Symplectic groups over noncommutative algebras

Abstract

We introduce the symplectic group ${{\,\mathrm{Sp}\,}}_2(A,\sigma )$ over a noncommutative algebra A with an anti-involution $\sigma$. We realize several classical Lie groups as ${{\,\mathrm{Sp}\,}}_2$ over various noncommutative algebras, which provides new insights into their structure theory. We construct several geometric spaces, on which the groups ${{\,\mathrm{Sp}\,}}_2(A,\sigma )$ act. We introduce the space of isotropic A-lines, which generalizes the projective line. We describe the action of ${{\,\mathrm{Sp}\,}}_2(A,\sigma )$ 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 $X_{{{\,\mathrm{Sp}\,}}_2(A,\sigma )}$, and construct different models of this space. Applying this to classical Hermitian Lie groups of tube type (realized as ${{\,\mathrm{Sp}\,}}_2(A,\sigma )$) 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.

Appendices

Appendix A: Classification of Hermitian algebras

The goal of this section is to classify all Hermitian algebras. To do this, we consider a more general class of algebras that we call pre-Hermitian, and classify them.

Let $(A,\sigma )$ be a ring with an anti-involution $\sigma$. As usual, we say that $a\in A$ is symmetric if $\sigma (a)=a$, denote the set of all symmetric elements in A by $A^\sigma$. Clearly, if $2\in A^\times$, then $A^\sigma$ is a (unital) Jordan ring under the operation $a\circ b=2^{-1}(ab+ba)$.

If A is an algebra over a commutative ring F and $\sigma$ is F-linear, then we will refer to $(A,\sigma )$ as an F-algebra.

Definition A.1

The Jacobson radical J(A) of a unital ring A is the set of all $x\in A$ such that $1+AxA\subseteq A^\times$. In particular, $1 + J(A)$ is a subgroup of $A^\times$.

It is well-known (see e.g., [17]) that J(A) is a nilpotent ideal for any (left or right) Artinian ring A. Moreover, such a ring is semisimple if and only if $J(A)=\{0\}$. In particular, this holds for finite dimensional algebras over any field.

Proposition A.2

J(A) is invariant under any anti-involution of any ring A.

Proof

Clearly, if $\sigma$ is any anti-involution of A then $1+A\sigma (x)A\subset A^\times$ for all $x\in J(A)$ hence $\sigma (x)\in J(A)$. $\square$

Definition A.3

We say that a ring $(A,\sigma )$ is pre-Hermitian if $A^\sigma \cap J(A)=\{0\}$.

If $J(A)=\{0\}$, A is sometimes called Jacobson semisimple (in particular, any $C^*$-algebra is Jacobson semisimple as a consequence of Gelfand-Naimark theorem). Also note that any ${\mathbb {R}}$-subalgebra of ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}})$ invariant under the Hermitian transposition is semisimple and, therefore, Jacobson semisimple (see [20], Exercise 18, p. 168). By definition, any Jacobson semisimple $(A,\sigma )$ is pre-Hermitian.

Definition A.4

We say that a ring $(A,\sigma )$ is Hermitian if $a^2+b^2=0$ for $a,b\in A^\sigma$ implies that $a=b=0$.

In particular, nonzero symmetric elements of Hermitian rings are not nilpotent.

Remark A.5

In contrast to the main part of this paper, we do not assume in this appendix that a Hermitian ring is an algebra over a real closed field.

Remark A.6

If $(A,\sigma )$ is a Hermitian ring such that $J(A)=\{0\}$, then, similarly to the Proposition 2.58, we can show that $-a^2\in A^\sigma _{\ge 0}$ (See definition 2.11) for all $a\in A^{-\sigma }$.

Remark A.7

Similarly to $A^\sigma _{\ge 0}$, for any ring $(A,\sigma )$ denote by $A^\sigma _{>0}$ the set of all sums $a_1^2+\cdots +a_n^2$, $n\ge 1$, where all $a_i$ are nonzero elements of $A^\sigma$. By definition, $A^\sigma _{>0}$ is an additive sub-semigroup of A, which may or may not contain 0. Clearly, $A^\sigma _{\ge 0}=A^\sigma _{>0}\cup \{0\}$. Also, it is immediate that if $0\notin A^\sigma _{>0}$, then $(A,\sigma )$ is Hermitian. It would be interesting to classify those rings in which the opposite implication holds.

Proposition A.8

Any Hermitian ring A with nilpotent J(A) is pre-Hermitian.

Proof

Assume, there exists $0\ne x\in J(A)\cap A^\sigma$. Since J(A) is a nilpotent ideal, there exist $x\in J(A)$ such that $a:=x^n\ne 0$ and $a^2=0$ for some $n>0$. Therefore, $a=0$, which is a contradiction. $\square$

Proposition A.9

Let $(A,\sigma )$ be a pre-Hermitian ring. Then $\sigma (x)=-x$ and $xy=-yx$ for any $x,y\in J(A)$. In particular, $2x^2=0$.

Proof

Let $x\in J(A)$. Since $x+\sigma (x)\in J(A)\cap A^\sigma =\{0\}$, $\sigma (x)=-x$. Furthermore, $-xy=\sigma (xy)=\sigma (y)\sigma (x)=yx$ for $x,y\in J(A)$. $\square$

Proposition A.10

If $(A,\sigma )$ is pre-Hermitian then:

• (a) $\sigma (x)=-x$ for all $x\in J(A)$ and $xa=\sigma (a)x$ for all $x\in J(A)$, $a\in A$.

• (b) $yx=-xy$ for all $x,y\in J(A)$ and $xyz=0$ for all $x,y,z\in J(A)$.

• (c) $(\sigma (a)-a)xy=xy(\sigma (a)-a)=0$ all $a\in A$, $x,y\in J(A)$

Proof

Prove (a). Since $\sigma (x)\in J(A)$ for all $x\in J(A)$ by Proposition A.2, $x+\sigma (x)\in J(A)\cap A^\sigma =\{0\}$. This proves the fist assertion. To prove the second assertion, using the fact that $xa\in J(A)$ for all $x\in J(A)$, $a\in A$, we obtain

$$\begin{aligned} xa=-\sigma (xa)=-\sigma (a)\sigma (x)=\sigma (a)x\ . \end{aligned}$$

This proves (a).

To prove (b) note that $xy=-\sigma (xy)=-yx$ for all $x,y\in J(A)$. To prove the second assertion, note that

$$\begin{aligned} \sigma (xyz)=-zyx=-yxz=xyz \end{aligned}$$

for all $x,y,z\in J(A)$ hence $xyz=0$. This proves (b).

To prove (c) note that on the one hand, $xya=x\sigma (a)y=axy$ and on the other hand, $(xy)a=\sigma (a)(xy)$ for all $a\in A$ and $x,y\in J(A)$. This proves (c).

The proposition is proved. $\square$

Corollary A.11

If $(A,\sigma )$ and $(B,\sigma ')$ are pre-Hermitian algebras over F, $char ~F \ne 2$ and $(A,\sigma )\otimes (B,\sigma ')$ is also pre-Hermitian, then either $J(A)=\{0\}$ or $J(B)=\{0\}$.

Proof

Indeed, If $x\in J(A)$, $y\in J(B)$, then $x\otimes y\in (A\otimes B)^{\sigma \otimes \sigma '}$ by Proposition A.10. $\square$

Example A.12

Let V be a finite-dimensional vector space over a field F, $char ~F\ne 2$, ${\hat{A}}=\Lambda (V)$ be the exterior algebra of V, and ${\hat{\sigma }}$ be the unique anti-involution of ${\hat{A}}$ such that ${\hat{\sigma }}(v)=-v$ for all $v\in V$. Denote by A the quotient of ${\hat{A}}$ by the ideal generated by $\Lambda ^3 V$, so that $A= F\oplus V\oplus \Lambda ^2 V$ as a vector space. Since ${\hat{\sigma }}$ preserves the ideal generated by $\Lambda ^3 V$, it induces a well-defined anti-involution $\sigma$ on A. Clearly, $J(A)=V\oplus \Lambda ^2 V$ and $\sigma (x)=-x$ for all $x\in J(A)$. Thus, $(A,\sigma )$ is pre-Hermitian with $A^\sigma =F$ and $A^\sigma _{> 0}=F_{> 0}$.

Proposition A.13

Let A be a pre-Hermitian ring. Then

$$\begin{aligned} A^\circ \cdot J(A)=J(A)\cdot A^\circ =0 \end{aligned}$$

where $A^\circ =A\cdot [A,A]=[A,A]\cdot A$ is the ideal of A generated by all commutators $[a,b]=ab-ba$, $a,b\in A$.

