D.A, E.R and A.W. acknowledge support from U.S. National Science Foundation grants DMS 1107452, 1107263, 1107367 "RNMS: GEometric structures And Representation varieties" (the GEAR Network). E.R. and A.W. were supported by the National Science Foundation under Grant No. 1440140 and the Clay Foundation (A.W.), while they were in residence at the Mathematical Sciences Research Institute in Berkeley, California, during the Fall Semester 2019. A.W acknowledges funding by the Deutsche Forschungsgemeinschaft within the Priority Program SPP 2026
“Geometry at Infinity”, by the European Research Council under
ERC-Consolidator grant 614733, and by the Klaus-Tschira-Foundation.
E.R. acknowledges funding by the Deutsche Forschungsgemeinschaft within the RTG 2229
“Asymptotic invariants and limits of groups and spaces” and the Priority Program SPP 2026 “Geometry at Infinity”, and thanks the Labex IRMIA of the Université de Strasbourg for support during the finishing of this project.
A.B. was partially supported by Simons Foundation Collaboration Grant  No. 636972. This work has been supported under Germany’s Excellence Strategy EXC-2181/1 - 390900948 (the Heidelberg STRUCTURES Cluster of Excellence).

We introduce the symplectic group $\Sp_2(A,\sigma)$ over a noncommutative algebra $A$ with an anti-involution $\sigma$.
We realize several classical Lie groups as $\Sp_2$ over various noncommutative algebras, which provides new insights into their structure theory.

We construct several geometric spaces, on which the groups $\Sp_2(A,\sigma)$ act.
We introduce the space of isotropic $A$-lines, which generalizes the projective line. We describe the action of $\Sp_2(A,\sigma)$ on isotropic $A$-lines, generalize the  Kashiwara-Maslov index of triples and the cross ratio of quadruples of isotropic $A$-lines as invariants of this action.
When the algebra $A$ is Hermitian or the complexification of a Hermitian algebra, we introduce the symmetric space $X_{\Sp_2(A,\sigma)}$, and construct different models of this space.

Applying this to classical Hermitian Lie groups of tube type (realized as $\Sp_2(A,\sigma)$) and their complexifications, we obtain different models of the symmetric space as noncommutative generalizations of models of the hyperbolic plane and of the three-dimensional hyperbolic space.

We also provide a partial classification of Hermitian algebras in Appendix A.

§ INTRODUCTION

The special linear groups $\SL_2(R)$, when $R$ is a commutative ring, are among the most important and best studied groups in mathematics. They arise in many different contexts: for example in number theory and arithmetic geometry and in the theory of finite simple groups. In the special case when $R = \R$ or $\CC$, they are ubiquitous in Lie theory and in representation theory. Via their geometric actions, they are fundamental objects in projective, conformal and hyperbolic geometry.

The main aim of this paper is to generalize these groups to the case when $R$ is a noncommutative ring. Unfortunately, for a noncommutative ring, the definition of $\SL_2(R)$ is tricky, see [5, 4] and the comments below.

We will thus slightly change our point of view, and notice that, for a commutative ring $R$, the special linear group $\SL_2(R)$ is isomorphic to the symplectic group $\Sp_2(R)$. In this article we develop the theory of symplectic groups $\Sp_2(A,\sigma)$ over possibly noncommutative algebras $A$ with an anti-involution $\sigma$.

We show that many aspects of the classical $\SL_2$ theory can be developed over such noncommutative algebras $(A,\sigma)$. By realizing several classical Lie groups as $\Sp_2$ over a noncommutative algebra, this also  provides new insights into the theory of some higher dimensional classical Lie groups.

* We define the group $\Sp_2(A,\sigma)$ and describe its action on a generalization of the projective line: the space of isotropic lines. We generalize the Kashiwara-Maslov index of triples and the  cross ratio of $4$-tuples of isotropic lines.

* We introduce Hermitian algebras $(A,\sigma)$ (see Definition <ref>), a special class of $\R$-algebras exhibiting positivity properties. In this case, we construct different models of the symmetric space associated to $\Sp_2(A,\sigma)$ and its complexification $\Sp_2(A_ \mathbb{C},\sigma_\mathbb{C})$. We obtain generalizations of several models of the hyperbolic plane and the three-dimensional hyperbolic space, which are the symmetric spaces associated to $\SL(2,\R)$ and $\SL(2,\CC)$ respectively.

* We show that classical Hermitian Lie groups of tube type, such as the standard real symplectic group $\Sp(2n,\mathbb{R})$ can be naturally realized as $\Sp_2(A,\sigma)$. This explains many aspects of the structure theory of Hermitian Lie groups of tube type.

* As an application we describe new explicit models of the symmetric spaces associated to the complex Lie groups $\Sp(2n,\CC), \GL(2n,\CC)$, and  $\OO(4n,\CC)$.

* A large part of the theory outlined in (2) is completely algebraic, and works even when $\R$ is replaced by any real closed field. This may be of interest for some applications.

In [5, 4] two of the authors started to develop the general theory of Lie groups and Lie algebras over noncommutative rings. This is a highly non-trivial theory. Some of the difficulties can be already be seen when trying to define the group $\SL_n$ over a noncommutative ring. It is immediate to define the group $\GL_n$ over a noncommutative ring $R$ as linear invertible maps from $R^n$ to $R^n$. But there is no appropriate definition for the group $\SL_n$ as a subgroup of $\GL_n$ because there is no canonical choice of a determinant, even though there is rich  theory of quasi-determinants [10].

Our approach is motivated by the notion of $\Theta$-positivity, a generalization of Lusztig's total positivity in the context of real Lie groups which are not necessarily split, introduced in [12, 11, 22]. Hermitian Lie groups of tube type admit a $\Theta$-positive structure.[In fact, for Hermitian Lie groups the $\Theta$-positive structure is related to the theory of Lie semigroups.]
The combinatorics of this $\Theta$-positive structure is governed by the Weyl group of a root system of type $A_1$, making Hermitian Lie groups of tube type “look like” $\SL_2$ theories over noncommutative algebras. Here we make this precise.

Our description of the different  models of the symmetric spaces for most of the Hermitian groups of tube type shows that the geometry of these symmetric space is a noncommutative version of the classical planar hyperbolic geometry. This is not surprising: for example a well-known model of the symmetric space of $\Sp(2n,\R)$ is the Siegel upper half space, and the formulae describing this model look like the formulae of planar hyperbolic geometry, with matrices replacing numbers. The theory developed here puts this observation into a more general theoretical framework.

Even more interesting is our description of different models of the  symmetric spaces of the complexifications of the groups discussed above, such as $\Sp(2n,\CC)$, $\GL(2n,\CC)$, $\OO(4n,\CC)$. Here we show that the geometry of their symmetric spaces is a noncommutative version of the classical $3$-dimensional hyperbolic geometry. This fact is more surprising.

We believe that this new point of view can be helpful and leads to new applications. An immediate application is that the construction of noncommutative coordinates for symplectic representations of fundamental groups of punctured surfaces developed in [2] and its relation to noncommutative cluster algebras introduced in [6] generalizes to representations into classical Hermitian Lie groups of tube type. This will appear in a forthcoming work.

We now describe the results of the paper in more detail.

§.§ Symplectic groups over involutive algebras

Let $\K$ be a field and consider an associative, possibly noncommutative unital finite-dimensional $\K$-algebra $A$. Let $\sigma: A \to A $ be an anti-involution, i.e. a $\K$-linear map with $\sigma(ab) = \sigma(b)\sigma(a)$ for all $a,b \in A$ and $\sigma^2 = \Id$. We denote the set of fixed points of $\sigma$ by $A^\sigma$.
Then we introduce the non-degenerate sesquilinear form $\omega: A^2 \times A^2 \to A$, defined by

\[\omega(x,y):= \sigma(x)^T \Omega y\,,\]

for all $x,y \in A^2$ and $\Omega =\Ome{1}$.
The symplectic group $\Sp_2(A, \sigma)$ is defined as

\[\Sp_2(A, \sigma) = \Aut(\omega)= \{f\in\Aut(V)\mid \forall x,y\in V:\omega(f(x),f(y))=\omega(x,y)\}\,.\]

Similar to the classical case, we look at natural spaces, on which the group $\Sp_2(A, \sigma)$ acts. The first such space is our noncommutative analog of the projective line. The group does not act transitively on the space of all lines in $A^2$, hence we restrict our attention to the lines that are isotropic for $\omega$:

\[\PP(\Is(\omega)) :=\{xA\mid \omega(x,x) = 0, \,  x\in A^2\text{ regular}\}\,,\]

where an element $x\in A^2$ is called regular if there exists $y \in A^2$ such that $x,y$ form a basis. Two isotropic lines $xA$ and $yA$ are said to be transverse if $x,y$ form a basis.

The action of $\Sp_2(A, \sigma)$ on $\PP(\Is(\omega))$ has some of the features we know from projective or Möbius geometry, e.g. the action on quadruples of points preserve an invariant that generalizes the cross-ratio:

* The group $\Sp_2(A,\sigma)$ acts transitively on $\PP(\Is(\omega))$

* The group $\Sp_2(A,\sigma)$ acts transitively on the set of pairs of transverse isotropic lines.

* The $\Sp_2(A,\sigma)$-orbits in the space of pairwise transverse triples of isotropic lines are in one-to-one correspondence with the orbits of the action of the group of invertible elements $A^\times$ on the set of invertible fixed points $(A^\sigma)^\times$ by

A^\times\times (A^\sigma)^\times & \mapsto & (A^\sigma)^\times\\
(a,b) & \mapsto & ab\sigma(a).
\end{matrix}$$

This gives rise to a generalization of the Kashiwara-Maslov index.

* The
$\Sp_2(A,\sigma)$-orbits in the space of pairwise transverse 4-tuples  of isotropic lines are in one-to-one correspondence with the orbits of the action of the group of invertible elements $A^\times$ on the set $A_0$ of products of invertible fixed points $(A^\sigma)^\times$, $A_0 =\{ bb'\, \mid \, b, b' \in (A^\sigma)^\times\}$:

A^\times\times A_0 & \mapsto & A_0\\
(a,b) & \mapsto & aba^{-1}.
\end{matrix}$$
The conjugacy class in $A_0$ is thus a  noncommutative cross-ratio of a four-tuple of pairwise transverse isotropic lines.

Of particular interest to us is a special class of involutive algebras that we call Hermitian algebras. We describe the geometry of the symmetric spaces for $\Sp_2(A,\sigma)$, when $(A,\sigma)$ is Hermitian or the complexification of an Hermitian algebra.

Hermitian algebras are  algebras $(A,\sigma)$ over a real closed field $\K$  where the following holds: if $a,b\in A^\sigma$, then $a^2 + b^2 = 0$ if and only if $a = b = 0$. Readers not familiar with real closed fields can simply think about the case where $\K=\R$, since this is the most interesting case for our applications.
Involutive algebras are closely related to Jordan algebras, and Hermitian algebras give rise to formally real Jordan algebras see Section <ref>. A key feature of Hermitian algebras is the existence of a proper convex cone $A^\sigma_+\subset A^\sigma$.

In the main body of the paper, we focus on semisimple Hermitian algebras. So, for the rest of the introduction,  whenever we say Hermitian algebra, we mean semisimple Hermitian algebra. In Appendix <ref> we consider a more general class of rings with anti-involution, which we call pre-Hermitian, investigate the theory of general Hermitian algebras, and classify semisimple Hermitian algebras.

There are different ways in which we can construct new Hermitian algebras out of old ones: Matrix algebras over Hermitian algebras are Hermitian, and complexifications (and quaternionifications) of Hermitian algebras provide new Hermitian algebras.
Given a Hermitian algebra $(A,\sigma)$ we consider the complexification $A_\CC$ of $A$ and extend the involution complex linearly to an involution $\sigma_\CC$ and complex anti-linearly to an involution $\bar\sigma_\CC$. Then $(A_\CC, \bar\sigma_\CC)$ is an Hermitian algebra, but $(A_\CC, \sigma_\CC)$ is not, see Section <ref>.

For Hermitian algebras and their complexifications we can refine Theorem <ref>.

For this we make the following definition: we say that a triple of pairwise transverse isotropic lines is positive, if its $\Sp_2(A,\sigma)$-orbits corresponds to an element in $A^\sigma_+$ by the identification given in Theorem <ref>.

If $(A,\sigma)$ is Hermitian, then $\Sp_2(A,\sigma)$ acts transitively on the space of positive triples of pairwise transverse isotropic lines.

Assume $\K=\R$ and let $(A,\sigma)$ be a Hermitian algebra or its complexification. Then the space of isotropic lines is compact.

§.§ Symmetric spaces associated to Sp2(A,sigma)

Recall that the symmetric space associated to $\Sp_2(\R)$ is $X_{\Sp_2(\R)} =\Sp_2(\R)/ \UU_1(\CC)$, where $\UU_1(\CC)$ is a maximal compact subgroup. It admits many explicit models. It can be realized as the space of compatible complex structures on $\R^2$, this is the space
$\mathfrak C:=\left\{J\text{ complex structure on $\R^2$}\mid \omega(J \cdot, \cdot) \text{ is an inner product}
\right\}.$ We call this the complex structure model.
To obtain the upper hemisphere model, we complexify $\R^2$ to $\CC^2$ and extend the symplectic form $\omega$ complex linearly to a symplectic from $\omega_\CC$. Similarly we extend a complex structure $J$ complex linearly to $\CC^2$. Over $\CC$ now the complex structure is a diagonalizable endomorphism and we can associate to every compatible complex structure its $+i$-eigenspace in $\CC^2$. This provides an embedding of $\mathfrak C$ into the space of isotropic lines $\PP(\Is(\omega_\CC)) = \PP(\CC^2)$, whose image we denote by $\mathfrak P$ and call the projective model.

The Poincaré disk model $D = \{ z\in \CC\mid \overline{z}z <1\}$ and the upper half plane model $\mathfrak U = \{ z\in \CC \mid \Imm(z) >0\}$ arise from projective model $\mathfrak{P}$ naturally by picking specific affine charts in $\PP(\CC^2)$. We call the disk model the precompact model.
Taking its closure in $\CC$, or the closure of $\mathfrak P$ in $\PP(\CC^2)$, we obtain a compactification of the symmetric space, in which the space of (isotropic) lines in $\R^2$ arises as the boundary.

We show that these constructions can be appropriately generalized to symplectic groups $\Sp_2(A,\sigma)$ over real Hermitian algebras $(A,\sigma)$.
In order to introduce the maximal compact subgroup we observe that
given $(A,\sigma)$ the algebra $\Mat_n(A)$ of $n\times n$-matrices over $A$ can be endowed with the anti-involution $\sigma^T$, which applies $\sigma$ to each entry and then the transpose. When $(A,\sigma)$ is a Hermitian algebra, then $(\Mat_n(A), \sigma^T)$ is Hermitian as well. In that case the subgroup $\UU_n(A,\sigma) = \{ M \in \Mat_n(A) \, \mid \, \sigma(M)^T M = \Id_n\}$ is compact.
This allows us to define the maximal compact subgroups
$\KSp_2(A,\sigma) = \Sp_2(A,\sigma)\cap \UU_2(A,\sigma)  \subset \Sp_2(A,\sigma)$, and $\KSp^c_2(A_\CC, \sigma_\CC)= \Sp_2(A_\CC,\sigma_\CC)\cap \UU_2(A_\CC,\bar\sigma_\CC)  \subset \Sp_2(A_\CC,\sigma_\CC)$.
We can thus consider the symmetric space associated to $\Sp_2(A,\sigma)$,
$X_{\Sp_2 (A,\sigma)}=  \Sp_2(A, \sigma)/ \KSp_2(A, \sigma)$, and the symmetric space
$X_{\Sp_2 (A_\CC,\sigma_\CC)}=  \Sp_2(A_\CC, \sigma_\CC)/ \KSp^c_2(A_\CC, \sigma_\CC)$.
We develop different explicit models for the symmetric spaces
$X_{\Sp_2 (A,\sigma)}$ and $X_{\Sp_2 (A_\CC,\sigma_\CC)}$.

Let $(A,\sigma)$ be a real Hermitian algebra and $(A_\CC, \sigma_\CC, \bar\sigma_\CC)$ its complexification. Then the symmetric space $X_{\Sp_2 (A,\sigma)}=  \Sp_2(A, \sigma)/ \KSp_2(A, \sigma)$ admits

* a complex structure model
$\mathfrak C:=\left\{J\text{ complex structure on $A^2$}\mid \omega(J\cdot,\cdot)\text{ is a $\sigma$-inner product}
\right\}.$

* a projective model $\mathfrak P:=\{vA_\CC\mid v\in\Is(\omega_\CC),\; i \omega_\CC(\overline{v},v)\in (A_\CC^{\bar\sigma})_+\},$ where $(A_\CC^{\bar\sigma})_+$ is the proper convex cone in $(A_\CC^{\bar\sigma})$.

* a precompact model $\mathring{D}(A^{\sigma_\CC}_\CC,\bar\sigma_\CC):=\{c\in A^{\sigma_\CC}_\CC\mid 1-\bar cc\in (A^{\bar\sigma_\CC}_\CC)_+\}$.

* an upper half-space model $\mathfrak U:=\{z\in A_\CC^{\sigma_\CC}\mid \Imm(z)\in A^\sigma_+\}$, where $A^\sigma_+$ is the proper convex cone in $A^\sigma$.

Furthermore, there are natural maps between these different models.
The closure of the precompact model gives rise to a compactification of $X_{\Sp_2 (A,\sigma)}$ in which the space of isotropic lines $\PP(\Is(\omega))$ appears as the closed $\Sp_2(A,\sigma)$-orbit.

For the group $\Sp_2(\CC)$ there is a similar construction of explicit models of the symmetric space, which is less well known.
The symplectic group $\Sp_2(\CC)$ acts on $(\CC^2, \omega)$, which we view as the complexification of $\R^2$.
A quaternionic structure on $\CC^2$ is an additive map $J: \CC^2 \to \CC^2$ such that $J^2=-\Id$ and such that  $J(xa)=J(x)\bar a$ for all $x\in
\CC^2$, $a\in \CC$.
Then $\Sp_2(\CC)$ acts naturally on the space of compatible quaternionic structures on $\CC^2$, which is
$\mathfrak C:=\{J\text{ quaternionic structure on $\CC^2$}\mid \omega( J \cdot, \cdot) \text{ is a Hermitian inner product}\}.$ We call $\mathfrak C$ the quaternionic structure model of $X_{\Sp_2(\CC)}$.
Analogously to the above construction, where we complexified $\R^2$ we can now quaternionify $\CC^2$ and extend the symplectic form $\omega$ as well as the quaternionic structure. One has to be a bit more cautious working with the quaternions $\HH$, but
one gets a projective model $\mathfrak P \subset \PP(\HH^2)$, a precompact model $D = \{x+yi+zj \in \HH\mid 1-(x^2+y^2+z^2)>0\}  \subset \HH$ and a upper half space model $\mathfrak U =\{x+iy+kj\in \HH\mid x,y,z\in\R,\;z>0\}  \subset \HH$.
These quaternionic models for the three-dimensional hyperbolic space are for example described in [1, 19].

We prove that an analogous construction can be made for the complexified symplectic groups $\Sp_2(A_\CC, \sigma_\CC)$, using quaternionic extensions $A_\HH$ of the Hermitian algebra $A$ with two appropriate quaternionic extensions $\sigma_0$ and $\sigma_1$ of the anti-involution $\sigma$. In particular, we obtain

Let $(A_\CC,\sigma_\CC)$ be the complexification of a real Hermitian algebra $(A, \sigma)$. Then the symmetric space $X_{\Sp_2 (A_\CC,\sigma_\CC)}=  \Sp_2(A_\CC, \sigma_\CC)/ \KSp^c_2(A_\CC, \sigma_\CC)$ admits

* a quaternionic structure model

$\mathfrak C:=\{J\text{ quaternionic structure on $A_\CC^2$}\mid \omega( J \cdot, \cdot) \text{ is a
$\bar\sigma$-inner product}\}$

* a projective model $\mathfrak P\subset \Is(\omega_\HH)$,

* a precompact model $\mathring{D}(A_\HH^{\sigma_0},\sigma_1):=\{c\in A^{\sigma_0}_\HH\mid 1-\sigma_1(c)c\in (A^{\sigma_1}_\HH)_+\}$, where $(A^{\sigma_1}_\HH)_+$ is the proper convex cone in the quaternionification of $A$.

* an upper half-space model $\mathfrak U:=\{z_0+z_1j\in A_\HH^{\sigma_0}\mid z_0\in A^{\sigma_\CC}_{\CC},\;z_1\in (A_\CC^{\bar\sigma})_+\}$.

