I$ into the respective subspaces of symmetric and antisymmetric matrices,
 we can see that $(\cdot,\cdot)|_{T_ICompl}$
has signature $(4n^2-2n, 4n^2+2n)$. As $G$ acts transitively on $Compl$ we see that the restriction of $(\cdot,\cdot)$
to $T_{I_1}Compl$ for every $I_1 \in Compl$ has the same signature, thus $(\cdot,\cdot)$ determines a pseudo-riemannian metric
on $Compl$. Note that for a tangent vector $J \in T_ICompl$, $JI=-IJ$ and $J^2=-Id$ we have that $(J,J)=-tr(J^2)>0$,
thus the tangent 2-planes to twistor lines are positive  with respect to this pseudo-riemannian metric.


The angles $\sphericalangle J_kJ_l$ between vectors $J_k$ and $J_l$ are  determined by their cosines 
$\cos\, \sphericalangle J_kJ_l=\frac{(J_k,J_l)}{\sqrt{(J_k,J_k)}\sqrt{(J_l,J_l)}}=\frac{-Tr \, J_kJ_l}{\sqrt{-Tr\, J^2_k }\sqrt{-Tr\, J^2_l }}$ (for the correct definition of the angle we certainly need to justify why $(\cdot,\cdot)|_{\langle J_k,J_l \rangle_\RR}$ is positively definite, this will be shown
in Section \ref{Algebraic-section}, see Remark \ref{Validity-cosine-remark}). In terms of our pseudometric on the tangent bundle 
$TCompl$, the angle $\sphericalangle J_kJ_l$ is the length of one of two arcs of the great circle in $S(J_k,J_l)$
through $J_k$ and $J_l$. This is easy to see using the parametrization of the great circle $t \mapsto e^{tJ}J_ke^{-tJ}$, where
$J \in S$ is a complex structure orthogonal to the plane $\langle J_k,J_l \rangle_\RR \subset End\, V_\RR$ (that is, $J \in S$ anticommutes
with both $J_k,J_l$). 

Now, we can ask ourselves if the lengths of the sides of the triangle we are dealing with satisfy the triangle inequality, that is,
for example, if
$\arccos(-\gamma) \leqslant \arccos(-\alpha)+\arccos(-\beta)$ holds. Assuming that both the angle on the left and the sum of the angles
on the right are between 0 and $\pi$ we take the reversed inequality for cosines of the left and the right side:
$$-\gamma \geqslant (-\alpha)(-\beta)-\sqrt{1-\alpha^2}\sqrt{1-\beta^2}.$$
This inequality implies easily that $$\alpha^2+\beta^2+\gamma^2+2\alpha\beta\gamma \leqslant 1.$$
Thus the triangle inequality is controlled by the earlier introduced function $f(\alpha,\beta,\gamma)$.

As one can easily see, the function $f$ not only verifies if for a given $\triangle J_1J_2J_3$ the triangle formed by the arcs joining the 
$J_1,J_2,J_3$ on the respective twistor spheres can be isometrically mapped to a (geodesic) triangle
lying on a round sphere of radius 1, but it actually checks if these vertices lie on an actual sphere in $Compl$ (which is not a twistor sphere,
just a real-analytic submanifold in $Compl$).
Indeed, the complex structures in the vector space $\langle J_1,J_2,J_3\rangle_\RR$
are given by the equation $(aJ_1+bJ_2+cJ_3)^2=-Id$, 
which can be rewritten as $-a^2-b^2-c^2+2\alpha ab+2\beta ac+2\gamma bc+1=0$. 
The left side of the latter equation is a quadratic form in $a,b,c$ with the matrix
$$Q=\left(\begin{array}{ccc}
-1 & \alpha & \beta \\
\alpha & -1 & \gamma \\
\beta & \gamma & -1
\end{array}\right),$$
which, due to Sylvester criterion and the fact that $|\alpha|<1$ ,
is negative definite if and only if the determinant $det\, Q =f(\alpha,\beta,\gamma)<0$.
\end{comment}


\begin{comment}
Each triangle $\triangle J_1J_2J_3$  determines the triple $T(\triangle J_1J_2J_3)=(\cos\, \sphericalangle J_1J_2, \cos\, \sphericalangle J_2J_3, \cos\, \sphericalangle J_3J_1)$. 
We will also need to associate to a triangle an extended set of its metric invariants, namely the quadruple 
$Q(\triangle J_1J_2J_3)=(\cos\, \sphericalangle J_1J_2, \cos\, \sphericalangle J_2J_3, \cos\, \sphericalangle J_3J_1, (J_1J_2,J_3))$ (here $J_1J_2$ need not be a point in
one of the three twistor lines, even need not be a complex structure, so that we just write the inner product $(J_1J_2,J_3)$, without appealing to any ``distance/angle" interpretation).

Denote the cosines of $\sphericalangle J_1J_2, \sphericalangle J_1J_3$ and $ \sphericalangle J_2J_3$ by $-\alpha, -\beta$ and $-\gamma$ respectively
(we introduce the minus sign for further notational convenience).
Introduce  the following metric invariant of the triangle $\triangle J_1J_2J_3$, $$|\triangle J_1J_2J_3|=\alpha^2+\beta^2+\gamma^2+2\alpha\beta\gamma.$$
We will call a triangle $\triangle J_1J_2J_3$ {\it thin}, if $|\triangle J_1J_2J_3|>1$, {\it thick}, if $|\triangle J_1J_2J_3|<1$
and {\it cylindrical} if  $|\triangle J_1J_2J_3|=1$.
\end{comment}


\begin{comment}
\section{Algebraic characterization of compact twistor spheres in $Compl$}
\label{Algebraic-section}
In this section we give an algebraic characterization of twistor lines, which is then
applied to twistor triangles to
construct the triple invariant $T$.
Let $I,J,K=IJ \in End\, V_\RR$ be complex structures satisfying the quaternionic identities
and $$S=\{aI+bJ+cK|a^2+b^2+c^2=1\}$$ be the corresponding twistor sphere, and $J_1\neq \pm J_2$ be some complex structures in $S$. Then the plane 
$\langle J_1,J_2\rangle_\RR \subset End\,V_\RR$ intersects $S$ in the
circle $S \cap \langle J_1,J_2\rangle_\RR$, which contains a complex structure which anticommutes with
$J_1$.
Then there exists a real number $\alpha$ such that $\alpha J_1+J_2$ determines,
after a scalar normalization, a complex structure in $S$  anticommuting with $J_1$,


that is (even without the normalization), $$(\alpha J_1+J_2)J_1+J_1(\alpha J_1+J_2)=0,$$
which results in the relation 
$$
J_1J_2+J_2J_1=2\alpha\cdot Id.
$$
The fact that $\alpha J_1+J_2$ is proportional to a complex structure
brings certain restrictions on $\alpha$, that is,
$$(\alpha J_1+J_2)^2=a\cdot Id, a<0.$$
The left side is equal to $$-(1+\alpha^2)Id+\alpha\cdot (J_1J_2+J_2J_1)=(\alpha^2-1)Id,$$
and thus the condition $a=\alpha^2-1<0$ is simply the condition $|\alpha|<1$.
Thus the necessary and sufficient condition that nonproportional $J_1,J_2$
belong to the same twistor sphere $S$ is that there exists $\alpha \in \RR$ such that
\begin{equation}
J_1J_2+J_2J_1=2\alpha\cdot Id, \, |\alpha|<1.
\end{equation}
\begin{rem}
\label{Validity-cosine-remark}
The latter equality justifies that the restriction of our indefinite metric to the plane $\langle J_1,J_2\rangle_\RR$
is positive definite, thus the definition of  $\cos \, \sphericalangle J_1J_2$ as $\frac {(J_1,J_2)}{\sqrt{(J_1,J_1)}\sqrt{(J_2,J_2)}}$ is indeed valid.
\end{rem}

As we said in the introduction there are generalized (noncompact) twistor lines defined for the cases
$|\alpha|=1$ and $|\alpha|>1$. We will discuss the definitions together with their 
compactifications in Section \ref{Generalized-twistor-lines}.
\end{comment}

\begin{comment}
As we said previously, the twistor triangle $\triangle J_1J_2J_3$ gives rise to the subalgebra $\mathcal H(J_1,J_2,J_3)$ in $End\, V_\RR$ generated by $J_1,J_2,J_3$.
The relations $J_kJ_l+J_lJ_k \in \RR Id$ guarantee that the dimension of our algebra is at most 8. Indeed,
as a vector space it is spanned by (possibly dependent) vectors $Id,J_1,J_2,J_3,J_1J_2,J_1J_3,J_2J_3, J_1J_2J_3$.
Next, this algebra can be considered as a vector space over quaternions $I,J,K \in S(J_1,J_2)$
and so the real dimension is either 4 or 8. In the case the dimension is 4 we must have that $J_3 \in S(J_1,J_2)$,
thus our twistor triangle is degenerate.  If the dimension is 8, the triangle is nondegenerate.
We want to classify the (real) irreducible representations of our algebra depending
on the trichotomy with respect to $|\triangle|<0, |\triangle|=0,|\triangle|>0$. This will be a crucial step
for the proof of Theorem \ref{Main-theorem}. For the classification purpose we need
to come up with a specific generating set for our algebra, whose construction we are explaining below.

\subsection{Reduction to a simpler triangle}
\label{reduction-subsection}

Let $\triangle J_1J_2J_3$ be an arbitrary twistor triangle, which we will denote for short by $\triangle$.


Let the relations, corresponding to the sides of our triangle be $$J_1J_2+J_2J_1=2\alpha Id, J_1J_3+J_3J_1=2\beta Id, J_2J_3+J_3J_2=2\gamma Id.$$


Assume that  none of $\alpha,\beta,\gamma$ is zero.
Consider the point $X=\gamma J_1- \beta J_2 \in End \, V_\RR$. Such $X$ anticommutes with $J_3$ and $X^2=-cId$, where $c=\beta^2+\gamma^2+2\alpha\beta\gamma$. Assume here that $c>0$ 
(this is the case, for example, if $|\alpha|<1$ and some of $\beta,\gamma$ is nonzero).

 Thus $X$, after scaling, determines a point $J$ in $S(J_1,J_2)$, anticommuting with $J_3$. Thus, the original
triangle $\triangle J_1J_2J_3$, is subdivided by the twistor line $S(J,J_3)$ into two twistor triangles 
$\triangle JJ_3J_1$ and $\triangle JJ_3J_2$,
with the angle $\sphericalangle JJ_3$ equal $\frac{\pi}{2}$. 


\vspace*{0.7cm}

\input{triangle2.pic}
\vspace*{-15.5cm}
\begin{center}
Picture 2: subdividing a triangle
\end{center}

\bigskip

If $c>0$ then we normalize $X$ to a reflection operator $R=\frac{1}{\sqrt{-c}}X$,
giving rise to the twistor line which we denote by $S(R,J_3)$.

If $c=0$ then $X$ is a nilpotent operator $N$, giving rise to the twistor line which we denote by  $S(N,J_3)$.

In the above picture the result $J$ of normalization of $X$ itself belongs to $S(J_1,J_2)$.
It is not true in general, say, for $c\leqslant 0$, as neither of the reflection or the nilpotent
operator is a complex structure. 


Denote the formed
subtriangles by $\triangle^{\prime}$ and $\triangle^{\prime\prime}$. 
Note that $\triangle^{\prime}$ and $\triangle^{\prime\prime}$ are determined by
the order of vertices of $\triangle J_1J_3J_2$.
It is clear that if $g(\triangle_1)=\triangle_2$, where $\triangle_1, \triangle_2$
are some twistor triangles and $g \in G$, then $g(\triangle_1^{\prime})=\triangle_2^{\prime}$
and $g(\triangle_1^{\prime \prime})=\triangle_2^{\prime \prime}$. On the opposite, if, for example,
$g(\triangle_1^{\prime})=\triangle_2^{\prime}$ then $g(\triangle_1)=\triangle_2$. Next, 
if we have $T(\triangle_1)=T(\triangle_2)$ then, trivially, $T(\triangle_1^\prime)=T(\triangle_2^\prime)$
and if $Q(\triangle_1)=Q(\triangle_2)$ then, slightly less trivially, $Q(\triangle_1^\prime)=Q(\triangle_2^\prime)$ 
(in order to show the equality of the additional 4-th invariant we need to actually use that
$J,J_1,J_2,J_3$ are complex structures).


This construction of switching from $\triangle=\triangle J_1 J_2 J_3$ to 
$\triangle^{\prime}=\triangle J_1 J J_3$ shows
that the transitivity of the action of $G$ on a certain set of twistor triangles $\triangle$ with fixed arbitrary angles is equivalent to the transitivity
of $G$ on the set of corresponding triangles $\triangle^{\prime}$. 


Note that while switching from the original triangle $\triangle$ to the right triangle $\triangle^{\prime}$ might not preserve the invariant $|\cdot|$, it still
preserves the case of trichotomy from which we started, which is the content of the following lemma,
\begin{lem}
\label{preservation-invariant}
 If under the above conditions on $\alpha,\beta\gamma$ we have $|\triangle|>0$ 
then $|\triangle^{\prime}|>0$, if $|\triangle|<0$ then $|\triangle^{\prime}|<0$ and  if $|\triangle|=0$ then $|\triangle^{\prime}|=0$ (it is preserved in the case of a cylindrical triangle).
\end{lem}

The proof of the lemma is obvious, given that
the matrices of the quadratic forms $Q=q|_\langle J_1,J_2,J_3\rangle$ and $Q^{\prime}=q|_\langle J_1,J,J_3\rangle$ are related as $Q^{\prime}=T^t Q T$ for an easily calculated
matrix $T$, so that the determinants of the forms are of the same sign. Below we give a proof by an explicit
calculation from scratch.


\begin{proof}[Proof of Lemma \ref{preservation-invariant}]
Let  $T(\triangle J_1J_2J_3)=(\alpha,\beta,\gamma)$ and $|\triangle J_1J_2J_3|=f(\alpha,\beta,\gamma)
=\alpha^2+\beta^2+\gamma^2+2\alpha\beta\gamma-1$. Denote $\triangle=\triangle J_1J_2J_3$
Let the triangle $\triangle^{\prime}$ have vertices $J_1,\frac{\gamma}{\sqrt{c}} J_1- 
\frac{\beta}{\sqrt{c}} J_2, J_3,$ where $c=\beta^2+\gamma^2+2\alpha\beta\gamma$.
Then $T(\triangle^{\prime})=(-\frac{\gamma+\alpha\beta}{\sqrt{c}},\beta,0)$ and
$$f\left(-\frac{\gamma+\alpha\beta}{\sqrt{c}},\beta,0\right)=\frac{\gamma^2+\alpha^2\beta^2+2\alpha\beta\gamma}{\gamma^2+\beta^2+2\alpha\beta\gamma}+\beta^2-1=\frac{\beta^2}{\gamma^2+\beta^2+2\alpha\beta\gamma}(f(\alpha,\beta,\gamma)-1).$$ 
Given that $\beta \neq 0$ this proves the lemma.
\end{proof}

Further we will always be dealing with the ``right'' triangles in the above described sense. So from now we fix notations, 
considering a triangle $\triangle XJ_2J_3$ such that $XJ_2=-J_2X$, $XJ_3+J_3X=2\alpha Id, J_2J_3+J_3J_2=2\beta Id$, where $X=J$ or $X=R$, or $X=N$ as above. This is the convenient generating set mentioned in the introduction.

\begin{rem}
In fact, as we will see it later, and this follows from Theorem \ref{Algebraic-theorem}, being hyperbolic, or spherical, or cylindrical can be considered as a property of the algebra $\mathcal H=\mathcal H(J_1,J_2,J_3)$, and does not depend on
the choice of the triangle, whose vertices generate this algebra.
\end{rem}

\end{comment}


\section{When is $\mathcal H_{\alpha,\beta,\gamma}$ a quaternionic algebra?}
\label{When-section}
In this section we prove Theorem \ref{Algebraic-theorem}. 


Let $\mathcal H=\mathcal H_{\alpha,\beta,\gamma}$ be the algebra over $\RR$ given by generators and defining relations $$\langle e_1,e_2,e_3\,|\, 
e^{2}_1=e^2_2=e^2_3=-1, e_1e_2+e_2e_1=2\alpha,
e_2e_3+e_3e_2=2\beta, e_1e_3+e_3e_1=2\gamma\rangle.$$


\begin{comment}
It is easy to study the decomposition of the regular (left or right) representation of the algebra $\mathcal H$
into irreducible representations in the considered case when 
one of $|\alpha|,|\beta|,|\gamma|$ is strictly less than 1,
because in that case $\mathcal H$ contains the algebra of quaternions $\HH$, 

so that the only proper $\mathcal H$-submodules of $\mathcal H$ (giving rise
to the subrepresentations of $\rho_{reg}$) are the left (or right) ideals in $\mathcal H$,
which are cyclic modules over $\HH$. 
\end{comment}
\begin{comment}
Moreover, we have seen
that we can even give a nice description of the geometry of such left (right) ideals:
depending on the sign of $\det\,Q_{\alpha,\beta,\gamma}$

 it is either a 2-sphere,
or one or two points (corresponding to two-sided ideals of $\mathcal H$).
\end{comment}
\begin{comment}
At the same time if we look at  $\mathcal H_{\alpha,\beta,\gamma}$ for general $\alpha,\beta,\gamma$, it is possible that $\mathcal H_{\alpha,\beta,\gamma}$ does not contain 
the algebra of quaternions, and so describing the irreducible representations, and the geometry 
of the left (right) ideals may be noticeably harder than in the quaternionic case.

We want thus to find necessary and  sufficient conditions on general 
$\alpha,\beta,\gamma$ in order for $\mathcal H=\mathcal H_{\alpha,\beta,\gamma}$ to contain the algebra $\HH$.
\end{comment}

In the introduction we defined
the bilinear form $q$ on $\mathcal H$, $$q(u,v)=\frac{1}{\dim_\RR \mathcal H}Tr(\rho_{reg}(uv)),
\dim_\RR \mathcal H=8,$$
and, thus, the associated quadratic form $q(v,v)$, which we will also denote by $q$.
Due to the relations of $\mathcal H$, the form $q$ has a lot in common with the vector-valued
quadratic form $Sq$ on $\mathcal H$ that squares the elements of $\mathcal H$, 
$Sq \colon v\mapsto v^2 \in \mathcal H$.