Proof

It follows from Proposition A.10 that

$$\begin{aligned} abx=x\sigma (ab)=x\sigma (b)\sigma (a)=bx\sigma (a)=bax \end{aligned}$$

for all $a,b\in A$, $x\in J(A)$ hence $[a,b]x=0$. Also, $x[a,b]=[\sigma (b),\sigma (a)]x=0$.

The proposition is proved. $\square$

Example A.14

$A^\circ =A$ if A is simple noncommutative and $({{\,\mathrm{Mat}\,}}_n (A))^{\circ }={{\,\mathrm{Mat}\,}}_n (A)$ if $A^{\circ }=A$.

The following definition is motivated by Proposition A.10.

Definition A.15

We say that a (unitless) ring J is nilpotent pre-Hermitian if:

• $2x=0$ implies that $x=0$ (this is relevant only in characteristic 2).

• $yx=-xy$ and $xyz=0$ for all $x,y,z\in J$.

Clearly, for any nilpotent pre-Hermitian ring J with $0\notin 2(J\setminus \{0\})$, the assignments $j\mapsto -j$ define an anti-involution on J with no fixed points in $J\setminus \{0\}$.

Let J be a ring and K be a commutative ring that acts on J from the left. We denote the action by:

$$\begin{aligned} \begin{matrix} {\triangleright }:&{} K\times J &{} \rightarrow &{} J\\ &{} (k,j) &{} \mapsto &{} k{\triangleright }j. \end{matrix} \end{aligned}$$

Similarly, if K acts in J from the right, we denote the action by:

$$\begin{aligned} \begin{matrix} {\triangleleft }:&{} J\times K &{} \rightarrow &{} J\\ &{} (j,k) &{} \mapsto &{} j{\triangleleft }k. \end{matrix} \end{aligned}$$

We say that a ring J is a left K-algebra if J is a left K-module with respect to ${\triangleright }$ and

$$\begin{aligned} k{\triangleright }(jj')=(k{\triangleright }j) j' \end{aligned}$$

for all $j,j'\in J$, $k\in K$ (so we will sometimes denote it simply by $k{\triangleright }jj'$).

We say that a ring J is a right K-algebra if J is a right K-module with respect to ${\triangleleft }$ and

$$\begin{aligned} (jj'){\triangleleft }k=j(j'{\triangleleft }k) \end{aligned}$$

for all $j,j'\in J$, $k\in K$ (so we will sometimes denote it simply by $jj'{\triangleleft }k$).

Definition A.16

Let K be a commutative unital ring with an involution ${\overline{\cdot }}$, J be a unitless K-algebra and $\gamma :K\rightarrow J$ be a homomorphism of abelian groups. We say that J is a $(K,{\overline{\cdot }},\gamma )$-algebra if:

$$\begin{aligned} \begin{aligned} j(\gamma (kk')-k{\triangleright }\gamma (k')-k'{\triangleright }\gamma (k)-\gamma (k)\gamma (k'))=0,\\ j(k{\triangleright }j')=({{\overline{k}}}{\triangleright }j-j\gamma (k))j',(k-{{\overline{k}}}){\triangleright }jj'=j\gamma (k)j' \end{aligned} \end{aligned}$$

(1.3)

for all $j,j'\in J$, $k,k'\in K$.

When J satisfies $J^3=0$, e.g., when J is pre-Hermitian, the conditions (1.3) simplify to

$$\begin{aligned} \begin{aligned} j(\gamma (kk')-k{\triangleright }\gamma (k')-k'{\triangleright }\gamma (k))=0,\\ j(k{\triangleright }j')={{\overline{k}}}{\triangleright }jj'=k{\triangleright }jj' \end{aligned} \end{aligned}$$

(1.4)

for all $j,j'\in J$, $k\in K$.

Proposition A.17

Let K be a commutative unital ring and J be a $(K,{\overline{\cdot }},\gamma )$-algebra. Then:

(a) $J\oplus K$ has a structure of an associative unital ring with the multiplication given by

$$\begin{aligned} (j+k)(j'+k')=jj'+k{\triangleright }j'+\overline{k'}{\triangleright }j-j\gamma (k')+kk' \end{aligned}$$

(1.5)

for all $j,j'\in J$, $k,k'\in K$ (we denote this ring by $J\rtimes _\gamma K$ and refer to as semidirect sum of J and K over $\gamma$).

(b) J is a two-sided ideal in $J\rtimes _\gamma K$, moreover, the projection to the second factor is a surjective homomorphism $J\rtimes _\gamma K \twoheadrightarrow K$ of rings whose kernel is J.

(c) Suppose that $2\in K^\times$ and $\gamma (kk')={{\overline{k}}}{\triangleright }\gamma (k')+k'{\triangleright }\gamma (k)-\gamma (k)\gamma (\overline{k'})+\frac{\gamma (k)\gamma (k')}{2}$ for all $k,k'\in K$. Then the assignments $k\mapsto \iota (k):={{\overline{k}}}+\frac{\gamma ( k)}{2}$ define an injective ring homomorphism $\iota :K\hookrightarrow J\rtimes _\gamma K$. Suppose additionally that $j\gamma (k)=2j\gamma ({{\overline{k}}})+\gamma ({{\overline{k}}})j$ for all $j\in J$, $k\in K$. Then $J\rtimes _\gamma K=J\rtimes _\mathbf{0} \iota (K)$.

Proof

Define a map ${\triangleleft }:J\times K\rightarrow J$ by

$$\begin{aligned} j{\triangleleft }k:={{\overline{k}}}{\triangleright }j-j\gamma (k)\ . \end{aligned}$$

Lemma A.18

${\triangleleft }$ is an action of K on J commuting with ${\triangleright }$ and $(jj'){\triangleleft }k=j(j'{\triangleleft }k)$ for all $j,j'\in J$, $k\in K$, (i.e., J is a right K-algebra).

Proof

First, show that ${\triangleleft }$ commute with ${\triangleright }$. Indeed,

$$\begin{aligned} (k{\triangleright }j){\triangleleft }k'=\overline{k'}{\triangleright }(k{\triangleright }j) - (k{\triangleright }j)\gamma (k')=k {\triangleright }(\overline{k'}{\triangleright }j) -k{\triangleright }j\gamma (k') =k{\triangleright }(j{\triangleleft }k') \end{aligned}$$

for all $j\in J$, $k,k'\in K$ because J is a left K-algebra.

Furthermore, $(j{\triangleleft }k){\triangleleft }k'=\overline{k'}{\triangleright }(j{\triangleleft }k)-(j{\triangleleft }k)\gamma (k')=\overline{kk'} {\triangleright }j-\overline{k'}{\triangleright }j\gamma (k)-(j{\triangleleft }k)\gamma (k')$

$$\begin{aligned} \overline{kk'} {\triangleright }j-(j{\triangleleft }k')\gamma (k)-j\gamma (k)\gamma (k')-(j{\triangleleft }k)\gamma (k') =\overline{kk'}{\triangleright }j-j\gamma (k'k)=j{\triangleleft }(kk') \end{aligned}$$

for all $k,k'\in K$, $j\in J$ by the commutation of ${\triangleleft }$ with ${\triangleright }$ and the first condition of (1.3).

Since $J\gamma (1)=\{0\}$ by the first condition (1.3), i.e., ${\triangleleft }1=Id_J$, this proves that J is a K-module under ${\triangleleft }$.

Finally,

$$\begin{aligned} j(j'{\triangleleft }k)= & {} j({{\overline{k}}}{\triangleright }j'-j'\gamma (k))=k{\triangleright }j j'-j\gamma ({{\overline{k}}})j'\\&-jj'\gamma (k)={{\overline{k}}}{\triangleright }jj'-jj'\gamma (k)=(jj'){\triangleleft }k \end{aligned}$$

for all $k,k'\in K$, $j\in J$ by the second and third conditions of (1.3). This proves that J is a right K-algebra under ${\triangleleft }$.

The lemma is proved. $\square$

Note that the identity

$$\begin{aligned} (j{\triangleleft }k)j'=j(k{\triangleright }j') \end{aligned}$$

(1.6)

for all $k\in K$, $j,j'\in J$ is equivalent to the second condition (1.3).

The following is immediate and well-known.