Furthermore, there are natural maps between these different models.
The closure of the precompact model gives rise to a compactification of $X_{\Sp_2(A_\CC,\sigma_\CC)}$ in which the space of isotropic lines $\PP(\Is(\omega_\CC))$ appears as the closed $\Sp_2(A_\CC,\sigma_\CC)$-orbit.

§.§ Hermitian Lie groups of tube type

As we mentioned before, Hermitian Lie groups of tube type look a lot like groups of type $\Sp_2$ over real involutive noncommutative algebras. We illustrate it on the following example:

We consider the $\R$-algebra $A=\Mat(n,\R)$ of real $n\times n$-matrices. A natural anti-involution on $A$ is the transposition that we denote by $\sigma$. Then by definition, $M\in\Sp_2(A,\sigma)$ if and only if $\sigma(M)\Omega M=M^T\Omega M-\Omega$ where $\Omega=\Om$. Here we used the standard identification of $\Mat(2n,\R)$ and $\Mat_2(\Mat(n,\R))$. That means, $M$ is a symplectic $2n\times 2n$-matrix and the group $\Sp_2(A,\sigma)$ agrees with $\Sp(2n,\R)$.

With the theory of symplectic groups over noncommutative involutive algebras, we can make the correspondence between Hermitian Lie groups of tube type and symplectic groups over semisimple Hermitian algebras very precise, at least for classical Hermitian Lie group of tube type.

The following classical Hermitian Lie groups of tube type can be realized as symplectic groups over Hermitian algebras:

* $\Sp(2n,\R) = \Sp_2(A,\sigma)$, where $A=\Mat(n,\R)$ is the algebra of $n\times n$ matrices over $\R$ with involution $\sigma: A \to A$ given by $\sigma(r)=r^T$.

* $\UU(n,n) = \Sp_2(A,\sigma)$, where $A=\Mat(n,\CC)$ is the algebra of $n\times n$ matrices over $\CC$ with involution $\sigma: A \to A$ given by $\sigma(r)=\bar r^T$.

* $\SO^*(4n) =  \Sp_2(A,\sigma)$, where $A=\Mat(n,\HH)$  is the algebra of $n\times n$ matrices over $\HH$ with involution $\sigma: A \to A$ given by $\sigma: A \to A$ is $\sigma(r)=\bar r^T = \bar{r_1}^T - \bar{r_2}^T j$ for $r=r_1+r_2j$ and $r_1,r_2\in \Mat(n,\CC)$.

The other classical Hermitian Lie group of tube type, $\SO_0(2,n)$, cannot be realized in the same way as $\Sp_2(A,\sigma)$. However its double cover $\Spin_0(2,n)$) can be realized as $\Sp_2$ over a slightly more complicated object $B_n$,
which is a Jordan subalgebra of an appropriate Clifford algebra $(A,\sigma)$. We will discuss this in a forthcoming article.

The exceptional Hermitian Lie group of tube type cannot be realized as $\Sp_2(A,\sigma)$, see Remark <ref>.

There are other Lie groups that can be realized as $\Sp_2(A,\sigma)$ over involutive algebras $(A,\sigma)$ that are not Hermitian, see Section <ref>.

For these Hermitian Lie groups of tube type, the above models for the symmetric space $X_{\Sp_2(A, \sigma)}$ give the well known explicit models of Hermitian symmetric spaces. The precompact model is the bounded symmetric domain model, and the space of isotropic lines identifies with the Shilov boundary of the bounded symmetric domain model.

In view of Theorem <ref> and Theorem <ref> classical Hermitian symmetric spaces of tube type can be thought of as hyperbolic planes over noncommutative involutive algebra $(A,\sigma)$.

As a Corollary of Theorem <ref> we can realize the complexifications of Hermitian Lie groups of tube type $\Sp_2(A,\sigma)$  as symplectic groups over complexifications of Hermitian algebras.

The following complex Lie groups can be realized as symplectic groups over involutive algebras:

* $\Sp(2n,\CC) = \Sp_2(A_\CC,\sigma_\CC)$, where $A=\Mat(n,\R)$ is the algebra of $n\times n$ matrices over $\R$ with involution $\sigma: A \to A$ given by $\sigma(r)=r^T$.

* $\GL(2n,\CC)= \Sp_2(A_\CC,\sigma_\CC)$, where $A=\Mat(n,\CC)$ is the algebra of $n\times n$ matrices over $\CC$ with involution $\sigma: A \to A$ given by $\sigma(r)=\bar r^T$.

* $\OO(4n,\CC) =  \Sp_2(A_\CC,\sigma_\CC)$, where $A=\Mat(n,\HH)$  is the algebra of $n\times n$ matrices over $\HH$ with involution $\sigma: A \to A$ given by $\sigma: A \to A$ is $\sigma(r)=\bar r^T = \bar r_1^T - r_2^T j$ for $r=r_1+r_2j$ and $r_1,r_2\in \Mat(n,\CC)$.

In particular, Theorem <ref> applies and we obtain explicit new realizations of models for the symmetric spaces associated to $\Sp(2n,\CC)$, $\GL(2n,\CC)$, and $\OO(4n,\CC)$.
We illustrate here the upper half-space model for $\Sp(2n,\CC)$ and refer the reader to Section <ref> for an explicit description of all the models for $\Sp(2n,\CC)$, $\GL(2n,\CC)$, and $\OO(4n,\CC)$.

For $\Sp(2n,\CC)$, we consider again $A=\Mat(n,\R)$ with the anti-involution $\sigma$ given by transposition. Then $A_\CC=\Mat(n,\CC)$, $\sigma_\CC$ is the transposition and $\Sp_2(A_\CC,\sigma_\CC)=\Sp(2n,\CC)$. The upper half-space model of the symmetric space of $\Sp_2(A_\CC,\sigma_\CC)$ is then (see  Theorem <ref>(4))

$$\mathfrak{U}=\{z_1+z_2 j\mid z_1\in A_\CC^{\sigma_\CC},\; z_2\in (A_\CC^{\bar\sigma_\CC})_+\}=\{z_1+z_2 j\mid z_1\in \Sym(n,\CC),\; z_2\in \Herm^+(n,\CC)\}.$$

Structure of the paper: In Section <ref> we discuss algebras $A$ with anti-involution, and introduce the notion of Hermitian algebra. A more general notion of pre-Hermitian algebras and a classification of Hermitian algebras is given in Appendix <ref>.
In Section <ref> we introduce the symplectic groups $\Sp_2(A,\sigma)$ over noncommutative rings and give examples of classical Lie groups that are realized as $\Sp_2(A,\sigma)$. In Section <ref> we investigate the action of $\Sp_2(A,\sigma)$ on the space of isotropic lines.
We construct the various models of the symmetric space $X_{\Sp_2(A,\sigma)}$ in Section <ref>, and of the symmetric space $X_{\Sp_2(A_\CC,\sigma_\CC)}$ in Section <ref>.
In Section <ref> we spell this construction out for the complexifications of the Hermitian Lie groups of tube type, giving explicit models for the symmetric spaces of $\Sp(2n,\CC)$, $\GL(2n,\CC)$, and $\OO(4n,\CC)$.

§ ALGEBRAS WITH ANTI-INVOLUTION

In this section we consider algebras with anti-involutions and introduce basic notions which play an important role throughout the paper.

§.§ Main definitions

Let $\K$ be a field and $A$ a unital associative possibly noncommutative finite-dimensional $\K$-algebra. If $\K$ is a topological field, $A$ has a well defined topology.

An anti-involution on $A$ is a $\K$-linear map $\sigma\colon A\to A$ such that

* $\sigma(ab)=\sigma(b)\sigma(a)$;

* $\sigma^2=\Id$.

An involutive $\K$-algebra is a pair $(A,\sigma)$, where $A$ is a $\K$-algebra and $\sigma$ is an anti-involution on $A$.

Sometimes in the literature the maps that satisfy  Definition <ref> are called just involutions. We add the prefix “anti” in order to emphasise that they exchange the factors.

Notice, since the algebra $A$ is unital, we always have the canonical copy of $\K$ in $A$, namely $\K\cdot 1$ where $1$ is the unit of $A$. We will always identify $\K\cdot 1$ with $\K$. Moreover, since $\sigma$ is linear, for all $k\in \K$, $\sigma(k\cdot 1)=k\sigma(1)=k\cdot 1$, i.e. $\sigma$ preserves $\K\cdot 1$.

An element $a\in A$ is called $\sigma$-normal if $\sigma(a)a=a\sigma(a)$. An element $a\in A$ is called $\sigma$-symmetric if $\sigma(a)=a$. An element $a\in A$ is called $\sigma$-anti-symmetric if $\sigma(a)=-a$. We denote

$$A^{\sigma}:=\Fix_A(\sigma)=\{a\in A\mid \sigma(a)=a\},$$

$$A^{-\sigma}:=\Fix_A(\sigma)=\{a\in A\mid \sigma(a)=-a\}.$$

Typical examples of involutive algebras are given by matrix algebras: If $\K$ is a field, then the space of $\K$-valued $n\times n$-matrices $\Mat(n,\K)$ with  anti-involution  given by the transposition is an involutive algebra.

If additionally $\K$ admits an involution $\delta\colon \K\to \K$, then $\Mat(n,\K)$ with the anti-involution $\delta\circ\sigma$ is again an example of an involutive $\K$-algebra.

Semisimple involutive finite-dimensional algebras over perfect fields with the additional assumption $A^\sigma=\K$ can be classified. To state the classification, we need the following well-known definitions:

A field $\K$ is called perfect if every irreducible polynomial over $\K$ has distinct roots in its splitting field.

Notice that every field of characteristic zero is perfect.

A non-zero algebra is called simple if it has no two-sided ideal besides the zero ideal and itself. A finite-dimensional algebra is called semisimple if it is isomorphic to a product of simple algebras.

Let $\K$ be a perfect field with $\ch(\K)\neq 2$ and $(A,\sigma)$ be a semisimple finite-dimensional involutive algebra over $\K$ such that $A^\sigma=\K$. Then either $A$ is a division algebra over $\K$ of dimension $1, 2$ or $4$, or $A = \K \oplus \K$ and the anti-involution exchanges the two summands.

A proof of this statement will be given in Proposition <ref> and Theorem <ref>.

We denote by $A^\times$ the group of all invertible elements of $A$. If $V \subset A$ is a vector subspace, we denote

$$V^\times = A^\times \cap V\,,$$

the set of invertible elements in $V$. We consider the following map $\theta$, that will play the role of a norm on $A$

\theta\colon & A & \to & A^\sigma\\
& a & \mapsto & \sigma(a)a
\end{matrix}$$

The closed subgroup
$$U_{(A,\sigma)}=\{a\in A^\times\mid \theta(a)=1\}$$
of $A^\times$ is called the unitary group of $A$. The Lie algebra of $U_{(A,\sigma)}$ agrees with $A^{-\sigma}$.

Let $(A,\sigma)$ be an algebra with an anti-involution. We define the set of $\sigma$-positive elements by

$$A^\sigma_+:=\left\{\ \sum_{i=1}^k a_i^2\ \midwd\ a_i\in (A^\sigma)^\times,\; k\in\N\ \right\},$$

and the set of $\sigma$-non-negative elements by

$$A^\sigma_{\geq 0}:=\left\{\ \sum_{i=1}^k a_i^2\ \midwd\ a_i\in A^\sigma,\; k\in\N\ \right\}.$$

If $(A,\sigma)$ is an algebra over an ordered field $\K$, then $A^\sigma_{\geq 0}$ is the topological closure of $A^\sigma_+$. This follows from Corollary <ref>.

We will now give the definition of a Hermitian algebra that will be a key notion in this paper. Before this, we remind the well-known definition of a real closed field. A real closed field is an ordered field $\K$ which is of index two in its algebraic closures. Equivalently, it is an  ordered field $\K$ in which every positive element of $\K$ is a square and every odd degree polynomial has at least one zero.

Let $\K$ be a real closed field. A unital associative $\K$-algebra with an anti-involution $(A,\sigma)$ is called Hermitian if for all $x,y\in A^\sigma$, $x^2+y^2=0$ implies $x=y=0$.

Let $\K$ be a real closed field. Slightly abusing our notation, we denote the algebraic closure of $\K$ by $\K_\CC$. Since $\K$ is real closed, $\K_\CC=\K[i]$ where $i$ is a square root of $-1$ called also the imaginary unit of $\K_\CC$. The field $\K_\CC$ is called the complexification of $\K$. The involution $\bar\cdot\colon \K_\CC\to\K_\CC$ defined as $\bar 1=1$, $\bar i=-i$ is called the complex conjugation. With this (anti-)involution, $\K_\CC$ becomes a Hermitian algebra over $\K$ of dimension $2$. If $\K=\R$ then as usual we denote $\CC=\R_\CC$.

For a real closed field $\K$ there exist a generalized quaternion skew-field $\K_\HH$ defined by the following presentation:

$$\K_\HH=\{x_0+x_1i+x_2j+x_3k\mid i^2=j^2=-1,~ij=-ji=k\} .$$

If $\K=\R$, then $\R_\HH$ is the classical quaternion skew-field that we will denote as usual by $\HH$.
We call $\K_\HH$ the quaternionic extension of $\K$. With the anti-involution $\bar\cdot$ defined as $\bar i=-i$, $\bar j=-j$ and called the quaternionic conjugation, $\K_\HH$ becomes a Hermitian algebra over $\K$ of dimension $4$. The elements $i,~j,~k$ are called imaginary units of $\K_\HH$.

Also the generalized octonionic algebra $\K_\Oc$ can be defined over any real closed field $\K$. This is the 8-dimendional non-associative noncommutative algebra that is generated (as a $\K$-vector space) by the unit $1\in\K$ and seven imaginary units $i,~j,~k,~E,~I,~J,~K$. The multiplication rule of the imaginary units is the same as in the classical octonionic $\Oc$ algebra over $\R$

Although in the notation of complexification and quaternionification we use symbols $\CC$, $\HH$ and $\Oc$, we do not assume that $\CC$ is a subfield of $\K_\CC$, $\HH$ is a subalgebra of $\K_\HH$ and $\Oc$ is a subalgebra of $\K_\Oc$. In the Appendix <ref>, we consider other division algebras over fields (not necessarily over real closed fields). In the notation of Appendix <ref>, $\K_\HH=\HH_{-1,-1}$ and $\K_\Oc=\Oc_{-1,-1}$.

The main examples of real closed fields are $\R$ and the subfield of algebraic reals. An example of real closed field not contained into $\R$ is given by the field of Puiseux series with real coefficients.  Another example is the field of hyperreals. The field $\Q$ is not real closed, since for example $2$ is not a square.

Every real closed field has characteristic zero and contains the field of algebraic reals as a subfield.

For real closed fields the following theorem holds:
[Generalized Frobenius Theorem]
Let $\K$ be a real closed field. Every associative division algebra over $\K$ is isomorphic to $\K$, $\K_\CC$ or $\K_\HH$. The only non-associative noncommutative division algebra over $\K$ is $\K_\Oc$.

In case $\K=\R$ this theorem is the classical Frobenius theorem. The general version of this theorem follows from the classical one using the Tarski–Seidenberg transfer principle (see <cit.>).

A subalgebra of a Hermitian algebra that is closed under the anti-involution is Hermitian.

The following matrix algebras provide examples of Hermitian $\R$-algebras: $\Mat(n,\R)$ with the transposition and $\Mat(n,\CC)$ with the transposition composed with the complex conjugation.

Semisimple finite-dimensional algebras over real closed fields with the additional property $A^\sigma=\K$ are all Hermitian and can be classified as follows:

Let $(A,\sigma)$ be a semisimple finite-dimensional algebra over $\K$ with $A^\sigma=\K$. Then one of the following cases holds:

* $A=\K$, the anti-involution $\sigma$ acts trivially;

* $A=\K\oplus\K$, the anti-involution $\sigma$ permutes copies of $\K$, i.e. $\sigma(x,y)=(y,x)$;

* $A=\K_\CC$, the anti-involution $\sigma$ acts by complex conjugation;

* $A=\K_\HH$, the anti-involution $\sigma$ acts by quaternionic conjugation.

In every case $(A,\sigma)$ is Hermitian. The bilinear form

\beta\colon & A^{-\sigma}\times A^{-\sigma} &\to&\K\\
& (a,a') & \mapsto & -aa'-a'a
\end{matrix}$$
is an inner product on $A^{-\sigma}$.

This follows from the Theorems <ref> and <ref>.

The property to be Hermitian can also be defined in the same way for algebras with an anti-involution over any field. In this paper, we are discussing only Hermitian algebras over real closed fields. In the Appendix <ref>, a larger class of Hermitian and pre-Hermitian rings is considered and classified.

Notice, if $\K$ is a real closed field and $(A,\sigma)$ is an involutive algebra over $\K$, then the addition and the multiplication in $A$ as well as the anti-involution $\sigma$ are semi-algebraic maps and the spaces $A$, $A^\sigma$, $A^{-\sigma}$, $A^\sigma_+$, $A^\sigma_{\geq 0}$, $U_{(A,\sigma)}$ are semi-algebraic sets.

If $\K$ is a real closed field and $(A,\sigma)$ is a Hermitian algebra over $\K$, then $A^\sigma_+=\{a^2\mid a\in (A^\sigma)^\times\}$ and $A^\sigma_{\geq 0}=\{a^2\mid a\in A^\sigma\}$ are proper convex cones in $A^\sigma$. For a proof in case $\K=\R$ see <cit.>. In the general case, the proof is identical.

§.§ Some properties

Let $\K$ be a field and $A$ be a unital associative $\K$-algebra of finite dimension $n$ over $\K$. Then $A$ is isomorphic to a subalgebra of $\Mat(n,\K)$.

For every $x \in A$, consider the linear map $L_x:A \rightarrow A$ defined by

$$L_x(y) = x y \,. $$

Consider the map

$$A \ni x \rightarrow L_x \in \Mat(n,\K) \,.$$

This is an injective $\K$-algebra homomorphism (there is no kernel because $A$ is unital).

Let $A$ be a unital associative $\K$-algebra, which is finite-dimensional over $\K$. The non-invertible elements are all zero-divisors.

Using the previous proposition, given $x \in A$, $x$ is invertible if and only if $L_x$ is surjective. If $L_x$ is not surjective, it has a kernel, hence $x$ is a zero-divisor.

Let $\K$ be a field, $A$ be a unital associative finite dimensional $\K$-algebra and $a\in A$. An element $\lambda\in\K$ is called eigenvalue of $a$ if $a-\lambda\cdot 1$ is not invertible.

Notice that this definition of an eigenvalue works for non-associative algebras as well.

Let $\K$ be a field and $A$ be a unital associative finite dimensional $\K$-algebra. Then $a\in A$ is not invertible if and only if $a$ is a zero divisor in $\K[a]$.

Since $A$ is of finite dimension over $\K$, for every $a\in A$ there exists $k\geq 0$ such that set $\{1,a,a^2,\dots,a^{k-1}\}$ is linearly independent over $\K$ and $\{1,a,a^2,\dots,a^k\}$ is linearly dependent over $\K$. This means that there exists a non-trivial polynomial $p\in\K[X]$ such that $\deg(p)=k$ and $p(a)=0$. Without lost of generality, assume $p=\sum_{i=0}^{k} c_iX^i$, $c_i\in \K$ and $c_k=1$.

First we note that if $a$ is non-invertible, then $p(0)=c_0=0$. Indeed, assume $p(0)=c_0\in\K^\times$. Then

$$p(a)=\sum_{i=0}^{k} c_ia^i=(\sum_{i=0}^{k-1} c_{i}a^{i-1})a+c_0=0.$$

This means that $a$ is invertible and $a^{-1}=-c_0^{-1}(\sum_{i=0}^{k-1} c_{i}a^{i-1})$.

Now, since $c_0=0$, we write $p(X)=q(X)X$ where $q\in\K[X]$ is a non-trivial polynomial of degree $k-1$. Hence, $q(a)\neq 0$ and we obtain $q(a)a=0$. That means, $a$ is a zero divisor in $\K[a]$

Let $\K$ be a field and $A$ be a unital associative finite dimensional $\K$-algebra. Let $a\in A$ and $p\in\K[X]$ such that $p(a)=0$, then $p(\lambda)=0$ for all eigenvalues $\lambda$ of $a$. In particular, there are only finitely many eigenvalues of $a$.