Introduce the subspaces $V=\langle e_1,e_2,e_3\rangle$, $\widetilde{V}=\langle  \beta-e_2e_3, \gamma-e_3e_1, \alpha-e_1e_2 \rangle$ and $c=\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3$. 
Set $Q_{\alpha,\beta,\gamma}=q|_V$ and $\widetilde{Q}_{\alpha,\beta,\gamma}=q|_{\widetilde{V}}$ and identify these restrictions with their matrices
in the specified bases of the respective subspaces.


We summarize the properties of algebra $\mathcal H$, in particular, the relation between the form $q$
and the square form $Sq$ on $\mathcal H$ 

 in the following proposition.

\begin{prop}
\label{Prop-properties-of-H}
\medskip

\noindent 1) $Sq|_{V}=q|_{V}$ and $\large Sq|_{\widetilde{V}}=q|_{\widetilde{V}}$;

\medskip

\noindent 2) The matrix $\widetilde{Q}_{\alpha,\beta,\gamma}$ is minus the adjugate of the matrix $Q_{\alpha,\beta,\gamma}$;

\medskip

\noindent 3) We have the $q$-orthogonal decomposition 
$\mathcal H=\RR\cdot 1  \oplus V \oplus \widetilde{V}\oplus \RR\cdot c$;

\medskip


\noindent  4) The element $c=\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3$
belongs to  the center of $\mathcal H$.

If $|\alpha|=|\beta|=|\gamma|=1$ and $\gamma=-\alpha\beta$ or, what is the same,  $rk\, Q_{\alpha,\beta,\gamma}=1$ $\iff$ the signature of $Q_{\alpha,\beta,\gamma}$ is $(0,1,2)$, 
the center is 3-dimensional, 
$Z(\mathcal H)=\langle 1,c,z\rangle$ where $z= -\gamma(\alpha-e_1e_2)+(\beta-e_2e_3)-\alpha(\gamma-e_3e_1)$. 


If  the signature of $Q_{\alpha,\beta,\gamma}$ is different from $(0,1,2)$,  
 then  the center is 2-dimensional, $Z(\mathcal H)=\langle 1,c \rangle$; 

\medskip

\noindent 5)  $c^2=
-\det Q_{\alpha,\beta,\gamma} \in \RR\cdot 1 \hookrightarrow \mathcal H$;


\medskip
\noindent  6) We have inclusions between subspaces $cV\subset \widetilde{V}$, $c\widetilde{V}\subset V$.
If $\det\,Q_{\alpha,\beta,\gamma}\neq 0$, then these inclusions become equalities and $c$ acts as an involution permuting
these subspaces, and hence also the larger subspaces $\langle 1,V \rangle$, $\langle c, \widetilde{V} \rangle$;


\medskip

\noindent 7) The pairs of elements $\{\alpha-e_1e_2,\gamma-e_3e_1\}, \{\alpha-e_1e_2,\beta-e_2e_3\},
\{\beta-e_2e_3,\gamma-e_3e_1\}$ anticommute, respectively, with $e_1$, $e_2$ and $e_3$;

\medskip
\noindent  8) The form $Sq|_{V\oplus \widetilde{V}}$ has as its range the subspace 
$\langle 1,c \rangle \subset  Z(\mathcal H)$.


\end{prop}
The equality of the restrictions of quadratic forms $q$ and $Sq$, stated in part 1, assumes that we identify $\RR$
with $\RR\cdot 1 \subset \mathcal H$. 
The first half of part 1 easily follows from the relations of algebra $\mathcal H$. 
The second half is proved in Section \ref{Appendix}.

In part 3 we easily have $1\perp V$, as $e_i$ are imaginary units, hence $Tr(\rho_{reg}(e_i\cdot 1)=0$
and  we also easily have $1\perp \widetilde{V}$.
The directly verifiable part 7 together with the orthogonality relation $c\perp 1$ (which is equivalent to
$Tr(\rho_{reg}(c))=Tr(\rho_{reg}(e_1e_2e_3))=0$, shown also in Section \ref{Appendix})
and the explained relations $1\perp V, 1 \perp \widetilde{V}$
imply most of the orthogonality relations in part 3,  the remaining ones are verified in Section \ref{Appendix}.

For the proofs of parts 4, 5, 6 and 8 we refer to Section \ref{Appendix}.


The property 2) can be directly verified using the calculations for  1)  in Section \ref{Appendix}
 and writing down the matrices 

of the forms $Q_{\alpha,\beta,\gamma}$ and $\widetilde{Q}_{\alpha,\beta,\gamma}$,

\[
Q_{\alpha,\beta,\gamma}=\left(\begin{array}{ccc}
-1 & \alpha & \gamma \\
\alpha & -1 & \beta \\
\gamma & \beta & -1 \\
\end{array}\right),
\widetilde{Q}_{\alpha,\beta,\gamma}=
-\left(\begin{array}{ccc}
1-\beta^2 & \beta\gamma+\alpha & \alpha\beta+\gamma \\
\beta\gamma+\alpha & 1-\gamma^2 & \alpha\gamma+\beta \\
\alpha\beta+\gamma & \alpha\gamma+\beta  & 1-\alpha^2
\end{array}\right).
\]
we see that $\widetilde{Q}_{\alpha,\beta,\gamma}$ is minus the adjugate matrix of $Q_{\alpha,\beta,\gamma}$. We will further use the shorter notations  $Q=Q_{\alpha,\beta,\gamma}$ and $\widetilde{Q}=\widetilde{Q}_{\alpha,\beta,\gamma}$.


\begin{thm}
\label{Signature-theorem}
 If the form $q$ is non-degenerate, then its signature is one of the three
$(2,6), (6,2)$ or $(4,4)$. 
In the cases of signature $(2,6)$ and $(4,4)$ the algebra $\mathcal H$
contains $\HH$ as a subalgebra. In the case of signature $(6,2)$ $\mathcal H$
does not contain $\HH$.
If $q$ is degenerate, its signature is $(1,3,4)$, $(3,1,4)$ or $(1,1,6)$.
For  a  degenerate $q$ the algebra $\mathcal H$ contains $\HH$ only
in the case of signature $(1,3,4)$.
\end{thm}

Note that the part of the statement about $\mathcal H$ not containing $\HH$
in the case of $q$ of signatures $(6,2)$, $(3,1,4)$ and $(1,1,6)$ is trivial: indeed, if there is $\HH \subset \mathcal H$ 
then the restriction $q|_{\HH}$  must have signature $(1,3)$, which is not possible
in the specified cases.

\begin{proof}[Proof of Theorem \ref{Signature-theorem}]
From the definition of $Q$ and the orthogonality relation $1 \perp V$, which is contained in part 3 of 
Proposition \ref{Prop-properties-of-H}, we  have that
the matrix of the restriction of $q$ to the 4-subspace $\langle 1,V\rangle = \langle 1, e_1,e_2,e_3\rangle$
in the basis $1, e_1,e_2,e_3$ is
\[
\left(\begin{array}{cc}
1 & \mathbb{0}_{1\times 3}\\
\mathbb{0}_{3\times 1} & Q \\
\end{array}\right). 


\]


 The definition of $\widetilde{Q}$, the orthogonality relation $c\perp \widetilde{V}$ contained in part 3 of Proposition \ref{Prop-properties-of-H} and part 5 of this proposition  allow us to write down the matrix of the restriction of $q$ 
to the subspace $\langle \widetilde{V},c\rangle$ in the respective basis, 

\[
\left(\begin{array}{cc}
\widetilde{Q} & \mathbb{0}_{3\times 1}\\
\mathbb{0}_{1\times 3} & -\det Q
\end{array}\right).


\] 


As $\widetilde{Q}$  is minus the adjugate matrix of  $Q$, 

the signature of $q$ is completely determined by  the signature of $Q$. 
When $Q$ is non-degenerate, we have the relation $$\widetilde{Q}=-(\det Q)Q^{-1},$$
which shows that $q$ is non-degenerate as well and makes it easy to determine the signature of $\widetilde{Q}$ and of $q$. 

\medskip

Let us now get to classifying all possible signatures of $Q$ and of $\widetilde{Q}$. 
Introducing 
\[
T=\left(\begin{array}{ccc}
1 & \alpha & \gamma \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right),
\mbox{ we get }
T^tQT=\left(\begin{array}{ccc}
-1 & 0 & 0 \\
0 & \alpha^2-1 & \alpha\gamma+\beta \\
0 &  \alpha\gamma+\beta & \gamma^2-1
\end{array}\right).
\]
Introduce the matrix
\[A=
\left(\begin{array}{cc}
\alpha^2-1 & \alpha\gamma+\beta \\
\alpha\gamma+\beta & \gamma^2-1
\end{array}\right),\det\,A=-\det\,Q.
\]
\noindent {\it Case 1.} $A$ has signature $(++)$ if and only if $|\alpha|>1$ and $\det Q<0$ (and
then, as the signature of $Q$ is determined by that of $A$, by choosing appropriate $T$'s we can
see that $|\beta|, |\gamma|>1$ as well). 
Then the signature of $Q$ is $(++-)=(2,1)$ and the signature of $q$ is $(6,2)$.
\medskip

\noindent {\it Case 2.} $A$ has signature $(--)$ if and only if $|\alpha|<1$ and $\det Q <0$ (again, then automatically  
$|\beta|, |\gamma|<1$). The signature of $Q$ in this case is $(---)=(0,3)$ and the signature of $q$ is $(2,6)$.

\medskip

\noindent {\it Case 3.} $A$ has signature $(+-)$ if and only if  $\det Q>0$.
In this case the signature of $Q$ is $(+--)=(1,2)$ and the signature of $q$ is $(4,4)$.
\medskip

In Case 1, as we discussed above, the signature $(6,2)$ guarantees that $\mathcal H$
does not contain $\HH$.


In Case 2 a subalgebra $\HH$ in $\mathcal H$ arises from a pair of anticommuting imaginary units
that can be taken already in the subspace $\langle e_1,e_2,e_3\rangle$. Indeed, if, 
for example, $|\alpha|<1$, then $e_1$ and $e_2$ generate a subalgebra in $\mathcal H$,  
isomorphic to $\HH$, 
as the imaginary unit $\frac{1}{\sqrt{1-\alpha^2}}(\alpha e_1+e_2)$ anticommutes with
$e_1$.

In Case 3 we consider, for example, the plane $P=\langle \alpha-e_1e_2, \gamma-e_3e_1\rangle
\subset \langle e_1,e_2,e_3 \rangle^{\perp}$ and note that actually both of
$\alpha-e_1e_2$ and $\gamma-e_3e_1$ anticommute with $e_1$ (and, of course, one could 
similarly choose analogous anticommuting planes for $e_2$ and $e_3$ as well).
Next, we want to show that $P$ contains an imaginary unit, which,  together with $e_1$,  would give 
us a quaternionic subalgebra $\HH \hookrightarrow \mathcal H$. 
For that we consider the square of a general element of $P$, 
$(x(\alpha -e_1e_2) +y(\gamma-e_3e_1))^2=x^2(\alpha^2-1)-2xy(\alpha\gamma+\beta)
+y^2(\gamma^2-1) \in \RR\cdot 1 \hookrightarrow \mathcal H$.
This is precisely the value $q(x(\alpha -e_1e_2) +y(\gamma-e_3e_1))$, which can be verified
directly or follows by part 1 of Proposition \ref{Prop-properties-of-H}. 
The matrix of $q|_P$
\[\left(\begin{array}{cc}
\alpha^2-1 & -(\alpha\gamma+\beta)\\
-(\alpha\gamma+\beta) & \gamma^2-1
\end{array}\right)
\]
has the determinant equal to $\det A=-\det Q<0$.
So the form $q|_P$ has signature $(+-)$ and it is possible to find a $q$-negative vector
$v=x(\alpha -e_1e_2) +y(\gamma-e_3e_1) \in P, x,y \in \RR$, 
such that $v^2=q(v)=-1 \in \RR\cdot 1 \hookrightarrow \mathcal H$.
  Then the anticommuting pair
$\langle e_1, v \rangle$ determines an embedding 
$\HH \hookrightarrow \mathcal H$. 

\bigskip

If $q$ is degenerate then, as above, we need to consider several cases for $A$.

\medskip

{\it Case 4. $A$ has signature  $(+0)$}. Then $|\alpha|,|\gamma|\geqslant 1$
and they cannot be both equal 1, so that there is at least one of them strictly greater than 1.
Assume, say $|\alpha|>1$ (the subcase when  $|\gamma|>1$ is ruled out in a similar way). We need to consider now $T^t\widetilde{Q}T$,
\[T=
\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0\\
\frac{\alpha\beta +\gamma}{\alpha^2-1} & \frac{\alpha\gamma +\beta}{\alpha^2-1} & 1\\
\end{array}\right),
T^t\widetilde{Q}T=\frac{1}{\alpha^2-1}
\left(\begin{array}{ccc}
-\det Q & -\alpha \det Q & 0\\
-\alpha \det Q & -\det Q & 0\\
0 & 0 & (\alpha^2-1)^2
\end{array}\right),
\]
which, given that $\det Q=0$ amounts to the signature $(+00)=(1,0,2)$ of $\widetilde{Q}$, which, together with the signature 
$(+-0)=(1,1,1)$ of $Q$ and the signature $(+0)=(1,0,1)$ of $q|_{\langle 1,c\rangle}$ gives the signature of $q$ being $(3,1,4)$. 

\medskip

{\it Case 5. $A$ has signature  $(-0)$}. Then $|\alpha|, 
 |\gamma|\leqslant 1$
and if we have both equalities, then the condition $det\, A=0$ means that $\alpha\gamma+\beta=0$,
so that $A=0$, which is impossible in the current case. So in this case at least one of the absolute values
$|\alpha|,|\gamma|$ is strictly less than 1. 
 We just repeat the arguments above and get that
the signature of $T^t\widetilde{Q}T$ is $(0,1,2)$, which, together with 
the signature $(0,2,1)$ of $Q$ and the signature $(1,0,1)$ of $q|_{\langle 1,c\rangle}$, gives the signature of $q$
being $(1,3,4)$. Note that in this case the fact, that some of $|\alpha|, |\gamma|$
must be strictly less than 1 guarantees that $\HH \hookrightarrow \mathcal  H$.

{\it Case 6. $A=0$}. In this case $|\alpha|=|\gamma|=1$ and $\beta=-\alpha\gamma$ (so that 
$|\beta|=1$ as well), and so we have that

$\alpha\beta+\gamma=\alpha\gamma+\beta=\beta\gamma+\alpha=0$. This means that $\widetilde{Q}$
is the zero matrix and in this case the signature of $Q$ is $(0,1,2)$ and  the signature of $q$  is $(1,1,6)$.

We have seen that if $q$ is degenerate then only in the case of signature $(1,3,4)$ we 
actually get that $\HH$ embeds into $\mathcal H$ and so the proof is now complete.
\end{proof}
\begin{rem}
\label{Universality-remark}
Note that we could use the argument, establishing the embedding $\HH \hookrightarrow \mathcal H$,  
in Case 3 for Case 2 as well,
because $\det A=-\det Q>0$ and the condition $|\alpha|<1$ that we have in Case 2 gives us that $q|_P$ is a negatively definite form.
It was illustrative, however, to emphasize that in Case 2 the embedding $\HH \hookrightarrow \mathcal H$
can be provided by the means of finding an anticommuting pair among the basis elements 
of $V=\langle e_1,e_2,e_3\rangle$ already, without
referring to its orthogonal complement.
\end{rem}

In the course of the proof of Theorem \ref{Signature-theorem}
we have seen that all possible signatures of the restriction $Q=q|_{\langle e_1,e_2,e_3\rangle}$,
determined by the choice of a standard generating set $e_1,e_2,e_3$ for our $\mathcal H$  are in one-to-one correspondence with the signatures of our (independent of choice of generators) form $q$: 
$(2,1)\leftrightarrow (6,2)$, $(0,3)\leftrightarrow (2,6)$, $(1,2)\leftrightarrow (4,4)$,
$(1,1,1)\leftrightarrow (3,1,4)$, $(0,2,1)\leftrightarrow (1,3,4)$, $(0,1,2)\leftrightarrow (1,1,6)$.