Lemma A.19

Let R and S be associative ring, R is unitless, S is unital and R is an S-bimodule such that

$$\begin{aligned} s{\triangleright }(rr')=(s{\triangleright }r)r',~(rr'){\triangleleft }s=r(r'{\triangleleft }s),~(r{\triangleleft }s)r'=r(s{\triangleright }r') \end{aligned}$$

(1.7)

for all $r,r'\in R$, $s,s'\in S$. Then $A:=R\oplus S$ is a unital associative ring with the product given by

$$\begin{aligned} (r+s)(r'+s')=rr'+s{\triangleright }r'+r{\triangleleft }s'+ss' \end{aligned}$$

for all $r,r'\in R$, $S,s'\in S$.

Thus, (1.6) and Lemma A.18 guarantee that all assumptions of Lemma A.19 hold for $R=J$, $S=K$, therefore, $J\oplus K$ is a unital associative ring. This proves (a).

Part (b) is obvious.

Prove (c). Indeed, $\iota (k)\iota (k')=({{\overline{k}}}+\frac{\gamma (k)}{2})(\overline{k'}+\frac{\gamma (k')}{2})=\overline{kk'}+\frac{{{\overline{k}}}{\triangleright }\gamma (k')+\gamma (k){\triangleleft }\overline{k'}}{2}+\frac{\gamma (k)\gamma (k')}{4}$

$$\begin{aligned} =\overline{kk'}+\frac{{{\overline{k}}}{\triangleright }\gamma (k')+k'{\triangleright }\gamma (k)-\gamma (k)\gamma (\overline{k'})}{2}+\frac{\gamma (k)\gamma (k')}{4} =\overline{kk'}+\frac{\gamma (k'k)}{2}=\iota (kk') \end{aligned}$$

for all $k,k'\in K$ by the first relation (1.3). Therefore, $\iota$ is a homomorphism of rings. Its injectivity follows because $\iota$ splits the canonical homomorphism from (b). Finally,

$$\begin{aligned} j\iota (k)=j({{\overline{k}}}+\frac{\gamma (k)}{2})=kj-j\gamma ({{\overline{k}}})+\frac{j\gamma (k)}{2}=k j+\frac{\gamma ({{\overline{k}}})}{2}j=\iota ({{\overline{k}}})j \end{aligned}$$

for all $j\in J$, $k\in K$. This proves that J is a $(\iota (K),{\widetilde{\cdot }},\mathbf{0})$-algebra with the trivial $\gamma =\mathbf{0}$ and:

$\bullet$ The involution ${\widetilde{\cdot }}$ defined by $\widetilde{\iota (k)}=\iota ({{\overline{k}}})$ for all $k\in K$.

$\bullet$ The left action of $\iota (k)$ on J by the left multiplication in $J\rtimes _\gamma K$.

In particular, $J\rtimes _\gamma K=J\rtimes _\mathbf{0} \iota (K)$. Part (c) is proved.

The proposition is proved. $\square$

Given a commutative ring $(K,{\overline{\cdot }})$ with anti-involution, we say that a left K-algebra J is a $(K,{\overline{\cdot }})$-algebra if

$$\begin{aligned} {{\overline{k}}}{\triangleright }jj'=k{\triangleright }jj'=j(k{\triangleright }j') \end{aligned}$$

for all $j,j'\in J$, $k\in K$. Clearly, $(K,{\overline{\cdot }})$-algebras are same as $(K,{\overline{\cdot }},\mathbf{0})$-algebras. Also, in view of (1.4), any nilpotent pre-Hermitian $(K,{\overline{\cdot }},\gamma )$-algebra is automatically a $(K,{\overline{\cdot }})$-algebra.

Proposition A.20

Let $(K,{\overline{\cdot }})$ be any commutative ring with anti-involution and let J be any nilpotent pre-Hermitian ring. Suppose that J is a $(K,{\overline{\cdot }})$-algebra and let $\gamma :K\rightarrow J$ be any homomorphism of abelian groups such that: $\gamma (kk')={{\overline{k}}}{\triangleright }\gamma (k')+k'{\triangleright }\gamma (k)$ for all $k,k'\in k$. Then:

• (a) J is a $(K,{\overline{\cdot }},\gamma )$-algebra.

• (b) Suppose additionally that $\gamma ({{\overline{k}}})=\gamma (k)$ for all $k\in K$. Then the assignments $j+k\mapsto \gamma (k)-j+{{\overline{k}}}$ define an anti-involution $\sigma$ on $J\rtimes _\gamma K$. Moreover, if K is semisimple (i.e., is a direct sum of fields), then $(J\rtimes _\gamma K,\sigma )$ is pre-Hermitian.

Proof

Prove (a). Indeed, the conditions (1.4) hold automatically for this choice of $\gamma$ and because J is $(K,{\overline{\cdot }})$-algebra. This proves (a).

Prove (b). First, let us verify that $\sigma$ is an anti-involution. Indeed,

$$\begin{aligned} \sigma (\sigma (j+k))\!=\!\sigma (\gamma (k)-j+{{\overline{k}}})\!=\!j-\gamma (k)+\sigma ({{\overline{k}}})\!=\!j-\gamma (k)+k+\gamma ({{\overline{k}}})=j+k \end{aligned}$$

for all $j\in J$, $k\in K$. That is, $\sigma ^2=1$.

Furthermore, by definition (1.5), $j k={{\overline{k}}}j-j\gamma (k)={{\overline{k}}}j+\gamma (k)j=k\sigma (k)j$ hence $kj=j\sigma (k)$ for all $j\in J$, $k\in K$. Then

$$\begin{aligned} \sigma (kk')= & {} \overline{kk'}+\gamma (kk')=\overline{kk'}+{{\overline{k}}}\gamma (k')+k'\gamma (k)=\overline{kk'}+{{\overline{k}}}(\sigma (k')-{{\overline{k}}}')+k'\gamma (k) \\= & {} {{\overline{k}}}\sigma (k')+\gamma (k)\sigma (k')=\sigma (k)\sigma (k') \end{aligned}$$

for all $k,k'\in K$. Clearly,

$$\begin{aligned} \sigma (jj')=-jj'=j'j=\sigma (j')\sigma (j) \end{aligned}$$

for all $j,j'\in J$. Also,

$$\begin{aligned} \sigma (kj)= & {} \sigma (k {\triangleright }j)=-k {\triangleright }j=-kj=-j\sigma (k)=\sigma (j)\sigma (k) \\ \sigma (jk)= & {} \sigma (\sigma (k)j)=\sigma (j)\sigma (\sigma (k)=-jk=-\sigma (k)j=\sigma (k)\sigma (j) \end{aligned}$$

for all $j\in J$, $k\in K$.

This proves the first assertion. To prove the second assertion, note that semisimplicity of K and Proposition A.17(b) imply that Jacobson radical of $J\rtimes _\gamma K$ is J. This finishes proof of (b).

The proposition is proved. $\square$

The following is an immediate corollary of Proposition A.20.

Corollary A.21

Let $(A,\sigma )$ be any pre-Hermitian ring and K be any commutative unital subring of A such that $K\cap J(A)=\{0\}$ and $\sigma (K)\subset K+J(A)$. Then

• (a) J(A) is both a $(K,{\overline{\cdot }})$-algebra and a $(K,{\overline{\cdot }},\gamma )$-algebra, where:

  • J(A) is a K-algebra via left multiplication.

  • ${\overline{\cdot }}:K\rightarrow K$ and $\gamma :K\rightarrow J(A)$ are determined by $\sigma (k)=\gamma (k)+{{\overline{k}}}$ for all $k\in K$.

• (b) The subring of A generated by K and J(A) is naturally isomorphic to $J(A)\rtimes _\gamma K$.

Recall that F is a perfect field if every irreducible polynomial over F has distinct roots. In particular, all fields of characteristic zero and all finite fields are perfect.

Theorem A.22

(Wedderburn-Mal’cev theorem (see [20], Exercise 18, p. 191)) Let R be a finite dimensional algebra over a perfect field F. Then

• (a) There is a splitting $\iota$ of the short exact sequence $J(R)\rightarrow R\rightarrow S:=R/J(R)$, e.g., $R=\iota (S)\oplus J(R)$.

• (b) The images of all splittings $\iota :S\hookrightarrow R$ are conjugate in R by $1+J(R)$.

Remark A.23

In fact, the multiplication in R in Theorem A.22 is as in Lemma A.19 since J(R) is naturally a bimodule over $S\iota (S)$.

The following is immediate.

Lemma A.24

In the assumptions of Theorem A.22 suppose that $\sigma (\iota (S))=\iota (S)$ for an anti-involution $\sigma$ of R and a splitting $\iota :S\hookrightarrow R$. Then $R^\sigma =\iota (S)^\sigma \oplus J(R)^\sigma$.