Let $\lambda$ be an eigenvalue of $a$. Therefore, by previous Proposition, there exists $0\neq b\in\K[a]$ with $(a-\lambda\cdot 1)b=0$. In particular, $ab=\lambda b$ and by induction $a^ib=a^{i-1}(ab)=a^{i-1}(\lambda b)=\lambda^i b$ for all $i\in\N$.

If $p\in\K[X]$ be such that $p(a)=0$, then $p(a)b=0$. Since $a^ib=\lambda^i b$ for all $i\in\N$, we obtain $0=p(a)b=p(\lambda)b$. But $p(\lambda)\in \K$ and $b\neq0$, therefore, $p(\lambda)=0$.

We will say that a  topological field has a non-trivial topology if its topology is not discrete nor indiscrete.

The condition that the topology on a field is non-trivial is remarkably strong. Such topologies are automatically $T_0$, because the intersection of all the neighborhoods of $0$ must be an ideal, hence it is $\{0\}$. This implies that the topology is Hausdorff, from the properties of topological groups. Fields with non-trivial topologies must be infinite: in fact, for finite $\K$, every one-point set is closed means that every one-point set is open as complement of finite-point (closed) set, i.e. the topology is discrete. An important example is an ordered field endowed with the order topology.

Let $\K$ be a topological field with a non-trivial topology. Let $A$ be a unital associative finite-dimensional $\K$-algebra, and let $V$ be a vector subspace of $A$ that contains at least one invertible element. Then $V^\times$ is an open dense subset of $V$.

Notice that $x\in A$ is invertible in $A$ if and only if the linear map $L_x$ is surjective, which holds if and only if $L_x$ is invertible in $\Mat(n,\K)$. Since $0\in\K$ is closed, the set of all invertible elements $\GL(n,\K)=\Mat(n,\K)\bs \det^{-1}(0)\subset \Mat(n,\K)$ is open. Hence, the intersection of $\GL(n,\K)$ with $V$ is open in $V$ as well.

To see the density, consider an invertible element $u \in V$. The set $u^{-1} \cdot V$ is again a vector subspace of $A$ and contains the unit $1$. It suffices to show that density holds for $u^{-1} \cdot V$. If $x \in u^{-1} \cdot V$ is not invertible, consider $y_\epsilon = x + \epsilon \cdot 1$. By the previous proposition, there are only finitely many values of $\epsilon$ such that $y_\epsilon$ is not invertible, namely the opposites of eigenvalues of $x$.

Since $\K$ is infinite and non-discrete with closed points, there exists a (maybe punctured) neighborhood $U$ of $0\in\K$ that does not contain eigenvalues of $x$. Now every open neighborhood of $x$ intersects non-trivially the set $\{x + \epsilon \cdot 1\mid \epsilon\in U\}$. To see this, we choose a basis $(e_1,\dots,e_n)$ of $u^{-1}\cdot V$ containing $1=e_1$. Then by definition of the product topology for every neighborhood $W$ of $x$ there exist a neighborhood $W'$ of $0\in\K$ such that $W$ contains the following open neighborhood $\{x+\lambda_1e_1+\dots+\lambda_ne_n\mid \lambda_i\in W'\}$. Since $U\cap W'\neq \emptyset$, $W\cap \{x + \epsilon \cdot 1\mid \epsilon\in U\}\neq \emptyset$. This shows that $V^\times$ is dense in $V$.

For algebras $(A,\sigma)$ satisfying Corollary <ref>,
$A^\times$ is open and dense in $A$, $(A^\sigma)^\times$ is open and dense in $A^\sigma$.

A closed half-space in $V$ is the set

$$\{x\in V \mid f(x) \geq 0\}\,, $$

for some linear functional $f$. Similarly, the set

$$\{x\in V \mid f(x) > 0\}$$

is an open half-space.

For a subset $D \subset V$, $C(D)$ is proper if and only if $\overline{D} \setminus \{0\}$ is contained in an open half-space.

First, notice that $\overline{C(D)}=C(\overline D)$.
One direction is clear, if $\overline{D} \setminus \{0\}$ is contained in an open half-space, i.e. there exists a linear functional $f$ such that for all $x\in \overline{D} \setminus \{0\}$, $f(x)>0$. Then for all $x,y\in \overline{D} \setminus \{0\}$ and for all $\lambda,\mu\in\R_+$, $f(\lambda x+\mu y)=\lambda f(x)+\mu f(y)>0$, i.e. $C(\overline D)\bs \{0\}\subseteq\{x\in V\mid f(x)>0\}$. Then for $x\in (-C(\overline D))\setminus \{0\}$, $f(x)<0$. Therefore, $C(\overline D)\cap-C(\overline D)=\{0\}$, i.e. $C(D)$ is proper.

Show the other direction. Let now $C(D)$ is proper. We fix some norm $\|\cdot\|$ on $V$. Let $S(r):=\{x\in V\mid \|x\|=r\}$ the sphere of radius $r$ with the center at the origin. We take the following set $\overline{C(D)}\cap S(1)$. We denote by $C$ its convex hull. It is clear that $C\subset \overline{C(D)}$, $C$ is compact and $0\neq C$ otherwise there are $x,y\in \overline{C(D)}\cap S(1)$ such that $x+y=0$. This contradicts to properness of the $C(D)$.

Further, by the Hahn-Banach separation theorem, there exists a linear functional $f$ such that $f(0)=0$ and $f(x)>0$ for all $x\in C$. Then, for every $x\in \overline{C(D)}\bs\{0\}$ there exists $y\in \overline{C(D)}\cap S(1)\subseteq C$ and $\lambda>0$ such that $x=\lambda y$. Therefore, $f(x)=\lambda f(y)>0$. Therefore,

$$\overline D\bs\{0\}\subseteq \overline{C(D)}\bs\{0\}\subset\{x\in V\mid f(x)>0\}.$$

When $\K$ is an ordered field, we will assume that it is endowed with the order topology, which is always non-trivial. Recall that a subset $C \subset V$ of a $\K$-vector space is a cone if it is stable under multiplication by a strictly positive scalar. A cone is convex if it is stable by sums of its elements.
If $C$ is a convex cone, its closure $\overline{C}$ and its interior $\mathring{C}$ are still convex cones. The set of the opposites of the elements of $C$, denoted by $-C$, is still a convex cone.
A convex cone $C$ is proper if

$$\overline{C} \cap - \overline{C} = \{0\}\,.$$

If $C$ is a cone in some algebra over $\K$, then we denote by $C^\times$ the subset of all invertible elements of $C$.

Similarly to Proposition <ref> can be proven:

Let $\K$ be an ordered field. Let $A$ be a unital associative finite-dimensional $\K$-algebra, and let $C$ be a convex cone of $A$ that contains at least one invertible element. Then $C^\times$ is an open dense subcone of $C$.

As we have seen in Remark <ref>, for Hermitian algebras, $A^\sigma_+$ and $A^\sigma_{\geq 0}$ are proper convex cones in $A^\sigma$. Proposition <ref> implies that $A^\sigma_+$ is open and dense in $A^\sigma_{\geq 0}$.

§.§ Jordan algebras and spectral theorem

In this section we recall the definitions of a Jordan algebra and formally real Jordan algebra and prove some properties of formally real Jordan algebra that we will need later. We also show how Jordan algebras arise from associative algebras.

Let $(V,\circ)$ be an possibly non-associative unital algebra over some field $\K$. $(V,\circ)$ said to be a Jordan algebra if for all $x,y\in V$

* $x\circ y=y\circ x$;

* $(x\circ y)\circ(x\circ x)=x\circ(y\circ(x\circ x))$  (Jordan identity).

A Jordan algebra $(V,\circ)$ over a real closed field is called formally real if for all $x,y\in V$, $x^2+y^2=0$ implies $x,y=0$.

The following is immediate:

Let $(A,\sigma)$ be a finite dimensional algebra with anti-involution, then the following hold:

* The algebra $(A^\sigma,\circ)$ is a Jordan algebra where

$$x\circ y= \frac{xy+yx}{2}.$$

* If $(A,\sigma)$ is Hermitian, the Jordan algebra $(A^\sigma,\circ)$ is formally real.

Jordan algebras that arise form associative algebras as in <ref> are called special Jordan algebras. All other Jordan algebras are called exceptional Jordan algebras.

In this section we establish some properties of formally real Jordan algebras. We do not assume that they are necessarily special.

Let $(V,\circ)$ be a formally real Jordan algebra over a real closed field $\K$. We use the following notation:

$$V_+=\left\{\sum_{i=1}^na_i^2\midwd a_i\in V^\times,\;n\in\N\right\},$$

$$V_{\geq 0}=\left\{\sum_{i=1}^na_i^2\midwd a_i\in V,\;n\in\N\right\}.$$

An element $c\in V$ is called an idempotent if $c^2=c$.

* Two idempotents $c,c'\in V$ are called orthogonal if $c\circ c'=0$.

* A tuple $(c_1,\dots,c_k)$ of pairwise orthogonal idempotents is called a complete orthogonal system of idempotents if $c_1+\dots+c_k=1$.

Note that every idempotent $c$ satisfies $c\in V_{\geq 0}$. Moreover, if $c\neq 1$, then $c\in V_{\geq 0}\bs V_+$

[Spectral theorem, first version]
Let $V$ be a formally real Jordan algebra over a real closed field $\K$. For every $b\in V$, there exist a unique $k\in\N$, unique elements $\lambda_1,\dots,\lambda_k\in\K$, all distinct, and a unique complete system of orthogonal idempotents $c_1,\dots,c_k\in \K[b]\subseteq V$ such that

We call this the spectral decomposition of $b$.

For a proof of this theorem for real Jordan algebras see <cit.>. A proof for general real closed fields is identical.

We collect some direct consequences of Theorem <ref>

Let $(V,\circ)$ be a formally real Jordan algebra and $c_1,c_2\in V$ be two orthogonal idempotents. There exists an $a\in V$ such that $c_1,c_2\in \K[a]$. If $V=A^\sigma$ for an associative Hermitian algebra $(A,\sigma)$, then $c_1c_2=0$.

We consider $a:=c_1-c_2$. This is the spectral decomposition of $a$. By the spectral theorem, $c_1,c_2\in \K[a]$. Let $V=A^\sigma$ for an Hermitian algebra. Since $c_1,c_2\in\K[a]$, they commute with respect to the product of $A$. Therefore, $0=c_1\circ c_2=c_1c_2$.

For $b\in V_{\geq 0}$, we have $\lambda_1,\dots,\lambda_k\geq 0$. If $b\in V_+$, we have moreover, $\lambda_1,\dots,\lambda_k > 0$.

The set of all invertible elements $V^\times$ of $V$ consists of elements such that in the spectral decomposition we have $\lambda_i\neq 0$ for all $i$. If all $\lambda_i\neq 0$, then

For every (continuous/smooth) function $f\colon \K \to \K$, the (continuous/smooth) map

$$\hat f\colon V \to V$$

can be defined: if

$$\hat f(b):=\sum_{i=1}^k f(\lambda_i)c_i.$$

This map is well defined because the spectral decomposition is unique. Analogously, for any function $f\colon \K_{\geq 0} \to \K$ or $f\colon \K_+ \to \K$, $\hat f\colon V_{\geq 0} \to V$ resp. $\hat f\colon V_+ \to V$ can be defined.

In particular, for every $b\in V_{\geq 0}$, the element $b^t\in V_{\geq 0}$ for $t\in\Q_+$ is well-defined (If $\K=\R$ it is even well-defined for all $t\in\R_+$). This definition is compatible with integer powers of elements.

* The space $V_+$ is
semi-algebraically contractible. The set $\{1\}$ is a semi-algebraic deformation retract of $V_+$. In particular, $V_+$ is semi-algebraically connected.

* The space $V_{\geq 0}$ is
semi-algebraically contractible. The set $\{1\}$ is a semi-algebraic deformation retract of $V_{\geq 0}$. In particular, $V_{\geq 0}$ is semi-algebraically connected.

We consider the following continuous semi-algebraic map: $H(t,a)=t\cdot a+(1-t)\cdot 1$ where $t\in [0,1]=\{s\in\K\mid 0\leq s\leq 1\}\subset\K$, $a\in  V_+$. Since, $H(t,a)=\sum_{i=1}^k (\lambda_i t+(1-t))c_i$ where $(c_i)$ is a complete system of orthogonal idempotents in the spectral decomposition of $a$. Since all $\lambda_i$ are positive, the convex combination $\lambda_i t+(1-t)$ is positive for all $t\in [0,1]$. Therefore, $H(t,a)\in V_+$ for all $t\in [0,1]$, $a\in V_+$. Moreover, $H(1,a)=a$ and $H(0,a)=1$, i.e. $H$ is a contraction and $\{1\}$ is a semi-algebraic deformation retract of $V_+$.

The same contraction works also for $V_{\geq 0}$.

Let $\K=\R$.
The space $V_+$ is (non-semi-algebraically) homeomorphic to $V$. In particular, $V_+$ is open in $V$ and contractible, and  $\{1\}\subset V_+$ is a deformation retract of $V_+$.

The map $f(t)=\log(t)$ gives a homeomorphism between $V_+$ and $V$. The rest follows from <ref>.

We now state the second version of the spectral theorem. For this we need to give some additional definitions:

An idempotent $0\neq c\in V$ is called primitive if it cannot be written as a sum of two orthogonal non-zero idempotents.

* A complete orthogonal system of primitive idempotents $(c_1,\dots,c_k)$ is called a Jordan frame.

* The maximal number of elements in a Jordan frame is called the rank of a Jordan algebra.

[Spectral theorem, second version]
Let $V$ be a formally real Jordan algebra over a real closed field $\K$. Suppose that $V$ has rank $n$. For every $b\in V$ there exist a Jordan frame $(e_1,\dots,e_n)$ and a unique $n$-tuple of elements  $(\lambda_1(b),\dots,\lambda_n(b))\in\K^n$ such that $\lambda_1(b)\geq\dots\geq\lambda_n(b)$ and

The elements $\lambda_1(b),\dots,\lambda_n(b)\in\K$ (with their multiplicities) are called the  eigenvalues of $b$ are uniquely determined by $b$. In particular, they do not depend (up to permutations) on the Jordan frame $e_1,\dots,e_n\in V$.

For a proof of this theorem for real Jordan algebras see <cit.>. The proof for a general real closed field is identical.

A Jordan frame $e_1,\dots,e_n\in V$ associated to the element $b\in V$ by Theorem <ref> is in general not unique and elements $e_1,\dots,e_n\notin \K[b]$, in contrast to the complete system of orthogonal idempotents from  Theorem <ref>.

In Definition <ref> we already defined  eigenvalues for elements of an associative algebras. This definition works also for non-associative algebras (see Remark <ref>). If $V$ is a formally real Jordan algebra, then the eigenvalues defined by the spectral theorem agree with the eigenvalues defined in Definition <ref>.

The signature of an element $b\in V$ is a pair $(p,q)$ with $p,q\in\N\cup\{0\}$ such that $b$ has $p$ positive eigenvalues and $q$ negative eigenvalues in the spectral decomposition defined in <ref>.

Notice that for an element $b$ of signature $(p,q)$, $p+q\leq n$, moreover, $p+q=n$ if and only if $b$ is invertible.

Let $b\in V$ and $\lambda_1(b),\dots,\lambda_n(b)$ are all its eigenvalues with multiplicities. We define
the eigenvalue map:

\lambda\colon & V & \mapsto & \K^n\\
& b & \to & (\lambda_1(b),\dots,\lambda_n(b)),
\end{matrix}$$

and the trace and the determinant maps:

$$\tr(b):=\sum_{i=1}^n \lambda_i(b),\; \det(b):=\prod_{i=1}^n \lambda_i(b).$$

Since the eigenvalues defined by the spectral theorem and eigenvalues defined in <ref> agree, the eigenvalue map is semi-algebraic. The eigenvalue map is continuous. For a proof when $\K = \R$ see <cit.>. For a general real closed field the proof is identical.

The function

\beta\colon & V\times V & \to & \K\\
& (b_1,b_2) & \mapsto & \tr\left(b_1\circ b_2\right)
\end{matrix}$$

is an inner product on $V$, where $V$ is intended as a $\K$-vector space.

* The set $V_+$ is an open and closed proper convex cone in $V^\times$. In particular, $V_+$ is a semi-algebraic connected component of $V^\times$.

* The set $V_{\geq 0}$ is a closed proper convex cone in $V$.

From <ref> we already know that $V_+$ and $V_{\geq 0}$ are proper convex cones in $V^\times$ and in $V$ respectively.

The set $V_+=\lambda^{-1}(\K_+^n)$ is open and closed in $V^\times=\lambda^{-1}((\K\bs\{0\})^n)$. In particular, $V_+$ is a semi-algebraic connected component of $V$ because by <ref> $V_+$ is semi-algebraically connected.

The set $V_+=\lambda^{-1}(\K^n_{\geq 0})$ is closed in $V$.

It is a cone since for any $a\in  A^\sigma_+$ and $\lambda>0$, $\lambda a\in  A^\sigma_+$ because $\lambda a=(\sqrt{\lambda}x)^2$ where $x^2=a$, $x\in  A^\sigma$.

The cone $ A^\sigma_+$ is proper because if $a,-a\in  A^\sigma_{\geq 0}=\overline{A^\sigma_+}$, then there exist $x,y\in A^\sigma$ such that $a=x^2$, $-a=y^2$. Then $0=x^2+y^2$, but since $A$ is Hermitian this implies $x=y=a=0$.

To see that the cone $ A^\sigma_+$ is convex, let us take $a,b\in  A^\sigma_+$. We have to show that $a+b\in  A^\sigma_+$.
Since $ A^\sigma_+$ is a connected component of $(A^\sigma)^\times$, it is enough to show that $a+b$ can be connected with $1\in  A^\sigma_+$ by a semi-algebraic path completely contained in $(A^\sigma)^\times$. We take the spectral decompositions of:

$$a=\sum_{i=1}^{n}\lambda_i c_i;\;b=\sum_{i=1}^{n}\mu_i c'_i$$

for two Jordan frames $\{c_i\}_{i=1}^n$ and $\{c'_i\}_{i=1}^n$. Since $\lambda_i>0$ and $\mu_i>0$ for all $i\in\{1,\dots,n\}$, $\lambda_i(t):=\lambda_it+\frac{(1-t)}{2}>0$ and $\mu_i(t):=\mu_it+\frac{(1-t)}{2}>0$ for all $i\in\{1,\dots,n\}$ and $t\in[0,1]$. Therefore,

$$x(t):=\sum_{i=1}^{n}\lambda_i(t) c_i+\sum_{i=1}^{n}\mu_i(t) c'_i$$

is a well defined semi-algebraic path in $(A^\sigma)^\times$ with $x(0)=a+b$ and $x(1)=1$.

Consider first $1+x^{-1}y^2x^{-1}=1+\sigma(yx^{-1})yx^{-1}$. The element $\sigma(yx^{-1})yx^{-1}$ is in  $A^\sigma_+$. We consider the spectral decomposition

where all $\lambda_i>0$, and $\{c_i\}$ is a complete system of orthogonal idempotents. Then

Since all $1+\lambda_i>0$, $1+x^{-1}y^2x^{-1}\in A^\sigma_+$, i.e. there exists $z\in (A^\sigma)^\times$ such that $1+x^{-1}y^2x^{-1}=z^2$. Then $a+b=xz^2x=\sigma(zx)zx\in A^\sigma_+$ because $zx\in A^\times$.

§.§ Classification of simple formally real Jordan algebras

In this section, we recall the well-known classification of simple formally real Jordan algebras over $\R$ and generalize it to all real closed fields.

Every simple formally real Jordan algebra over a real closed field $\K$ is isomorphic to one of the following Jordan algebras:

* Symmetric real matrices $(\Sym(n,\K),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Sym(n,\K)$;

* Hermitian complex matrices $(\Herm(n,\K_\CC),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Herm(n,\K_\CC)$;

* Hermitian quaternionic matrices $(\Herm(n,\K_\HH),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Herm(n,\K_\HH)$;

* $(B_n,\circ)$ where $B_n=\Span_\K(1,x_1,\dots, x_n)$ with $1\circ x_i=x_i$, $x_i\circ x_j=0$ for $i\neq j$, $x_i\circ x_i=1$ for all $i,j\in\{1,\dots,n\}$;

* Hermitian octonionic $3\times 3$ matrices $(\Herm(3,\K_\Oc),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Herm(3,\K_\Oc)$.

In case $\K=\R$ this theorem agrees with the classification of formally real Jordan algebras over $\R$ that is proven in [13, 9]. The general version of this classification follows from the classification in case $\K=\R$ using the Tarski–Seidenberg transfer principle (see <cit.>).