That is, indeed the signature of $Q$ does not depend on the choice of a standard generating set and so is an isomorphism invariant of $\mathcal H$. This completes the 
proof of the part of the statement of Theorem \ref{Algebraic-theorem} regarding the signature of $Q$.


\begin{cor}
If $Q$ has any of signatures $(0,3)$, $(0,2,1)$, $(1,2)$, or, what is the same,
$\mathcal H$ contains $\HH$, then  $\mathcal H\cong \mathcal H(\varepsilon)
=\HH\oplus \HH c$ for the central element $c$, proportional to 
$\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3$, $c^2=\varepsilon$, $\varepsilon=1,0,-1$
respectively.
\end{cor}

In order to verify the statement of the corollary one just needs to observe 
the (trivial) fact that the (nonzero) central element does not belong to $\HH\subset \mathcal H$,
so, given that $\dim_\RR\mathcal{H}=8$, we have that $\mathcal H=\HH\oplus \HH c$.
 This 
 completes the proof of the part of the statement of Theorem \ref{Algebraic-theorem},
regarding the isomorphism classes of $\mathcal H \supset \HH$. 
\begin{comment}
The properties 1) and 2) give that we have an octahedron formed by generalized twistor lines
in $Gr(2n,V_\CC)$:  $S(e_1,e_2),S(e_2,e_3),S(e_1,e_3)$, 
$S(e_1,\alpha-e_1e_2)$, $S(e_2,\alpha-e_1e_2)$,
$S(e_2,\beta-e_2e_3)$, $S(e_3,\beta-e_2e_3)$
$S(e_3,\gamma-e_1e_3)$, $S(e_1,\gamma-e_1e_3)$,
$S(\alpha-e_1e_2,\beta-e_2e_3)$,
$S(\alpha-e_1e_2,\gamma-e_1e_3)$,
$S(\beta-e_2e_3,\gamma-e_1e_3)$.
\end{comment}

\begin{rem}

When $Q=Q_{\alpha,\beta,\gamma}$ is of signature $(0,3)$,  the triangle 
$\triangle e_1 e_2 e_3$
is formed by 2-spheres (which can be considered as twistor spheres in $\mathcal H$
spanned by anticommuting elements,
or, under the regular representation, as twistor spheres in $End\, \RR^8$). 
Then, as we know, $\alpha=-\cos\,\sphericalangle e_1 e_2, 
\beta=-\cos\,\sphericalangle e_2 e_3,\gamma=-\cos\,\sphericalangle e_3 e_1$.
Normalizing the respective basis of $\widetilde{V}$ as 
$f_1=\frac{\beta-e_2e_3}{\sqrt{1-\beta^2}}, f_2=\frac{\gamma-e_3e_1}{\sqrt{1-\gamma^2}},
f_3=\frac{\alpha-e_1e_2}{\sqrt{1-\alpha^2}}$ so as to have $f_1^2=f_2^2=f_3^2=-1$ and observing that 
$\cos\, \sphericalangle f_1 f_2=-q(f_1,f_2)=-\frac{1}{2}(f_1f_2+f_2f_1)=\frac{\alpha+\beta\gamma}{\sqrt{1-\beta^2}\sqrt{1-\gamma^2}}=
-\frac{(-\alpha)-(-\beta)(-\gamma)}{\sqrt{1-\beta^2}\sqrt{1-\gamma^2}}=
-\frac{\cos\,\sphericalangle e_1e_2-\cos\,\sphericalangle e_2e_3 \cdot
\cos\,\sphericalangle e_3 e_1}{\sin\,\sphericalangle e_2 e_3\cdot
\sin\,\sphericalangle e_3 e_1}$, which is equal, by the spherical cosine law, 
to $\cos\, (\pi- \angle e_1e_3e_2)$,
and similarly for the pairs $f_1,f_3$ and $f_2,f_3$, so that the triangle  $\triangle f_1 f_2 f_3$
is also compact and is {\it polar} with respect to $\triangle e_1 e_2 e_3$,
that is, the distances between its vertices are equal $\pi - \angle e_1e_3e_2, 
\pi - \angle e_1e_2e_3, \pi - \angle e_3e_1e_2$, here 
the angles are taken between the geodesic segments lying on the corresponding 2-spheres 
forming the sides of
$\triangle e_1 e_2 e_3$,
and the distances between the vertices of $\triangle f_1f_2f_3$ are measured in the corresponding spheres, forming the sides of $\triangle f_1f_2f_3$. 
Here we extend the classical terminology for triangles  on a unit 2-sphere 
(see, for example, 
\cite[p. 49]{Rat}) to our twistor triangles.

\end{rem}

\section{The representation theory of $\mathcal H$}
\label{Regular-section}
This section is devoted to 
completing the proof of Theorem \ref{Algebraic-theorem} by classifying the irreducible representations of 
$\mathcal H$ and 
 proving Theorem \ref{Main-theorem}. 


Let us reproduce the respective part of the statement of Theorem \ref{Algebraic-theorem}
as a separate proposition.

\begin{prop}
\label{prop-regular}
The algebra $\mathcal H(1)=\HH\oplus \HH c, c^2=1,$ contains exactly two nontrivial two-sided 
ideals $\HH(1+c), \HH(1-c)$. 


The regular representation $\rho_{reg}$ decomposes as $\rho_1\oplus\rho_2$, where the non-faithful representations $\rho_1,\rho_2$ 
are the only irreducible (non-faithful) 4-representations of $\mathcal H$, $\rho_1,\rho_2$ are given by restrictions
of the regular representations
$\rho_{1}(h)=\rho_{reg}(h)|_{\HH(1+c)},\rho_2(h)=\rho_{reg}(h)|_{\HH(1-c)}$ for all $h \in \mathcal H$, and are isomorphic,
respectively, to the regular representation $\rho_{\HH}$ precomposed with   
 the quotient maps  $\mathcal H\rightarrow \mathcal H/\HH(1-c)\cong\HH, 
\mathcal H\rightarrow \mathcal H/\HH(1+c)\cong \HH$.


\medskip

The algebra $\mathcal H(0)=\HH\oplus \HH c, c^2=0$, contains exactly one nontrivial two-sided 
ideal $\HH c$.  
The 8-representation $\rho_8=\rho_{reg}$ is reducible but not completely reducible. There
are no faithful 4-representations of $\mathcal H$ and there exists exactly one non-faithful
4-representation $\rho_4(h)=\rho_{reg}(h)|_{\HH c}$, isomorphic to the regular
representation $\rho_{\HH}$ precomposed with the quotient map 
$\mathcal H\rightarrow \mathcal H/\HH c \cong \HH$.

\medskip

 The algebra $\mathcal H(-1)=\HH\oplus \HH\cdot c, c^2=-1$ 
 has no nontrivial two-sided ideals.
All proper left ideals are subspaces of dimension 4 in $\mathcal H$, that are of the form $\HH(w+c), w\in S^2\subset\HH, w^2=-1$.


The regular representation $\rho_{reg}$ decomposes as $\rho_0\oplus \rho_0$, where 
$\rho_0$ is the unique, up to isomorphism, irreducible (faithful) 4-representation of $\mathcal H$,
arising from any proper left ideal $\HH(w+c)$, $\rho_0(h)=\rho_{reg}(h)|_{\HH(w+c)}, h\in \mathcal H$.
\end{prop}

The proof of Proposition \ref{prop-regular} is given in subsections \ref{Spherical-irreps}, \ref{Hyperbolic-irreps}, 
\ref{Cylindrical-irreps}. 


\subsection{The case of spherical $\mathcal H=\mathcal H (1)$.}
\label{Spherical-irreps}
A proper  left ideal in $$\mathcal H (1)=\HH \oplus \HH\cdot c, c^2=1,$$
is an $\HH$-submodule of real dimension 4. Denote the generator of such an ideal by $w+c$
for $w\in\HH$.  Then the fact that $\mathcal H (w+c)=\HH(w+c)$
means that  $c(w+c)=q(w+c)$ for some $q \in \HH$. Then $c(w+c)=1+wc=q(w+c)=qw+qc$
means that $q=w$, $qw=w^2=1$, which means $w=\pm 1$ and so we have 
exactly two proper left ideals in $\mathcal H$: $\HH(1+c)\cong \RR^4$ and $\HH(1-c)\cong \RR^4$, with zero intersection, each of which is a two-sided ideal in $\mathcal H$. Clearly $c$ acts on the generator $1+c$ as the identity, so that 
$\rho_{reg}(c)|_{\HH(1+c)}=Id_{\RR^4}$, similarly,  $\rho_{reg}(c)|_{\HH(1-c)}=-Id_{\RR^4}$.

The rest of the statements about the irreducible 4-representations now follows easily. 


\subsection{The case of hyperbolic $\mathcal H=\mathcal H (-1)$.}
\label{Hyperbolic-irreps}
A proper  left ideal in $$\mathcal H(-1)=\HH \oplus \HH\cdot c, c^2=-1,$$
is an $\HH$-submodule of real dimension 4. Denote the generator of such an ideal by $w+c$
for $w\in\HH$.  As above, the fact that $\mathcal H (w+c)=\HH(w+c)$
means that  $c(w+c)=q(w+c)$ for some $q \in \HH$. Then $c(w+c)=-1+wc=q(w+c)=qw+qc$
means that $q=w$, $qw=w^2=-1$, which means $w=xi+yj+zk\in S^2\subset \HH, x^2+y^2+z^2=1$. 

 So we have a sphere $S^2$ of (distinct) left ideals $\HH(w+c),w\in S^2$. 
All representations $h\mapsto \rho_{reg}(h)|_{\HH(w+c)}$ are equivalent,
as the right action of the group of unit quaternions $S^3\subset \HH$ on the set of our
left ideals is isomorphic to the conjugation action of $S^3$ on the sphere of imaginary quaternions $S^2$: 
$\HH(w+c)h=\HH(wh+ch)=\HH h(h^{-1}wh+c)=\HH(h^{-1}wh+c)$,
$w \mapsto h^{-1}wh, h\in S^3, w\in S^2$, and the latter action is transitive. 


As the above calculation shows, $c$ acts on $v:=w+c$, $w=xi+yj+zk$, on the left by 
the left multiplication by $q=w=xi+yj+zk$.
The $c$-invariant subspace $\HH v \subset \mathcal H$ is spanned over $\RR$ by vectors $v,iv,jv,kv$.
We have $cv=qv$, $$c\cdot iv=icv=iqv=(-x\cdot 1-zj+yk)v,$$ 
$$c\cdot jv=jcv=jqv=(-y\cdot 1 + zi - xk )v,$$
and $$c\cdot kv=kcv=kqv=(-z\cdot 1 - yi + xj)v.$$
In the basis $v,iv,jv,kv$ the operator of the left multiplication by $c$ has the following matrix
\[


\left(
\begin{array}{cc|cc}
0  &  -x &           -y         & -z \\
x  &   0           & z           & -y \\
\hline
y  &  -z            & 0           & x \\
z   &  y            & -x           & 0 
\end{array}\right).

\]
Set $\rho_0(h)=\rho_{reg}(h)|_{\HH(w+c)}$ for $h \in \mathcal H$. Then,
as $\mathcal H=\HH\oplus \HH c$ and $\rho_0(c),\rho_0(1),\rho_0(i),\rho_0(j),\rho_0(k)$
are linearly independent over $\RR$, we see that $\rho_0$ is faithful.

\subsection{The case of $\mathcal H (0)$.} 
\label{Cylindrical-irreps}
Arguing similarly to the above it is easy to see that $\mathcal H c=\HH c$ is the only proper left ideal in 
$\mathcal H$ 
 (which is also a two-sided ideal). 
In this case we do not have faithful 4-dimensional representations,
the only 4-dimensional representation comes from 

the regular representation  $\rho_\HH$ of $\mathcal H/\mathcal H c \cong \HH$ precomposed with
 the quotient map $\mathcal H\rightarrow \mathcal H/\mathcal H c$. 

 For the  operator $L_c$, acting on $\mathcal H=\HH\oplus \HH c, c^2=0,$ by the left multiplication by $c$, we have the equality
$Ker\,L_c= Im\, L_c$ of its kernel and image. 


 For example, in the $\RR$-base
$c,ic,jc,kc, 1, i,j,k$ our operator $L_c$ has the matrix 
$$\left(
\begin{array}{cc}
\mathbb 0_{4\times 4}  &  \mathbb 1_{4\times 4} \\
\mathbb 0_{4\times 4}  &  \mathbb 0_{4\times 4} \\
\end{array}\right),$$
where 
$\mathbb 1_{4\times 4}$ is the $4\times 4$-identity matrix and $\mathbb 0_{4\times 4}$
is the $4\times 4$-zero matrix.

So, summarizing our observations for this case we conclude that there is exactly one, up to isomorphism, irreducible representation for each of the dimensions 4 and 8, which we call $\rho_4$ and $\rho_8=\rho_{reg}$, 
and only the latter is faithful. 

\bigskip

\begin{comment}
{\it Case of $Q$ of signature $(1,2)$.} 

In this case $\HH$ is generated by anticommuting imaginary units 
$I=e_1$ and $J=x(\alpha-e_1e_2)+y(\gamma-e_3e_1)$
for certain $x,y \in \RR$. Next, we would like to find the analog of the $J_3$
in the previous case, so as to have  $IJ_3+J_3I, JJ_3+J_3J \in \RR\cdot Id$.
We take as our $J_3$ the element $R=\frac{1}{\sqrt{\alpha^2-1}}(\alpha-e_1e_2)$.
Clearly $R^2=Id$, $R$ anticommutes with $I$ and $JR+RJ \in \RR\cdot Id$.
Let us from now set $\delta=\frac{1}{2\cdot 8}Tr(JR+RJ)$.
We can  study the cyclic $\HH$-submodules in $\mathcal H=\HH\oplus \HH\cdot R$
in a similar way to how it was done in the case of signature $(2,6)$. The only difference will
be the absence of the two-sided ideals in $\mathcal H$, the left ideals always form a 2-sphere
of  a nonzero radius, on which $\HH$ acts transitively, establishing isomorphisms of all
irreducible $4$-representations. Indeed, let $v=w+R\in \mathcal H$ be such element that
$Rv=qv$ for $q\in \HH$. Let $Rw=w_1+w_2R$ for $w_1,w_2 \in \HH$. Then
$Rv=(w_1+1)+w_2R=qw+qR$ so that $q=w_2=q$ and $qw=w_2w=w_1+1$
from where we write down $$w_2w-w_1-1=0.$$
Next, $Rw=R(xId+aI+bJ+cK)=xR-aIR+b(-JR+2\delta Id)-cI(-JR+2\delta Id)
=(2\delta bId-2\delta cI)+(xId-aI-bJ+cK)R$.
Now $w_2w=(xId-aI-bJ+cK)(xId+aI+bJ+cK)=(x^2+a^2+b^2-c^2)Id-2bc I+2ac J + 2cx K =w_1+1=
(2\delta b+1)Id-2\delta cI$.
Now, equating the respective coefficients  we immediately get
$$x^2+a^2+(b-\delta)^2=1+c^2+\delta^2, bc=\delta c, ac=0, cx=0,$$
which implies that either $c=0$ and we get a 2-sphere of solutions
$$x^2+a^2+(b-\delta)^2=1+\delta^2,$$ or $c\neq 0 $ and then $x=a=0,b=\delta$,
so that we get $0=1+c^2+\delta^2$ implying that the system has no solutions.
The fact that the representations corresponding to the points of the 2-sphere
are isomorphic, is proven similarly to the previous case.
\end{comment}

\begin{comment}
We will be looking at the elements of the form $J_3=ae_2+be_3$ for  $a,b \in \RR$.
Then $IJ_3+J_3I\in \RR \cdot Id$ is automatically satisfied,
and we need to satisfy the second condition. 
We can write the relations 

$$(\alpha-e_1e_2)(ae_2+be_3)+(ae_2+be_3)(\alpha-e_1e_2)=2\beta b e_1-2\gamma b e_2
+2\alpha b e_3-2be_1e_2e_3,$$
$$(\gamma-e_3e_1)(ae_2+be_3)+(ae_2+be_3)(\gamma-e_3e_1)=2\beta a e_1-2\gamma a e_2
+2\alpha a e_3-2ae_1e_2e_3,$$
so that the combination $JJ_3+J_3J$ is obtained by taking the first relation with coefficient $x$
and the second one with coefficient $y$,
$$JJ_3+J_3J=2\beta(bx+ ay) e_1-2\gamma (bx+ ay) e_2
+2\alpha (bx+ ay) e_3-2(bx+ ay)e_1e_2e_3.$$

Now setting, say, $a=x,b=-y$ we get $JJ_3+J_3J=0$. What is left to check is
if the defined $J_3$ normalizes to an imaginary unit, or to a reflection, or nilpotent operator.
For this we consider $$J^2_3=(xe_2-ye_3)^2=
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
-1 & -\beta \\
-\beta & -1
\end{array}\right)
\left(\begin{array}{c}
x \\ 
y
\end{array}\right) Id.$$

We recall that our $x$ and $y$ were chosen so that the
\[
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
\alpha^2-1 & -(\alpha\gamma+\beta) \\
-(\alpha\gamma+\beta) & \gamma^2-1
\end{array}\right)
\left(\begin{array}{c}
x \\ 
y
\end{array}\right)<0.
\]

Noting that

\[
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
\alpha^2-1 & -(\alpha\gamma+\beta) \\
-(\alpha\gamma+\beta) & \gamma^2-1
\end{array}\right)
\left(\begin{array}{c}
x \\ 
y
\end{array}\right)=\]
\[
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
\alpha^2 & -\alpha\gamma \\
-\alpha\gamma & \gamma^2
\end{array}\right)
\left(\begin{array}{c}
x \\ 
y
\end{array}\right)+
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
-1 & -\beta \\
-\beta & -1\\
\end{array}\right)
\left(\begin{array}{c}
x \\ 
y
\end{array}\right)=\]
\[=
(x\alpha-y\gamma)^2+
\left(\begin{array}{cc}
x & y
\end{array}\right)
\left(\begin{array}{cc}
-1 & -\beta \\
-\beta & -1\\
\end{array}\right)
\left(\begin{array}{c}
x \\ 
y
\end{array}\right)<0,
\]
we conclude that $J_3$ normalizes to an imaginary unit.
So $\mathcal H=\HH\oplus \HH J_3$ and we can apply the 
regular representation study from the previous case.
Moreover, for a normalized $J_3$ we see that $IJ_3+J_3I=2(x\alpha-y\gamma)$
and as $$(x\alpha-y\gamma)^2< - \left(\begin{array}{cc}
-1 & -\beta \\
-\beta & -1\\
\end{array}\right)
\left(\begin{array}{c}
x \\ 
y
\end{array}\right)=1,$$ we conclude
that it is precisely the subcase $c<1$
of the Case of signature $(2,6)$. No! No! No! Easy to see that $xe_2-ye_3 \in \HH$,
where $\HH$ is generated by $I$ and $J$!!!
\end{comment}


Thus, the proof of Proposition \ref{prop-regular} is now complete and so is the proof of Theorem \ref{Algebraic-theorem}.

\subsection {General representations of the cylindrical $\mathcal H=\mathcal H(0)$.} For proving Theorem
\ref{Main-theorem} below, we need to show that a general representation of 
$\mathcal H\cong \mathcal H(0)$ is a sum of irreducible representations that arise from its regular representation.
\begin{comment}
This statement is trivial for the above considered $\mathcal H(\varepsilon)$,
 when $\varepsilon=\pm 1$, as the irreducible representations in that case are 4-dimensional
and, as the corresponding representation spaces are cyclic  $\HH$-modules,  every representation
of $\mathcal H$ is a direct sum of the irreducible ones.