The following is well-known (cf. [20], Theorem 25C.17).

Theorem A.25

In the assumptions of Theorem A.22, there exists a faithful n-dimensional representation $\rho$ of R into the algebra of $\mathbf{n}$-block upper triangular matrices (for some partition $\mathbf{n}$ of n) such that $\rho (J(R))$ is in the block-strictly upper triangular part of $Mat (\mathbf{n}, F)$ and the image under $\rho$ of at least one splitting $\iota :S\hookrightarrow R$ is in the block-diagonal part of $Mat (\mathbf{n}, F)$.

We will use Theorem A.22 to finish the classification of finite-dimensional pre-Hermitian algebras over perfect fields as follows.

Theorem A.26

Let $(A,\sigma )$ be a finite-dimensional pre-Hermitian algebra over a perfect field F, denote by K the maximal abelian ideal of the semisimple quotient $S=A/J(A)$ and by B its complement so that $S=B\oplus K$. Then there is a unique copy of B in A splitting the canonical homomorphism $\pi :A\twoheadrightarrow S$, that is, $A= B\oplus {\tilde{K}}$, $\sigma (B)=B$, $\sigma ({\tilde{K}})={\tilde{K}}$, where ${\tilde{K}}=\pi ^{-1}(K)$. More precisely, ${\tilde{K}}=J(A)$ if $K= \{0\}$ and ${\tilde{K}}\cong J(A)\rtimes _\gamma K$ otherwise (in the notation of Corollary A.21).

Proof

Indeed, $B^\circ =B$ because each simple component C of B satisfies $C^\circ =C$ since $[C,C]\ne 0$. Furthermore, in the notation of Theorem A.22(a), fix a splitting $\iota :S\hookrightarrow A$ of homomorphism $\pi :A\twoheadrightarrow S$. Then $\iota (B)=\iota (B^\circ )=\iota (B)^\circ \subset A^\circ$ hence $\iota (B)J(A)=J(A)\iota (B)=\{0\}$ by Proposition A.13. To prove the first assertion note that, by Theorem A.22(b), the images of B under any splitting $S\hookrightarrow A$ are conjugate to $\iota (B)$ by $1+ J(A)$. Since $(1+J(A))\iota (b)=\iota (b)(1+J(A))=\iota (b)$ for all $b\in B$, we see that $\iota (B)$ is a unique copy of B in A. In particular, $\sigma (\iota (B))=\iota (B)$.

Furthermore, by definition, $\sigma ({\tilde{K}})={\tilde{K}}$ and ${\tilde{K}}=\iota (K)+J(A)$ as a vector space over F. Since $\iota (B) \iota (K)= \iota (K)\iota (B)=\{0\}$, we see that $\iota (B){\tilde{K}}={\tilde{K}} \iota (B)=\{0\}$. Therefore, $A=B\oplus {\tilde{K}}$, as an algebra.

If $K=\{0\}$, then, clearly, ${\tilde{K}}=J(A)$. Suppose that $K\ne \{0\}$. Since $\sigma (\iota (K))\subset {\tilde{K}}=\iota (K)+J(A)$, we see that $\iota (K)$ satisfies the hypotheses of Corollary A.21. Therefore, ${\tilde{K}}=J(A)\rtimes _\gamma \iota (K)$.

The theorem is proved. $\square$

The following is immediate.

Corollary A.27

In the assumptions of Theorem A.26, one has

• (a) If $K=\{0\}$ then A is Hermitian if and only if B is Hermitian. Otherwise, A is Hermitian if and only if both B and ${\tilde{K}}$ are Hermitian.

• (b) If A is Hermitian, then

  $A^\sigma _{\ge 0}={\left\{ \begin{array}{ll} B^\sigma _{\ge 0} &{} \text {if }\tilde{K}=\{0\}\\ B^\sigma _{\ge 0}\oplus {\tilde{K}}^\sigma _{\ge 0} &{}\text {otherwise} \end{array}\right. }$ and $A^\sigma _{+}={\left\{ \begin{array}{ll} B^\sigma _{+} &{} \text {if }\tilde{K}=\{0\}\\ B^\sigma _{+}\oplus K^\sigma _+ &{}\text {otherwise} \end{array}\right. }$.

We discuss now anti-involutions on simple rings.

Clearly, if $(A,\sigma )$ is a Hermitian algebra over a field F, then so is F, i.e., F is a formally real field. In particular, if $A^\sigma =F$, then $(A,\sigma )$ is Hermitian if and only if F is a formally real.

The following is an immediate consequence of the Skolem-Noether theorem.

For any ring $(A,\tau )$ with an anti-involution $\tau$ denote by $A^{[\tau ]}$ the set of all $v\in A^\times$ such that $\tau (v)=v z$, for some $z\in Z(A)^\times$ such that $\tau (z)=z^{-1}$ (In particular, if each element of the center Z(A) is fixed under $\tau$, then $\tau (v)\in \{v,-v\}$, i.e., ). Clearly, the assignments $(w,v)\mapsto w{\triangleright }v:=w v\tau (w)$ define an action of the multiplicative group $A^\times$ on $A^{[\tau ]}$.

Lemma A.28

Let $(A,\tau )$ be a simple Artinian ring. Then

• (a) For any anti-involution $\sigma :A\rightarrow A$ there is an element $v\in A^{[\tau ]}$ (unique up to multiplication by elements of the field Z(A)) such that $\sigma (a)=v\tau (a)v^{-1}$ for all $a\in A$.

• (b) For any $v\in A^{[\tau ]}$ the assignments $a\mapsto v\tau (a)v^{-1}$ define an involution $\sigma _v$ on A.

• (c) For any $w\in A^\times$ the assignments $a\mapsto waw^{-1}$ define an isomorphism of rings with anti-involution $(A,\sigma _v){\widetilde{\rightarrow }} (A,\sigma _{w{\triangleright }v})$.

Definition A.29

We say that an algebra $(A,\sigma )$ over F is thin if $A^\sigma =F$.

The first example of a thin algebra is given by Example A.12. Another one is a (generalized) quaternion algebra (see Remark A.33 below).

Clearly every thin algebra $(A,\sigma )$ over F is pre-Hermitian and the direct sum of two F-algebras is never thin because $(A\oplus A')^{\sigma \oplus \sigma '}=A^\sigma \oplus {A'}^{\sigma '}$. Moreover, the following is an immediate consequence of Theorem A.26.

Lemma A.30

If a finite-dimensional algebra $(A,\sigma )$ over a perfect field F with $char(F)\ne 2$ is thin then it is either simple noncommutative, or direct sum of two copes of a simple algebra interchanged by $\sigma$, or pre-Hermitian whose semisimple quotient A/J(A) is a field extension of F.

Proposition A.31

Let F be a perfect field with $char(F)\ne 2$ and $(A,\sigma )$ be a thin semisimple finite-dimensional algebra over F. Then:

• (a) $A=F\oplus A^{-\sigma }$ where $A^{-\sigma }=\{a\in A:\sigma (a)=-a\}$ is the space of skew-symmetric elements.

• (b) $A^{-\sigma }$ admits a unique nonsingular symmetric bilinear form $\beta$ such that $aa'+a'a=-\beta (a,a')$ for all $a\in A^{-\sigma }$.

• (c) If a is a nilpotent element in A then $a\in A^{-\sigma }$ and $a^2=0$.

• (d) $A^{-\sigma }$ is a Lie algebra with respect to the commutator bracket $[a,a']=aa'-a'a=2aa'+\beta (a,a')$

• (e) $2a''a'a-2aa'a''=\beta ([a,a'],a'')=\beta (a,[a',a''])$ and $\beta ([a,a'],[a,a'])=\beta (a,a)\beta (a',a')-\beta (a,a')^2$ for $a,a',a''\in A^{-\sigma }$.

Proof

Part (a) is obvious because $A^\sigma =F$. To prove (b) note that ${\hat{\beta }}(a,a'):=-aa'-a'a$ is fixed by $\sigma$ and thus belongs to F for all $a,a'\in A$. This is obviously a symmetric bilinear form on A, denote by $\beta$ its restriction to $A^{-\sigma }$. If z is in the radical of $\beta$, then $za+az=0$ for all $a\in A^{-\sigma }$, which implies that $z=0$ by Lemma  A.30. This proves (b).