Every simple formally real Jordan algebra is isomorphic to one of the following Jordan algebras:

* Symmetric real matrices $(\Sym(n,\R),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Sym(n,\R)$;

* Hermitian complex matrices $(\Herm(n,\CC),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Herm(n,\CC)$;

* Hermitian quaternionic matrices $(\Herm(n,\HH),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Herm(n,\HH)$;

* $(B_n,\circ)$ where $B_n=\Span_\R(1,x_1,\dots, x_n)$ with $1\circ x_i=x_i$, $x_i\circ x_j=0$ for $i\neq j$, $x_i\circ x_i=1$ for all $i,j\in\{1,\dots,n\}$;

* Hermitian octonionic $3\times 3$ matrices $(\Herm(3,\Oc),\circ)$ where $a\circ b=\frac{ab+ba}{2}$ for $a,b\in \Herm(3,\Oc)$.

The Jordan algebras (1)–(3) from <ref> can be seen as $A^\sigma$ for certain simple algebras $(A,\sigma)$. Let $\K$ be a real closed field.

* Symmetric real matrices: $A=\Mat(n,\K)$, $\sigma(r):=r^T$ is an algebra with an anti-involution. Then $A^\sigma=\Sym(n,\K)$ space of all symmetric matrices. The algebra $(A,\sigma)$ is Hermitian with $A^{\sigma}_+=\Sym^+(n,\K)$ real symmetric positive definite matrices.

* Hermitian complex matrices: $A=\Mat(n,\K_\CC)$, $\bar\sigma(r):=\bar r^T$ is a Hermitian algebra with $A^{\bar\sigma}=\Herm(n,\K_\CC)$ complex Hermitian matrices and $A^{\bar\sigma}_+=\Herm^+(n,\K_\CC)$ complex Hermitian positive definite matrices.

* Hermitian quaternionic matrices: $A=\Mat(n,\K_\HH)$, $\bar\sigma(r):=\bar r^T$, is a Hermitian algebra  with $A^{\bar\sigma}=\Herm(n,\K_\HH)$ quaternionic Hermitian matrices and $A^{\bar\sigma}_+=\Herm^+(n,\K_\HH)$ quaternionic Hermitian positive definite matrices.

There is another anti-involution on $A=\Mat(n,\K_\HH)$, namely $\sigma_1(r):=\sigma(r_1)+\bar\sigma(r_2)j$ where $r_1,r_2\in\Mat(n,\K_\CC)$. This algebra $(A,\sigma_1)$ is not Hermitian.

The Jordan algebra $(\Herm(3,\K_\Oc),\circ)$ is exceptional. This means that there is no associative real algebra $A$ that contains $\Herm(3,\K_\Oc)$ as a Jordan subalgebra.

The Jordan algebra $(B_n,\circ)$ can be embedded as a Jordan subalgebra into the even Clifford algebra $\Cl_{even}(1,n)$ for some appropriate anti-involution $\sigma$, but $\Cl(1,n)^\sigma$ is strictly bigger than $B_n$ for $n>2$. In the case $n=2$, $\Cl(1,2)$ is isomorphic to $\Mat(2,\K)$ as an associative algebra, and $B_2$ is isomorphic to $\Sym(2,n)$ as a Jordan algebra.

§.§ Positivity of the norm

Let $\K$ be a real closed field and $(A,\sigma)$ be a Hermitian semisimple algebra over $\K$. The goal of this subsection is to show that $\theta(a)=\sigma(a)a\in A^\sigma_{\geq 0}$ for all $a\in A$. To do it, first, we prove some technical propositions.

Let $(A,\sigma)$ be a Hermitian semisimple algebra, then $-a^2\in A^\sigma_{\geq 0}$ for every $a\in A^{-\sigma}$.

We prove it by induction on the rank of $A$. If its rank is one, then the statement follows from the Corollary <ref>.

Let now rank of $A$ be equal $n>1$ and we assume that for every Hermitian semisimple algebra $A'$ of rank smaller then $n$, the Proposition holds. Take $a\in A^{-\sigma}$, then $a^2\in A^\sigma$. We consider a spectral decomposition:

$$a^2=\sum_{i=1}^n \lambda_i e_i$$

where $(e_i)_{i=1}^n$ is a Jordan frame. We denote

Then $E+e=1$, $eE=Ee=0$, and algebras $(EAE,\sigma)$ and $(eAe,\sigma)$ are Hermitian semisimple subalgebras of $A$ of rank strictly smaller then rank of $A$. Therefore,

\begin{align*}
a^2 = & (EaE+Eae+eaE+eae)^2=EaEaE+eaeae+(eaEae+EaeaE)\\
& +(Eaeae+EaEae)+(eaeaE+eaEaE).
\end{align*}

Since $EaEaE=(EaE)^2$ and $EaE\in (EAE)^{-\sigma}$, we have $-EaEaE\in (EAE)^\sigma_{\geq 0}\subseteq A^\sigma_{\geq 0}$. The same holds for $-eaeae$. Further,

$$-(eaEae+EaeaE)=(exE-Exe)^2\in A^\sigma_{\geq 0}$$

because $exE-Exe\in A^\sigma$.

beacsue of the spectral decomposition of $a^2$ and $eE=Ee=0$. The same holds for $eaeaE+eaEaE$.

Let $(A,\sigma)$ be a Hermitian semisimple algebra and $x\in A$ satisfy the property $\sigma(x)x=0$, then $x=0$.

We denote: $A_0:=\{x\in A\mid \sigma(x)x=0\}$.
First, note that $\sigma(x)x=0$ if and only if $x\sigma(x)=0$. Indeed, let $\sigma(x)x=0$, then $x\sigma(x)\in A^\sigma$ and $(x\sigma(x))^2=x\sigma(x)x\sigma(x)=0$. Therefore, since $A$ is Hermitian, $x\sigma(x)=0$.

Further, write $x=x^s+x^a$ where $x^s\in A^\sigma$, $x^a\in A^{-\sigma}$. Then

Therefore, $x^sx^a-x^ax^s=0$ and $(x^s)^2-(x^a)^2=0$. Since $(x^s)^2,-(x^a)^2\in A^{\sigma}_{\geq 0}$, $x^s=0$ and $(x^a)^2=0$. Therefore $1+x^a=1+x$ is invertible and $(1+x)^{-1}=1-x$.

Let $r\in A$, consider $y=xr$. Then $\sigma(y)y=\sigma(r)\sigma(x)xr=0$, i.e. $y\in A_0$. Further, take $r'\in A$ and consider $z=r'y$, then $z\sigma(z)=0$. Therefore, as we have seen, $\sigma(z)z=0$ and $z\in A_0$. That means, for every $r,r'\in A$, $1+r'xr\in A^\times$, i.e. $x\in J(A)$ where $J(A)$ is the Jacobson radical of $A$ (see Definition <ref>). Because $A$ is semisimple, $J(A)=\{0\}$.

Let $\K$ be a real closed field. As before we denote by $\K_\CC$ the algebraic closure of $\K$.

Let $(A,\sigma)$ be a Hermitian semisimple algebra, then the complexified algebra $(A_\CC,\bar\sigma_\CC)$, where $A_\CC=A\otimes_\K \K_\CC$ and $\bar\sigma_\CC$ the complex anti-linear extension of $\sigma$ (i.e. $\bar\sigma(x+iy)=\sigma(x)-i\sigma(y)$ for $x,y\in A$), is Hermitian as an involutive algebra over $\K$ ans semisimple.

The algebra $(A_\CC,\bar\sigma_\CC)$ is clearly unital and associative. We check now that, if $x^2+y^2=0$, then $x=y=0\in A^{\bar\sigma_\CC}_\CC$.

For $x\in A^{\bar\sigma_\CC}_\CC$ we write $x=x_1+ix_2$, $x_1,x_2\in A$. Since
$x_1\in A^\sigma$, $x_2\in A^{-\sigma}$. Further,

Since $x_1^2,y_1^2,-x_2^2,-y_2^2\in A^\sigma_{\geq 0}$, we have $x_1^2=y_1^2=-x_2^2=-y_2^2=0$. Therefore, by Proposition <ref>, $x_1=x_2=y_1=y_2=0$, i.e. $x=y=0$.

Assume now that $A_\CC$ is not semisimple, i.e. the Jacobson radical $J(A_\CC)$ of $A_\CC$ is non-trivial. Let $0\neq x\in J(A_\CC)$, then by <ref>, $\bar\sigma_\CC(x)=-x$ and $x^2=0$. Therefore, $\bar\sigma_\CC(ix)=ix$, i.e. $ix\in A^{\bar\sigma_\CC}_\CC$ and $(ix)^2=0$. This contradicts to the property to be Hermitian for $(A_\CC,\bar\sigma_\CC)$.

Let $Y$ be a finite dimensional $\K_\CC$-algebra, $V\subseteq Y$ be a $\K_\CC$-vector subspace. Then $V^\times$ is semi-algebraically connected.

If $V^\times=\emptyset$ then $V^\times$ is semi-algebraically connected.

Assume now that $1\in V^\times\neq\emptyset$. As we have seen in the Proposition <ref>, $Y$ can be embedded as a subalgebra into $\Mat(r,\K_\CC)$ for some $r\in \N$. We identify $Y$ as a subalgebra of $\Mat(r,\K_\CC)$.

Let $a\in V^\times\subseteq\GL(n,\K_\CC)$. We consider

$$S^1_{\K_\CC}:=\{x+iy\in\K_\CC\mid x,y\in\K, x^2+y^2=1\}.$$

Notice, that $S^1_{\K_\CC}$ is a semi-algebraically connected set. In particular, it has infinitely many points.
Since $a$ has only finitely many eigenvalues, and $0$ is not one of them, there is a point $z\in S^1_{\K_\CC}\subset\K_\CC$ such that the $\K$-line $\{tz\mid t\in\K\}$ in $\K_\CC$ through the origin containing $z$ does not intersect any of the eigenvalues of $a$. Now, consider the $\K$-path $f(t)=at+z(1-t)\Id$, $t\in[0,1]=\{s\in\K\mid 0\leq s\leq 1\}\subset \K$. It lies completely in $V$ because it is a $\CC$-vector space. This has determinant $0$ if and only if $z(t-1)$ is an eigenvalue of $at$, which happens if and only if $z(1-t)/t$ is an eigenvalue of $a$ (this does not work when $t=0$, but then it is clear that the determinant is non-zero). By construction, it is not the case for any $t\in[0,1]\subset\K$, so this defines a path form $a$ to $z\Id$.
Now, there is a path in $\K_\CC$ not passing through $0$ from $z$ to $1$, and, since $\{z\Id\mid z\in S^1_{\K_\CC}\}\subset V^\times$, this gives rise to a semi-algebraic path in $A_\CC^\times$ from $z\Id$ to $\Id$, and so concatenating these two paths, we get a path from $a$ to $\Id$ that lies in $V^\times$, showing that $V^\times$ is semi-algebraically connected.

Finally, if $1\notin V^\times$ but there exists $v\in V^\times$, then $V^\times$ is semi-algebraically connected if and only if $(v^{-1}V)^\times=v^{-1}(V^\times)$ is semi-algebraically connected. Moreover, $v^{-1}V$ is also a $\K_\CC$-vector space and $1\in (v^{-1}V)^\times$. So $(v^{-1}V)^\times$ is semi-algebraically connected and, therefore, $V^\times$ is semi-algebraically connected as well.

Now we are ready to prove the following

Let $(A,\sigma)$ be a Hermitian semisimple algebra, then $\sigma(a)a\in A^\sigma_{\geq 0}$ for all $a\in A$. In particular, $A^\sigma_+=\theta(A^\times)$ and $A^\sigma_{\geq 0}=\theta(A)$.

First, we embed $(A,\sigma)$ into its complexification $(A_\CC,
\bar\sigma_\CC)$ that is Hermitian as well. Let $a\in A^\times\subset A_\CC^\times$. Since $A_\CC^\times$ is semi-algebraically connected, we take a semi-algebraic path $\gamma\colon [0,1]\to A^\times_\CC$ such that $\gamma(0)=a$, $\gamma(1)=1\in A$. The eigenvalue map:
$\lambda(\sigma(\gamma(t))\gamma(t))$ takes values in the semi-algebraic connected component of $(1,\dots,1)\in(\K\bs\{0\})^n$, i.e. in $\K_+^n$. In particular, $\lambda(\sigma(a)a)=\lambda(\sigma(\gamma(0))\gamma(0))$ is positive meaning that $a\in A^\sigma_+\subset (A^{\bar\sigma_\CC}_\CC)_+$.

Since $A$ is the closure of $A^\times$, the eigenvalue map takes values in the closure of the connected component of $(1,\dots,1)\in(\R\bs\{0\})^n$, i.e. all eigenvalues of $\sigma(a)a$ are non-negative for $a\in A$.

We conjecture that the Theorem <ref> holds also for non-semisimple Hermitian algebras.

Let $V$ be a formally real Jordan algebra over a real closed field $\K$ and $V_+$ be the cone of positive elements of $V$. The group of all linear transformations of $V$ (as $\K$-vector space) which preserve $V_+$ is called structure group of $V$ and denoted by $G(V)$.

The group $A^\times$ acts on $A^\sigma$ preserving $A^\sigma_+$ in the following way:

\psi\colon & A^\times\times A^\sigma & \mapsto & A^\sigma\\
& (a,b) & \to & \sigma(a)ba.
\end{matrix}$$

In other words, $\psi(A^\times)$ is a subgroup of $G(A^\sigma)$. Moreover, the restriction of this action to $A^\sigma_+$ is transitive.

§.§ Polar decomposition

Let $(A,\sigma)$ to be a Hermitian algebra. From now on, we always assume the algebra $A$ to be semisimple.

[Polar decomposition, first version]
The map

\pol\colon & U_{(A,\sigma)}\times A^\sigma_+ & \to & A^\times \\
& (u,b) & \mapsto & ub
\end{matrix}$$

is a semi-algebraic homeomorphism. In particular, for every $g\in A^\times$ there exist unique $b\in A^\sigma_+$ and $u\in U_{(A,\sigma)}$ such that $g=ub$.

The map $\pol$ is well-defined because $A^\sigma_+\subseteq A^\times$. First, we prove the surjectivity. Take $g\in A^\times$, then $\sigma(g)g\in A^\sigma_+$ by Proposition <ref>. Take $b:=(\sigma(g)g)^{\frac{1}{2}}$, then $u:=g(\sigma(g)g)^{-\frac{1}{2}}\in U_{(A,\sigma)}$. Indeed,

Now, we prove the injectivity. Let $g=ub=u'b'$ where $u,u'\in U_{(A,\sigma)}$, $b,b'\in A^\sigma_+$. Then $\sigma(g)g=(b')^2=b^2\in A^\sigma_+$. We take the spectral decompositions of $b$ and $b'$:

$$b=\sum_{i=1}^k\lambda_i c_i,\;b'=\sum_{i=1}^{k'}\lambda_i' c_i'$$

where all $k,k'\in \N$, $\lambda_i,\lambda_i'>0$ and $\{c_i\}$, $\{c'_i\}$ are complete orthogonal systems of idempotents of $A^\sigma$. Then

$$b^2=\sum_{i=1}^k\lambda_i^2 c_i=\sum_{i=1}^{k'}(\lambda_i')^2 c_i'=(b')^2.$$

Because of the uniqueness of the spectral decomposition, $k=k'$ and, up to reordering, all $\lambda_i^2=(\lambda_i')^2$, $c_i=c_i'$. But all $\lambda_i>0$, therefore, $\lambda_i=\lambda_i'$, i.e. $b=b'$ and $u=gb^{-1}=g(b')^{-1}=u'$.

Finally, by definition, $\pol$ is continuous. Moreover,

is continuous as well and contains only semi-algebraic operations. Therefore, $\pol$ is a semi-algebraic homeomorphism.

The map

A^\sigma_+ \times U_{(A,\sigma)} & \to & A^\times \\
(b,u) & \mapsto & bu
\end{matrix}$$

is a semi-algebraic homeomorphism. In particular, for every $g\in A^\times$ there exist unique $b\in A^\sigma_+$ and $u\in U_{(A,\sigma)}$ such that $g=bu$.

The group $U_{(A,\sigma)}<A^\times$ is a semi-algebraic deformation retract of $A^\times$.

This follows from the polar decomposition <ref> and from <ref>.

The following $\K$-bilinear form $\beta\colon A\times A\to\K$:

on $A$ is positive definite. The group $U_{(A,\sigma)}$ acts on $A$ by the left and right multiplication preserving $\beta$.

The map $\beta$ is $\K$-bilinear on $A$. Moreover, $\beta(a,a)=\tr(\sigma(a)a)\geq 0$ for all $a\in A$ because $\sigma(a)a\in A^\sigma_{\geq 0}$.

If $\beta(a,a)=0$, then all eigenvalues of $\sigma(a)a$ are zero, i.e. $\sigma(a)a=0$ and, therefore, because of semisimplicity of $A$ we have $a=0$ (see Proposition <ref>).

The form $\beta$ is invariant under the left and right action by multiplication of $U_{(A,\sigma)}$ because the trace is invariant under conjugation.

The adjoint action of $U_{(A,\sigma)}$ on $A$ preserves $\beta$. In particular, the Lie algebra $A^{-\sigma}$ of $U_{(A,\sigma)}$ is compact.

With respect to the norm induced by the inner product $\beta$ on $A$, the following subsets of $A$ are closed, bounded and semi-algebraic:

* the group $U_{(A,\sigma)}$;

* $D(A,\sigma):=\{a\in A\mid 1-\sigma(a)a\in A^\sigma_{\geq 0}\}$.

In particular, for every vector subspace $V$ of $A$, the set

$$D(V,\sigma):=\{a\in V \mid 1-\sigma(a)a\in A^\sigma_{\geq 0}\}=D(A,\sigma)\cap V$$

closed, bounded and semi-algebraic.

We remind the definition of a derivation of a Jordan algebra:

Let $V$ be a Jordan algebra over some field $\K$. A $\K$-linear map $D\colon V\to V$ is called a derivation if $D(xy)=D(x)y+xD(y)$ for all $x,y\in V$. The space of all derivations of $V$ is denoted by $\Der(V)$.

The space $\Der(V)$ is a subspace of the space of all $\K$-vector space endomorphisms $\End(V)$ of $V$. Moreover, $\Der(V)$ is a Lie algebra with respect to the commutator and it agrees with the Lie algebra of the group of all Jordan algebra automorphisms $\Aut(V)$ of $V$.

Let $(A,\sigma)$ be a semisimple associative algebra over some field $\K$. The map $\ad\colon A^{-\sigma}\to \Der(A^\sigma)$, $\ad(u)a:=ua-au$ for $u\in A^{-\sigma}$, $a\in A^\sigma$ is a surjective homomorphism of Lie algebras.

As we have seen before, for all $u\in A^{-\sigma}$ and for all $a\in A^\sigma$, $\ad(u)a=ua-au\in A^\sigma$. Moreover, it is easy to see that $\ad(u)$ is a derivation for all $u\in A^{-\sigma}$ and that $\ad$ is a Lie algebra homomorphism. Therefore, $\ad(A^{-\sigma})\subseteq \Der(A^\sigma)$.

Since $A$ is semisimple, $A^\sigma$ is a semisimple Jordan algebra.  Every derivation of a semisimple Jordan algebra is inner (see <cit.>), i.e. for every derivation $D\in\Der(A^\sigma)$ there exist $a_1,a_2\in A^\sigma$ such that for all $a\in A^\sigma$, $D(a)=\ad([a_1,a_2])a=[a_1,a_2]a-a[a_1,a_2]$, where $[\cdot,\cdot]$ is the commutator on $A$. Finally, notice that $[a_1,a_2]\in A^{-\sigma}$ for $a_1,a_2\in A^\sigma$. That means that $\ad(A^{-\sigma})=\Der(A,\sigma)$.

Let $(A,\sigma)$ be a semisimple associative algebra over a real closed field $\K$.
The semi-algebraic connected component of the identity $\Aut_0(A^\sigma)$ of the automorphism group $\Aut(A^\sigma)$ of the Jordan algebra $A^\sigma$ agrees with the semi-algebraic  connected component of the identity of the group $\Ad(U_{(A,\sigma)})$ where $\Ad$ is the adjoint Lie group action on its Lie algebra.

This statement follows from the fact that $\Lie(U_{(A,\sigma)})=A^{-\sigma}$, $\Lie(\Aut(A^\sigma))=\Der(A^\sigma)$ and the derivative of the adjoint action $\Ad$ is $\ad$.

Let $(A,\sigma)$ be a semisimple Hermitian algebra over a real closed field $\K$ such that the Jordan algebra $A^\sigma$ is simple. Then $U_{(A,\sigma)}$ acts transitively on the set of all Jordan frames of $A^\sigma$.