This statement requires a little of explanation when $\varepsilon=0$, as there is an 
irreducible 8-representation which is not a cyclic $\HH$-module, so the above argument 
does not work in precisely the same way. Let us consider this case.
\end{comment}


\begin{comment}
We have $\mathcal H =\HH \oplus \HH c$, where 
$c^2=0$ is our central nilpotent element. 

\begin{lem}
\label{Lemma-kernel}
The subspace 
$\HH c=L_c(\mathcal H)$ is exactly the kernel of the operator $L_c\colon \mathcal H \rightarrow \mathcal H$.
This subspace is the only proper $\mathcal H$-submodule in $\mathcal H$.
\end{lem}

\begin{proof}
The first part of the statement of the lemma is obvious.
Let us prove the second part of the statement. Indeed,  if we have an $\mathcal H$-submodule containing an element $q_1+q_2c$ with $q_1\neq 0$,
then the submodule must also contain $c(q_1+q_2c)=q_1c+q_2cc=q_1c$,
hence it contains $c$ and $q_1=(q_1+q_2c)-q_2c$ and so must coincide with the whole of $\mathcal H$.
\end{proof}
\end{comment}


\begin{comment}
So, assume we have a real vector space $V$ and a representation $\rho\colon \mathcal H \rightarrow 
End(V)$. We will further omit the letter $\rho$ and instead of $\rho(h)y$, $y\in V$, we will write simply $hy$.


Choose a nonzero $x \in V$.
Set $V_1=\mathcal Hx$. Then $V_1$, as a  subrepresentation of $\rho$, is isomorphic to
$\rho_4$ or $\rho_8$.
Assume now that we have defined a subspace $V_k\subset V$ for a certain $k\geqslant 1$
such that $V_k$ is a direct sum of some number of representations $\rho_4$ and $\rho_8$.
Choose $y \notin V_k$ and set $V_{k+1}=V_k+\mathcal Hy$. 
\end{comment}


\begin{comment}
Then we have two cases: 

{\it Case 1.} The intersection $V_k \cap \mathcal Hy$
is zero, in which case we have constructed a subrepresentation  $V_{k+1}=V_k\oplus \mathcal Hy \subset V$, which also decomposes as a sum of representations $\rho_4$ and $\rho_8$.

 {\it Case 2.} The intersection $V_k \cap \mathcal Hy$ is nonzero. Then either $V_k \subset \mathcal Hy$,
so that $V_{k+1}=\mathcal Hy$ which is $\rho_4$ or $\rho_8$, or $V_k \cap \mathcal Hy$
is a proper $\mathcal H$-submodule in both $V_k$ and $\mathcal Hy$,
so that necesarily
$\dim\, \mathcal Hy=8$. It follows that
$V_k\cap \mathcal Hy=c\mathcal Hy\subset V_k= m\rho_4\oplus n\rho_8$ for some $m,n\geqslant 0$,


so that $cy=w_1+w_2, w_1\in W_1= m\rho_4, w_2\in W_2 = n\rho_8$, moreover, by Lemma \ref{Lemma-kernel} 
$w_2$ must be of the form $cw$ for a certain $w \in W_2 \subset V_k$.
If $w_1=0$ then we have the direct sum decomposition $V_{k+1}=V_k\oplus \mathcal H(y-w)$,
where $\mathcal H(y-w)$ is isomorphic to $\rho_4$ and thus the representation $V_{k+1}$
is again a sum of representations $\rho_4$ and $\rho_8$.

If $w_1\neq 0$ then we can write $W_1= m\rho_4= (m-1)\rho_4\oplus \mathcal Hw_1$,
where obviously $\mathcal Hw_1= \rho_4$. Next, $V_k+\mathcal Hy=V_k+\mathcal H(y-w)$,
where now $c(y-w)=w_1$, and we can write $W_1+\mathcal H(y-w)= ((m-1)\rho_4\oplus \mathcal Hw_1)
+\mathcal H(c-w)= (m-1)\rho_4\oplus \rho_8$. Thus $V_{k+1}= (m-1)\rho_4\oplus (n+1)\rho_8$
in this case.

So in every case the subrepresentation $V_{k+1}$ that we construct is again 
a sum of a certain number of representations $\rho_4$ and $\rho_8$.
\end{commen}

\begin{comment}
 and is thus
a proper $\mathcal H$-submodule in both (8-dimensional) $\mathcal Hx$ and $\mathcal Hy$.
It is obviously an $\HH$-submodule, hence it has dimension 4. As the mappings 
$h \mapsto hx, h \mapsto hy$ are bijective, we see that 
$\mathcal Hx \cap \mathcal Hy=\mathcal Hvx=\mathcal Hvy\subset Ker\, L_v|_{V_1}$. Next, 
for an apropriate nonzero $q\in \HH$ we have that $v(y-qx)=0$. This implies that
on the (non-direct) sum $\mathcal Hx+\mathcal Hy=\mathcal Hx+\mathcal H(y-qx)$ the operator $L_v$ in the basis 
$\HH vx, \HH x, \HH(y-qx)$ has the matrix 
\[
\left(
\begin{array}{cc|c}
\mathbb 0_{4\times 4}  &  B & \mathbb 0_{4\times 4}\\
\mathbb 0_{4\times 4}  &  \mathbb 0_{4\times 4} & \mathbb 0_{4\times 4}\\

\hline

\mathbb 0_{4\times 4} & \mathbb 0_{4\times 4} & \mathbb 0_{4\times 4}\\
\end{array}\right),
\]
where, as earlier, $B$ is a 4$\times$4-matrix appearing in the irreducible 8-representation for 
$\mathcal H$. Thus $V_2=\mathcal H x + \mathcal H y$ splits as $\rho_8\oplus \rho_4$.

\end{comment}
\begin{comment}
Next, let us choose $z \in V, z\notin V_2$. Then again, either 
$\mathcal Hz \cap V_2=\{0\}$, in which case we set $V_3=V_2\oplus \mathcal Hz$,
where $\mathcal Hz=\rho_8$ or $\rho_4$,
or there is a nonzero intersection, which must be again $\mathcal Hvz\subset Ker\,L_v|_{V_2}=\mathcal Hz \cap V_2$.

For  $V_2$ arising from the Case 1 above we have that that $Ker\, L_v|_{V_2}=
\mathcal Hvx \oplus \mathcal Hvy$, so that $\mathcal Hvz \subset \mathcal Hvx \oplus \mathcal Hvy$. Hence
for appropriate $q_1,q_2 \in \HH$ we have that $z-q_1x-q_2y \in Ker\,L_v$
and we can write $V_3=V_2\oplus \HH(z-q_1x-q_2y)=\rho_8\oplus \rho_8\oplus \rho_4$
or $\rho_4\oplus\rho_4\oplus \rho_4$, depending on what kind of decomposition held for $V_2$.

For $V_2$ arising from the Case 2 above we have that $\mathcal Hvz\subset \mathcal Hvx=\mathcal Hvy$,
and so there are $q_1,q_2 \in \HH$, such that $y-q_1x, z-q_2x \in Ker\, L_v$. Then $V_3=V_2+\mathcal Hz=(\HH vx\oplus \HH x) \oplus \HH (y-q_1x) \oplus \HH (z-q_2x)=\rho_8\oplus \rho_4\oplus \rho_4$.
\end{comment}
\begin{comment}
 Continuing in the same way we get a chain of 
strictly increasing subspaces $V_1\subset V_2\subset V_3\subset \dots V$, so that
after finitely many steps we end up with the representation $\rho$ of $\mathcal H$, $\rho=m\rho_8\oplus n\rho_4$
for appropriate integers $m,n\geqslant 0$ (obviously $8m+4n=\dim\, V$).
Let us formulate this fact as a lemma.
\end{comment}
\begin{prop}
\label{Prop-special-representation}
For every representation $\rho \colon \mathcal H(0) \rightarrow End\,V_\RR$ 

 we have $\rho=k\rho_8\oplus l\rho_4$
for appropriate integers $k,l\geqslant 0$. 
\end{prop}

\begin{proof}
Let us write $\mathcal H=\HH\oplus \HH c$, where $c \in Z(\mathcal H),c^2=0$. 
Then we have $Im\,\rho(c)\subset Ker\,\rho(c)$. Both subspaces $Im\,\rho(c), Ker\,\rho(c)\subset V_\RR$
are $\HH$-invariant, so, choosing an $\HH$-invariant complement $U\subset V_\RR$ to $Ker\,\rho(c)$
we get that $\rho(c)$ induces an isomorphism $U \cong Im\,\rho(c)$. Set $4l=\dim\, U$.
Similarly, there is an
$\HH$-invariant complement to $Im\,\rho(c)$ in $Ker\,\rho(c)$, of dimension $4k$ for an appropriate $k$.
 Hence  

we can write $\rho=k\rho_4\oplus l\rho_8$. 

\end{proof}


\subsection{The proof of Theorem \ref{Main-theorem}}
\label{Main-theorem-subsection}
Let us deal first with the uniqueness statements.
The part of the statement of Theorem \ref{Main-theorem} for hyperbolic $\mathcal H\cong \mathcal H(-1)$
follows from the statement of Proposition \ref{prop-regular}, that there is exactly one irreducible representation $\rho_0$ of $\mathcal H$,
which is a cyclic $\HH$-module.  
 Hence an arbitrary $4n$-representation $\rho$ of $\mathcal H$ is isomorphic to $n\rho_0=\bigoplus^n \rho_0$.

The part of the statement for spherical $\mathcal H=\mathcal H(1)$ follows from the statement
of Proposition \ref{prop-regular}, that there are exactly two irreducible 4-representations $\rho_1,\rho_2$
of $\mathcal H$, which are non-faithful and correspond to 
factoring $\mathcal H \rightarrow \HH$ with respect to each of the two ideals found in this case. Hence, an arbitrary representation can be written as $\rho=
k\rho_1\oplus (n-k)\rho_2$ 
and the fact that $\rho_1(c)=Id_{\RR^4}$ and $\rho_2(c)=-Id_{\RR^4}$ explained in 
\ref{Spherical-irreps} tells us that 

$Tr(\rho_1(c))=4$ 
and $Tr\,\rho_2(c)=-4$, so that $Tr\,\rho(c)=4(2k-n)$, which uniquely identifies $k$ and hence multiplicities of both $\rho_1$
and $\rho_2$  in the decomposition of $\rho$. 

\begin{comment}
Now, as $\det\, Q_{\alpha,\beta,\gamma} \neq 0$, 
by Proposition \ref{Prop-properties-of-H},
$Z(\mathcal H)=\langle 1, c\rangle$, $1^2=1$, $c^2=-1$.
The plane $\rho(Z(\mathcal H))\subset End\,V_\RR$ is endowed with the  metric $(A,B)=-\frac{1}{4n}Tr(AB)$,
introduced in \ref{subsec-pseudometric}.
The isometry class of the plane $(\rho(\mathcal Z(\mathcal H)),(\cdot,\cdot))$ is uniquely determined by $(\rho(1),\rho(1)), (\rho(c),\rho(c))$ and $(\rho(c),\rho(1))$, and so 
is, actually, uniquely determined by
$(\rho(c),\rho(1))=-Tr\,\rho(c)$. 
As $c$ is proportional to $\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3$, and we have  $Tr(\rho(e_j))=0,j=1,2,3$, 
so that $Tr(\rho(\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3))=-Tr\,\rho(e_1e_2e_3)$,
the trace  $Tr\,\rho(e_1e_2e_3)$ also identifies both the isometry class of
$(\rho(\mathcal Z(\mathcal H)),(\cdot,\cdot))$ and the multiplicity $k$ uniquely (and is uniquely determined by any of those, as well).
\end{comment}

Note that $k=0,\dots,n$, where the extremal cases $k=0,n$
correspond to  non-faithful representations, so that there are total of $n+1$ non-equivalent 
representations of the spherical algebra $\mathcal H$.


In the case $\mathcal H\cong \mathcal H(0)$, by Proposition \ref{prop-regular}, the irreducible representations of the cylindrical algebra $\mathcal H$
are the 8-representation $\rho_8$ and the (non-faithful, factoring through $\HH$) 4-representation $\rho_4$. 

 Proposition \ref{Prop-special-representation} tells us 
that an arbitrary representation $\rho$ of $\mathcal H(0)$ is isomorphic to a sum of 
these representations, $\rho=k\rho_8 \oplus l\rho_4$, where $8k+4l=4n$. The number of all such possible non-equivalent representations, including the
trivial one, $\mathcal H \rightarrow \HH$, is $\lfloor \frac{n}{2}\rfloor+1$. It is also clear that
$rk\, \rho(c)=4k$, which uniquely identifies the multiplicities $k$ and $l$.

The existence part now follows for the case $a)$ $\mathcal H_{\alpha,\beta,\gamma}\cong\mathcal H(-1)$ 
from existence and faithfulness of the 4-dimensional representation $\rho_0$, 
and in the case $b)$ 


it follows from  
the obvious faithfulness of $\rho_{reg}\colon \mathcal H \rightarrow End\,\RR^8$.


\section{Proof of Theorem \ref{Theorem-relation}}
\label{Proof of Theorem-relation}

Let us sketch the plan of the proof.  Let $S\subset Compl \subset End\,V_\RR$
be a compact twistor line and $I_1,I_2,I_3 \in S$ be linearly independent 
complex structure operators, $G_\HH=G_{I_1}\cap G_{I_2} \cap G_{I_3}=G_{I_1}\cap G_{I_2}$. 
Set $(\alpha,\beta,\gamma){=}T(\triangle I_1 I_2 I_3)$. 
 As it was explained in the introduction,
given $(g_1,g_2,g_3)\in m^{-1}(G_\HH)$ we can construct a (possibly degenerate) twistor triangle 
$\triangle I_1 I_2J_3,J_3= g_2(I_3)=g_1^{-1}(I_3),$ formed by the lines $S=S(I_1,I_2),g_2(S),g_2g_3(S)$, $g_2(I_3)\in g_2(S) \cap g_2g_3(S)$. As $Tr(I_2I_3)=Tr(g_2(I_2I_3))=Tr(g_2(I_2)g_2(I_3))=Tr(I_2J_3)$
and, similarly, $Tr(I_1I_3)=Tr(g_1^{-1}(I_1I_3))=Tr(I_1J_3)$, we have $T(\triangle I_1 I_2 J_3)=(\alpha,\beta,\gamma)$,


so that there is a mapping $$\tau\colon m^{-1}(G_\HH) \rightarrow \mathcal T_{\alpha,\beta,\gamma},$$
$$(g_1,g_2,g_3)\mapsto \triangle I_1 I_2 g_2(I_3),$$


where $\mathcal T_{\alpha,\beta,\gamma}=\{\mbox{triangles }\triangle I_1I_2J_3\subset Compl \mbox{ with }T(\triangle I_1I_2J_3)=(\alpha,\beta,\gamma)\}$. Here $I_1,I_2$ are fixed. 

First, we can easily see that $\mathcal T_{\alpha,\beta,\gamma}$ is naturally a union of $G_\HH$-orbits,  
which are in one-to-one correspondence with the $n+1$
equivalence classes of representations 
$\rho\colon \mathcal H_{\alpha,\beta,\gamma}\rightarrow End\,V_\RR$ of the spherical algebra
$\mathcal H_{\alpha,\beta,\gamma}$. There are exactly two orbits in $\mathcal T_{\alpha,\beta,\gamma}$,
that are one-element sets, each of which consists of a degenerate triangle, supported on $S$, these orbits correspond to the two classes of nonfaithful representations. 


Second, we explicitly determine the fibers of $\tau$ over the two $G_\HH$-inequivalent degenerate triangles in the image of $\tau$,
supported on $S$:  these fibers are the trivial and the $SO(3)$-type
components of $m^{-1}(G_\HH)$ listed in the statement of Theorem \ref{Theorem-relation}. 
\begin{comment}
In fact, further we develop a general description of the components of $m^{-1}(G_\HH)$, that includes, in particular, these two, but we find it illustrative,  at this point, to give a concrete example.
\end{comment}

Third, we show that $\tau$ is onto, in particular, the connected components of 
$m^{-1}(G_\HH)$ reduce to those of the fibers $\tau^{-1}(G_\HH \cdot \triangle I_1 I_2 J_3)$ 
of $\tau$ 
over the $G_\HH$-orbits in $\mathcal T_{\alpha,\beta,\gamma}$. 

Fourth, we calculate the fibers of $\tau$ over individual nondegenerate triangles in $\mathcal T_{\alpha,\beta,\gamma}$.
Each such fiber  is shown to be diffeomorphic to $G_\HH\times G_\HH\times G_\HH$, being thus connected. 


Fifth, 

 we calculate the fibers of $\tau$ over the $G_\HH$-orbits $G_\HH \cdot \triangle I_1 I_2 J_3$ in $\mathcal T_{\alpha,\beta,\gamma}$, 
these fibers, being connected, are the connected components of $m^{-1}(G_\HH)$ described in Theorem \ref{Theorem-relation}. 

Finally, we do a calculation showing that for any triangle in the orbit, corresponding to the 
equivalence class of the balanced representation of $\mathcal H_{\alpha,\beta,\gamma}$,   
its angles, up to taking complements to $\pi$, are equal to the respective angles of 
the spherical triangle $\triangle I_1 I_2 I_3$.