Prove (c). Let $a\in A$ be a nilpotent and suppose that $a\notin A^{-\sigma }$. Then there exists $c\in F^\times$ such that the nilpotent element $b:=ca$ satisfies $\sigma (b)=1-b$. Therefore, $(1-b)^n=0$ for some $n\ge 2$. This contradicts to the non-invertibility of b. Thus, $\sigma (a)=-a$ and $a^2$ is a nilpotent in $A^\sigma =F$, that is, $a^2=0$.

Part (d) is obvious because $\sigma ([a,a'])=-[a,a']$ for all $a,a'\in A^\sigma$.

Prove (e). Indeed,

$$\begin{aligned} \beta ([a,a'],a'')= & {} -[a,a']a''-a''[a,a']=-(2aa'+\beta (a,a'))a''\\&-a''(-2a'a-\beta (a',a)) \\= & {} 2a''a'a-2aa'a''=-\beta ([a'',a'],a) \end{aligned}$$

for all $a,a',a''\in A^\sigma$. This proves the first assertion. To prove the second one, compute

$$\begin{aligned} \beta ([a,a'],[a,a'])= & {} 2(aa'-a'a)a'a+2aa'(a'a-aa') \\= & {} -2(aa'+a'a)a'a-2aa'(a'a+aa')+4a{a'}^2a+4a'a^2a' \\= & {} 2\beta (a,a')(aa'+a'a)-2a^2\beta (a',a')\\&-2{a'}^2\beta (a,a)=\beta (a,a)\beta (a',a') -\beta (a,a')^2 \end{aligned}$$

for all $a,a'\in A^{-\sigma }$. This proves (d).

The proposition is proved. $\square$

Theorem A.32

Let F be a perfect field with $char(F)\ne 2$ and $(A,\sigma )$ be a thin semisimple finite-dimensional algebra over F. Then A is either a division algebra over F with $\dim _F A\in \{1,2,4\}$ or $A=F\oplus F$ (so that $\sigma$ is the permutation of summands).

Proof

Indeed, any simple component of A is ${{\,\mathrm{Mat}\,}}_n(D)$, where D is a division algebra over F. Note that if $n\ge 3$, then A contains an element $e=e_{12}+e_{23}$ which contradicts to Proposition A.31(c). Thus, $n\le 2$ and $A\in \{D,D\oplus D,{{\,\mathrm{Mat}\,}}_2(D),{{\,\mathrm{Mat}\,}}_2(D)\oplus {{\,\mathrm{Mat}\,}}_2(D)\}$ by Lemma A.30. Note, however, that the last two algebras are not thin. If $A=D\oplus D$, then $(x,\sigma (x))\in A^\sigma$ for all $x\in D$, i.e. we always have a copy of D in $A^\sigma =F$, that is, $D=F$.

The theorem is proved. $\square$

For any ring F and $\alpha ,\beta \in F$ denote by ${\mathbb {H}}_{\alpha ,\beta }$ the F-algebra with a presentation:

$$\begin{aligned} {\mathbb {H}}_{\alpha , \beta }=\langle i,j\ |\ i^2=\alpha ,~j^2=\beta ,~ji=-ij\rangle \ . \end{aligned}$$

Clearly, it admits a unique anti-involution ${\overline{\cdot }}$ such that ${{\overline{i}}}=-i$, ${{\overline{j}}}=-j$. By construction, $a{{\overline{a}}}={{\overline{a}}} a\in F$ for all $a\in {\mathbb {H}}_{\alpha , \beta }$. In particular, ${\mathbb {H}}_{\alpha , \beta }$ is a division algebra if and only if F is a field and $a {\overline{a}} \in F^\times$ for all nonzero $a\in {\mathbb {H}}_{\alpha , \beta }$.

Remark A.33

Any 4-dimensional division algebra D over a field F with ${{\,\mathrm{char}\,}}F\ne 2$ is isomorphic to ${\mathbb {H}}_{\alpha ,\beta }$ for some non-squares $\alpha ,\beta \in F^\times$. Clearly, it is thin with the above anti-involution ${\overline{\cdot }}$. Note that if both $-\alpha$ and $-\beta$ are complete squares in F, then ${\mathbb {H}}_{\alpha ,\beta } \cong {\mathbb {H}}_{-1,-1}$ is the ordinary algebra of quaternions.

Remark A.34

Suppose that ${{\,\mathrm{char}\,}}F\ne 2$ and the algebraic closure ${{\overline{F}}}$ of F is a quadratic extension of F. Then any division algebra over F is isomorphic to either ${{\overline{F}}}$ or to some ${\mathbb {H}}_{\alpha ,\beta }$.

Generalizing the above observations, we construct some twisted group algebras of abelian groups with anti-involutions. Recall that, given a group G and its linear action $\triangleright$ on a commutative ring K, a map $\chi :G\times G\rightarrow K$ is called a 2-cocycle on G (in K) if

$$\begin{aligned} (g\triangleright \chi (g',g''))\chi (g,g'g'')=\chi (g,g')\chi (gg',g'') \end{aligned}$$

for all $g,g',g''\in G$.

Furthermore, denote by $K_\chi G$ the $\chi$-twisted group algebra of G, i.e., a K-algebra with a free K-basis $\{[g], g\in G\}$ and the multiplication table:

$$\begin{aligned}{}[g][g']=\chi (g,g')\cdot [gg'], [g]k=g\triangleright k\cdot [g] \end{aligned}$$

for all $g,g'\in G$, $k\in K$. It is well-known that any central simple (e.g., a division) algebra A over a field $F=K^G$ is isomorphic to some $K_\chi G$ so that K is a Galois field extension of F.

The following is immediate.

Lemma A.35

Let ${\overline{\cdot }}$ be an involution on K. Then the following statements are equivalent for a 2-cocycle $\chi :G\times G\rightarrow K$:

• (a) The assignments $k\cdot [g]\mapsto [\sigma (g)]\cdot {{\overline{k}}}$ for $g\in G$, $k\in K$ define an anti-involution on $K_\chi G$.

• (b) $\sigma (g)\triangleright {{\overline{k}}}=\overline{g^{-1}{\triangleright }k}$ and $\overline{\chi (g,g')}=\overline{gg'}\triangleright \chi (\sigma (g'),\sigma (g))$ for all $g,g'\in G$, $k\in K$.

In particular, if the G-action $\triangleright$ is trivial, the cocycle conditions simplifies:

$$\begin{aligned} \chi (gg',g'')\chi (g,g')=\chi (g,g'g'')\chi (g',g'') \end{aligned}$$

for all $g,g',g''\in G$ (e.g., $\chi$ is a bicharacter of G).

The following is immediate.

Lemma A.36

Let G be an abelian group, trivially acting on K, ${\overline{\cdot }}$ be an anti-involution on K and $\varepsilon :G\rightarrow K^\times$, $g\mapsto \varepsilon _g$ be a map. Then the following are equivalent:

• (a) The assignments $k\cdot [g]\mapsto {{\overline{k}}}\varepsilon _g\cdot [g]$ for $g\in G$, $k\in K$ define an anti-involution $\sigma _\varepsilon$ on $K_\chi G$.

• (b) $\overline{\varepsilon _g}=\varepsilon _g^{-1}$ and $\chi (g',g)=\overline{\chi (g,g')}\varepsilon _{gg'}\varepsilon _g^{-1}\varepsilon _{g'}^{-1}$ for all $g,g'\in G$.

In particular, $(K_\chi G)^{\sigma _\varepsilon }=\bigoplus \limits _{g\in G}K_g\cdot [g]$, where $K_g=\{k\in K:{{\overline{k}}}=k\cdot \varepsilon _g\}$ for all $g\in G$.

Now we classify anti-involutions on 4-dimensional division algebras.

Proposition A.37

In the notation as above, suppose that ${\mathbb {H}}_{\alpha ,\beta }$ is a noncommutative division algebra (over a field F). Then any anti-involution on ${\mathbb {H}}_{\alpha ,\beta }$ is either ${\overline{\cdot }}$ or is given by

$$\begin{aligned} \sigma (a)= v{{\overline{x}}}v^{-1} \end{aligned}$$

for some nonzero imaginary v, i.e., such that ${{\overline{v}}}=-v$ (in what follows, we denote the latter anti-involution by $\sigma _v$).