This statement follows from the fact that $\Aut_0(A^\sigma)$ acts transitively on the set of all Jordan frames of $A^\sigma$ (see <cit.> for a proof over $\R$, general case follows identically).

Let $(A,\sigma)$ be a semisimple Hermitian algebra over a real closed field $\K$ such that the Jordan algebra $A^\sigma$ is simple. The action

\psi\colon & A^\times\times A^\sigma & \mapsto & A^\sigma\\
& (a,b) & \to & \sigma(a)ba.
\end{matrix}$$

of the group $A^\times$ on $A^\sigma$ preserves the signature of elements. The orbits of this action are precisely the sets of all elements of fixed signature in $A^\sigma$.

Let $k$ be the rang of $A^\sigma$. We fix a Jordan frame $(e_i)_{i=1}^k$ and denote: $o_{p,q}:=\sum_{i=1}^p e_i-\sum_{i=1}^qe_{k-i+1}$ for $p+q\leq k$.
First notice that for every element $b\in A^\sigma$ of signature $(p,q)$ there is an element $g\in A^\times$ such that $o_{p,q}=\sigma(g)b_1g$. To see this, first, we take a spectral decomposition $b=\sum_{i=1}^k \lambda_ic_i$ for a Jordan frame $(c_i)_{i=1}^k$ and $\lambda_1\geq\dots\geq\lambda_k$. Let $u\in U_{(A,\sigma)}$ such that $u$ maps the Jordan frame $(c_i)_{i=1}^k$ to $(e_i)_{i=1}^k$. Therefore: $\sigma(u)bu=\sum_{i=1}^k \lambda_ie_i$. Finally, we take $a:=\sum_{i=1}^k \mu_ie_i$ where $\mu_i=\lambda_i^{-1}$ if $\lambda_i\neq 0$ and $\mu_i=1$ otherwise. Then $\sigma(ua)b(ua)=o_{p,q}$.

The structure group $G(A^\sigma)$ acts on $A^\sigma$ preserving the signature (see <cit.>). Therefore, $A^\times$ does it as well.

§.§ Matrix algebra over a Hermitian algebra

One way to construct new semisimple Hermitian algebras is to consider a matrix algebra over a semisimple Hermitian algebra. For this, we assume $(A,\sigma)$ to be a semisimple Hermitian algebra. We consider the following anti-involution on the algebra $\Mat_n(A)$ of $n\times n$-matrices over $A$:

\sigma^T\colon & \Mat_n(A) & \to & \Mat_n(A)\\
& M & \mapsto & \sigma(M)^T
\end{matrix}$$

where $\sigma(M)$ means applying $\sigma$ componentwise to elements of the matrix $M\in\Mat_n(A)$. We denote

$$\Sym_n^{\geq 0}(A,\sigma):=\{M\in\Sym_n(A,\sigma)\mid \sigma(x)^TMx\in A^\sigma_{\geq 0}\text{ for all }x\in A^n\};$$

$$\Sym_n^+(A,\sigma):=\{M\in\Sym_n(A,\sigma)\mid \sigma(x)^TMx\in A^\sigma_+\text{ for all regular }x\in A^n\};$$

$$\UU_n(A,\sigma):=U_{(\Mat_n^\times(A),\sigma^T)}=\{M\in \Mat_n(A)\mid \sigma(M)^TM=\Id_n\}.$$

For a Hermitian semisimple algebra $(A,\sigma)$, the algebra $(\Mat_n(A),\sigma^T)$ is Hermitian and semisimple.

First, we have to check that for every two $M,N\in\Sym_n(A,\sigma)$, if $M^2+N^2=0$, then $M=N=0$.

Note that for every $x\in A^n$, if $\sigma(x)^Tx=0$, then $x=0$. Indeed, take $x=(x_1,\dots,x_n)^T$ such that $\sigma(x)^Tx=\sigma(x_1)x_1+\dots+\sigma(x_n)x_n=0$. Since all $\sigma(x_i)x_i\in A^\sigma_{\geq 0}$, all $\sigma(x_i)x_i=0$. Because of semisimplicity of $A$, all $x_i=0$.

Now, take $x\in A^n$ and consider $Mx,Nx\in A^n$. Assume $M^2+N^2=0$, then

$$0=\sigma(x)^T(M^2+N^2)x=\sigma(x)^TM^2x+\sigma(x)^TN^2x=\sigma(Mx)^T Mx+\sigma(Nx)^T Nx.$$

Since for every $y\in A^2$, $\sigma(y)^Ty\in A^{\sigma}_{\geq 0}$, $\sigma(Mx)^T Mx,\sigma(Nx)^T Nx=0$, By semisimplicity of $A$, $Mx=Nx=0$ for all $x\in A^n$. Therefore, $M=N=0$.

Finally, we have to check that $\Mat_n(A)$ is semisimple. We do it by induction. Assume $\Mat_k(A)$ is semisimple for all $k<n$.

Take $X\in J(\Mat_n(A))\cap \Sym_n(A,\sigma)$. Assume $X\neq 0$. Since $X\in J(\Mat_n(A))$, there exists $n>0$ such that $X^{2n}=0$ and $X^n\neq 0$. We write $X^n$ as a block matrix:

x & a \\
\sigma(a)^T & y
\end{pmatrix}$$

where $x\in\Sym_{n-1}(A,\sigma)$, $a\in A^{n-1}$, $y\in A^\sigma$. Then

x^2+a\sigma(a) & xa+ay \\
\sigma(a)^Tx+y\sigma(a)^T & \sigma(a)^Ta+y^2
\end{pmatrix}=0.$$

Therefore,  $\sigma(a)^Ta+y^2=\sigma(a)^Ta+\sigma(y)y=0$ and $\sigma(a)^Ta,\sigma(y)y\in A^\sigma_{\geq 0}$. That means $\sigma(a)a=\sigma(y)y=0$ and, because of semisimplicity of $A$, $a=0$, $y=0$. Further, $x^2+a\sigma(a)=x^2=0$. Since $\Mat_{n-1}(A)$ is Hermitian, $x=0$, i.e. $X^n=0$. That contradicts to the assumption $X\neq 0$. We obtain $J(\Mat_n(A))\cap \Sym_n(A,\sigma)=\{0\}$.

Take $X\in J:=J(\Mat_n(A))$ in the Jacobson radical of $\Mat_n(A)$ (see Definition <ref>), we write $X$ as a block matrix:

x & a \\
b & y
\end{pmatrix}$$

where $x\in\Mat_{n-1}(A)$, $a,b^T\in A^{n-1}$, $y\in A$. Since $X\in J$ by Proposition <ref>, $\sigma^T(X)=-X$ and $X^2=0$, i.e. $\sigma(x)^T=x$, $b=-\sigma(a)^T$, $y\in A^{-\sigma}$,

x^2-a\sigma(a)^T & xa+ay \\
-\sigma(a)^Tx-y\sigma(a)^T & -\sigma(a)^Ta+y^2
\end{pmatrix}=0.$$

Since $0=\sigma(a)^Ta-y^2=\sigma(a)^Ta+\sigma(y)y$ and $\sigma(a)^Ta,\sigma(y)y\in A^\sigma_{\geq 0}$, $\sigma(a)^Ta=\sigma(y)y=0$. Because of semisimplicity of $A$, $a=0$, $y=0$. Moreover, since $x^2-a\sigma(a)^T=x^2=0$ and $\Mat_{n-1}(A)$ is semisimple, $x=0$, i.e. $X=0$. Therefore, $J=\{0\}$ and $\Mat_n(A)$ is semisimple.

§.§ Complex extensions of algebras

Let $A$ be an algebra over a real closed field $\K$. We denote $A_\CC:=A\otimes_\K\K_\CC$ and call $A_\CC$ the complexification of $A$. We extend $\sigma$ “complex anti-linearly” to an  anti-involution $\bar\sigma_\CC$ on $A_\CC$:

We embed $A_\CC$ into $\Mat_2(A)$ in the following way:

\begin{equation}\label{Upsilon}
\begin{matrix}
\Upsilon \colon& A_\CC & \to & \Mat_2(A)\\
& x+yi & \mapsto & \begin{pmatrix}
x & y \\
-y & x
\end{pmatrix}.
\end{matrix}
\end{equation}

This map is an injective homomorphism of $\K$-algebras. Moreover, the anti-involution $\bar\sigma$ corresponds to $\sigma^T$ under this embedding. From Corollary <ref> we obtain:

Let $(A,\sigma)$ be a Hermitian algebra. The algebra $(A_\CC,\bar\sigma_\CC)$ is Hermitian if and only if $A$ is semisimple. If $A$ is semisimple then $A_\CC$ is semisimple.

If $(A,\sigma)$ is Hermitian and semisimple, then $(A_\CC,\sigma)$ is Hermitian and semisimple by <ref>.

If $A$ is not semisimple, i.e. the Jacobson radical $J(A)$ of $A$ is not trivial, then by <ref>, for every $0\neq x\in J(A)$, $x^2=0$ and $\sigma(x)=-x$. Therefore, $\bar\sigma_\CC(ix)=ix\neq 0$, i.e. $ix\in A^{\bar\sigma_\CC}_\CC$ and $(ix)^2=0$. This contradicts to the property to be Hermitian for $(A_\CC,\bar\sigma_\CC)$.

Let $(A,\sigma)$ be Hermitian and semisimple.

* The group $U_{(A_\CC,\bar\sigma_\CC)}=\{z\in A_\CC\mid\bar\sigma_\CC(z)z=1\}$ is semi-algebraically connected (as a semi-algebraic deformation retract of a semi-algebraically connected space $A^\times_\CC$).

* If $\K=\R$, then the group $U_{(A_\CC,\bar\sigma_\CC)}$ is a maximal compact subgroup of $A_\CC^\times$.

There is another anti-involution on $A_\CC$, namely the complex linear extension $\sigma_\CC$ of $\sigma$. Together with this anti-involution $A_\CC$ is never Hermitian because
$$0=1-1=\sigma_\CC(1)\cdot 1+\sigma_\CC(i)\cdot i.$$

Similarly to spectral theorems for $\sigma$-symmetric elements of a semisimple, Hermitian algebra $(A,\sigma)$, the spectral theorem for $\bar\sigma_\CC$-normal elements of $(A_\CC,\bar\sigma_\CC)$ can be proven.

[Spectral theorem for normal elements, first version]
Let $(A_\CC,\bar\sigma_\CC)$ be Hermitian. For every $\bar\sigma_\CC$-normal element $b\in A_\CC$, there exist a unique $k\in\N$, unique elements $\lambda_1,\dots,\lambda_k\in\K_\CC$, all distinct, and a unique complete system of orthogonal idempotents $c_1,\dots,c_k\in \K_\CC[b]\cap A_\CC^{\bar\sigma_\CC}$ such that

We call this the spectral decomposition of $b$.

[Spectral theorem for normal elements, second version]
Let $(A_\CC,\bar\sigma_\CC)$ be Hermitian. Suppose that the Jordan algebra $A_\CC^{\bar\sigma_\CC}$ has rank $n$. For every $\bar\sigma_\CC$-normal element $b\in A_\CC$ there exist a Jordan frame $(e_1,\dots,e_n)$ of $A_\CC^{\bar\sigma_\CC}$ and an $n$-tuple of elements  $(\lambda_1(b),\dots,\lambda_n(b))\in\K^n_\CC$ such that

The tuple $(\lambda_1(b),\dots,\lambda_n(b))$ is uniquely defined by $b$ up to permutations. The elements $\lambda_1(b),\dots,\lambda_n(b)\in\K$ (with their multiplicities) are called the eigenvalues of $b$. In particular, they do not depend on the Jordan frame $e_1,\dots,e_n\in A_\CC^{\bar\sigma_\CC}$.

A Jordan frame $e_1,\dots,e_n\in A_\CC^{\bar\sigma_\CC}$ associated to a $\bar\sigma_\CC$-normal element $b\in A_\CC$ by Theorem <ref> is in general not unique and elements $e_1,\dots,e_n\notin \K_\CC[b]$, in contrast to the complete system of orthogonal idempotents from  Theorem <ref>.

Let $(A,\sigma)$ be semisimple and Hermitian. Let $a_1,a_2\in A^\sigma$ that commute. Then there exist a complete orthogonal system of idempotents $c_1,\dots c_k\in A^\sigma$ and elements $\lambda_1,\dots,\lambda_k,\mu_1,\dots,\mu_k\in \K$ such that

Moreover, there exist an $a\in A^\sigma$ such that $c_1,\dots,c_k\in\K[a]$. In particular, $a_1,a_2\in \K[a]$.

If $a\in A_\CC$ is $\bar\sigma_\CC$-normal, then $\sigma(a)\in\K[a]$.

Let $(A_\CC,\bar\sigma_\CC)$ be Hermitian. If $a\in A^{-\bar\sigma_\CC}$, then all the eigenvalues of $a$ are purely imagine. If $a\in U_{(A_\CC,\bar\sigma_\CC)}$, then all the eigenvalues are in $S^1_\K=\{z\in\K_\CC\mid |z|=1\}$.

For every $u\in U_{(A,\sigma)}$, there exist unique $r\in\N\cup\{0\}$, $s\in\{0,1,2\}$ with $r+s>0$, unique systems of idempotents $c_1,\dots,c_r\in A^{\bar\sigma_\CC}_\CC$ and $c_1',\dots,c_s'\in A^{\sigma}$ such that $c_1,\dots,c_r,\bar c_1,\dots,\bar c_r,c_1',\dots,c_s'$ is a complete orthogonal system of idempotents of $A^{\bar\sigma_\CC}_\CC$ and unique elements $\zeta_1,\dots,\zeta_r\in S^1_\K$ with $\Imm(\zeta_i)>0$ for all $i\in\{1,\dots,r\}$ all distinct and $\varepsilon_1,\dots\varepsilon_s\in\{1,-1\}$ all distinct such that

$$u=\sum_{j=1}^r (\zeta_j c_j+\bar\zeta_j\bar c_j)+\sum_{j=1}^s\varepsilon_jc_j'.$$

For an element $u\in U(A_\CC,\bar\sigma_\CC)$, $u\in U(A,\sigma)$ if and only if $u=\bar u$. We take the spectral decompositions of $u$ and $\bar u$:

$$u=\sum_{j=1}^k \zeta_j c_j,\;\;\bar u=\sum_{j=1}^k \bar\zeta_j\bar c_j$$

where all $\zeta_j\in S_1$, $(c_j)$ is a complete orthogonal system of idempotents of $A^{\bar\sigma_\CC}_\CC$.
Notice, all $\bar c_1,\dots,\bar c_k$ is a complete orthogonal system of idempotents of $A^{\bar\sigma_\CC}_\CC$ because $\bar c_i\bar c_j=\overline{c_ic_j}$.
If $u=\bar u$, then, because of uniqueness of the spectral decomposition, for every $j\in\{1,\dots, k\}$ there exists $j'\in\{1,\dots, k\}$ such that $\zeta_j c_j=\bar\zeta_j \bar c_{j'}$. There can be to cases:

* $j=j'$ then $\zeta_j\in\R$ i.e. $\zeta_j\in\{1,-1\}$.

* $j\neq j'$ then $c_{j'}=\bar c_j$ and $\zeta_j\notin \R$, i.e. $\Imm(\zeta_j)\neq 0$.

Because all $\zeta_j$ are distinct, there can be at most one $j$ such that $\zeta_j=1$ and at most one $j$ with $\zeta_j=-1$. For such $j$, $c_j\in A^\sigma$. So we obtain

$$u=\sum_{j=1}^r (\zeta_j c_j+\bar\zeta_j\bar c_j)+\sum_{j=1}^s\varepsilon_jc_j'.$$

for appropriate $r,s\in\N\cup\{0\}$, $s\leq 2$. Finally, we can rename the idempotents so that $\Imm(\zeta_j)>0$ for all $j\in\{1,\dots r\}$.

The determinant map on $U_{(A_\CC,\bar\sigma_\CC)}$ is given by:

$$\det(u)=\prod_{j=1}^n \zeta_j\in S^1_\K\subset \K_\CC$$

where $u=\sum_{j=1}^n \zeta_je_j$ for some Jordan frame $e_1,\dots,e_n\in A^{\bar\sigma_
\CC}_\CC$.

Let $\K=\R$. The fundamental group of $U_{(A_\CC,\bar\sigma_\CC)}$ contains a copy of $(\Z,+)$.

The determinant map

$$\det\colon U_{(A_\CC,\bar\sigma_\CC)}\to S^1$$
is continuous and surjective. It induces the homomorphism of fundamental groups:

$$(\det)_*\colon\pi_1(U_{(A_\CC,\bar\sigma_\CC)}, 1)\to \pi_1(S^1,1).$$

This homomorphism is surjective because the curve $u(t)=e^{it}e_1+\sum_{j=2}^ne_j$ for $t\in[0,2\pi]$ for some Jordan frame $e_1,\dots,e_n$ maps to the curve $\det_*u(t)=e^{it}$ which generates $\pi_1(S^1,1)$. Therefore $u(t)$ generates in $\pi_1(U_{(A_\CC,\bar\sigma_\CC)}, 1)$ a subgroup that is isomorphic to $(\Z,+)$.

Let $(A,\sigma)$ be a semisimple Hermitian algebra and $(A_\CC,\sigma_\CC)$ be its complexification. Assume that the Jordan algebra $A_\CC^{\bar\sigma_\CC}$ is simple.
The following action of $A_\CC^\times$ on $(A_\CC^{\sigma_\CC})^\times$

\psi_\CC\colon & A_\CC^\times\times (A_\CC^{\sigma_\CC})^\times & \to & (A_\CC^{\sigma_\CC})^\times \\
& (a,b) & \mapsto & \sigma(a)ba
\end{array}$$

is transitive.

We will show that for every $b\in (A_\CC^{\sigma_\CC})^\times$ there exists an $a\in A_\CC^\times$ such that $\psi_\CC(a,1)=b$. We take $b\in (A_\CC^{\sigma_\CC})^\times$ and consider its polar decomposition of $b=ub'$ where $u\in U_{(A_\CC,\bar\sigma_\CC)}$, $b'\in (A_\CC^{\bar\sigma_\CC})^\times$. Using the spectral theorem we obtain:
where $(c_i)_{i=1}^k$ is a Jordan frame of $A_\CC^{\bar\sigma_\CC}$ and $\lambda_i>0$ for all $i\in\{1,\dots, k\}$.

We fix a Jordan frame $(x_i)_{i=1}^k$ of $A^\sigma$. This is also a Jordan frame of $A_\CC^{\bar\sigma_\CC}$. By <ref>, the group $U_{(A_\CC,\bar\sigma_\CC)}$ acts transitively on Jordan frames of $A_\CC^{\bar\sigma_\CC}$. Therefore, there exists $u'\in U_{(A_\CC,\bar\sigma_\CC)}$ such that $x_i=\bar\sigma(u')c_i u'$ for all $i\in\{1,\dots,k\}$. Further

where $u''=\sigma_\CC(u')u\bar\sigma_\CC(u')^{-1}\in U_{(A_\CC,\bar\sigma_\CC)}$, $b''=\sum_{i=1}^k\lambda_ix_i\in A^\sigma$. Since $u''b''=\sigma(u')bu'\in A_\CC^{\sigma_\CC}$,
$$u''b''=\sigma_\CC(u''b'')=b''\sigma_\CC(u'')=b''(\bar u'')^{-1}.$$
Therefore, $b''=u''b''\bar u''$. By induction, we obtain $b''=(u'')^n b'' (\bar u'')^n$ for all $n\in\Z$, or equivalently $(u'')^nb''=b'' (\bar u'')^{-n}$.

Now assume $\K=\R$. Since $(u'')^nb''=b'' (\bar u'')^{-n}$ holds for every $n\in\Z$, the following equality $f(u'')b''=b''f((\bar u'')^{-1})$ holds for every function $f$ that can be expressed as a convergent Taylor series at $u''$. Since $u''\in U_{(A_\CC,\bar\sigma_\CC)}$, $u''\neq 0$. By <ref>, there exists $w\in U_{(A_\CC,\bar\sigma_\CC)}$ such that $(u'')=w^2$. So we can take as $f$ the branch of the square root such that $f(u'')=w$. Then we obtain $wb''=b''(\bar w)^{-1}$. Since any branch of the square root is a semi-algebraic function, using the Tarski–Seidenberg transfer principle (see <cit.>), this holds for every real closed field $\K$.