\subsection{ The $G_\HH$-orbit structure of  $\mathcal T_{\alpha,\beta,\gamma}$}


\begin{comment}
If we have a twistor triangle $\triangle J_1J_2J_3$ with $T(\triangle J_1J_2J_3)=T(\triangle I_1I_2I_3)$
then, using the transitivity of the (pseudo-isometric) $G$-action on twistor lines we can assume that
actually $J_1=I_1$ and $J_2=I_2$, and so that our twistor triangle is $\triangle I_1I_2J_3$.
\end{comment}

\begin{comment}
For a triangle $\tau(g_1,g_2,g_3)=\triangle I_1I_2J_3, J_3=g_2(I_3)$ (not necessarily non-degenerate) formed by lines
$S(I_1,I_2)=S, S(I_2,J_3)=g_{2}(S), S(I_1,J_3)=g_{2}g_{3}(S)=g_1^{-1}(S)$, 

 where $g_2(I_3)=J_3$, we have  $T(\triangle I_1 I_2 J_3)=T(\triangle I_1 I_2 I_3)=(\alpha,\beta,\gamma)$. 
\end{comment}
By definition, for every triangle $\triangle I_1I_2J_3 \in \mathcal T_{\alpha,\beta,\gamma}$
we have $T(\triangle I_1I_2J_3)=T(\triangle I_1I_2I_3)=(\alpha,\beta,\gamma)$. 
Writing $T(\triangle I_1 I_2 I_3)=\frac{1}{4n}(Tr\, I_1I_2,Tr\, I_2I_3,Tr\, I_3I_1)=
(-\cos \sphericalangle I_1I_2,-\cos \sphericalangle I_2I_3,-\cos \sphericalangle I_3I_1)$, by the spherical
cosine law we have that $\cos\,\angle I_2I_1I_3=\frac{\cos \sphericalangle I_2I_3-
\cos \sphericalangle I_1I_2 \cos \sphericalangle I_1I_3}
{\sin \sphericalangle I_1I_2 \sin \sphericalangle I_1I_3}=-\frac{\beta+\alpha\gamma}
{\sqrt{1-\alpha^2}\sqrt{1-\gamma^2}}$, where by the $\angle I_2I_1I_3$ we mean the angle formed by the geodesic segments $I_1I_2$ and $I_1I_3$. 
Now the fact that the ratio on the right side is a cosine of a certain angle, 
the angle is not equal to $0$ or $\pi$,
means that $\left|\frac{\beta+\alpha\gamma}
{\sqrt{1-\alpha^2}\sqrt{1-\gamma^2}}\right|<1$ which is precisely the requirement that 
$det\, Q_{\alpha,\beta,\gamma}<0$. As, for example, 
$|\alpha|<1$, 
we see that $\mathcal H_{\alpha,\beta,\gamma}$ contains $\HH$,  

thus, the signature of $Q_{\alpha,\beta,\gamma}$ can only be $(0,3)$. Thus, 

 the algebra $\mathcal H_{\alpha,\beta,\gamma}$  is spherical. 

Next, Theorem \ref{Main-theorem} tells us that there are $n+1$ classes of $G$-equivalent representations
of $\rho\colon \mathcal H_{\alpha,\beta,\gamma} \rightarrow End\,V_\RR$, among which there are
 2 non-faithful representations, factoring through $\mathcal H_{\alpha,\beta,\gamma}\rightarrow \HH$ and $n-1$ faithful representations. Fix an embedding of $\HH\hookrightarrow End\,V_\RR$
which corresponds to the subalgebra in $End\,V_\RR$ generated by $I_1,I_2$,
$\HH\cong \langle Id,I_1,I_2,I_1I_2\rangle \subset End\,V_\RR$, and define the set
$$Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})=
\{\mbox{representations }
\rho\colon \mathcal H_{\alpha,\beta,\gamma}\rightarrow End\,V_\RR,\\ \rho(e_1)=I_1,\rho(e_2)=I_2\}.$$
The group $G_\HH$ acts on $Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$
via the adjoint action 
and, as there are total
of $n+1$ $G$-equivalence classes of general representations $\rho \colon \mathcal H_{\alpha,\beta,\gamma} \rightarrow End\,V_\RR$, we see that $Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$
is a union of $n+1$ $G_\HH$-orbits, that are $G_\HH$-equivalence classes. 


\begin{comment}
where
$\rho(e_1)=I_1,\rho(e_2)=I_2$ (each such $\rho$ is determined by the $\rho(c)$ for the central element $c, c^2=1$, of $\mathcal H_{\alpha,\beta,\gamma}=\HH\oplus \HH c$,
or, equivalently,  by the choice of $\rho(e_3)=J_3\in Compl$). In fact, two such representations are 
$G$-equivalent if and only if they are $G_\HH$-equivalent.  
\end{comment}


\begin{comment}
Here we note that two triangles $\triangle I_1 I_2 J_3$
and $\triangle I_1 I_2 K_3$ are $G$-equivalent if and only if they are $G_\HH$-equivalent.
\end{comment}

Once we make a choice of a central element $c,c^2=1$, each $\rho \in Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$ is uniquely determined by 
the image $\rho(c)$ 

or, equivalently,  by  the  image $\rho(e_3)=J_3\in Compl$. For example,
if we set $c=\frac{1}{r}(\beta e_1 -\gamma e_2 +(\alpha-e_1e_2)e_3),r=\sqrt{-\det\,Q_{\alpha,\beta,\gamma}}$, then $c$ is such a normalized central element and 
$$\rho(e_3)=\rho\left( (\alpha-e_1 e_2)^{-1}(rc - \beta e_1 + \gamma e_2)\right)
=(\alpha Id-I_1 I_2)^{-1}(r\rho(c) - \beta I_1 + \gamma I_2) \in S.$$ 

Thus, there is a $G_\HH$-equivariant bijection between the sets $Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$ and $ \mathcal T_{\alpha,\beta,\gamma}$ 
given by $\rho\mapsto \triangle I_1 I_2\rho(e_3)$.

Two $G_\HH$-equivalence classes of nonfaithful representations lead to two
 degenerate triangles supported on the same sphere $S$, and are determined by the choice 
$\rho(c)=\pm Id$ 

 or, equivalently, by
the choice of $\rho(e_3)=(\alpha Id-I_1 I_2)^{-1}(\pm r Id - \beta I_1 + \gamma I_2) \in S,$
one of which is our fixed $I_3$, we denote the other one by $I_3^\prime$. The relation of $I_3^\prime$ to $I_3$ will be explained below. 
The $G_\HH$-equivalence classes of $n-1$ faithful representations lead to $n-1$ 
$G_\HH$-equivalence classes of non-degenerate triangles $\triangle I_1 I_2 J_3 \in \mathcal T_{\alpha,\beta,\gamma}$. 


\subsection{The trivial component $\tau^{-1}(\triangle I_1 I_2 I_3)$}
For any $(g_1,g_2,g_3) \in \tau^{-1}(\triangle I_1 I_2 I_3)$ we must have $I_3=g_2(I_3)$,
that is, $g_2\in G_{I_2} \cap G_{I_3}=G_\HH$. Next, the requirement $g_1g_2g_3 \in G_\HH$
tells us that $I_3=g_1g_2g_3(I_3)=g_1g_2(I_3)=g_1(I_3)$, that is $g_1\in G_{I_1}\cap G_{I_3}=G_\HH$
and then, clearly, $g_3 \in G_\HH$. That is, we have the inclusion $\tau^{-1}(\triangle I_1 I_2 I_3)
\subset G_\HH\times G_\HH \times G_\HH$. The inclusion $G_\HH\times G_\HH \times G_\HH
\subset \tau^{-1}(\triangle I_1 I_2 I_3)$ is obvious, so that we have 
$\tau^{-1}(\triangle I_1 I_2 I_3)=G_\HH\times G_\HH \times G_\HH$, the most trivial component of 
$m^{-1}(G_\HH)$. 


\begin{comment}
\subsection{The three types of components.} First of all, 
we want to  show that $m^{-1}(G_\HH)$ is contained in the union $G_\HH^3 \cup 
\Pi_{j=1}^3 (G_{I_j,S}\setminus G_\HH)  \cup
\Pi_{j=1}^3 (G_{I_j}\setminus G_{I_j,S})$,
where $G_\HH^3=G_\HH\times G_\HH \times G_\HH$.  
We trivially have that $G_\HH^3 \subset m^{-1}(G_\HH)$, this is the
most trivial component of the fiber.

If any of the $g_i$'s in the question satisfying $g_1g_2g_3 \in G_\HH$ belongs to $G_\HH$, then
actually all of them must be in $G_\HH$, so that
$m^{-1}(G_\HH) \subset G_\HH^3 \cup
(G\setminus G_\HH)^3$. 
Indeed, if $g_1g_2g_3 \in G_\HH$ and, 
for example,  $g_2\in G_\HH$, then $I_3\in g_3(S)=(g_1g_2)^{-1}(S)\ni I_1$,
implying  $g_3(S)=(g_1g_2)^{-1}(S)=S$, so that
 $g_1,g_3\in G_{I_1,S}\cap G_{I_3,S}=G_\HH$ (see Lemma \ref{Two-point-lemma}) the latter, other cases are analogous. 


 Let us now consider
 the part of the fiber in $(G\setminus G_\HH)^3$.


The study of the part of the fiber $m^{-1}(G_\HH)\cap \Pi_{j=1}^3
(G_{I_j,S}\setminus G_\HH)$ 
is about study of the relations of the kind $g_1g_2g_3\in G_\HH$ in the subgroup  $G_{I_1,S}G_{I_2,S}G_{I_3, S}\subset G, G_{I_1,S}G_{I_2,S}G_{I_3, S}
\cong SO(3)\times G_\HH$, 
which, modulo $G_\HH$,
is just $SO(3)$ and there is a clear geometric picture behind three independent rotations in $SO(3)$
whose product is the identity element, which we will describe below. 


If any of the elements $g_1,g_3$ in the product $g_1g_2g_3$, satisfying $g_1g_2g_3\in G_\HH$,
takes $S$, as a set, to itself (which means that this $g_j$ is in $G_{I_j,S}$) then each of the three takes $S$,
as a set, to itself.
Indeed, if, for example, $g_3$ is in $G_{I_3,S}$ then $g_3^{-1}(I_2)\in S$ and
$g_3^{-1}(I_2)=g_1g_2g_3(g_3^{-1}(I_2))=g_1(I_2)$, so that $g_1$ takes $I_2$ to 
$g_3^{-1}(I_2)$, therefore, the latter cannot be equal $\pm I_1$. Besides that,  $g_1$ takes, by definition, $I_1$ to $I_1$. But then, by Lemma \ref{Two-point-lemma} $g_1$ takes $S=S(I_1,I_2)=S(I_1,g_3^{-1}(I_2))$ to itself, so that $g_1 \in G_{I_1,S}$
and finally $g_2$ must also take $S$ to itself. 

 So, in that case, we are again dealing with the part of the fiber described in part 2) of Theorem  \ref{Theorem-relation}. If we initially assume that $g_1\in G_{I_1,S}$ then, analogously, we arrive
to the conclusion that $g_3 \in G_{I_3,S}$. 

If $g_1,g_2,g_3$ satisfy $g_1g_2g_3\in G_\HH$ and $g_2 \in G_{I_2,S}$, in particular, $g_2(I_3)\in S$, then,
as we must have $I_3=g_1g_2g_3(I_3)=g_1(g_2(I_3))$, so that $g_2(I_3) \neq \pm I_1$
and, therefore, the element $g_1$ takes $S=S(I_1,g_2(I_3))=S(I_1,I_3)$ to itself. Finally,
$g_3$ takes $S$ to itself as well. 


This finally justifies the inclusion $m^{-1}(G_\HH) \subset G_\HH^3 \cup 
\Pi_{j=1}^3 (G_{I_j,S}\setminus G_\HH)  \cup
\Pi_{j=1}^3 (G_{I_j}\setminus G_{I_j,S})$.


\end{comment}

\subsection{The $SO(3)$-type component $\tau^{-1}(\triangle I_1 I_2 I_3^\prime)$.}
For a triple $(g_1,g_2,g_3) \in \tau^{-1}(\triangle I_1 I_2 I_3^\prime)$
we must have $g_2(I_3)=I_3^\prime \in S$, that is, $g_2$ takes 
$S=S(I_2,I_3)$ to $S=S(I_2,I_3^\prime)$, so that  $g_2\in G_{I_2,S}$, moreover,  
$g_2$ is determined uniquely, up to an element in $G_\HH$.
Then writing $g_1g_2g_3=h \in G_\HH$ we see that $I_3=g_1g_2g_3(I_3)=g_1(I_3^\prime)$,
that is, $g_1$ takes $I_3^\prime \in S$ to $I_3\in S\setminus \{\pm I_1\}$. This, together with
$g_1(I_1)=I_1$  implies that $g_1$ takes $S=S(I_1,I_3)$ to $S=S(I_1,I_3^\prime)$,
that is, $g_1 \in G_{I_1,S}$. Again, such $g_1$ is determined uniquely, up to an element in $G_\HH$.
Finally, $g_3$ must also take $S$ to $S$, so that $g_3\in G_{I_3,S}$ and again, due to the relation 
$g_1g_2g_3 \in G_\HH$, it is determined uniquely, 
up to an element in $G_\HH$.

 


The element $g_j$ as above acts as a rotation of $S$ about the corresponding axis $\{\pm I_j\},j=1,2,3$. 
The subgroups of (isometric) rotations $\langle e^{tI_1}|\,t\in\RR\rangle \subset G_{I_1}, 
\langle e^{tI_2}|\, t\in \RR \rangle \subset G_{I_2},\langle e^{tI_3}|\,t\in\RR \rangle  \subset G_{I_3}$
generate a subgroup in $G_S$ isomorphic to $SO(3)$. 
It is clear that fixing the (unique) rotations $e^{t_1I_1}, e^{t_2I_2}$, such that
$I_3^{\prime}=e^{t_1I_1}(I_3)=e^{t_2I_2}(I_3)$, we get that $e^{-t_1I_1}e^{t_2I_2}$ is a rotation in $SO(3)\subset G_S$ 
about the axis $\{\pm I_3\}$. Then we can find
a unique $e^{t_3I_3}$, such that $e^{-t_1I_1}e^{t_2I_2}e^{t_3I_3}=1 \in G_\HH$.

\vspace*{1cm}
\hspace*{4cm}
\input{SO3-corr.pic}
\vspace*{-12.5cm}
\begin{center}
Picture 2: Obtaining a relation among rotations around $I_1,I_2,I_3$.
\end{center}

Thus, allowing $g_1,g_2,g_3$ to be defined up to elements in $G_\HH$, we get
that $\tau^{-1}(\triangle I_1 I_2 I_3^\prime)=e^{-t_1I_1}G_\HH \times e^{t_2I_2}G_\HH
\times e^{t_3I_3}G_\HH$. This is the $SO(3)$-type component from the statement
 of Theorem \ref{Theorem-relation}. 


\begin{comment}
\subsection{The fibers of $\tau$ over non-degenerate triangles.}
It was, basically,  shown above that for $(g_1,g_2,g_3)\in m^{-1}(G_\HH)$
the assumption that $g_2 \in G_\HH$ (respectively, $g_2\in G_{I_2,S}$) implies
that $g_1,g_3\in G_\HH$ (respectively, $g_1\in G_{I_1,S},g_3\in G_{I_3,S}$).
Similarly one shows that if for $(g_1,g_2,g_3)\in m^{-1}(G_\HH)$ we assume that 
$g_1$ or $g_3$ is in $G_\HH$ (respectively,  $g_1\in G_{I_1,S}$ or $g_3\in G_{I_3,S}$),
then actually all $g_j\in G_\HH, j=1,2,3$ (respectively, $g_j\in G_{I_j,S},j=1,2,3$). 
That is, we have  the inclusion $m^{-1}(G_\HH) \subset G_\HH^3 \cup 
\Pi_{j=1}^3 (G_{I_j,S}\setminus G_\HH)  \cup
\Pi_{j=1}^3 (G_{I_j}\setminus G_{I_j,S})$, where $G_\HH^3:=G_\HH\times G_\HH\times G_\HH$. 
So the fibers of the mapping $\tau$ over the nondegenerate triangles are all contained
in $\Pi_{j=1}^3 (G_{I_j}\setminus G_{I_j,S})$. 
\medskip

 Now let us get to the existence and the structure of the most non-trivial components of the fiber 
in part 3) of Theorem \ref{Theorem-relation}.
First of all, we explain the part of the statement of Theorem \ref{Theorem-relation} about the geometry
of triangles that we will use in our search for these components of $m^{-1}(G_\HH)$.  
\end{comment}

\subsection{The surjectivity of $\tau \colon m^{-1}(G_\HH) \to \mathcal T_{\alpha,\beta,\gamma}$}


  


Given a triangle $\triangle I_1 I_2 J_3 \in \mathcal T_{\alpha,\beta,\gamma}$ with
$T( \triangle I_1 I_2 J_3)=T(\triangle  I_1 I_2 I_3)=(\alpha,\beta,\gamma)$, 
we need to find a triple  $(g_1,g_2,g_3) \in m^{-1}(G_\HH)$, such that 
$\tau(g_1,g_2,g_3)= \triangle I_1 I_2 J_3$. 


\begin{comment}
We know from the introduction that an arbitrary twistor triangle with a side $S$ 
gives rise  to (non-unique) triples $I_1,I_2,I_3$ of points in $S$ and to elements  $g_1,g_2,g_3$, $g_j\in G_{I_j}, I_j\in S, j=1,2,3$, 

such that $g_1g_2g_3(S)=S$, that is $g_1g_2g_3\in G_S$.
\end{comment}
As we discussed in the introduction, we can find $g_1\in G_{I_1},g_2\in G_{I_2}$ 
such that 

$g_2(S)=S(I_2,J_3)$ and $g_1^{-1}(S)=S(I_1,J_3)$. 
As $G_{J_3}$ acts transitively on the set of twistor lines containing 
$\pm J_3$ (see \cite{Twistor-lines}), we can take $f_3\in G_{J_3}$, such that $f_3(S(I_2,J_3))=S(I_1,J_3)$
(here we do not assume that $f_3$ takes $I_2$ to $I_1$, it is merely an equality of sets), and set $g_3=g_2^{-1}f_3g_2 \in 
G_{g_2^{-1}(J_3)}$. Then, obviously $g_1g_2g_3(S)=S$. 


Here, certainly, $g_2^{-1}(J_3)$ need not be equal the initially fixed $I_3$ and while the product
$g_1g_2g_3$ belongs to $G_S$, it need not be in $G_\HH$. We want to modify the triple $(g_1,g_2,g_3)$
so as to satisfy conditions $g_1g_2g_3 \in G_\HH$, $g_j\in G_{I_j},j=1,2,3$, and 
$\tau(g_1,g_2,g_3)=\triangle I_1 I_2 J_3$. 