Proof

By Skolem-Noether theorem, any anti-involution $\sigma$ on ${\mathbb {H}}_{\alpha ,\beta }$ can be written as $\sigma (x)=v{{\overline{x}}}v^{-1}$ where $v\in {\mathbb {H}}_{\alpha ,\beta }^{\times }$. Then $\sigma ^2(a)=uau^{-1}$ where $u=v{{\overline{v}}}^{-1}$. Since $\sigma ^2=1$, we have $a=uau^{-1}$ for all $a\in$, i.e. u is central in ${\mathbb {H}}_{\alpha ,\beta }$. That is $u=c\in F^\times$. This implies that ${{\overline{v}}}=c^{-1} v$ hence either $c=1$, $v\in F^\times$ or $c=-1$, ${{\overline{v}}}=-v$. $\square$

Proposition A.38

In the assumptions of Proposition A.37, for any nonzero imaginary $v\in {\mathbb {H}}_{\alpha ,\beta }$ there exist imaginary $w,w'$ such that $[w,w']=v$ and $\{1,w,w',v\}$ is a basis of ${\mathbb {H}}_{\alpha ,\beta }$. In particular, $vw=-wv$, $vw'=-w'v$, $ww'+w'w\in F$, $({\mathbb {H}}_{\alpha ,\beta })^{\sigma _v}=F+Fw+Fw'$ and $({\mathbb {H}}_{\alpha ,\beta })^{-\sigma _v}=F\cdot v$.

Proof

Denote by V the set of all imaginary elements of ${\mathbb {H}}_{\alpha ,\beta }$. For an imaginary v define an F-linear map $f_v:V\rightarrow {\mathbb {H}}_{\alpha ,\beta }$ by $f_v(a):= va+av$. Clearly, the range of $f_v$ is F, in particular, $Ker~f_v$ is a 2-dimensional subspace of the (3-dimensional) space V. Then

$$\begin{aligned} \sigma _v(a)=v{{\overline{a}}}v^{-1}=-vav^{-1}=avv^{-1}=a \end{aligned}$$

for any $a\in Ker~f_v$. That is, $F+Ker~f_v\subset ({\mathbb {H}}_{\alpha ,\beta })^{\sigma _v}$. Since $v\in ({\mathbb {H}}_{\alpha ,\beta })^{-\sigma _v}$, we see that $F+Ker~f_v= ({\mathbb {H}}_{\alpha ,\beta })^{\sigma _v}$ and $({\mathbb {H}}_{\alpha ,\beta })^{-\sigma _v}=Fv$.

Therefore, $[w,w']\in F^\times v$ for any basis $\{w,w'\}$ of $Ker~f_v$.

The proposition is proved. $\square$

Note that if ${\mathbb {H}}={\mathbb {H}}_{-1,-1}$ is the ordinary quaternion algebra over ${\mathbb {R}}$, then $w^2=-w{{\overline{w}}}<0$ for any imaginary $w\in {\mathbb {H}}$, hence $(\sqrt{-w^2})^2+w^2=0$ and we obtain the following immediate corollary of Proposition A.38.

Corollary A.39

The algebra $({\mathbb {H}}, \sigma )$ is Hermitian if and only if $\sigma ={\overline{\cdot }}$.

We conclude the section with recalling the Cayley-Dickson construction of (generalized) octonions ${{\mathbb {O}}}_{\alpha ,\beta }:={\mathbb {H}}_{\alpha ,\beta }\oplus {\mathbb {H}}_{\alpha ,\beta }$, as a free module over a ring F with the following multiplication table

$$\begin{aligned} (a,b)(a',b')=(aa'-\overline{b'}b,b'a+b\,\overline{a'})\ . \end{aligned}$$

This is a non-associative F-algebra with the anti-involution given by

$$\begin{aligned} \overline{(a,b)}=({\overline{a}},-b) \end{aligned}$$

so that $\overline{(a,b)}(a,b)=(a,b)\overline{(a,b)}=(a{{\overline{a}}}+\overline{b}b,0)$ for all $a,b\in {{\mathbb {O}}}_{\alpha ,\beta }$. In particular, if ${\mathbb {H}}_{\alpha ,\beta }$ is a division algebra, then ${{\mathbb {O}}}_{\alpha ,\beta }$ is a non-associative division algebra as well.

Appendix B: Isomorphisms and embeddings of matrix algebras

Some tensor products and direct products of matrix algebras are related in a way we want to discuss in this section. The described in here isomorphisms and embeddings are used in the main part of the paper as a tool to construct symmetric spaces and study their properties.

As usual, for every algebra A and (anti-)involution $\sigma$, we denote by $A^\sigma$ the set of fixed points of $\sigma$ in A.

We also remind our standard notation of anti-involutions on matrix algebras.

• (1) If A is ${{\,\mathrm{Mat}\,}}(n,{\mathbb {R}})$ or ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}})$, then the anti-involution given by the transposition of the matrix is denoted by $\sigma$.

• (2) If A is ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}})$, then by ${\bar{\sigma }}$ is denoted the anti-involution given by $\sigma$ composed with the complex conjugation.

• (3) If A is ${{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})$, then the anti-involution $\sigma _0:A\rightarrow A$ is given by the rule $\sigma _0(r_1+r_2j)=\sigma (r_1)+{\bar{\sigma }}(r_2)j$ where $r_1,r_2\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$. Another anti-involution $\sigma _1:A\rightarrow A$ is given by the rule $\sigma _1(r_1+r_2j)={\bar{\sigma }}(r_1)-\sigma (r_2)j$ where $r_1,r_2\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$.

If $n=1$, we identify the algebra of $1\times 1$ matrices with the corresponding (skew-)field and use the same notation for the (anti-)involution in a (skew-)field as in the matrix algebra over this (skew-)field, in particular, in this case $\sigma$ is the identity map on ${\mathbb {R}}$ or ${\mathbb {C}}$, ${\bar{\sigma }}$ is the complex conjugation on ${\mathbb {C}}$ and $\sigma _1$ is the quaternionic conjugation on ${\mathbb {H}}$.

1.1 Three isomorphisms of matrix algebras

In this section, we describe three well-known matrix algebras isomorphisms.

1.1.1 ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}})\otimes _{\mathbb {R}}{\mathbb {C}}$ and ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}})$

Fact B:1

The following map is an isomorphism of ${\mathbb {C}}\{i\}$-algebras:

$$\begin{aligned} \begin{matrix} \chi :&{} {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes _{\mathbb {R}}{\mathbb {C}}\{i\} &{} \rightarrow &{} {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\\ &{} a+bI &{} \mapsto &{} (a+bi , a-bi) \end{matrix} \end{aligned}$$

where $a,b\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$. In particular,

$$\begin{aligned} \chi ({{\,\mathrm{Id}\,}}I\otimes 1)=(i,-i),\;\chi ({{\,\mathrm{Id}\,}}\otimes i)=(i,i). \end{aligned}$$

The anti-involution induced by $\sigma \otimes {{\,\mathrm{Id}\,}}$

$$\begin{aligned} \chi \circ (\sigma \otimes {{\,\mathrm{Id}\,}})\circ \chi ^{-1} \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$ acts in the following way:

$$\begin{aligned} (m_1,m_2)\mapsto (m^T_1,m^T_2). \end{aligned}$$

The anti-involution induced by ${\bar{\sigma }}\otimes {{\,\mathrm{Id}\,}}$

$$\begin{aligned} \chi \circ ({\bar{\sigma }}\otimes {{\,\mathrm{Id}\,}})\circ \chi ^{-1} \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$ acts in the following way:

$$\begin{aligned} (m_1,m_2)\mapsto (m^T_2,m^T_1). \end{aligned}$$

The involution induced by ${{\,\mathrm{Id}\,}}\otimes {\bar{\sigma }}$

$$\begin{aligned} \chi \circ ({{\,\mathrm{Id}\,}}\otimes {\bar{\sigma }})\circ \chi ^{-1} \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$ acts in the following way:

$$\begin{aligned} (m_1,m_2)\mapsto ({\bar{m}}_2,{\bar{m}}_1). \end{aligned}$$

Therefore:

$$\begin{aligned} \chi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes _{\mathbb {R}}{\mathbb {C}}\{i\})^{{\bar{\sigma }}\otimes {{\,\mathrm{Id}\,}}})= & {} \{(m,m^T)\mid m\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\}, \\ \chi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes _{\mathbb {R}}{\mathbb {C}}\{i\})^{{\bar{\sigma }}\otimes {\bar{\sigma }}})= & {} {{\,\mathrm{Herm}\,}}(n,{\mathbb {C}}\{i\})\times {{\,\mathrm{Herm}\,}}(n,{\mathbb {C}}\{i\}),\\ \chi ({{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\}))= & {} \chi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes _{\mathbb {R}}{\mathbb {C}}\{i\})^{{{\,\mathrm{Id}\,}}\otimes {\bar{\sigma }}})\\= & {} \{(m,{\bar{m}})\mid m\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\}. \end{aligned}$$