This implies:

$$u''b''=wb''\bar w^{-1}=(wb^{\frac{1}{2}})b^{\frac{1}{2}}\bar w^{-1}=\sigma_\CC(b^{\frac{1}{2}}\bar w^{-1})b^{\frac{1}{2}}\bar w^{-1}=\psi_\CC(b^{\frac{1}{2}}\bar w^{-1},1).$$

$b=\sigma_\CC((u')^{-1})u''b''(u')^{-1}=\psi_\CC(b^{\frac{1}{2}}\bar w^{-1}(u')^{-1},1)$.

§.§ Quaternionic extensions of algebras

If $A$ is an algebra over a real closed field $\K$, then we call $A_\HH=A\otimes_\K\K_\HH$ the quaternionification of $A$.

Sometimes, to make a construction more precise, we will
write $\K_\HH\{i,j,k\}$ to emphasize the imaginary units. The multiplication rule is then $ij=-ji=k$. Sometimes, we will also write $\K_\CC\{\kappa\}$ to emphasize the imaginary unit $\kappa$ of the complexified field $\K_\CC$.

If $A_\CC=A\otimes_\K\K_\CC$ is the complexification of some real algebra $A$, then it can be
embedded into $A\otimes_\K\K_\HH$ in many different ways. If we write $A_\CC:=A\otimes_\K\K_\CC\{i\}$,
$A_\HH:=A\otimes_\K\K_\HH\{i,J,K\}$, it means that $A_\CC$ is embedded into $A_\HH$ by the map induced by
the identification $A_\CC\ni i\mapsto i\in A_\HH$.

Let $A$ be a $\K_\CC$-algebra, $A_0\subset A$ be $\K$-subalgebra of $A$ with the property that there is a central element $I\in Z(A)$ such that
$I^2=-1$ and $A=A_0\oplus A_0I$. Then we say that $A_0$ is a real locus of $A$ with respect to the
imaginary unit $I$. In this case, $A$ is isomorphic to $A_0\otimes_\K\K_\CC\{I\}$ as $\K_\CC\{I\}$-algebra. We
take the following $\K_\HH$-algebra:

The algebra $A$ sits inside $\HH[A,A_0,I,J]$ as described above.

We call $\HH[A,A_0,I,J]$ the quaternionification of $A$ with respect to the real locus
$A_0$ and the imaginary unit $I$.

Assume $\Phi\colon B=B_\R\otimes_R\CC[i]\to B'$ to be an isomorphism of $\CC$-algebras and
$\Phi(1\otimes i)=I\in B'$. Then $B'=\Phi(B_\R)\oplus I\Phi(B_\R)$, i.e. we can see $B'$ as
$\Phi(B_\R)\otimes_\R\CC[I]$. We take the free extension $B'\langle J\rangle$ of $B'$ by an element
$J\notin B'$ and then divide the ideal

$$\mathcal I:=(J^2+1, IJ+JI, Ja-aJ\mid a\in\Phi(B_\R)).$$

The quotient algebra $B'\langle J\rangle/\mathcal I$, denoted by $\HH[B',I,J]$, is isomorphic as
$\HH$-algebra to $\Phi(B_\R)\otimes_\R\HH[i,j,k]$ by the map induced by the identification
$I\mapsto i$, $J\mapsto j$. In this case, we will talk about quaternionification of $B'$ with
respect to the imaginary unit $I$.

The algebra $A_\HH$ can be embedded into $\Mat_2(A_\CC)$ in the following way:

\begin{equation}\label{Upsilon_H}
\begin{matrix}
\Upsilon_\HH \colon& A_\HH & \to & \Mat_2(A_\CC)\\
& x+yj & \mapsto & \begin{pmatrix}
x & y \\
-\bar y & \bar x
\end{pmatrix}.
\end{matrix}
\end{equation}

This map is an injective homomorphism of $\CC\{i\}$-algebras. Moreover, the anti-involution $\sigma_1$ on $A_\HH$ defined as follows:

for $x,y\in A_\CC$, corresponds under this embedding to $\bar\sigma^T$. By Corollary <ref>, we obtain:

Let $(A,\sigma)$ be Hermitian and semisimple.

* The algebra $(A_\HH,\sigma_1)$ is Hermitian and semisimple.

* If $\K=\R$, then the group $U_{(A_\HH,\sigma_1)}=\{z\in A_\CC\mid \sigma_1(z)z=1\}$ is a maximal compact subgroup of $A_\HH^\times$.

There is another anti-involution $\sigma_0$ on $A_\HH$ defined as follows:

for $x,y\in A_\CC$. The algebra $(A_\HH,\sigma_0)$ is never Hermitian.

§.§ Some properties of real Hermitian algebras

In this section, we assume that $(A,\sigma)$ is a semisimple real Hermitian algebra. We will state some properties that depend on the fact that $\R$ is locally compact, contrary to the other real closed fields.

The group $U_{(A,\sigma)}$ is a maximal compact subgroup of $A^\times$.

The group  $U_{(A,\sigma)}\subset\Isom(\beta)$ is compact by <ref>. By <ref>, $U_{(A,\sigma)}$ is a deformation retract of $A^\times$. Hence, it is a maximal compact subgroup of $A^\times$.

$\UU_n(A,\sigma)$ is a maximal compact subgroup of $\Mat_n^\times(A)$

The following map:

\bpol\colon & U_{(A,\sigma)}\times A^\sigma_{\geq 0} & \to & A \\
& (u,b) & \mapsto & ub.
\end{matrix}$$

is proper, surjective.

Since the group $U_{(A,\sigma)}$ is compact, we can take a closure in the polar decomposition and the map stays surjective. Let $K\subset A$ be a compact subset, then

$$\bpol^{-1}(K)\subseteq\{(\sigma(a)a)^{\frac{1}{2}}\mid a\in K\}\times U_{(A,\sigma)}$$

is compact.

If $A$ is Hermitian, the map $\theta\colon A\to A^\sigma_{\geq 0}$ is proper.

Let $K\subset A^\sigma_{\geq 0}$ be a compact subset. Then

$$\theta^{-1}(K)=\{ub_+^{\frac{1}{2}}\mid u\in U_{(A,\sigma)}, b_+\in K\}=
\bpol(U_{(A,\sigma)}\times K).$$

Since $U_{(A,\sigma)}\times K$ is compact in $U_{(A,\sigma)}\times A^\sigma_{\geq 0}$ and $\bpol$ is continuous, $\theta^{-1}(K)$ is compact.

Let $(A,\sigma)$ be Hermitian. The domain

$$D(A,\sigma):=\{a\in A\mid 1-\sigma(a)a\in A^\sigma_{\geq 0}\}$$

is compact.

In particular, for every vector subspace $V$ of $A$, the domain

$$D(V,\sigma):=\{a\in V \mid 1-\sigma(a)a\in A^\sigma_{\geq 0}\}=D\cap V$$

is compact.

This follows directly from <ref>

[Spectral theorem for $U_{(A_\CC,\bar\sigma_\CC)}$, first version]
For every $u\in U_{(A_\CC,\bar\sigma_\CC)}$, there exists a unique $r\in\N$, a unique complete system of orthogonal idempotents $c_1,\dots,c_r\in A^{\bar\sigma_\CC}_\CC$ and unique elements $\zeta_1,\dots,\zeta_r\in S^1$ all distinct such that

$$u=\sum_{j=1}^r \zeta_jc_j.$$

We call this the spectral decomposition of $u$.

Consider the Lie algebra $A^{-\bar\sigma_\CC}_\CC$ of $U_{(A_\CC,\bar\sigma_\CC)}$. The following map:

\psi\colon& A^{\bar\sigma_\CC}_\CC & \mapsto & A^{-\bar\sigma_\CC}_\CC\\
& a & \mapsto & ia
\end{matrix}$$

is injective. Moreover, because
$\dim_\R(A^{\bar\sigma_\CC}_\CC)=\dim_\R A=\dim_\R(A^{-\bar\sigma_\CC}_\CC),$
this map is an isomorphism of $\R$-vector spaces. By the first version of the spectral theorem, for every $a\in A^{\bar\sigma}_\CC$, there exists a unique complete system of orthogonal idempotents $c_1,\dots,c_r\in A^{\bar\sigma_\CC}_\CC$ of $A_\CC^{\bar\sigma_\CC}$ and a unique distinct elements $\theta_1,\dots,\theta_r\in\R$ such that for all $i\neq j$, $\theta_i\neq \theta_j$ and

$$a=\sum_{j=1}^r \theta_j c_j.$$

$$\exp(ia)= \sum_{j=1}^r e^{i\theta_j} c_j\in U_{(A_\CC,\bar\sigma_\CC)}.$$

Notice that $U_{(A_\CC,\bar\sigma_\CC)}$ is connected and compact by <ref>. The exponential map is surjective for compact connected Lie groups, so every element of $U_{(A_\CC,\bar\sigma_\CC)}$ admit such decomposition. We denote $\zeta_j:=e^{i\theta_j}\in S^1$.

We can always assume that all $\zeta_j$ are pairwise distinct. Indeed, if there are $j,k\in\{1,\dots,r\}$ such that $j\neq k$ but $\zeta_j=\zeta_k$, then $\zeta_jc_j+\zeta_kc_k=\zeta_j(c_j+c_k)$ and $c_j+c_k$ is an idempotent which is orthogonal to all other idempotents $c_l$, $l\neq j,k$. So we obtain a decomposition of $u$ where the number of idempotents is less then before. We can go on until all $\zeta_j$ become distinct.

Assume, there are two such decompositions for an element $u\in U_{(A_\CC,\bar\sigma_\CC)}$:

$$u=\sum_{j=1}^r \zeta_j c_j=\sum_{j=1}^{r'} \zeta'_j c'_j$$

for some $r,r'\in\N$, $(c_j)$, $(c'_j)$ complete systems of orthogonal idempotents and $(\zeta_j)$, $(\zeta'_j)$ in $S^1$ such that all $\zeta_j$ are pairwise distinct and all $\zeta'_j$ are pairwise distinct.

We multiply $u$ by $c_k$ from the left and by $c'_l$ from the right:

$$c_kuc'_l=\zeta_k c_kc'_l=\zeta'_lc_kc'_l.$$

So either $c_kc'_l=0$ or $\zeta_k=\zeta'_l$, i.e.

$$\{\zeta_k\mid k\in\{1,\dots,r\}\}=\{\zeta'_l\mid l\in\{1,\dots,r'\}\},$$

in particular $r=r'$. We can reorder indices so that $\zeta'_k=\zeta_k$ for all $k\in\{1,\dots r\}$. Therefore:

We get $c_k=c_kc'_k=c'_k$ for all $k\in\{1,\dots,r\}$. So we obtain that the spectral decomposition of $u$ is unique.

[Spectral theorem for $U_{(A,\sigma)}$]
For every $u\in U_{(A,\sigma)}$, there exist unique $r\in\N\cup\{0\}$, $s\in\{0,1,2\}$ with $r+s>0$, unique systems of idempotents $c_1,\dots,c_r\in A^{\bar\sigma_\CC}_\CC$ and $c_1',\dots,c_s'\in A^{\sigma}$ such that $c_1,\dots,c_r,\bar c_1,\dots,\bar c_r,c_1',\dots,c_s'$ is a complete orthogonal system of idempotents of $A^{\bar\sigma_\CC}_\CC$ and unique elements $\zeta_1,\dots,\zeta_r\in S^1$ with $\Imm(\zeta_i)>0$ for all $i\in\{1,\dots,r\}$ all distinct and $\varepsilon_1,\dots\varepsilon_s\in\{1,-1\}$ all distinct such that

$$u=\sum_{j=1}^r (\zeta_j c_j+\bar\zeta_j\bar c_j)+\sum_{j=1}^s\varepsilon_jc_j'.$$

For an element $u\in U(A_\CC,\bar\sigma_\CC)$, $u\in U(A,\sigma)$ if and only if $u=\bar u$. We take the spectral decompositions of $u$ and $\bar u$:

$$u=\sum_{j=1}^k \zeta_j c_j,\;\;\bar u=\sum_{j=1}^k \bar\zeta_j\bar c_j$$

where all $\zeta_j\in S_1$, $(c_j)$ is a complete orthogonal system of idempotents of $A^{\bar\sigma_\CC}_\CC$.
Notice, all $\bar c_1,\dots,\bar c_k$ is a complete orthogonal system of idempotents of $A^{\bar\sigma_\CC}_\CC$ because $\bar c_i\bar c_j=\overline{c_ic_j}$.
If $u=\bar u$, then, because of uniqueness of the spectral decomposition, for every $j\in\{1,\dots, k\}$ there exists $j'\in\{1,\dots, k\}$ such that $\zeta_j c_j=\bar\zeta_j \bar c_{j'}$. There can be to cases:

* $j=j'$ then $\zeta_j\in\R$ i.e. $\zeta_j\in\{1,-1\}$.

* $j\neq j'$ then $c_{j'}=\bar c_j$ and $\zeta_j\notin \R$, i.e. $\Imm(\zeta_j)\neq 0$.

Because all $\zeta_j$ are distinct, there can be at most one $j$ such that $\zeta_j=1$ and at most one $j$ with $\zeta_j=-1$. For such $j$, $c_j\in A^\sigma$. So we obtain

$$u=\sum_{j=1}^r (\zeta_j c_j+\bar\zeta_j\bar c_j)+\sum_{j=1}^s\varepsilon_jc_j'.$$

for appropriate $r,s\in\N\cup\{0\}$, $s\leq 2$. Finally, we can rename the idempotents so that $\Imm(\zeta_j)>0$ for all $j\in\{1,\dots r\}$.

Suppose, $A^{\bar\sigma_\CC}_\CC$ has rank $n$. For every $u\in U_{(A_\CC,\bar\sigma_\CC)}$, there exist a Jordan frame $e_1,\dots,e_n\in A^{\bar\sigma}_\CC$ and an $n$-tuple of elements $(\zeta_1,\dots,\zeta_n)\in (S^1)^n$ and  such that

The elements $\zeta_1,\dots,\zeta_n$ (with their multiplicities) are called eigenvalues of $u$. They are uniquely defined by $u$ up to permutations. In particular, they do not depend on the Jordan frame $e_1,\dots,e_n\in A^{\bar\sigma_\CC}_\CC$.

The determinant map on $U_{(A_\CC,\bar\sigma_\CC)}$ is given by:

$$\det(u)=\prod_{j=1}^n \zeta_j\in S^1\subset \CC$$

where $u=\sum_{j=1}^n \zeta_je_j$ for some Jordan frame $e_1,\dots,e_n\in A^{\bar\sigma_
\CC}_\CC$.

The fundamental group of $U_{(A_\CC,\bar\sigma_\CC)}$ contains a copy of $(\Z,+)$.

The determinant map

$$\det\colon U_{(A_\CC,\bar\sigma_\CC)}\to S^1$$
is continuous and surjective. It induces the homomorphism of fundamental groups:

$$(\det)_*\colon\pi_1(U_{(A_\CC,\bar\sigma_\CC)}, 1)\to \pi_1(S^1,1).$$

This homomorphism is surjective because the curve $u(t)=e^{it}e_1+\sum_{j=2}^ne_j$ for $t\in[0,2\pi]$ for some Jordan frame $e_1,\dots,e_n$ maps to the curve $\det_*u(t)=e^{it}$ which generates $\pi_1(S^1,1)$. Therefore $u(t)$ generates in $\pi_1(U_{(A_\CC,\bar\sigma_\CC)}, 1)$ a subgroup that is isomorphic to $(\Z,+)$.

Let $(A,\sigma)$ be a semisimple Hermitian algebra over $\R$ and $(A_\CC,\sigma_\CC)$ be the complexification. Assume that the Jordan algebra $A_\CC^{\bar\sigma_\CC}$ is simple.
The following action of $A_\CC^\times$ on $(A_\CC^{\sigma_\CC})^\times$

\psi_\CC\colon & A_\CC^\times\times (A_\CC^{\sigma_\CC})^\times & \to & (A_\CC^{\sigma_\CC})^\times \\
& (a,b) & \mapsto & \sigma(a)ba
\end{array}$$

is transitive.

We will show that for every $b\in (A_\CC^{\sigma_\CC})^\times$ there exists an $a\in A_\CC^\times$ such that $\psi_\CC(a,1)=b$. We take $b\in (A_\CC^{\sigma_\CC})^\times$ and consider its polar decomposition of $b=ub'$ where $u\in U_{(A_\CC,\bar\sigma_\CC)}$, $b'\in (A_\CC^{\bar\sigma_\CC})^\times$. Using the spectral theorem we obtain:
where $(c_i)_{i=1}^k$ is a Jordan frame of $A_\CC^{\bar\sigma_\CC}$ and $\lambda_i>0$ for all $i\in\{1,\dots, k\}$.

We fix a Jordan frame $(x_i)_{i=1}^k$ of $A^\sigma$. This is also a Jordan frame of $A_\CC^{\bar\sigma_\CC}$. By <ref>, the group $U_{(A_\CC,\bar\sigma_\CC)}$ acts transitively on Jordan frames of $A_\CC^{\bar\sigma_\CC}$. Therefore, there exists $u'\in U_{(A_\CC,\bar\sigma_\CC)}$ such that $x_i=\bar\sigma(u')c_i u'$ for all $i\in\{1,\dots,k\}$. Further

where $u''=\sigma_\CC(u')u\bar\sigma_\CC(u')^{-1}\in U_{(A_\CC,\bar\sigma_\CC)}$, $b''=\sum_{i=1}^k\lambda_ix_i\in A^\sigma$. Since $u''b''=\sigma(u')bu'\in A_\CC^{\sigma_\CC}$,
$$u''b''=\sigma_\CC(u''b'')=b''\sigma_\CC(u'')=b''(\bar u'')^{-1}.$$
Therefore, $b''=u''b''\bar u''$. By induction, we obtain $b''=(u'')^n b'' (\bar u'')^n$ for all $n\in\Z$, or equivalently $(u'')^nb''=b'' (\bar u'')^{-n}$. Since this holds for every $n\in\Z$, the following equality $f(u'')b''=b''f((\bar u'')^{-1})$ holds for every function $f$ that can be expressed as a convergent Taylor series at $u''$. Since $u''\in U_{(A_\CC,\bar\sigma_\CC)}$, $u''\neq 0$. By <ref>, there exists $w\in U_{(A_\CC,\bar\sigma_\CC)}$ such that $(u'')=w^2$. So we can take as $f$ the branch of square root such that $f(u'')=w$. Then we obtain $wb''=b''(\bar w)^{-1}$. This implies:

$$u''b''=wb''\bar w^{-1}=(wb^{\frac{1}{2}})b^{\frac{1}{2}}\bar w^{-1}=\sigma_\CC(b^{\frac{1}{2}}\bar w^{-1})b^{\frac{1}{2}}\bar w^{-1}=\psi_\CC(b^{\frac{1}{2}}\bar w^{-1},1).$$

$b=\sigma_\CC((u')^{-1})u''b''(u')^{-1}=\psi_\CC(b^{\frac{1}{2}}\bar w^{-1}(u')^{-1},1)$.

First, we need the following Lemma:

Let $C$ be a closed proper convex cone in some finite-dimensional $\R$-vector space $V$. Then for every $c\in V$, the set $K:=C\cap(c-C)$ is compact.

Assume $K$ is not compact. We fix some norm $\|\cdot\|$ on $V$. There exists a sequence $(x_n)$ such that $\|x_n\|\to\infty$. Since $y_n:=\frac{x_n}{\|x_n\|}\in S^1$ and for finite-dimensional $V$, $S_1$ is compact, there exists a limit point $y$ of $(y_n)$. Since $C$ is a closed cone, $y\R_+\subseteq C\cap(c-C)$ and, therefore $c-y\R_+\subseteq C\cap(c-C)$. Analogously, $c+y\R_+\subseteq C\cap(c-C)$ and, therefore $-y\R_+\subseteq C\cap(c-C)$. That means, $y\R\in C\cap(-C)$, so $y=0$. This contradicts that $y\in S^1$. Therefore, $K$ is compact.

By the Lemma <ref>, the set

$$K:=\{x\in A^\sigma_{\geq 0}\colon 1-x\in A^\sigma_{\geq 0}\}=A^\sigma_{\geq 0}\cap(1-A^\sigma_{\geq 0})$$

is compact. Since $\theta^{-1}(K)=D$ and $\theta$ is proper, $D$ is compact.

The domain $D(V,\sigma)$ is compact because $D$ is compact and $V$ is closed in $A$.

§ SYMPLECTIC GROUPS OVER NONCOMMUTATIVE ALGEBRAS

In this section we introduce symplectic groups over unital rings with anti-involution.

§.§ Sesquilinear forms on A-modules and their groups of symmetries

Let $A$ be a unital associative ring with an anti-involution $\sigma$.