Set $\widetilde{I_3}=g_2^{-1}(J_3)$ and set $\widetilde{\widetilde{I_3}}=g_1(J_3)$.
We clearly have that $\arccos(-\beta)=d(I_2,I_3)=d(I_2,\widetilde{I_3})\mbox{ and }
\arccos(-\gamma)=d(I_1,I_3)=d(I_1,\widetilde{\widetilde{I_3}}),$
where $d(\cdot,\cdot)$ denotes the (spherical) distance between the points of $S$, see the picture below.

Now let us choose $e^{t_2I_2}\in G_{I_2,S}$ and $e^{t_1I_1} \in G_{I_1,S}$ such that $e^{t_2I_2}(\widetilde{I_3})=I_3 \mbox{ and }e^{t_1I_1}(\widetilde{\widetilde{I_3}})=I_3.$ Then the modified elements $e^{t_1I_1}g_1\in G_{I_1},g_2e^{-t_2I_2}\in G_{I_2}$
still  lead to the same triple of consecutive twistor lines as $g_1,g_2$ did: $(e^{t_1I_1}g_1)^{-1}(S)=g_1^{-1}(e^{-t_1I_1}(S))=g_1^{-1}(S)=S(I_1,J_3)$
and $g_2e^{-t_2I_2}(S)=g_2(S)=S(I_2,J_3)$. Now we have that $e^{t_2I_2}g_3e^{-t_2I_2}\in G_{I_3}$,
indeed: $e^{t_2I_2}g_3e^{-t_2I_2}(I_3)=e^{t_2I_2}g_3(\widetilde{I_3})=e^{t_2I_2}(\widetilde{I_3})=I_3$.
 Obviously the element $e^{t_1I_1}g_1 \cdot g_2e^{-t_2I_2}\cdot e^{t_2I_2}g_3e^{-t_2I_2}$ 
takes $S$ to itself,
moreover,  
$e^{t_1I_1}g_1 \cdot g_2e^{-t_2I_2}\cdot e^{t_2I_2}g_3e^{-t_2I_2}(I_3)=
e^{t_1I_1}g_1 \cdot g_2e^{-t_2I_2}(I_3)=
e^{t_1I_1}g_1 \cdot g_2(\widetilde{I_3})=e^{t_1I_1}(\widetilde{\widetilde{I_3}})=I_3$. This means
that our product $e^{t_1I_1}g_1 \cdot g_2e^{-t_2I_2}\cdot e^{t_2I_2}g_3e^{-t_2I_2}=e^{t_3I_3}h
= he^{t_3 I_3} \in G_{I_3,S}$ for an appropriate $h \in G_\HH$. 
 So for the ``corrected'' elements $e^{t_1I_1}g_1 \in G_{I_1},g_2e^{-t_2I_2} \in G_{I_2},
e^{t_2I_2}g_3e^{-t_2I_2}e^{-t_3I_3}\in G_{I_3}$ their product   $(e^{t_1I_1}g_1) \cdot (g_2e^{-t_2I_2})\cdot (e^{t_2I_2}g_3e^{-t_2I_2}e^{-t_3I_3})=h$ belongs to $G_\HH$
and they map under $\tau$ to the twistor triangle $\triangle I_1 I_2 J_3$. 

That is, for an arbitrary twistor triangle $\triangle I_1I_2J_3 \in \mathcal T_{\alpha,\beta,\gamma}$ 

 we can find $(g_1,g_2,g_3)\in m^{-1}(G_\HH)$ such that $\tau(g_1,g_2,g_3)=\triangle I_1I_2J_3$. 


\vspace*{0.2cm}
\hspace*{2cm}
\input{triangle-relation-2.pic}

\vspace*{-10.50cm}
\begin{center}
Picture 3: modifying the triple $(g_1,g_2,g_3)$ in order to get $g_1g_2g_3\in G_\HH$.
\end{center}

\begin{comment}
The map $g_{1}$ takes isometrically 
the side $g_{2}(g_{3}(S))$ onto the side $S$ so that $\sphericalangle I_1g_{2}(J_3)=
\sphericalangle I_1I_3=\sphericalangle I_1g_{1}(g_{2}(I_3))$, so that there is a rotation
$e^{tI_1}$ such that $e^{tI_1}(g_{1}(g_{2}(I_3)))=I_3$.  Now replacing 
$g_{1}$ with $e^{tI_3}g_{1}$ for an appropriate $t$ we can assume that 
actually $g_{1}g_{2}g_{3}$ fixes $I_3$ and so replacing $g_{3}$ with 
$g_{3}e^{sI_3}$ for an appropriate $s$ we get that $g_{1}g_{2}g_{3}$
induces the identity mapping of $S$ to itself, thus the product $g_{1}g_{2}g_{3}$
belongs to $G_\HH$. 
\end{comment}


\subsection{The $G_\HH^3$-structure of  the fiber $\tau^{-1}(\triangle I_1 I_2 J_3)$.} 
\label{Subsec-individual-fiber}
Given an arbitrary 

nondegenerate triangle $\triangle I_1 I_2 J_3 \in \mathcal T_{\alpha,\beta,\gamma}$ we want to describe $\tau^{-1}(\triangle I_1 I_2 J_3)$. 


As we already know, $\tau^{-1}(\triangle I_1 I_2 J_3)$ is non-empty, so let us choose
$(f_1,f_2,f_3)\in \tau^{-1}(\triangle I_1 I_2 J_3)$. 
Let us now show that $$\tau^{-1}(\triangle I_1 I_2 J_3)=\{(h_1f_1, f_2h_2^{-1}, h_2f_3h_3^{-1})|h_1,h_2,h_3\in G_\HH\}\cong G_\HH^3,$$
the orbit under the action of $G_\HH^3$ defined by 
$(h_1,h_2,h_3)\cdot (f_1,f_2,f_3)=(h_1f_1, f_2h_2^{-1}, h_2f_3h_3^{-1})$. 
The inclusion `$\supset$' is obvious, so we only need to show the inclusion `$\subset$',
that is, given $(g_1,g_2,g_3)\in \tau^{-1}(\triangle I_1 I_2 J_3)$ we need to find $h_1,h_2,h_3 \in G_\HH$
such that $(g_1,g_2,g_3)=(h_1f_1, f_2h_2^{-1}, h_2f_3h_3^{-1})$. 


We recall that, as every twistor line is uniquely identified by its any two non-proportional points,
the condition that $(f_1,f_2,f_3),(g_1,g_2,g_3)\in \tau^{-1}(\triangle I_1 I_2 J_3)$, implying that
$f_2(I_3)=g_2(I_3)$ tells us that we have set-theoretic equalities $f_2(S)=g_2(S), f_2(f_3(S))=g_2(g_3(S))$. 
We note that as $f_1^{-1}(S)=f_2(f_3(S))=g_2(g_3(S))=g_1^{-1}(S)$, 
we have that $g_1f_1^{-1}(S)=S$ and as $g_1f_1^{-1} \in G_{I_1}$ we have that
$g_1=h_1e^{t_1I_1}f_1=h_1f_1e^{t_1I_1}$ for $h_1\in G_\HH$ and an appropriate $t_1 \in \RR$. 

Next, we certainly have that $f_2(S)=g_2(S)$, 
which analogously gives that  $g_2^{-1}f_2(S)=S$, hence $g_2^{-1}f_2=e^{t_2I_2}h_2$
for an appropriate $h_2\in G_\HH$ and $t_2\in \RR$,                
so that $g_2=f_2h_2^{-1}e^{-t_2I_2}$.
At the same time, as $\tau(f_1,f_2,f_3)=\tau(g_1,g_2,g_3)$ and so $f_2(I_3)=g_2(I_3)$, 

we see that $g_2^{-1}f_2(I_3)= I_3$,


and then $g_2^{-1}f_2 \in G_{I_2}\cap G_{I_3}=G_\HH$, that is, $t_2=0$ and $g_2=f_2h_2^{-1}$,

\begin{comment}
or $f_2^{-1}g_2(I_3)=- I_3$ which is only possible when $I_2$ and $I_3$ anticommute ($\beta=0$) and then
it specifies $t_2=\pm\frac{\pi}{2}$ and then $g_2=f_2I_2h_2$ (the different choices of $\pm$ for $I_2$ can be identified as $\pm I_2$ differ by $-Id \in End\,V_\RR$, which acts trivially on $Compl$).
\end{comment}

Now let us figure how much freedom the choice of $g_3$ has with respect to $f_3$.
For that we look at the set-theoretic equality $f_2(f_3(S))=g_2(g_3(S))$
and use that $g_2=f_2h_2^{-1}$.  


The equality  $g_2(g_3(S))=f_2h_2^{-1}(g_3(S))=f_2(f_3(S))$
implies that $g_3^{-1}h_2f_3= e^{t_3I_3}h_3\in G_{I_3,S}$ for appropriate $t_3 \in \RR$ and $h_3\in G_\HH$, so that $g_3=h_2f_3h_3^{-1}e^{-t_3I_3}$. 
The condition $(f_1,f_2,f_3),(g_1,g_2,g_3)\in m^{-1}(G_\HH)$ 
translates into the equality of mappings $Id_S=g_1g_2g_3=h_1f_1e^{t_1I_1}\cdot f_2h_2^{-1} \cdot h_2f_3h_3^{-1}e^{-t_3I_3}=h_1f_1e^{t_1I_1}\cdot f_2f_3h_3^{-1}e^{-t_3I_3}$. 
Now, as $f_1f_2f_3=h\in G_\HH$ we have that $f_2f_3=f_1^{-1}h$
so that replacing $f_2f_3$ with this expression in the formula for $g_1g_2g_3$
we get that $g_1g_2g_3=h_1f_1e^{t_1I_1}\cdot f_1^{-1}hh_3^{-1}\cdot e^{-t_3I_3}
=h_1hh_3^{-1}e^{t_1I_1}e^{-t_3I_3}=Id_S$. From this we see that
we must already have $e^{t_1I_1}e^{-t_3I_3}=Id_S$, so that 
$e^{t_1I_1}=Id_S,e^{t_3I_3}=Id_S$, which immediately specifies $t_1=t_3=0$. 
So finally $(g_1,g_2,g_3)=(h_1f_1, f_2h_2^{-1}, h_2f_3h_3^{-1})$, 
concluding that the set $\tau^{-1}(\triangle I_1 I_2 J_3)$ is diffeomorphic to 
$G_\HH \times G_\HH \times G_\HH$. 
\begin{comment}
Now Theorem \ref{Main-theorem}
tells us that there are $n+1$, up to $G$-equivalence,  representations $\rho\colon \mathcal H \rightarrow 
End\, V_\RR$, among which there are $n-1$ classes of faithful representations.
Thus, given that we fix $I_1$ and $I_2$ determining the side $S(I_1,I_2)$, we get $n-1$ different,
up to $G$-action,  choices for $J_3 \notin S(I_1,I_2)$ to form $\triangle I_1I_2J_3$.
These $n-1$ triangles lead to $n-1$ disjoint families of the above specified type.
\end{comment}
\begin{comment}
\medskip
{\it Case $g_2=f_2I_2h_2$: $I_2$ and $I_3$ anticommute.}

Besides the general solution that we obtained above for general $I_2$ and $I_3$
there exists what looks like an additional solution,  
in the case when $g_2=f_2I_2h_2$. As earlier, requiring that 
$f_2(f_3(S))=g_2(g_3(S))= f_2I_2h_2(g_3(S))$ 
we get $f_3^{-1}(I_2(h_2(g_3(S))))=S$ and the latter mapping clearly takes 
$I_3$ to $-I_3$ and so clearly 

$I_2 f_3^{-1} I_2  h_2  g_3 (I_3)= I_3$
so that $I_2f_3^{-1}I_2h_2g_3 = e^{t_3I_3}h_3$, allowing to write
$g_3=h_2^{-1}I_2f_3I_2e^{t_3I_3}h_3$.
Finally writing $Id_S=g_1g_2g_3
=h_1f_1e^{t_1I_1}\cdot f_2I_2h_2 \cdot h_2^{-1}I_2f_3I_2e^{t_3I_3}h_3=-h_1f_1e^{t_1I_1}\cdot f_2f_3I_2e^{t_3I_3}h_3=-
h_1f_1e^{t_1I_1}\cdot f_1^{-1}hI_2e^{t_3I_3}h_3$
and we get the relation $e^{t_1I_1}I_2e^{t_3I_3}=Id_S$.
The latter mapping maps $I_3$ to $e^{t_1I_1}(-I_3)$ so we must have that 
$e^{t_1I_1}(-I_3)=I_3$ which means that $I_1I_3=-I_3I_1$ ($\gamma=0$)
and then $t_1=\pm \frac{\pi}{2}$. As $I_3$ anticommutes with $I_1,I_2$,
the periods $I_1,I_2$ lie in the equatorial plane orthogonal to the axis $\{I_3,-I_3\}$. 
As for $e^{t_1I_1}I_2e^{t_3I_3}=Id_S$ we must also have that
$e^{t_1I_1}I_2e^{t_3I_3}(I_1)=I_1$ so that $I_2e^{t_3I_3}(I_1)=I_1$, 
and we also  know  
 that $e^{t_3I_3}(I_1)$ lies on the  large circle 
containing $I_1,I_2$. Now $I_2$ rotates around $\{I_2,-I_2\}$ by $\pi$ and takes
$e^{t_3I_3}(I_1)$ back to $I_1$. This means that $d(I_1,I_2)=d(e^{t_3I_3}(I_1),I_2)$ which allows to 
determine $e^{t_3I_3}$ uniquely.

So in the case $\beta=\gamma=0$ given a triple $(f_1,f_2,f_3), f_1f_2f_3=G_\HH,$ giving rise 
to the triangle $\triangle I_1I_2J_3$, $J_3=f_2(I_3)$, we obtain, as above, a triple $(g_1,g_2,g_3) \in G_{I_1}\times G_{I_2} \times G_{I_3}$ such that $g_1g_2g_3\in G_\HH$.
This triple corresponds to the triangle $\triangle I_1I_2(-J_3)$, $-J_3=-f_2(I_3)=g_2(I_3)$. 
In fact, this triple is one of the earlier obtained $n-1$ solutions, that is the triangle 
$\triangle I_1I_2(-J_3)$ corresponds (as well as  the original $\triangle I_1I_2J_3$)
to one of $n-1$ non-equivalent faithful representations of $\mathcal H$.


As in  our case 
we have $\det\,Q_{\alpha,\beta,\gamma}<0$ 
and thus, by Theorem \ref{Algebraic-theorem} 
we have $\mathcal H = \HH \oplus \HH c$
for the central element $c$, $c^2=1$. 
Fix $c=\frac{1}{\sqrt{-\det\,Q_{\alpha,\beta,\gamma}}}(\beta e_1-\gamma e_2+\alpha e_3-e_1e_2e_3)
|_{\beta=\gamma=0}=\frac{1}{\sqrt{1-\alpha^2}}(\alpha e_3-e_1e_2e_3)$. 

 This algebra $\mathcal H$ has exactly two two-sided ideals generated, respectively,  by $1+c$
and $1-c$.


Now 
replacing the generators $e_1,e_2,e_3$ respectively with $I_1,I_2,J_3$ we get that the sign change $J_3\mapsto -J_3$
simply permutes these ideals, $1+\frac{1}{\sqrt{1-\alpha^2}}(\alpha J_3-I_1I_2J_3) \mapsto 
1-\frac{1}{\sqrt{1-\alpha^2}}(\alpha J_3-I_1I_2J_3)$. 
Thus if we had $\mathcal H(I_1,I_2,J_3)=\rho (\mathcal H)$ 
for some $\rho=k\rho_1\oplus l\rho_2$, then 
$\mathcal H(I_1,I_2,-J_3)=\widetilde{\rho} (\mathcal H)$ for $\widetilde{\rho}=l\rho_1\oplus k\rho_2$.
So indeed our ``new'' triangle $\triangle I_1I_2(-J_3)$ and the respective family of triples
$(g_1,g_2,g_3)$ are ones of the $n+1$ above described in the case of general $\alpha,\beta,\gamma$.