1.1.2 ${{\,\mathrm{Mat}\,}}(n,{\mathbb {H}})\otimes _{\mathbb {R}}{\mathbb {C}}$ and ${{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}})$

Fact B:2

The following map is an isomorphism of algebras:

$$\begin{aligned} \begin{matrix} \psi :&{} {{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})\otimes _{\mathbb {R}}{\mathbb {C}}\{I\} &{} \rightarrow &{} {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{i\})\\ &{} (q_1+q_2j) + (p_1+p_2j)I &{} \mapsto &{} \begin{pmatrix} q_1+p_1i &{} q_2+p_2i\\ -{\bar{q}}_2-{\bar{p}}_2i &{} {\bar{q}}_1+{\bar{p}}_1i \end{pmatrix}. \end{matrix} \end{aligned}$$

where $q_1,q_2,p_1,p_2\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$. This is a ${\mathbb {C}}\{I\}$-${\mathbb {C}}\{i\}$-isomorphism, i.e. $\psi (xI)=\psi (x)i$ for $x\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})\otimes _{\mathbb {R}}{\mathbb {C}}\{I\}$. In particular,

$$\begin{aligned} \chi ({{\,\mathrm{Id}\,}}i\otimes 1)= & {} \begin{pmatrix} {{\,\mathrm{Id}\,}}i &{} 0\\ 0 &{} -{{\,\mathrm{Id}\,}}i \end{pmatrix},\;\chi ({{\,\mathrm{Id}\,}}\otimes j)=\begin{pmatrix} 0 &{} {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}}&{} 0 \end{pmatrix}, \\ \chi ({{\,\mathrm{Id}\,}}k\otimes 1)= & {} \begin{pmatrix} 0&{} {{\,\mathrm{Id}\,}}i\\ {{\,\mathrm{Id}\,}}i &{} 0 \end{pmatrix},\;\chi ({{\,\mathrm{Id}\,}}\otimes I)={{\,\mathrm{Id}\,}}i. \end{aligned}$$

The anti-involution induced by $\sigma _0\otimes {{\,\mathrm{Id}\,}}$

$$\begin{aligned} \psi \circ (\sigma _0\otimes {{\,\mathrm{Id}\,}})\circ \psi ^{-1} \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}})$ acts in the following way:

$$\begin{aligned} m\mapsto \begin{pmatrix} {{\,\mathrm{Id}\,}}&{} 0\\ 0 &{} -{{\,\mathrm{Id}\,}}\end{pmatrix}m^T\begin{pmatrix} {{\,\mathrm{Id}\,}}&{}\quad 0\\ 0 &{}\quad -{{\,\mathrm{Id}\,}}\end{pmatrix}. \end{aligned}$$

The anti-involution induced by $\sigma _1\otimes {{\,\mathrm{Id}\,}}$

$$\begin{aligned} \psi \circ (\sigma _1\otimes {{\,\mathrm{Id}\,}})\circ \psi ^{-1} \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}})$ acts in the following way:

$$\begin{aligned} m\mapsto -\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}m^T\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}=\begin{pmatrix} 0 &{}\quad i\\ -i &{}\quad 0 \end{pmatrix}m^T\begin{pmatrix} 0 &{}\quad i\\ -i &{}\quad 0 \end{pmatrix}. \end{aligned}$$

The involution induced by ${{\,\mathrm{Id}\,}}\otimes {\bar{\sigma }}$

$$\begin{aligned} \psi \circ ({{\,\mathrm{Id}\,}}\otimes {\bar{\sigma }})\circ \psi ^{-1} \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}})$ acts in the following way:

$$\begin{aligned} m\mapsto -\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}{\bar{m}}\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}=\begin{pmatrix} 0 &{}\quad i\\ -i &{}\quad 0 \end{pmatrix}{\bar{m}}\begin{pmatrix} 0 &{}\quad i\\ -i &{}\quad 0 \end{pmatrix}. \end{aligned}$$

The anti-involution induced by $\sigma _0\otimes {\bar{\sigma }}$

$$\begin{aligned} \psi \circ (\sigma _0\otimes {\bar{\sigma }})\circ \psi ^{-1} \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}})$ acts in the following way:

$$\begin{aligned} m\mapsto \begin{pmatrix} 0 &{} {{\,\mathrm{Id}\,}}\\ {{\,\mathrm{Id}\,}}&{} 0 \end{pmatrix}{\bar{m}}^T\begin{pmatrix} 0 &{} {{\,\mathrm{Id}\,}}\\ {{\,\mathrm{Id}\,}}&{} 0 \end{pmatrix}. \end{aligned}$$

Therefore:

$$\begin{aligned}&\psi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})\otimes _{\mathbb {R}}{\mathbb {C}}\{I\})^{\sigma _1\otimes {{\,\mathrm{Id}\,}}}) \\&\quad =\left\{ m\in {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{i\})\,\Big \vert \,m=-\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}m^T\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}\right\} \\&\quad ={{\,\mathrm{{\mathfrak {o}}}\,}}\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}={{\,\mathrm{{\mathfrak {sp}}}\,}}(2n,{\mathbb {C}}), \\&\psi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})\otimes _{\mathbb {R}}{\mathbb {C}}\{I\})^{\sigma _0\otimes {\bar{\sigma }}}) \\&\quad =\left\{ m\in {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{i\})\,\Big \vert \,m=\begin{pmatrix} 0 &{} {{\,\mathrm{Id}\,}}\\ {{\,\mathrm{Id}\,}}&{} 0 \end{pmatrix}{\bar{m}}^T\begin{pmatrix} 0 &{} {{\,\mathrm{Id}\,}}\\ {{\,\mathrm{Id}\,}}&{} 0\end{pmatrix}\right\} , \\&\psi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})\otimes _{\mathbb {R}}{\mathbb {C}}\{I\})^{\sigma _1 \otimes {\bar{\sigma }}})={{\,\mathrm{Herm}\,}}(2n,{\mathbb {C}}),\\&\psi ({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\}))=\psi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})\otimes _{\mathbb {R}}{\mathbb {C}}\{I\})^{{{\,\mathrm{Id}\,}}\otimes {\bar{\sigma }}}) \\&\quad =\left\{ m\in {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{i\})\,\Big \vert \,m=-\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}{\bar{m}}\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}\right\} \\&\quad =\left\{ \begin{pmatrix} q_1 &{} q_2\\ -{\bar{q}}_2 &{} {\bar{q}}_1 \end{pmatrix}\,\Big \vert \,q_1,q_2\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\right\} . \end{aligned}$$

1.1.3 ${{\,\mathrm{Mat}\,}}(n,{\mathbb {H}})\otimes _{\mathbb {R}}{\mathbb {H}}$ and ${{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}})$

Fact B:3

The following map:

$$\begin{aligned} \phi :{{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes _{\mathbb {R}}{\mathbb {H}}\{i,j,k\}\rightarrow {{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}}) \end{aligned}$$

defined on generators of $A_{\mathbb {H}}$ as follows:

$$\begin{aligned} \phi (a\otimes i)= & {} \begin{pmatrix} 0 &{} a &{} 0 &{} 0 \\ -a &{} 0 &{} 0 &{} 0 \\ 0 &{} \quad 0 &{}\quad 0 &{}\quad -a \\ 0 &{}\quad 0 &{} \quad a &{} \quad 0 \end{pmatrix},\; \phi (a\otimes j)=\begin{pmatrix} 0 &{}\quad 0 &{} \quad a &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad a \\ -a &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad -a &{} \quad 0 &{}\quad 0 \end{pmatrix}, \\ \phi (a\otimes k)= & {} \begin{pmatrix} 0 &{}\quad 0 &{} \quad 0 &{}\quad a \\ 0 &{}\quad 0 &{}\quad -a &{}\quad 0 \\ 0 &{}\quad a &{}\quad 0 &{}\quad 0 \\ -a &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix},\; \phi (aI\otimes 1)=\begin{pmatrix} 0 &{}\quad -a &{}\quad 0 &{}\quad 0 \\ a &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{} \quad 0 &{}\quad -a \\ 0 &{}\quad 0 &{}\quad a &{}\quad 0 \end{pmatrix}, \\ \phi (aJ\otimes 1)= & {} \begin{pmatrix} 0 &{}\quad 0 &{}\quad -a &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad a \\ a &{} \quad 0 &{} \quad 0 &{}\quad 0 \\ 0 &{}\quad -a &{}\quad 0 &{}\quad 0 \end{pmatrix},\; \phi (aK\otimes 1)=\begin{pmatrix} 0 &{} 0 &{} 0 &{} -a \\ 0 &{}\quad 0 &{}\quad -a &{}\quad 0 \\ 0 &{}\quad a &{}\quad 0 &{} \quad 0 \\ a &{} \quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix} \end{aligned}$$

where $a\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {R}})$ is an ${\mathbb {R}}$-algebra isomorphism.

The anti-involution $\sigma _1\otimes \sigma _0$ corresponds under $\phi$ to the following anti-involution

$$\begin{aligned} \phi \circ (\sigma _1\otimes \sigma _0)\circ \phi \end{aligned}$$

on ${{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}})$: $m\mapsto -\Xi m^T \Xi$ where

$$\begin{aligned} \Xi :=\begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 &{}\quad {{\,\mathrm{Id}\,}}_n \\ 0 &{}\quad 0 &{}\quad -{{\,\mathrm{Id}\,}}_n &{}\quad 0 \\ 0 &{}\quad {{\,\mathrm{Id}\,}}_n &{}\quad 0 &{}\quad 0 \\ -{{\,\mathrm{Id}\,}}_n &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}. \end{aligned}$$

The anti-involution $\sigma _1\otimes \sigma _1$ corresponds under $\phi$ to the transposition on ${{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}})$.

Therefore:

$$\begin{aligned}&\phi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes _{\mathbb {R}}{\mathbb {H}}\{i,j,k\})^{\sigma _1\otimes \sigma _0})\\&\quad = \left\{ m\in {{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}})\mid m=-\Xi m^T \Xi \right\} \\&\quad ={{\,\mathrm{{\mathfrak {o}}}\,}}(\Xi )\cong {{\,\mathrm{{\mathfrak {sp}}}\,}}(4n,{\mathbb {R}}), \\&\phi (({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{i,j,k\})\otimes _{\mathbb {R}}{\mathbb {H}}\{i,j,k\})^{\sigma _1\otimes \sigma _1})={{\,\mathrm{Sym}\,}}(4n,{\mathbb {R}}), \end{aligned}$$

The real locus ${{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})$ of ${{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes _{\mathbb {R}}{\mathbb {H}}\{i,j,k\}$ is mapped by $\phi$ to:

$$\begin{aligned}&\phi ({{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})) \\&\quad =\left\{ \begin{pmatrix} a &{} \quad -b &{}\quad -c &{}\quad -d \\ b &{}\quad a &{}\quad -d &{}\quad c \\ c &{}\quad d &{}\quad a &{}\quad -b \\ d &{}\quad -c &{}\quad b &{}\quad a \end{pmatrix}\,\Big \vert \,a,b,c,d\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {R}})\right\} . \end{aligned}$$

1.2 Embeddings between matrix algebras

In this section, we consider the following two embeddings:

$$\begin{aligned}&{{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes {\mathbb {C}}\{j\}\hookrightarrow {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes {\mathbb {H}}\{i,j,k\}, \\&{{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {C}}\{j\}\hookrightarrow {{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {H}}\{i,j,k\}. \end{aligned}$$

We use these embedding to see the symmetric space for a real group inside the symmetric space for a complexified group.

1.2.1 Embedding ${{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes {\mathbb {C}}\{j\}\hookrightarrow {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes {\mathbb {H}}\{i,j,k\}$

In the previous sections, we have seen the isomorphisms:

$$\begin{aligned} \begin{matrix} \chi :&{} {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes _{\mathbb {R}}{\mathbb {C}}\{j\} &{} \rightarrow &{} {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{j\})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{j\})\\ &{} a+bI &{} \mapsto &{} (a+bj , a-bj) \end{matrix} \end{aligned}$$

where $a,b\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{j\})$ and

$$\begin{aligned} \begin{matrix} \psi :&{} {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes _{\mathbb {R}}{\mathbb {H}}\{i,j,k\} &{} \rightarrow &{} {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{i\})\\ &{} (q_1+q_2j) + (p_1+p_2j)I &{} \mapsto &{} \begin{pmatrix} q_1+p_1i &{} q_2+p_2i\\ -{\bar{q}}_2-{\bar{p}}_2i &{} {\bar{q}}_1+{\bar{p}}_1i \end{pmatrix}. \end{matrix} \end{aligned}$$

where $q_1,q_2,p_1,p_2\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})$.

We want to describe the map

$$\begin{aligned} \psi \circ \iota \circ \chi ^{-1}:{{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{j\})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{j\})\rightarrow {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{i\}) \end{aligned}$$

where

$$\begin{aligned} \iota :{{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes {\mathbb {C}}\{j\}\hookrightarrow {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})\otimes {\mathbb {H}}\{i,j,k\} \end{aligned}$$

is the natural embedding. Let $(a,b):=(a_1+a_2j,b_1+b_2j)\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{j\})\times {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{j\})$ for $a_1,a_2,b_1,b_2\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {R}})$, then

$$\begin{aligned} \chi ^{-1}(a,b)=\frac{a+b}{2}+\frac{a-b}{2j}I=\frac{a_1+b_1+(a_2+b_2)j}{2}+\frac{a_2-b_2-(a_1-b_1)j}{2}I. \end{aligned}$$

Therefore,

$$\begin{aligned} \psi (\chi ^{-1}(a,b))=\frac{1}{2}\begin{pmatrix} a_1+b_1+(a_2-b_2)i &{} a_2+b_2-(a_1-b_1)i\\ -(a_2+b_2)+(a_1-b_1)i &{} a_1+b_1+(a_2-b_2)i \end{pmatrix}. \end{aligned}$$

We also describe the image of the map $\psi \circ \iota \circ \chi ^{-1}$:

$$\begin{aligned} {{\,\mathrm{Im}\,}}(\psi \circ \iota \circ \chi ^{-1})= & {} \left\{ \begin{pmatrix} q &{} p\\ -p &{} q \end{pmatrix}\,\Big \vert \,p,q\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{i\})\right\} \\= & {} \left\{ m\in {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{i\})\,\Big \vert \,m=-\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}m\begin{pmatrix} 0 &{}\quad {{\,\mathrm{Id}\,}}\\ -{{\,\mathrm{Id}\,}} &{}\quad 0 \end{pmatrix}\right\} . \end{aligned}$$

1.2.2 Embedding ${{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {C}}\{j\}\hookrightarrow {{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {H}}\{i,j,k\}$

In the previous sections, we have seen the isomorphisms:

$$\begin{aligned} \begin{matrix} \psi :&{} {{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes _{\mathbb {R}}{\mathbb {C}}\{j\} &{} \rightarrow &{} {{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{I\})\\ &{} (q_1+q_2J) + (p_1+p_2J)j &{} \mapsto &{} \begin{pmatrix} q_1+p_1I &{} q_2+p_2I\\ -{\bar{q}}_2-{\bar{p}}_2I &{} {\bar{q}}_1+{\bar{p}}_1I \end{pmatrix}. \end{matrix} \end{aligned}$$

where $q_1,q_2,p_1,p_2\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {C}}\{I\})$ and

$$\begin{aligned} \phi :{{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes _{\mathbb {R}}{\mathbb {H}}\{i,j,k\}\rightarrow {{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}}) \end{aligned}$$

defined as in the Sect. B.1.3.

We want to describe the image of the map

$$\begin{aligned} \phi \circ \iota \circ \psi ^{-1}:{{\,\mathrm{Mat}\,}}(2n,{\mathbb {C}}\{I\})\rightarrow {{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}}) \end{aligned}$$

where

$$\begin{aligned} \iota :{{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {C}}\{j\}\hookrightarrow {{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {H}}\{i,j,k\}, \end{aligned}$$

is the natural embedding.

Note that an element $x\in {{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {H}}\{i,j,k\}$ is contained in the subalgebra ${{\,\mathrm{Mat}\,}}(n,{\mathbb {H}}\{I,J,K\})\otimes {\mathbb {C}}\{j\}$ if and only if x commutes with $1\otimes j$. So we obtain:

$$\begin{aligned} {{\,\mathrm{Im}\,}}(\psi \circ \iota \circ \chi ^{-1})=\left\{ m\in {{\,\mathrm{Mat}\,}}(4n,{\mathbb {R}})\mid m= -\phi ({{\,\mathrm{Id}\,}}_n\otimes j)m\phi ({{\,\mathrm{Id}\,}}_n\otimes j) \right\} . \end{aligned}$$