A $\sigma$-sesquilinear form $\omega$ on a right $A$-module $V$ is a map

$$\omega\colon V\times V\to A$$

such that

We denote by
$$\Aut(\omega):=\{f\in\Aut(V)\mid \forall x,y\in V:\omega(f(x),f(y))=\omega(x,y)\}$$
the group of symmetries of $\omega$. We also define the corresponding Lie algebra:

$$\End(\omega):=\{f\in\End(V)\mid \forall x,y\in V:\omega(f(x),y)+\omega(x,f(y))=0\}$$

with the usual Lie bracket $[f,g]=fg-gf$.

We now set  $V=A^2$. We view $V$ as the set of columns and endow it with the structure of a right $A$-module.

$ $
We make the following definitions:

* A pair $(x,y)$ for $x,y\in A^2$ is called basis of $A^2$ if for every $z\in A^2$ there exist $a,b\in A$ such that $z=xa+yb$.

* The element $x\in A^2$ is called regular if there exists $y\in A^2$ such that $(x,y)$ is a basis of $A^2$.

* $l\subseteq A^2$ is called a line if $l=xA$ for a regular $x\in A^2$. We denote the space of lines of $A^2$ by $\PP(A^2)$.

* Two regular elements $x,y\in A^2$ are called linearly independent
if $(x,y)$ is a basis of $A^2$.

* Two lines $l,m$ are called transverse if $l=xA$, $m=yA$ for linearly independent $x,y\in A^2$.

* An element $x\in A^2$ is called isotropic with respect to $\omega$ if $\omega(x,x)=0$. The set of all isotropic regular elements of $(A^2,\omega)$ is denoted by $\Is(\omega)$.

* A line $l$ is called isotropic if $l=xA$ for an regular isotropic $x\in A^2$. The set of all isotropic lines of $(A^2,\omega)$ is denoted by $\PP(\Is(\omega))$.

$ $
We consider a form $\omega$ and say

* The form
$\omega$ is non-degenerate if for every regular $x\in A^2$ there exists a $y\in A^2$ such that $\omega(x,y)\in A^\times$.

* The form is $\sigma$-symmetric if $\omega(x_2,x_1)=\sigma(\omega(x_1,x_2))$ for all $x_1,x_2\in A^2$.

* The form is $\sigma$-skew-symmetric  if $\omega(x_2,x_1)=-\sigma(\omega(x_1,x_2))$ for all $x_1,x_2\in A^2$.

* When $A$ is  Hermitian, a $\sigma$-symmetric form is called $\sigma$-inner product if $\omega(x,x)\in A^\sigma_+$ for all regular $x\in A^2$.

We can now introduce the symplectic group $\Sp_2(A,\sigma)$ over $(A,\sigma)$.

Let $(A,\sigma)$ be a unital ring with anti-involution.
Let $\omega(x,y):=\sigma(x)^T\Omega y$ with $\Omega=\Ome{1}$. The form $\omega$ is called the standard symplectic form on $A^2$.
The group $\Sp_2(A,\sigma):=\Aut(\omega)$ is the symplectic group $\Sp_2$ over $(A,\sigma)$. Its Lie algebra is $\spp_2(A,\sigma):=\End(\omega)$.

We have

a & b \\
c & d
\end{pmatrix}\midwd \sigma(a)c,\,\sigma(b)d\in A^\sigma,\,\sigma(a)d-\sigma(c)b=1\right\}\subseteq \Mat_2^\times(A)$$

x & z \\
y & -\sigma(x)
\end{pmatrix}\midwd x\in A,\;y,z\in A^\sigma\right\}\subseteq \Mat_2(A).$$

From now on, we assume $\omega(x,y):=\sigma(x)^T\Omega y$ on $A^2$.

A basis $(x,y)$ of $A^2$ is called symplectic if $x,y$ are isotropic and $\omega(x,y)=1$.

For every basis $(x,y)$ of $A^2$ and for every $z\in A^2$ there exist unique $a,b\in A$ such that $z=xa+yb$. Moreover, for every regular $x\in A^2$, the map

A & \to & xA\\
a & \mapsto & xa
\end{matrix}$$

is an isomorphism of right $A$-modules.

Take a basis $(x,y)$ of $A^2$.
Consider the following $A$-homomorphism of right $A$-modules:

A^2 & \to & A^2\\
(a,b) & \mapsto & z= xa+yb.
\end{matrix}$$

This is also a surjective $\R$-homomorphism of vector spaces of the same dimension. Therefore, it is injective, i.e. $(a,b)$ is uniquely defines by $z$. The restriction of this homomorphism to $A\times\{0\}$ is an isomorphism $A\to xA$ of right $A$-modules.

The form $\omega$ is non-degenerate.

Let $x=(x_1,x_2)^T\in A^2$ regular. We want to find $y\in A^2$ such that $\omega(y,x)=1$. Since $x$ is regular, there exists $x'=(x_1',x_2')^T\in A^2$ such that $(x,x')$ is a basis. That means that the matrix
x_1 & x_1'\\
x_2 & x_2'
\end{pmatrix}$ is invertible, i.e. there exists the inverse matrix
a_1 & a_2\\
a_1' & a_2'
\end{pmatrix}$. Therefore, $a_1x_1+a_2x_2=1$. We take $y:=(\sigma(a_2),-\sigma(a_1))^T$, then

So $\omega$ is non-degenerate.

An element $x=(x_1,x_2)^T\in A^2$ is isotropic if and only if $\sigma(x_1)x_2\in A^\sigma$.

The proof follows by direct computation.

If $x,y\in A^2$ are isotropic and $\omega(x,y)=1$, then $(x,y)$ is a basis.

Let $x,y\in A^2$ are isotropic and $\omega(x,y)=1$. Consider the map

A^2 & \to & A^2\\
(a,b) & \mapsto & xa+yb.
\end{matrix}$$

To see that this map is an isomorphism, it is enough to check that it is injective. Assume $xa+yb=0$ for some $a,b\in A$, then

So $a=b=0$.

a & b \\
c & d
\end{pmatrix}\midwd \left(\Clm{a}{c},\Clm{b}{d}\right)\text{ is a symplectic basis}
\right\} $$

Let $x\in A^2$ regular isotropic, $y\in A^2$ and $\omega(x,y)\in A^\times$. Then $(x,y)$ is a basis of $A^2$. In particular, $y$ is regular.

To see that $(x,y)$ is a basis, it is enough to check that the map

A^2 & \to & A^2\\
(a,b) & \mapsto & xa+yb
\end{matrix}$$

is injective. Assume $xa+yb=0$ for some $a,b\in A$, then

Since $\omega(x,y)\in A^\times$, $b=0$.

The element $x\in A^2$ is regular, therefore, by Proposition <ref>, if $xa=0$, then $a=0$. So, we obtain $(a,b)=(0,0)$, i.e. the map above is an isomorphism.

For every regular isotropic $x\in A^2$, there exists an isotropic $y\in A^2$ such that $(x,y)$ is a symplectic basis.

Since $\omega$ is non-degenerate, there exists $y'\in A^2$ such that $\omega(x,y')\in A^{\times}$ and $(x,y')$ is a basis. We take $y'':=y'-\frac{x}{2}\omega(y',x)^{-1}\omega(y,y)$, then

Since $\omega(x,y')=\omega(x,y'')$, if we take $y:=y''\omega(x,y')^{-1}$, we obtain $\omega(x,y)=1$ and $x,y$ are isotropic, so $(x,y)$ is a symplectic basis.

The group $\Sp_2(A,\sigma)$ acts transitively on regular isotropic elements of $(A^2,\omega)$.

If $x=(x_1,x_2)^T\in A^2$ is regular isotropic, then there exists $y=(y_1,y_2)^T\in A^2$ regular isotropic such that $(x,y)$ is a symplectic basis. Then

x_1 & y_1 \\
x_2 & y_2
\end{pmatrix}\in \Sp_2(A,\sigma)$$

and $g(1,0)^T=x$.

Let now $\K$ be a real closed field and $(A,\sigma)$ be a Hermitian $\K$-algebra. We consider the following sesquilinear form $b\colon A^2\times A^2 \to A$, $b(x,y):=\sigma(x)^Ty$ for $x,y\in A^2$.

$b(x,x)=0$ for an element $x\in A^2$ if and only if $x=0$.

For $x=(x_1,x_2)^T\in A^2$, $b(x,x)=\sigma(x_1)x_1+\sigma(x_2)x_2$ with $\sigma(x_1)x_1,\sigma(x_2)x_2\in A^\sigma_{\geq 0}$. If $b(x,x)=0$, then $\sigma(x_1)x_1=\sigma(x_2)x_2=0$.

If $x\in A^2$ is regular then $b(x,x)\in A^\sigma_+$.

Let $a\in A^\sigma$ is not invertible, then there exists $c\in A^\sigma\bs\{0\}$ such that $ac=ca=0$.

Consider the spectral decomposition of $a=\sum_{i=1}^k \lambda_ic_i$ for some complete system of idempotents $(c_i)_{i=1}^k$ and $\lambda_i\in\R$. Since $a$ is not invertible, there exists $i$ such that $\lambda_i=0$. Then $ac_i=c_ia=0$.

Let $x=(x_1, x_2)^T$ and assume $b(x,x)=\sigma(x)x=\sigma(x_1)x_1+\sigma(x_2)x_2\in A^\sigma_{\geq 0}$ is not invertible. By the Lemma <ref>, there exists $c\in\R[a]\subseteq A^\sigma$ such that $0=cb(x,x)c=b(xc,xc)$. Therefore, $xc=0$ and so by Proposition <ref>, the map $A\to xA$, $a\mapsto xa$ is not invertible. In particular, $x$ is not regular.

Let $x\in A^2$ be regular, then there exists $a\in A$ and $y\in A^2$ such that $(xa,y)$ is an orthonormal with respect to $b$ basis of $A^2$, i.e. $b(xa,xa)=b(y,y)=1$, $b(xa,y)=0$.

Since $x$ is regular, $b(x,x)\in A^\sigma_+$. Take $a=b(x,x)^{-\frac{1}{2}}$, then $b(xa,xa)=1$. Moreover, there exists $z\in A^2$ such that $(x,z)$ is a basis of $A^2$. Consider $y':=z-xab(xa,z)$, obviously $(xa,y')$ is a basis as well. Then $b(xa,y')=b(xa,z-xab(xa,z))=b(xa,z)-b(xa,xa)b(xa,z)=0$. Since $y'$ is regular and so $b(y',y')\in A^\sigma_+$, we can take $y:=y'b(y',y')^{-1}$.

§.§ Maximal compact subgroup of Sp2(A,sigma) over Hermitian algebras

We now assume $\K=\R$ and $(A,\sigma)$ to be Hermitian, semisimple $\R$-algebra. We describe a maximal compact subgroup of $\Sp_2(A,\sigma)$.

We denote:

$$\UU_2(A,\sigma):=\{M\in\Mat_2(A)\mid \sigma(M)^TM=\Id\};$$

a & b \\
-b & a
\end{pmatrix}\left|\,\begin{matrix}
\sigma(a)a+\sigma(b)b=1\\
\sigma(a)b-\sigma(b)a=0
\end{matrix}\right.,\,a,b\in A\right\}.$

a & b \\
c & d
\end{pmatrix}\in\KSp_2(A,\sigma)$. On one hand, $M\in\Sp_2(A,\sigma)$, therefore,

$$M^{-1}= -\Om \sigma(M)^T\Om=\begin{pmatrix}
\sigma(d) & -\sigma(b) \\
-\sigma(c) & \sigma(a)
\end{pmatrix}.$$

On the other hand, $M\in\UU_2(A,\sigma)$, therefore,

\sigma(a) & \sigma(c) \\
\sigma(b) & \sigma(d)
\end{pmatrix}.$$

So we obtain, $a=d$ and $b=-c$.

The group
$\KSp_2(A,\sigma)$ is a maximal compact subgroup of $\Sp_2(A,\sigma)$.

The group $\KSp_2(A,\sigma)$ is a closed subgroup of the compact group $\UU_2(A,\sigma)$, so it is compact.

Now, we show that $\KSp_2(A,\sigma)$ is a maximal compact subgroup of $\Sp_2(A,\sigma)$. For this let $K$ be a compact subgroup containing $\KSp_2(A,\sigma)$ as a proper subgroup. We consider the following decomposition of $\spp_2(A,\sigma)$:

$$\spp_2(A,\sigma)=\ksp_2(A,\sigma)\oplus S$$

\begin{pmatrix}
a & b \\
-b & a
\end{pmatrix}
\midwd \sigma(a)=-a\in A, b\in A^\sigma
\right\},$$

\begin{pmatrix}
c & d \\
d & -c
\end{pmatrix}
\midwd c,d\in A^\sigma
\right\}.$$

By our assumption, $\Lie(K)$ contains $\ksp_2(A,\sigma)$ and has nontrivial intersection with $S$. Take some matrix

c & d \\
d & -c
\end{pmatrix}\in\Lie(K)\cap S,\;c,d\in A^\sigma.$$

The matrix

0 & d \\
-d & 0
\end{pmatrix}\in\ksp_2(A,\sigma)\subset\Lie(K),$$

c & 2d \\
0 & -c
\end{pmatrix}=\begin{pmatrix}
c & d \\
d & -c
\end{pmatrix}+\begin{pmatrix}
0 & d \\
-d & 0
\end{pmatrix}\in\Lie(K)\bs\ksp_2(A,\sigma).$$

Using the exponential map of $\spp_2(A,\sigma)$ restricted to $\Lie(K)$, we obtain that there exits a matrix $M:=\begin{pmatrix}
g & gx \\
0 & g^{-1}
\end{pmatrix}\in K\bs\KSp_2(A,\sigma)$ where $g=\exp(c)\in A^\sigma$, $x\in A^\sigma$. We consider the spectral decomposition of $g=\sum_{i=1}^k\lambda_ic_i$ with $\lambda_i>0$ and $(c_i)_{i=1}^k$ a complete orthogonal system of idempotents. Take a sequence $\{M^r\}\subseteq K$. Then

Assume that there exists $s\in\{1,\dots,k\}$ such that $\lambda_s\neq\pm 1$. Then either $0<|\lambda_s|<1$ or $0<|\lambda_s^{-1}|<1$. Without loss of generality, we may assume $0<|\lambda_s|<1$. Since $K$ is compact, $\{M^r\}\subseteq K$ has a convergent subsequence $\{M^{r_j}\}\subseteq K$:

$$\lim M^{r_j}_{11}=\lim\sum_{i=1}^k\lambda_i^{r_j}c_i=\sum_{i=1}^k\hat\lambda_ic_i$$

where $\hat\lambda_i=\lim\lambda_i^{r_j}$. But $\hat\lambda_s=\lim\lambda_s^{r_j}=0$ for any subsequence $\{r_j\}$. Therefore $\lim M^{r_j}_{11}$ is not invertible and so $\lim M^{r_j}$ is not invertible as well. Therefore, all $\lambda_i=\pm 1$ and $g^2=1$. The element $L:=\begin{pmatrix}
g & 0\\
0 & g^{-1}
\end{pmatrix}\in\KSp_2(A,\sigma)\subset K$. Then $ML=\begin{pmatrix}
1 & x \\
0 & 1
\end{pmatrix}\in K$. Take $(ML)^r=\begin{pmatrix}
1 & rx \\
0 & 1
\end{pmatrix}\in K$. This sequence does not have any convergent subsequence unless $x=0$. So we get $M=L\in\KSp_2(A,\sigma)$. This contradicts the assumption $M\notin\KSp_2(A,\sigma)$ and we obtain that $\KSp_2(A,\sigma)$ is a maximal compact subgroup of $\Sp_2(A,\sigma)$.

The embedding
$$\Upsilon \colon A_\CC  \to  \Mat_2(A)$$
(see equation <ref>) maps $U_{(A_\CC,\bar\sigma_\CC)}$ isomorphically to $\KSp_2(A,\sigma)$. In particular, the fundamental group of $\Sp_2(A,\sigma)$ is infinite.

§.§ Maximal compact subgroups of Sp2(A,sigma) over complexified algebras

Let $\K=\R$ and $(A,\sigma)$ be a Hermitian, semisimple $\R$-algebra. In Section <ref> we have seen that $(A_\CC,\bar\sigma_\CC)$ is also Hermitian and semisimple and $(A_\CC,\sigma_\CC)$ is never Hermitian.

We set

The group $\KSp^c_2(A_\CC,\sigma_\CC)$ is given by

a & b \\
-\bar b & \bar a
\end{pmatrix}\left|\;\begin{matrix}
\bar\sigma_\CC(a)a+\sigma_\CC(b)\bar b=1\\
\bar\sigma_\CC(a)b-\sigma_\CC(b)\bar a=0
\end{matrix}\right.,\;a,b\in A_\CC\right\}.$$

a & b \\
c & d
\end{pmatrix}\in\KSp^c_2(A_\CC,\sigma_\CC)$. On one hand, $M\in\Sp_2(A_\CC,\sigma_\CC)$, therefore,

$$M^{-1}= -\Om \sigma_\CC(M)^T\Om=\begin{pmatrix}
\sigma_\CC(d) & -\sigma_\CC(b) \\
-\sigma_\CC(c) & \sigma_\CC(a)
\end{pmatrix}.$$

On the other hand, $M\in\UU_2(A_\CC,\bar\sigma_\CC)$, therefore,

\bar\sigma_\CC(a) & \bar\sigma_\CC(c) \\
\bar\sigma_\CC(b) & \bar\sigma_\CC(d)
\end{pmatrix}.$$

So we obtain, $d=\bar a$ and $c=-\bar b$.

The group
$\KSp^c_2(A_\CC,\sigma_\CC)$ is a maximal compact subgroup of $\Sp_2(A_\CC,\sigma_\CC)$.

The proof follows the same strategy as the proof of Theorem <ref>.

By definition, $\KSp^c_2(A_\CC,\sigma_\CC)$ is closed subgroup of $\UU_2(A_\CC,\bar\sigma_\CC)$ which is compact, so $\KSp^c_2(A_\CC,\sigma_\CC)$ is compact as well.

To show that $\KSp^c_2(A_\CC,\sigma_\CC)$ is a maximal compact subgroup of $\Sp_2(A_\CC,\sigma_\CC)$, we assume $K$ to be a compact subgroup of $\Sp_2(A_\CC,\sigma_\CC)$ containing $\KSp^c_2(A_\CC,\sigma_\CC)$ as a proper subgroup. We consider the following decomposition of $\spp_2(A_\CC,\sigma_\CC)$:

$$\spp_2(A_\CC,\sigma_\CC)=\ksp^c_2(A_\CC,\sigma_\CC)\oplus S$$

\begin{pmatrix}
a & b \\
-\bar b & \bar a
\end{pmatrix}
\midwd \bar\sigma_\CC(a)=-a\in A_\CC, b\in A^{\bar\sigma}_\CC
\right\},$$

\begin{pmatrix}
c & d \\
\bar d & -\bar c
\end{pmatrix}
\midwd c,d\in A^{\bar\sigma}_\CC
\right\}.$$

By our assumption, $\Lie(K)$ contains $\ksp^c_2(A_\CC,\sigma_\CC)$ and has nontrivial intersection with $S$. So we can take a matrix

c & d \\
\bar d & -\bar c
\end{pmatrix}\in\Lie(K)\cap S,\; c,d\in A^{\bar\sigma}_\CC.$$

0 & d \\
-\bar d & 0
\end{pmatrix}\in\ksp^c_2(A_\CC,\sigma_\CC)\subset\Lie(K),$$

c & 2d \\
0 & -\bar c
\end{pmatrix}=\begin{pmatrix}
c & d \\
\bar d & -\bar c
\end{pmatrix}+\begin{pmatrix}
0 & d \\
-\bar d & 0
\end{pmatrix}\in\Lie(K)\bs\ksp^c_2(A_\CC,\sigma_\CC).$$

Using the exponential map of $\spp_2(A_\CC,\sigma_\CC)$ restricted to $\Lie(K)$, we obtain that there exits a matrix $M:=\begin{pmatrix}
g & gx \\
0 & \bar g^{-1}
\end{pmatrix}\in K\bs\KSp^c_2(A_\CC,\sigma_\CC)$ where $g=\exp(c)\in A^{\bar\sigma}_\CC$, $x\in A^{\bar\sigma}_\CC$. We now consider the spectral decomposition of $g=\sum_{i=1}^k\lambda_ic_i$ with $\lambda_i>0$ and $(c_i)_{i=1}^k$ a complete orthogonal system of idempotents. Take a sequence $\{M^r\}\subseteq K$. Then

M^r_{22}=\bar g^{-k}=\sum_{i=1}^k\lambda_i^{-r}c_i.$$

We assume, there exists $s\in\{1,\dots,k\}$ such that $\lambda_s\neq\pm 1$. Then either $0<|\lambda_s|<1$ or $0<|\lambda_s^{-1}|<1$. Without loss of generality, we assume $0<|\lambda_s|<1$. Since $K$ is compact, $\{M^r\}\subseteq K$ has a convergent subsequence $\{M^{r_j}\}\subseteq K$:

$$\lim M^{r_j}_{11}=\lim\sum_{i=1}^k\lambda_i^{r_j}c_i=\sum_{i=1}^k\hat\lambda_ic_i$$

where $\hat\lambda_i=\lim\lambda_i^{r_j}$. But $\hat\lambda_s=\lim\lambda_s^{r_j}=0$ for any subsequence $\{r_j\}$. Therefore $\lim M^{r_j}_{11}$ is not invertible and so $\lim M^{r_j}$ is not invertible as well. Therefore, all $\lambda_i=\pm 1$ and $g^2=1$. The element $L:=\begin{pmatrix}
g & 0\\
0 & g^{-1}
\end{pmatrix}\in\KSp^c_2(A_\CC,\sigma_\CC)\subset K$. Then $ML=\begin{pmatrix}
1 & x \\
0 & 1
\end{pmatrix}\in K$. Take $(ML)^r=\begin{pmatrix}
1 & rx \\
0 & 1
\end{pmatrix}\in K$. This sequence does not have any convergent subsequence unless $x=0$. So we get $M=L\in\KSp^c_2(A_\CC,\sigma_\CC)$. This contradicts the assumption $M\notin\KSp^c_2(A_\CC,\sigma_\CC)$ and we obtain that $\KSp^c_2(A_\CC,\sigma_\CC)$ is a maximal compact subgroup of $\Sp_2(A_\CC,\sigma_\CC)$.