\end{comment}


\begin{comment}
So in that case we have an additional family of solutions (a connected component of
the set of solutions in $G\times G\times G$)
of the kind $(g_1,g_2,g_3)=(g_{1,\HH}f_1I_1, f_2I_2g_{2,\HH}, g_{2,\HH}^{-1}I_2f_3I_2e^{t_3I_3}g_{3,\HH})$,
which looks like the previous set of solutions rotated by the corresponding action of $I_1$
and $I_2$ on the coordinate entries. It is clear that this action by $I_1$ and $I_2$
acts as involution on the two families of solutions: if we apply it one more time to  $(g_{1,\HH}f_1I_1, f_2I_2g_{2,\HH}, g_{2,\HH}^{-1}I_2f_3I_2e^{t_3I_3}g_{3,\HH})$ we get
$(g_{1,\HH}f_1I_1I_1, (f_2I_2g_{2,\HH})I_2, 
I_2(g_{2,\HH}^{-1}I_2f_3I_2e^{t_3I_3}g_{3,\HH})I_2e^{t_3I_3})=((-g_{1,\HH})f_1,
f_2\cdot (-g_{2,\HH}), (-g_{2,\HH})^{-1}f_3e^{-t_3I_3}e^{t_3I_3}(-g_{3,\HH}))$ which is
a solution of the first type. Now  collecting the connected components of
the set of solutions over $n+1$ distinct spherical triangles, we get the required $2(n+1)$
components of the specified type.
\end{comment}

\medskip

\subsection {The fiber $\tau^{-1}(G_\HH\cdot \triangle I_1 I_2 J_3)$}
For any $g \in G_\HH$ and any $\triangle I_1 I_2 J_3 \in \mathcal T_{\alpha,\beta,\gamma}$ 
we have $g\cdot \triangle I_1 I_2 J_3=\triangle g(I_1)g(I_2)g(J_3)
=\triangle I_1 I_2 g(J_3)$, 
so that the orbit $G_\HH\cdot \triangle I_1 I_2 J_3$ is the subset $\{\triangle I_1 I_2 g(J_3)\,|\,g \in G_\HH\}
\subset \mathcal T_{\alpha,\beta,\gamma}$. 
As we have seen in \ref{Subsec-individual-fiber}, the preimage of an individual triangle 
$\tau^{-1}(\triangle I_1 I_2 J_3)=
\{(h_1f_1, f_2h_2^{-1}, h_2f_3h_3^{-1})|h_1,h_2,h_3\in G_\HH\}$, where $(f_1,f_2,f_3)$ 
is an arbitrary point in $\tau^{-1}(\triangle I_1 I_2 J_3)$, has an obvious structure
of the homogeneous space diffeomorphic to $G_\HH^3$. 
The preimage  $\tau^{-1}(g\cdot \triangle I_1 I_2 J_3), g \in G_\HH,$ contains the point 
$g\cdot (f_1,f_2,f_3)=(gf_1g^{-1},gf_2g^{-1},gf_3g^{-1})$
and so, again by \ref{Subsec-individual-fiber}, we can write $$\tau^{-1}(g\cdot \triangle I_1 I_2 J_3)=\{(h_1gf_1g^{-1}, gf_2g^{-1}h_2^{-1}, h_2gf_3g^{-1}h_3^{-1})|h_1,h_2,h_3\in G_\HH\}=$$ $$=
\{(h_1f_1g^{-1}, gf_2h_2^{-1}, h_2f_3h_3^{-1})|h_1,h_2,h_3\in G_\HH\}.$$
Then we have $$\tau^{-1}(G_\HH\cdot \triangle I_1 I_2 J_3)=\underset{g\in G_\HH}{\bigcup}
\tau^{-1}(g\cdot \triangle I_1 I_2 J_3)=$$ $$
=\{(h_1f_1g^{-1}, gf_2h_2^{-1}, h_2f_3h_3^{-1})|g,h_1,h_2,h_3\in G_\HH\},$$
so that the preimage $\tau^{-1}(G_\HH\cdot \triangle I_1 I_2 J_3)$ has an obvious structure of
a homogeneous $G^4_\HH$-manifold,

namely the action is defined by $$(h_1,g,h_2,h_3)\cdot (f_1,f_2,f_3)=(h_1f_1g^{-1}, gf_2h_2^{-1}, h_2f_3h_3^{-1}).$$ Thus, up to renaming the entries of tuples in $G_\HH^4$, it is a connected subset in $m^{-1}(G_\HH)$ of the form specified in
Theorem \ref{Theorem-relation}. In order to
determine the diffeomorphism type of the orbit $G_\HH^4\cdot (f_1,f_2,f_3)$ for an arbitrary $(f_1,f_2,f_3)\in \tau^{-1}(G_\HH\cdot \triangle I_1 I_2 J_3)$ we need to find the stabilizer of that point.  
The stabilizer, by definition of the action,
consists of those 4-tuples $(h_1,g,h_2,h_3)\in G_\HH^4$ which satisfy
$(h_1,g,h_2,h_3)\cdot (f_1,f_2,f_3)=(h_1f_1g^{-1}, gf_2h_2^{-1}, h_2f_3h_3^{-1})=(f_1,f_2,f_3)$.

Equating the first entries, we get  $h_1f_1g^{-1}=f_1$ or $g=f_1^{-1}h_1f_1$, thus
$g \in G_\HH \cap f_1^{-1}G_\HH f_1$. As $f_1^{-1}(I_1)=I_1$, $f_1^{-1}(I_3)=f_2f_3(I_3)=f_2(I_3)=J_3$, we see that $g$ must pointwise stabilize $I_1,I_2$ and $J_3=f_1^{-1}(I_3)$,
so that in the end it stabilizes pointwise the subalgebra $\mathcal H(I_1,I_2,J_3)\subset End\,V_\RR$
under the $G$-action. 
Further we denote the pointwise $G$-action stabilizer of  $\mathcal H(I_1,I_2,J_3)$ by $G_{\mathcal H(I_1,I_2,J_3)}\subset G_\HH$. 
For any $g \in G_{\mathcal H(I_1,I_2,J_3)}$ we can determine, in a unique way, 
the respective $h_1=f_1gf_1^{-1}\in G_\HH$.

Equating the second entries we get $gf_2h_2^{-1}=f_2$, that is, $g\in G_\HH \cap f_2 G_\HH f_2^{-1}$. This condition is equivalent to $g$ stabilizing pointwise $I_1,I_2$ and $J_3=f_2(I_3)$,
that is, $g\in  G_{\mathcal H(I_1,I_2,J_3)}$, so that no further restriction is added, and, again, for any such $g$ we uniquely determine 
$h_2=f_2^{-1}g f_2 \in G_\HH$. 

Equating the third entries and using the previously determined $h_2$ we get that $h_3=f_3^{-1}h_2f_3=f_3^{-1}\cdot f_2^{-1}g f_2 \cdot f_3$.
As we know that $(f_1,f_2,f_3)\in m^{-1}(G_\HH)$, so that $f_1f_2f_3=h\in G_\HH$, we
get $f_2f_3=f_1^{-1}h$, and then, using the previously determined $h_1$,  
we get $h_3=h^{-1} \cdot f_1 g f_1^{-1}\cdot  h=h^{-1}h_1h\in G_\HH$. 

Thus, $Stab_{G_\HH^4}(f_1,f_2,f_3)=\{(g,f_1gf_1^{-1}, f_2^{-1}g f_2, f_3^{-1}f_2^{-1}g f_2f_3)\,|\, g \in G_{\mathcal H(I_1,I_2,J_3)}\} \cong G_{\mathcal H(I_1,I_2,J_3)}$  
and $\tau^{-1}(G_\HH\cdot \triangle I_1 I_2 J_3)\cong G_\HH^4/G_{\mathcal H(I_1,I_2,J_3)}$,
here we identify $G_{\mathcal H(I_1,I_2,J_3)}$ with its image in $G_\HH^4$
under the above specified isomorphism $G_{\mathcal H(I_1,I_2,J_3)}\cong Stab_{G_\HH^4}(f_1,f_2,f_3)$. 

Now $\mathcal T_{\alpha,\beta,\gamma}$ is a union of $n+1$ distinct orbits 
of the form $G_\HH\cdot \triangle I_1 I_2 J_3$. As $G_\HH$ is connected, 
and each of the fibers of $\tau$ over such an orbit is diffeomorphic to $ G_\HH^4/G_{\mathcal H(I_1,I_2,J_3)}$, we see that the fibers of $\tau$ are connected subsets of $m^{-1}(G_\HH)$. 

\subsection{The connected components of $m^{-1}(G_\HH)$}


First, we want to show  that the 
 (proved to be connected) fibers $\tau^{-1}(G_\HH \cdot \triangle I_1 I_2 J_3)$
are the connected components of $m^{-1}(G_\HH)$.  Second,
we will calculate the dimension of the connected components. 

For that it is sufficient to show that the $G_\HH$-orbits form the set of connected 
components of $\mathcal T_{\alpha,\beta,\gamma}$. This  
follows from the fact that $G_\HH$-orbits form the set of connected components of 
the topological space $Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$, which is 
$G_\HH$-equivariantly isomorphic to  $\mathcal T_{\alpha,\beta,\gamma}$, the isomorphism
$Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma}) \cong \mathcal T_{\alpha,\beta,\gamma}$ is
given by $\rho \mapsto \triangle I_1 I_2 \rho(I_3)$. 
Indeed,  each representation $$\rho \in Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma}), \rho\colon \mathcal H_{\alpha,\beta,\gamma} \rightarrow End\,V_\RR,   
\rho(e_1)=I_1, \rho(e_2)=I_2,$$ is uniquely determined by $\rho(e_3)$, or, which is equivalent, by
$\rho(c)=Id_{\RR^{4k}}\oplus -Id_{\RR^{4(n-k)}}$, see Theorem \ref{Main-theorem}, where $c$ is a the choice of a central element, satisfying $c^2=1$, 

$k=\frac{1}{8}(Tr(\rho(c))+4n)$ (such representation $\rho$ is $G_\HH$-isomorphic to $\rho_k$ defined in the introduction). 
Any two representations $\rho_1,\rho_2 \in Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$
are $G_\HH$-equivalent if and only if $Tr(\rho_1(c))=Tr(\rho_2(c))$. Thus 
the set $Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$ is a disjoint union of
its $n+1$ closed $G_\HH$-orbits $$\{\rho\in Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})\,|\,Tr(\rho(c))=4(2k-n)\},k=0,\dots,n.$$ Hence, as $G_\HH$ is connected, the $G_\HH$-orbits form the set of connected components
of $\mathcal T_{\alpha,\beta,\gamma}\cong Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma})$,
therefore the fibers $\tau^{-1}(G_\HH \cdot \triangle I_1 I_2 J_3)$ are the connected components of $m^{-1}(G_\HH)$. 
Besides that, from the identification 
$Rep_{I_1,I_2}(\mathcal H_{\alpha,\beta,\gamma}) \cong \mathcal T_{\alpha,\beta,\gamma}$ we obtain that each connected component of 
$m^{-1}(G_\HH)=\tau^{-1}(\mathcal T_{\alpha,\beta,\gamma})$ has the stated form $$
\{(f_1,f_2,f_3)\in m^{-1}(G_\HH)\,|\,f_2(I_3) \in G_\HH \cdot \rho_k(e_3)\}$$
for an appropriate $k$, $0\leqslant k \leqslant n$. 


The dimension of $ G_\HH^4/G_{\mathcal H(I_1,I_2,J_3)}$ is determined
from the fact that $g \in G$ stabilizes the subalgebra $\mathcal H(I_1,I_2,J_3) \subset End\, V_\RR$
if and only if it stabilizes 


the representation $\rho\colon \mathcal H_{\alpha,\beta,\gamma} \rightarrow End\,V_\RR$ given by  
$\rho(e_1)=I_1, \rho(e_2)=I_2,\rho(e_3) =  J_3$, or, equivalently, $g$ stabilizes $I_1,I_2$
together with  $\rho(c)=Id_{\RR^{4k}}\oplus -Id_{\RR^{4(n-k)}}$. 

The condition of centralizing $\rho(c)$ cuts out the subgroup $G_{\mathcal H(I_1,I_2,J_3)}= G_{\HH,\rho_k}$ of dimension $4k^2+4(n-k)^2$ in $G_\HH$. 


That is,  $\dim\,G_\HH^4/G_{\mathcal H(I_1,I_2,J_3)}=4 \cdot 4n^2-(4k^2+4(n-k)^2)=12n^2+8nk-8k^2$.
This dimension takes its smallest value $12n^2=\dim\, G_\HH^3$ exactly when $k=0$ or $k=n$,
that is, when $J_3=I_3$ or $J_3=I_3^\prime$. 

\bigskip
What is left now is the calculation comparing the angles of the triangle $\triangle I_1 I_2J_3$
to those of $\triangle I_1 I_2 I_3$. 
\subsection{Triangles: comparison of angles.}

Given a twistor triangle $\triangle I_1 I_2 J_3$
with $T(\triangle I_1 I_2 J_3)=(\alpha,\beta,\gamma)$, we have the relation between the 
corresponding complex structures: $I_1I_2+I_2I_1=2\alpha Id$, $I_1J_3+J_3I_1=2\gamma Id$, 
$I_2J_3+J_3 I_2=2\beta Id$. 
The complex structures $\frac{\alpha I_1+I_2}{\sqrt{1-\alpha^2}}\in S(I_1,I_2)$
and $\frac{\gamma I_1+J_3}{\sqrt{1-\gamma^2}}\in S(I_1,J_3)$ anticommute with
 $I_1$. Then we can write $$T_{I_1}S(I_1,I_2)=\left \langle \frac{\alpha I_1+I_2}{\sqrt{1-\alpha^2}}, 
I_1\frac{\alpha I_1+I_2}{\sqrt{1-\alpha^2}}\right \rangle, T_{I_1}S(I_1,J_3)=\left\langle \frac{\gamma I_1+J_3}{\sqrt{1-\gamma^2}}, 
I_1\frac{\gamma I_1+J_3}{\sqrt{1-\gamma^2}}\right\rangle.$$
This spaces obviously have trivial intersection. Set $W=T_{I_1}S(I_1,I_2) \oplus T_{I_1}S(I_1,J_3)$.
Consider the values of the form $q(x,y)=-\frac{1}{4n}Tr(x\cdot y) \colon W \times W \rightarrow \RR$
for the unit vectors in the tangent planes. Setting $c_t=\cos t \cdot \frac{\alpha I_1+I_2}{\sqrt{1-\alpha^2}}+ \sin t \cdot \frac{-\alpha Id+I_1I_2}{\sqrt{1-\alpha^2}}$
and $d_s=\cos s \cdot \frac{\gamma I_1+J_3}{\sqrt{1-\gamma^2}}+ \sin s \cdot 
\frac{-\gamma Id+I_1J_3}{\sqrt{1-\gamma^2}}$ we consider
$q(c_t, d_s),$
which is equal to the trace of the following
$$-\frac{1}{4n\sqrt{1-\alpha^2}\sqrt{1-\gamma^2}}\biggl(\cos t\cos s\cdot (-\alpha\gamma Id+\alpha I_1J_3$$
$$+\gamma I_2I_1 +I_2J_3)+$$
$$+\cos t\sin s (-\alpha\gamma I_1 -\alpha J_3-\gamma I_2+I_2I_1J_3)+$$
$$+\sin t\cos s(-\alpha\gamma I_1 -\alpha J_3+\gamma I_1 I_2 I_1 + I_1 I_2 J_3)+$$
$$+\sin t \sin s(\alpha\gamma Id-\alpha I_1 J_3 -\gamma I_1I_2+I_1 I_2 I_1 J_3)\biggr).$$


We know that traces of complex structures are zeroes and we know traces of all products of pairs
of distinct complex structures, like $I_1I_2, I_1J_3, I_2J_3$. The trace of $I_1I_2I_1=-I_1I_2I_1^{-1}$
is zero, and we have  $I_2I_1J_3=(-I_1I_2+2\alpha Id)J_3=-I_1I_2J_3+2\alpha J_3$, 
$I_1I_2I_1J_3=I_1(-I_1I_2+2\alpha Id)J_3=I_2J_3+2\alpha I_1 J_3$,
so what remains to calculate is the trace of $I_1 I_2 J_3$.

As in general the natural representation  $\rho\colon \mathcal H=\mathcal H(I_1,I_2,J_3) \rightarrow
End\,V_\RR$ decomposes, by Theorem \ref{Main-theorem}, as $\rho=k \rho_1\oplus l \rho_2$ for certain $k,l$, setting $m=4(l-k)$ and $c=\beta e_1-\gamma e_2 +\alpha e_3-e_1e_2e_3, r=\sqrt{-\det\,Q_{\alpha,\beta,\gamma}}$, so that $\rho_1(c)=rId_{\RR^4}, \rho_2(c)=-rId_{\RR^4}$, 
we have that
$Tr(I_1 I_2 J_3)=-Tr(\beta I_1-\gamma I_2 +\alpha J_3-I_1I_2J_3)=-Tr(k\rho_1(c)\oplus l\rho_2(c))=mr$.  


Now we calculate the above value of $q(c_t, d_s)$, $$q(c_t, d_s)=-\frac{1}{\sqrt{1-\alpha^2}\sqrt{1-\gamma^2}}
\left((\beta+\alpha\gamma)\cos(t-s)+ \frac{m}{4n}r\cdot\sin(t-s)\right).$$
Next,  if $m=0$, which is equivalent to $\rho$ being balanced, then $|q(c_t, d_s)|< 1$,
(so that the form $q$ is indeed positively definite on $W$) and the maximal value 
of $|q(c_t, d_s)|$ is attained for $t=s$,
it is equal to $\left|\frac{\beta+\alpha\gamma}{\sqrt{1-\alpha^2}\sqrt{1-\gamma^2}}\right|$, 
which, by the spherical cosine law, equals $\pm \cos \angle I_2I_1I_3$. The analogous computations can be done for the other two angles of $\triangle I_1I_2J_3$ showing that these angles are well defined and are equal to
the respective angles of $\triangle I_1 I_2 I_3$, up to taking complements to $\pi$.


\section{Appendix}
\label{Appendix}
In this section we prove the most nontrivial parts of the statement
of Proposition \ref{Prop-properties-of-H}.


\subsection{$S|_V=q|_V,  S|_{\widetilde{V}}=q|_{\widetilde{V}}$}


The part $S|_V=q|_V$ is really trivial and follows from the definition of $q$ and the relations
of the algebra $\mathcal H$.

Let us get to showing $S|_{\widetilde{V}}=q|_{\widetilde{V}}$. 
We easily see that 
$S(\alpha-e_1e_2)=\alpha^2-2\alpha e_1e_2+e_1 (e_2e_1)e_2=
\alpha^2-2\alpha e_1e_2+e_1 (-e_1e_2+2\alpha)e_2=\alpha^2-1=
q(\alpha-e_1e_2)$ and similarly for other basis elements of $\widetilde{V}$. 
We need to check the equality $S(v)=q(v)$ for the general elements $v\in \widetilde{V}$, for which now it  suffices
to check that mixed symmetric products of the kind $(\alpha-e_1e_2)(\beta-e_2e_3)+(\beta-e_2e_3)(\alpha-e_1e_2)$ land in $\RR\cdot Id$. Indeed,
$(\alpha-e_1e_2)(\beta-e_2e_3)+(\beta-e_2e_3)(\alpha-e_1e_2)=
2\alpha\beta-2\beta e_1e_2 -2 \alpha e_2 e_3 +e_1e_2\cdot e_2e_3+e_2e_3\cdot e_1e_2
=
2\alpha\beta-2\beta e_1e_2 -2 \alpha e_2 e_3 -e_1e_3+(-e_3e_2+2\beta) e_1e_2=
2\alpha\beta -2 \alpha e_2 e_3 -e_1e_3-e_3e_2e_1e_2=
2\alpha\beta -2 \alpha e_2 e_3 -e_1e_3-e_3(-e_1e_2+2\alpha)e_2=
2\alpha\beta -2 \alpha (e_2 e_3+e_3e_2) -e_1e_3-e_3e_1=2\alpha\beta-4\alpha\beta-2\gamma=
-2(\alpha\beta+\gamma) \in \RR$ and similarly for other pairs of basis elements.

\medskip

\subsection{ $Tr\, \rho_{reg}(e_1e_2e_3)=0$}
\label{subsec-triple-trace}

For the trace calculation we consider the basis $1,e_1,e_2,e_3,e_1e_2,e_2e_3,e_3e_1,e_1e_2e_3$ of $\mathcal H$.
Then 

$e_1e_2e_3\cdot 1=e_1e_2e_3$, 

$e_1e_2e_3\cdot e_1=e_1e_2(-e_1e_3+2\gamma)=-e_1(-e_1e_2+2\alpha)e_3+2\gamma e_1e_2
=-2\alpha e_1e_3 - e_2e_3+2\gamma e_1e_2$,

$e_1e_2e_3\cdot e_2=e_1e_2(-e_2e_3+2\beta)=e_1e_3+2\beta e_1e_2$,

$e_1e_2e_3\cdot e_3=-e_1e_2$,

$e_1e_2e_3\cdot e_1e_2=e_1e_2(-e_1e_3+2\gamma)e_2=-e_1(-e_1e_2+2\alpha)e_3e_2-2\gamma e_1
=-e_2e_3e_2-2\alpha e_1e_3e_2-2\gamma e_1=-(-e_3e_2+2\beta)e_2-
2\alpha e_1(-e_2e_3+2\beta)-2\gamma e_1=-(2\gamma+4\alpha\beta) e_1-2\beta e_2-e_3+2\alpha e_1e_2e_3$,

$e_1e_2e_3\cdot e_2e_3=e_1e_2(-e_2e_3+2\beta)e_3=-e_1+2\beta e_1e_2e_3$,

$e_1e_2e_3\cdot e_3e_1=-e_1(-e_1e_2+2\alpha)=-2\alpha e_1-e_2$,

$e_1e_2e_3\cdot e_1e_2e_3=e_1e_2(-e_1e_3+2\gamma)e_2e_3=-e_1(-e_1e_2+2\alpha)e_3e_2e_3
-2\gamma e_1e_3=-e_2(-e_2e_3+2\beta)e_3-2\alpha e_1(-e_2e_3+2\beta)e_3-2\gamma e_1e_3
= 1 -2\alpha e_1e_2-2\beta e_2e_3 -(2\gamma+4\alpha\beta)e_1e_3$,
which finally shows that for every element of our basis $x$ the result of the left multiplication  
$e_1e_2e_3\cdot x$ never contains a nonzero $x$-component, so that $Tr(\rho_{reg}(e_1e_2e_3))=0$.

\medskip

\subsection{\bf The orthogonal decomposition $\mathcal H=\RR\cdot 1 \oplus V \oplus 
\widetilde{V}  \oplus \RR\cdot c$.}
We recall that $c=\beta e_1-\gamma e_2 +\alpha e_3-e_1 e_2 e_3$.
From  \ref{subsec-triple-trace} we get $1\perp c$. The orthogonality of 1 to the rest is clear, moreover, for the anticommuting 
pairs of elements of bases of $V$ and $\widetilde{V}$ we get $e_1 \perp \alpha-e_1e_2,\gamma-e_3e_1$ and similarly for $e_2,e_3$. The fact that for non-anticommuting pairs 
like $e_1,\beta-e_2e_3$ their symmetric product $e_1(\beta-e_2e_3)+(\beta-e_2e_3)e_1$ lands in 
$\RR\cdot c$ (see an explicit calculation of that in \ref{subsec-square-map}) implies that even for non-anticommuting pairs we have the orthogonality, 
$e_1\perp \beta-e_2e_3$ and similarly for $e_2,e_3$. 

Next, the orthogonality $c\perp V$
means that we need to check that $Tr(\rho_{reg}(e_j\cdot c))=0$, for example,
$e_1(\beta e_1-\gamma e_2 +\alpha e_3-e_1e_2e_3)=-\beta-\gamma e_1e_2 +\alpha e_1e_3+e_2e_3$
and the corresponding trace of this element is zero. 

The orthogonality $c\perp \widetilde{V}$ means that we need to check
that $Tr(\rho_{reg}(\alpha-e_1e_2)\cdot c)=0$ etc. Here
$(\alpha-e_1e_2)\cdot (\beta e_1-\gamma e_2 +\alpha e_3-e_1e_2e_3)=
\alpha\cdot c-\beta e_1(-e_1e_2+2\alpha)-\gamma e_1 -\alpha e_1e_2e_3 +e_1e_2\cdot e_1e_2e_3
=\alpha \cdot c -\beta e_2-(2\alpha\beta +\gamma)e_1-\alpha e_1e_2e_3 +
e_1(-e_1e_2+2\alpha)e_2e_3=
\alpha \cdot c -\beta e_2-(2\alpha\beta +\gamma)e_1-e_3+\alpha e_1e_2e_3$
and the corresponding trace of the latter element is clearly zero.

\medskip

\subsection{\bf The centrality of $c$} First, in order to see that $c$ indeed belongs to the center $\mathcal Z(\mathcal H)$
it is necessary and sufficient to check that $ce_i=e_ic, i=1,2,3$. Let us check that
$ce_1=e_1c$, the other cases are done similarly.

We have $ce_1=(\beta e_1-\gamma e_2+\alpha e_3 -e_1e_2e_3)e_1=\beta e_1e_1-\gamma (-e_1e_2+
2\alpha)+\alpha (-e_1e_3+2\gamma)-e_1e_2e_3e_1
=e_1(\beta e_1+\gamma e_2-\alpha e_3-e_2e_3e_1)$.
Now $e_2e_3e_1=e_2(-e_1e_3+2\gamma)=-(-e_1e_2+2\alpha)e_3+2\gamma e_2
=e_1e_2e_3+2\gamma e_2-2\alpha e_3$, so that 
$ce_1=e_1(\beta e_1+\gamma e_2-\alpha e_3-(e_1e_2e_3+2\gamma e_2-2\alpha e_3))
=e_1(\beta e_1-\gamma e_2+\alpha e_3 -e_1e_2e_3)=e_1c$.


\medskip

\subsection{\bf The center of $\mathcal H$.}  We know already that, for $c=\beta e_1-\gamma e_2+\alpha e_3 -e_1e_2e_3$ as above, 
we have that $\langle 1,c\rangle \subset \mathcal Z(\mathcal H)$.
Now, for an element $z \in \mathcal H$, $z=x_0\cdot 1+x_1e_1+x_2e_2+x_3e_3+
y_{12}(\alpha-e_1e_2)+y_{23}(\beta-e_2e_3)+y_{31}(\gamma-e_3e_1)+x_4\cdot c$, 
$x_i,y_{ij} \in \RR$, to be in the center means $ze_i=e_iz$ for $i=1,2,3$.
Let us check when there exist such $z$.

As the part $x_0\cdot 1+x_4\cdot c$ is already in $ \mathcal Z(\mathcal H)$, we may assume
that $x_0=x_4=0$. Now the difference
$e_1z-ze_1=x_2(e_1e_2-e_2e_1)+x_3(e_1e_3-e_3e_1)+y_{12}(e_2+e_1e_2e_1)
+y_{23}(-e_1e_2e_3+e_2e_3e_1)+y_{31}(-e_1e_3e_1-e_3)=
 -2x_2(\alpha-e_1e_2)+2x_3(\gamma-e_3e_1)+2y_{12}(e_2+\alpha e_1)
+y_{23}(-e_1e_2e_3+e_2(-e_1e_3+2\gamma))-2y_{31}(e_3+\gamma e_1)=
-2x_2(\alpha-e_1e_2)+2x_3(\gamma-e_3e_1)+2y_{12}(e_2+\alpha e_1)
+y_{23}(-e_1e_2e_3-(-e_1e_2+2\alpha)e_3+2\gamma e_2)-2y_{31}(e_3+\gamma e_1)=
-2x_2(\alpha-e_1e_2)+2x_3(\gamma-e_3e_1)+2y_{12}(e_2+\alpha e_1)
+y_{23}(-e_1e_2e_3+e_1e_2e_3-2\alpha e_3+2\gamma e_2)-2y_{31}(e_3+\gamma e_1)=
-2x_2(\alpha-e_1e_2)+2x_3(\gamma-e_3e_1)+2y_{12}(e_2+\alpha e_1)
+2y_{23}(\gamma e_2-\alpha e_3)-2y_{31}(e_3+\gamma e_1)=
-2x_2(\alpha-e_1e_2)+2x_3(\gamma-e_3e_1)+2(y_{12} \alpha-y_{31}\gamma)e_1  
+2(y_{12}+y_{23}\gamma) e_2 - 2(y_{23}\alpha+y_{31}) e_3
$ equals zero if and only if $x_2=x_3=0$ and $y_{12}\alpha-y_{31}\gamma=0, y_{12}=-y_{23}\gamma$
and $y_{31}=-y_{23}\alpha$. Note that the last two equalities trivially imply the first of the three 
last equalities, so that we see that $z=x_1e_1+y_{23}(-\gamma(\alpha- e_1e_2)+(\beta-e_2e_3)-\alpha(\gamma-e_3e_1))$. Next, commutation with $e_2,e_3$ means that $x_1=0$
and that $z=y_{23}(-\gamma(\alpha- e_1e_2)+(\beta-e_2e_3)-\alpha(\gamma-e_3e_1))
=y_{31}(-\beta(\alpha-e_1e_2)-\alpha(\beta-e_2e_3)+(\gamma-e_3e_1))=
y_{12}((\alpha-e_1e_2)-\gamma(\beta-e_2e_3)-\beta(\gamma-e_3e_1))$,
that is, if $z\neq 0$ then we must have that the following matrix
\[\left(\begin{array}{ccc}
1 & -\gamma & -\beta\\
-\gamma & 1 & -\alpha \\
-\beta & -\alpha & 1
\end{array}\right)
\]
is of rank 1. This is precisely the case when the form $Q$ has signature $(0,1,2)$
($q$ has signature $(1,1,6)$), that is, $|\alpha|=|\beta|=|\gamma|=1$ and $\gamma=-\alpha\beta$.

In this case the center of $\mathcal H$ is spanned by $1,c$ and $z$.

\medskip


\subsection{\bf Equality $c^2=-\det\,Q_{\alpha,\beta,\gamma}$.} Now let us calculate $c^2=(\beta e_1-\gamma e_2+\alpha e_3 -e_1e_2e_3)(\beta e_1-\gamma e_2+\alpha e_3 -e_1e_2e_3)=
-\beta^2-\gamma^2-\alpha^2-\beta\gamma(e_1e_2+e_2e_1)-\alpha\gamma(e_2e_3+e_3e_2)
+\alpha\beta(e_1e_3+e_3e_1)-\beta(e_1\cdot e_1e_2e_3+e_1e_2e_3\cdot e_1)+\gamma(e_2\cdot e_1e_2e_3
+e_1e_2e_3\cdot e_2)-\alpha(e_3\cdot e_1e_2e_3+e_1e_2e_3\cdot e_3)+(e_1e_2e_3)^2=
-\alpha^2-\beta^2-\gamma^2-2\alpha\beta\gamma-\beta(-e_2e_3+(-e_2e_1+2\alpha)e_3e_1)
+\gamma((-e_1e_2+2\alpha)e_2e_3+e_1e_2(-e_2e_3+2\beta))-\alpha((-e_1e_3+2\gamma)e_2e_3-e_1e_2)+(e_1e_2e_3)^2=
-\alpha^2-\beta^2-\gamma^2-2\alpha\beta\gamma-\beta(-e_2e_3-e_2(-e_3e_1+2\gamma)e_1+2\alpha e_3e_1)
+\gamma(e_1e_3+2\alpha e_2e_3+e_1e_3+2\beta e_1e_2)-\alpha(-e_1(-e_2e_3+2\beta)e_3+2\gamma e_2e_3-e_1e_2)+(e_1e_2e_3)^2=
-\alpha^2-\beta^2-\gamma^2-2\alpha\beta\gamma-\beta(-2e_2e_3-2\gamma e_2e_1+2\alpha e_3e_1)
+\gamma(2e_1e_3+2\alpha e_2e_3+2\beta e_1e_2)-\alpha(-2e_1e_2-2\beta e_1e_3+2\gamma e_2e_3)+(e_1e_2e_3)^2=
-\alpha^2-\beta^2-\gamma^2-2\alpha\beta\gamma-\beta(-2e_2e_3-2\gamma (-e_1e_2+2\alpha)+2\alpha e_3e_1)
+\gamma(2(-e_3e_1+2\gamma)+2\alpha e_2e_3+2\beta e_1e_2)-\alpha(-2e_1e_2-2\beta (-e_3e_1+2\gamma)+2\gamma e_2e_3)+(e_1e_2e_3)^2=
-\alpha^2-\beta^2-\gamma^2-2\alpha\beta\gamma-\beta(-2e_2e_3+2\gamma e_1e_2-4\alpha\gamma+2\alpha e_3e_1)
+\gamma (-2e_3e_1+4\gamma+2\alpha e_2e_3+2\beta e_1e_2)-\alpha(-2e_1e_2+2\beta e_3e_1-4\beta\gamma+2\gamma e_2e_3)+(e_1e_2e_3)^2=
-\alpha^2-\beta^2-\gamma^2+6\alpha\beta\gamma+4\gamma^2+2\alpha e_1e_2+
2\beta e_2e_3-2(\gamma+2\alpha\beta) e_3e_1+(e_1e_2e_3)^2=
-\alpha^2-\beta^2-\gamma^2+6\alpha\beta\gamma+4\gamma^2+2\alpha e_1e_2+
2\beta e_2e_3-2(\gamma+2\alpha\beta) e_3e_1+e_1e_2(-e_1e_3+2\gamma)e_2e_3=
-\alpha^2-\beta^2-\gamma^2+6\alpha\beta\gamma+4\gamma^2+2\alpha e_1e_2+
2\beta e_2e_3-2(\gamma+2\alpha\beta) e_3e_1-e_1e_2e_1e_3e_2e_3-2\gamma e_1e_3=
-\alpha^2-\beta^2-\gamma^2+6\alpha\beta\gamma+2\alpha e_1e_2+
2\beta e_2e_3-4\alpha\beta e_3e_1-e_1(-e_1e_2+2\alpha)e_3e_2e_3=
-\alpha^2-\beta^2-\gamma^2+6\alpha\beta\gamma+2\alpha e_1e_2+
2\beta e_2e_3-4\alpha\beta e_3e_1-e_2e_3e_2e_3-2\alpha e_1e_3e_2e_3=
-\alpha^2-\beta^2-\gamma^2+6\alpha\beta\gamma+2\alpha e_1e_2+
2\beta e_2e_3-4\alpha\beta e_3e_1-e_2(-e_2e_3+2\beta)e_3-2\alpha e_1(-e_2e_3+2\beta)e_3=
1-\alpha^2-\beta^2-\gamma^2+6\alpha\beta\gamma
-4\alpha\beta (e_1e_3 +e_3e_1)=1-\alpha^2-\beta^2-\gamma^2-2\alpha\beta\gamma=-\det\,Q_{\alpha,\beta,\gamma}$, as 
was stated.

\medskip


\subsection{\bf Inclusions $cV\subset \widetilde{V},c\widetilde{V}\subset V$.} 

From the above shown, as $ce_1=e_1c=-\beta -\gamma e_1e_2 +\alpha e_1e_3+e_2e_3=
-(\beta-e_2e_3)+\gamma (\alpha-e_1e_2)-\alpha\gamma+\alpha (-e_3e_1+2\gamma)=
-(\beta-e_2e_3)+\alpha (\gamma-e_3e_1)+\gamma (\alpha-e_1e_2) \in \widetilde{V}$.

Analogously one shows that $ce_2,ce_3 \in \widetilde{V}$, that is $cV\subset \widetilde{V}$.


Now let us show that $c\widetilde{V}\subset V$. Again, we are going to give a computation for some base element
of $\widetilde{V}$, leaving to the reader the similar computations with the remaining base elements.

Consider $c(\beta-e_2e_3)=
(\beta e_1-\gamma e_2+\alpha e_3 -e_1e_2e_3)(\beta-e_2e_3)=
\beta^2 e_1 -\beta\gamma e_2+\beta \alpha e_3-\beta e_1e_2e_3
-\beta e_1e_2e_3-\gamma e_3-\alpha e_3e_2e_3+e_1e_2e_3e_2e_3=
\beta^2 e_1 -\beta\gamma e_2+(\alpha\beta-\gamma) e_3-2\beta e_1e_2e_3
-\alpha (-e_2e_3+2\beta)e_3+e_1e_2(-e_2e_3+2\beta)e_3=
(\beta^2 -1)e_1 -(\beta\gamma+\alpha) e_2-(\alpha\beta+\gamma) e_3 \in V$.

\medskip


\subsection{\bf The square map $S\colon v\mapsto v^2$ sends $V \oplus \widetilde{V}$ to $\langle 1,c \rangle \subset \mathcal Z(\mathcal H)$.} 
\label{subsec-square-map}
It is easy to see that each of $V$ and $\widetilde{V}$ is mapped to $\RR\cdot 1$,
moreover as we have that every basis element in $V$ $e_1,e_2,e_3$ anticommutes with the respective two elements of the basis $\alpha-e_1e_2,\beta-e_2e_3,\gamma-e_3e_1$, we see that
in order to prove that $S(V\oplus \widetilde{V}) \subset \langle 1,c\rangle$ we only need to 
consider the symmetric products of the non-anticommuting elements, for example, $e_1(\beta-e_2e_3)+(\beta-e_2e_3)e_1=
2\beta e_1-e_1e_2e_3-e_2e_3e_1=2\beta e_1-e_1e_2e_3-e_2(-e_1e_3+2\gamma)=
2\beta e_1-2\gamma e_2 -e_1e_2e_3+e_2e_1e_3=2\beta e_1-2\gamma e_2 -e_1e_2e_3+(-e_1e_2+2\alpha)e_3=2(\beta e_1-\gamma e_2 +\alpha e_3-e_1e_2e_3)=2c$
and similarly for other respective pairs of basis elements from $V$ and $\widetilde{V}$.

\medskip
{"config": "arxiv", "file": "1806.09794/Triangles-28-06-2018.tex"}