Let $(A,\sigma)$ be a real Hermitian algebra, $A_\CC$ be the complexification of $A$, and $\sigma_\CC$ be the complex linear extension of $\sigma$. The embedding

$$\Upsilon_\HH \colon A_\HH \to \Mat_2(A_\CC)$$
from <ref> maps $U_{(A_\HH,\sigma_1)}$ isomorphically to $\KSp^c_2(A_\CC,\sigma_\CC)$.

Notice, that the group $\KSp_2(A_\CC,\sigma_\CC)$ is never compact because it is a complexification of the real group $\KSp_2(A_\CC,\sigma_\CC)$.

§.§ Realization of classical Lie groups as Sp2(A,sigma)

In the case when $(A, \sigma)$ is a Hermitian algebra so that $(A^\sigma,\circ)$ is a Jordan algebra, the symplectic groups $\Sp_2(A,\sigma)$ are isomorphic to classical Hermitian Lie groups of tube type.
There are also other groups that can be realized as $\Sp_2(A,\sigma)$ for algebras with anti-involution which are not Hermitian.

* Real symplectic group $\Sp(2n,\R)$

In order to realize the real symplectic group $\Sp(2n,\R)$, we take
$A=\Mat(n,\R)$ to the be algebra of $n\times n$ matrices over $\R$ and consider the involution $\sigma: A \to A$ given by $\sigma(r)=r^T$, $r\in A$.

Then $\Sp_2(A,\sigma)$ is isomorphic to $\Sp(2n,\R)$.
The maximal compact subgroup is

$$\KSp_2(A,\sigma)=\Upsilon(\UU(n))\cong \UU(n).$$

* Indefinite unitary group $\UU(n,n)$

To realize the unitary group $\UU(n,n)$ of an indefinite Hermitian form of signature $(n,n)$ we consider $A=\Mat(n,\CC)$, and the involution $\bar\sigma: A \to A$ given by $\bar\sigma(r)=\bar r^T$.
Then $\Sp_2(A,\bar\sigma)$ is isomorphic to $\UU(n,n)$. To see this, we notice that the standard Hermitian form $h$ of signature $(n,n)$ on $\CC^{2n}$ is given by $h(x,y):=i\omega(xT,yT)$ where $T=\diag(\Id_n,-i\Id_n)$). The complexification $A_\CC$ is isomorphic to $\Mat(n,\CC)\times\Mat(n,\CC)$ (see Section <ref>). Therefore,

$$\KSp_2(A,\sigma)\cong \UU(n)\times\UU(n).$$

Note that we cannot realize the special unitary group $\mathrm{SU}(n,n)$ as $\Sp_2(A,\sigma)$.

* The group $\SO^*(4n)$

The groups $\SO^*(2n)$ are Hermitian Lie groups, but they are of tube type only if $n$ is even.
In order to realize $\SO^*(4n)$ as $\Sp_2(A,\sigma)$ we consider  $A=\Mat(n,\HH)$ and the involution $\sigma_1: A \to A$, given by $\sigma_1(r)=\bar r^T=\bar\sigma(r_1)-\sigma(r_2)j$ for $r=r_1+r_2j$ and $r_1,r_2\in \Mat(n,\CC)$.
Then $\Sp_2(A,\sigma_1)$ is isomorphic $\SO^*(4n)$ (some authors also use the notation $\OO(2n,\HH)$) considered as the group of isometries of the  quaternionic form $\beta$ on $\HH^{2n}$ defined by

$$\beta(x,y)=\sum_{i=1}^{2n} \bar x_i j y_i=\bar x^T(\Id_{2n}j)y.$$

To see this, we notice that

$$\Id_{2n} j=\sigma_1(T)\Ome{\Id_n} T$$

$$T=\frac{1}{\sqrt{2}}\begin{pmatrix}\Id_n & -\Id_n j \\-\Id_n j & \Id_n\end{pmatrix}.$$

In this case $A_\CC$ is isomorphic to $\Mat(2n,\CC)$ (see Section <ref>). Therefore.

$$\KSp_2(A,\sigma)\cong \UU(2n).$$

The Hermitian Lie group of tube type $\SO_0(2,n)$ cannot be realized in the same way as $\Sp_2(A,\sigma)$. The reason is that the Jordan algebra $B_n$ (see the classification <ref>) cannot be seen as $A^\sigma$ for an appropriate Hermitian algebra $(A,\sigma)$. Nevertheless, since $B_n$ can be seen as Jordan subalgebra of an appropriate Clifford algebra, the group $\SO_0(2,n)$ (or more precisely, its double cover $\Spin_0(2,n)$) can be realized as $\Sp_2$ over some more complicated object which we do not discuss in this paper.

The exceptional Hermitian Lie group of tube type $E_7$ cannot be realized as $\Sp_2(A,\sigma)$. The reason is that the Jordan algebra $\Herm(3,\Oc)$ (see the classification <ref>) cannot be embedded into any associative algebra, in particular it cannot be realized as $A^\sigma$ for a Hermitian algebra $(A,\sigma)$.

Some other Lie groups can also be realized as $\Sp_2(A,\sigma)$ for algebras with anti-involution $(A,\sigma)$ that is not Hermitian.

§.§.§ Other examples

There are other interesting cases of classical groups that can be realized as $\Sp_2(A,\sigma)$ where $(A,\sigma)$ is not Hermitian.

* To realize the symplectic group over any field $\K$ we  consider $A=\Mat(n,\K)$ with the anti-involution $\sigma(r):=r^T$. Then $\Sp_2(A,\sigma)$ is isomorphic to $\Sp(2n,\K)$.

If $\K=\CC$, $A$ is the complexification of $\Mat(n,\R)$, and the maximal compact subgroup is

$$\KSp^c_2(A,\sigma)\cong \Sp(n).$$

* The indefinite symplectic group can be realized as $\Sp_2(A,\sigma)$. For this consider $A=\Mat(n,\HH)$ with involution  $\sigma_0(r)=\sigma(r_1)+\bar\sigma(r_2)j$ for $r=r_1+r_2j$ and $r_1,r_2\in \Mat(n,\CC)$. Note that the algebra $(A,\sigma_0)$ is not Hermitian and also does not appear as a complexification of a Hermitian algebra.

Then $\Sp_2(A,\sigma_0)$ is isomorphic $\Sp(n,n)$ considered as the group of isometries of the quaternionic form $\omega$ on $\HH^{2n}$ given by

$$\omega(x,y)=\sum_{i=1}^{2n} \bar x_i y_i=\bar x^T\Dia{-\Id_n}{\Id_n}y=$$

$$=k\left(\sigma_0(x)\Dia{\Id_n k}{-\Id_n k}y\right).$$

To see this, we notice that

$$\Dia{\Id_n k}{-\Id_n k}=\sigma_0(T)\Ome{\Id_n} T$$

$$T=\frac{1}{\sqrt{2}}\begin{pmatrix}\Id_n & \Id_n k \\ \Id_n k & \Id_n\end{pmatrix}.$$

The maximal compact subgroup of $\Sp(n,n)$ is $\Sp(n)\times\Sp(n)$. One can see this using the machinery developed in the Section <ref>.

The subgroup $\KSp_2(A,\sigma_0)$ is isomorphic to $\GL_n(\HH)$ which is not compact because $(A,\sigma_0)$ is not Hermitian.

§ SPACE OF ISOTROPIC LINES

We assume $(A,\sigma)$ to be a unital ring with an anti-involution. We denote by $\PP(A^2)$ the space of lines in $A^2$, i.e.

$$\PP(A^2):=\{xA\mid x\in A^2\text{ regular}\}.$$

The group $\Mat_2^\times(A)$ acts on $\PP(A^2)$.

If $A^2$ is equipped with the standard symplectic form $\omega$, we denote by $\PP(\Is(\omega))$ the space of all isotropic (with respect to $\omega$) lines:

$$\PP(\Is(\omega))=\{xA\subset A^2\mid \omega(x,x)=0,\,x\text{ regular}\}.$$

This space is a closed subspace of $\PP(A^2)$. The group $\Sp_2(A,\sigma)$ acts on the space of isotropic lines.

§.§ Space of lines as a homogeneous space

Now assume $\K$ to be real closed field and $(A,\sigma)$ is an algebra with an anti-involution over $\K$. We show that $\PP(A^2)$ and $\PP(\Is(\omega))$ can be seen as homogeneous (even symmetric) spaces that is compact in case $\K=\R$.

Let $(A,\sigma)$ be Hermitian. The group $\UU_2(A,\sigma)$ acts continuously  and transitively on $\PP(A^2)$ with

$$\Stab_{\UU_2(A,\sigma)}\Clm{1}{0}A=\left\{\begin{pmatrix}u_1 & 0 \\ 0 & u_2\end{pmatrix}\midwd u_1,u_2\in U_{(A,\sigma)}\right\}\cong U_{(A,\sigma)}\times U_{(A,\sigma)}.$$

Let $xA$ be a line. Since $x\in A^2$ is regular, by Proposition <ref>, we can assume $b(x,x):=\sigma(x)^Tx=1$. By Proposition <ref>, there exists $y$ such that $(x,y)$ is an orthonormal basis, i.e. the matrix

x_1 & y_1\\
x_2 & y_2
\end{pmatrix}\in \UU_2(A,\sigma),$$

where $x=(x_1,x_2)^T$, $y=(y_1,y_2)^T$. Moreover

x_1 & y_1\\
x_2 & y_2
\end{pmatrix}\Clm{1}{0}=x.$$

That means $\UU_2(A,\sigma)$ acts transitively on $\PP(A^2)$.

Now compute the stabilizer of $(1,0)^TA$. Take $U:=\begin{pmatrix}
x_1 & y_1\\
x_2 & y_2
\end{pmatrix}\in \UU_2(A,\sigma).$ Since $U(1,0)=(a,0)$ for some $a\in A$, $x_2=0$. Further

$$\Id_2=\sigma(U)^TU= \begin{pmatrix}
\sigma(x_1) & 0 \\
\sigma(y_1) & \sigma(y_2)
\end{pmatrix}\begin{pmatrix}
x_1 & y_1\\
0 & y_2
\end{pmatrix}=\begin{pmatrix}
\sigma(x_1)x_1 & \sigma(x_1)y_1 \\
\sigma(y_1)x_2 & \sigma(y_1)y_1+\sigma(y_2)y_2
\end{pmatrix}.$$

Therefore, $\sigma(x_1)x_1=1$, i.e. $x_1\in U_{(A,\sigma)}$. Further, $\sigma(x_1)y_1=0$. Since $x_1\in U_{(A,\sigma)}$, it is invertible and so $y_1=0$. Finally, $\sigma(y_1)y_1+\sigma(y_2)y_2=\sigma(y_2)y_2=1$. Therefore, $y_2\in U_{(A,\sigma)}$.

The space $\PP(A^2)$ is homeomorphic to the quotient space:

$$\UU_2(A,\sigma)/U_{(A,\sigma)}\times U_{(A,\sigma)}$$

where the group $U_{(A,\sigma)}\times U_{(A,\sigma)}$ is embedded into $\UU_2(A,\sigma)$ diagonally.

In particular, if $\K=\R$, $\PP(A^2)$ is compact and $\PP(\Is(\omega))$ is compact for any sesquilinear form $\omega\colon A^2\times A^2\to A$.

For Hermitian algebras and complexifications of Hermitian algebras, we can describe the space $\PP(\Is(\omega))$ more precisely. Similarly to the Proposition <ref> and Corollary <ref>, one can prove the following two statements:

* Let $(A,\sigma)$ be Hermitian and $\omega$ be the standard symplectic form on $A^2$. Then $\KSp_2(A,\sigma)$ acts transitively on $\PP(\Is(\omega))$ with

$$\Stab_{\KSp_2(A,\sigma)}\Clm{1}{0}A=\left\{\begin{pmatrix}u & 0 \\ 0 & u\end{pmatrix}\midwd u \in U_{(A,\sigma)}\right\}\cong U_{(A,\sigma)}.$$

In particular, $\PP(\Is(\omega))$ is homeomorphic to the quotient space:

where the group $U_{(A,\sigma)}$ is embedded into $\KSp_2(A,\sigma)$ in the diagonal way.

* Let $(A_\CC,\sigma_\CC)$ be the complexification of a Hermitian algebra $(A,\sigma)$ and $\omega$ be the standard symplectic form on $A_\CC^2$. Then $\KSp^c_2(A_\CC,\sigma_\CC)$ acts transitively on $\PP(\Is(\omega))$ with

$$\Stab_{\KSp^c_2(A_\CC,\sigma_\CC)}\Clm{1}{0}A=\left\{\begin{pmatrix}u & 0 \\ 0 & u\end{pmatrix}\midwd u \in U_{(A_\CC,\bar\sigma_\CC)}\right\}\cong U_{(A_\CC,\bar\sigma_\CC)}.$$

In particular, $\PP(\Is(\omega))$ is homeomorphic to the quotient space:

where the group $U_{(A_\CC,\bar\sigma_\CC)}$ is embedded into $\KSp^c_2(A_\CC,\sigma_\CC)$ in the diagonal way.

$\Sp_2(A,\sigma)$ acts transitively on $\PP(\Is(\omega))$.

\begin{pmatrix}
x & xy \\
0 & \sigma(x)^{-1}
\end{pmatrix}
\midwd
x\in A^\times, y\in A^\sigma
\right\}$$

\begin{pmatrix}
x & 0 \\
zx & \sigma(x)^{-1}
\end{pmatrix}
\midwd
x\in A^\times, z\in A^\sigma
\right\}$$

$\Sp_2(A,\sigma)$ acts transitively on the space of isotropic lines since it acts transitively on $\Is(\omega)$.

We prove only the statement for the first stabilizer. The second one can be proved analogously.

x & a \\
b & t
\end{pmatrix}
\begin{pmatrix}
1 \\ 0
\end{pmatrix}=\begin{pmatrix}
x \\ b
\end{pmatrix},$$

$x\in A^\times$ and $b=0$. Furthermore,

x & a \\
0 & t
\end{pmatrix}\right)^T\Ome{1}
\begin{pmatrix}
x & a \\
0 & t
\end{pmatrix}=\begin{pmatrix}
0 & \sigma(x)t \\
-\sigma(t)x & -\sigma(t)a+\sigma(a)t
\end{pmatrix}=\Ome{1},$$

we obtain $t=\sigma(x)^{-1}$, $a=xy$ for $y\in A^\sigma$.

§.§ Action of Sp2(A,sigma) on pairs of isotropic lines

Let $(A,\sigma)$ be a unital ring with an anti-involution.

Two elements $u,v\in\Is(\omega)$ are linearly independent if and only if, up to action of $\Sp_2(A,\sigma)$, $u=(1,0)^T$, $v=(a,b)^T$ with $b\in A^\times$. Moreover, if $\omega(u,v)=1$, then $a\in A^\sigma$, $b=1$.

$\Sp_2(A,\sigma)$ acts transitively on $\Is(\omega)$, therefore, up to $\Sp_2(A,\sigma)$-action, we can assume that $u=(1,0)^T$. Since $u$ and $v$ are linearly independent, $b\in A^\times$. If $\omega(u,v)=1=b$, then $v=(a,1)^T$ is isotropic, i.e.

So $a\in A^\sigma$.

If $x,y\in\Is(\omega)$ linearly independent, then $\omega(x,y)\in A^\times$.

If $(x,y)$ is a symplectic basis then there exists the unique $g\in \Sp_2(A,\sigma)$ such that $g(1,0)^T=x$, $g(0,1)^T=y$. In particular, $\Sp_2(A,\sigma)$ acts transitively on symplectic bases.

We can assume, $x=(1,0)^T$, $y=(a,1)^T$ and $a\in A^\sigma$. Take $g:=\begin{pmatrix}1 & a \\ 0 & 1\end{pmatrix}$, then $gx=x$, $gy=(0,1)^T$.

Let $xA$, $yA$ be two transverse isotropic lines with $x,y\in\Is(\omega)$. Then there exist $M\in \Sp_2(A,\sigma)$ and $y'\in\Is(\omega)$ such that $y'A=yA$ and $Mx=(1,0)^T$, $My'=(0,1)^T$. In particular, $\omega(x,y')=1$.

$\Sp_2(A,\sigma)$ acts transitively on pairs of transverse isotropic lines.

\begin{pmatrix}
x & 0 \\
0 & \sigma(x)^{-1}
\end{pmatrix}
\mid
x\in A^\times\right\}\cong A^\times.$$

By the Corollary <ref>, every pair of transverse isotropic lines can be mapped to $((1,0)^TA,(0,1)^TA)$ by an element of $\Sp_2(A,\sigma)$. So $\Sp_2(A,\sigma)$ acts transitively on pairs of transverse isotropic lines.

By the Proposition <ref>,

$$=\Stab_{\Sp_2(A,\sigma)}\left(\Clm{1}{0}A\right)\cap \Stab_{\Sp_2(A,\sigma)}\left(\Clm{0}{1}A\right)=$$

\begin{equation*}
\begin{pmatrix}
x & 0 \\
0 & \sigma(x)^{-1}
\end{pmatrix}
\mid
x\in A\right\}.\qedhere
\end{equation*}

§.§ Action of Sp2(A,sigma) on triples of isotropic lines

Let $(A,\sigma)$ be a unital ring with an anti-involution. Let $(x_1A,x_3A,x_2A)$ be a triple of pairwise transverse isotropic lines where all $x_i\in \Is(\omega)$. Because of transversality of $x_1A$ and $x_2A$, we can assume $\omega(x_1,x_2)=1$. Up to action of $\Sp_2(A,\sigma)$, we can assume $x_1=(1,0)^T$, $x_2=(0,1)^T$. We can also normalize $x_3$ so that $\omega(x_1,x_3)=1$. Then $x_3=(b,1)^T$, $b=\omega(x_3,x_2)\in (A^\sigma)^\times$.

Orbits of the action of $\Sp_2(A,\sigma)$ on triples of pairwise transverse isotropic lines are in 1-1 correspondence with orbits of the following action of $A^\times$ on $(A^\sigma)^\times$:

\psi\colon & A^\times\times (A^\sigma)^\times & \mapsto & (A^\sigma)^\times\\
& (a,b) & \mapsto & ab\sigma(a).
\end{array}$$

Let $(l_1,l_3,l_2)$ be a triple pairwise transverse of isotropic lines.
Up to $\Sp_2(A,\sigma)$-action, we can assume $l_i=x_iA$ for $x_1=(1,0)^T$, $x_2=(0,1)^T$, $x_3=(1,b)^T$ with $b\in (A^\sigma)^\times$. The stabilizer $\Stab_{\Sp_2(A,\sigma)}((1,0)^TA,(0,1)^TA)\cong A^\times$ acts on $x_3$ in the following way: