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\begin{document}
\title{On Hermitian and skew-Hermitian matrix algebras over octonions}
\author{Arezoo Zohrabi}
\address{University of Ostrava, Ostrava, Czech Republic}
\email{azohrabi230@gmail.com}
\author{Pasha Zusmanovich}
\email{pasha.zusmanovich@gmail.com}
\date{First written December 6, 2019; last revised March 28, 2022}
\thanks{%
Journal of Nonlinear Mathematical Physics \textbf{28} (2021), no.1, 108--122}
\keywords{Matrix octonion algebras; Hermitian; skew-Hermitian;
$\delta$-derivations; associative forms}
\subjclass[2020]{17A36; 17A75; 17B20; 17B81; 17C20}
\begin{abstract}
We prove simplicity of algebras in the title, and compute their
$\delta$-derivations and symmetric associative forms.
\end{abstract}
\maketitle
\section*{Introduction}
We consider algebras of Hermitian and skew-Hermitian matrices over octonions.
While such algebras of matrices of low order are well researched and well
understood (the algebra of $3 \times 3$ Hermitian matrices being the famous
exceptional simple Jordan algebra), this is not so for higher orders; the case
of Hermitian matrices of order $4 \times 4$ appears in modern physics (string
theory, M-theory).
Derivation algebras of algebras of Hermitian and skew-Hermitian matrices over
octonions were recently computed in \cite{petyt}, and here we continue to study
these algebras. After the preliminary \S \ref{sec-1}, where we set notation and
remind basic facts about algebras with involution, we prove simplicity of the
algebras in question (\S \ref{sec-simp}), and compute their $\delta$-derivations (\S \ref{sec-der}) and symmetric
associative forms (\S \ref{sec-form}). The last \S \ref{sec-q} contains some
further questions.
\section{Notation, conventions, preliminary remarks}\label{sec-1}
\subsection{}
The ground field $K$ of characteristic $\ne 2,3$ is assumed to be arbitrary,
unless stated otherwise; $\overline K$ and $K^{q}$ denote the algebraic and the
quadratic closure of $K$, respectively. ``Algebra'' means an arbitrary algebra
over $K$, not necessary associative, or Lie, or Jordan, or satisfying any other
distinguished identity, unless specified otherwise. If $a$ is an element of an
algebra $A$, then $R_a$ denotes the linear operator of the right multiplication by $a$. All
unadorned tensor products and $\Hom$'s are over the ground field $K$. The symbol
$\dotplus$ denotes the direct sum of vector spaces, while $\oplus$ denotes the direct sum of algebras
or modules.
\subsection{Algebras with involution}
An \emph{involution} on a vector space $V$ is a linear map $j: V \to V$ such
that $j^2 = \id_V$. If $j$ is an involution on $V$, define
\begin{align*}
&\sym^{+}(V,j) = \set{x \in V}{j(x) = x} \\
\intertext{and}
&\sym^{-}(V,j) = \set{x \in V}{j(x) = -x} ,
\end{align*}
the subspaces of $j$-symmetric and $j$-skew-symmetric elements of $V$,
respectively.
For an arbitrary vector space with involution $j$, we have the direct sum
decomposition:
$$
V = \sym^+(V,j) \>\dotplus\> \sym^-(V,j) .
$$
An \emph{involution} on an algebra $A$ is a linear map $j: A \to A$ which is an
involution on $A$ as a vector space, and, additionally, is an antiautomorphism
of $A$, i.e., $j(xy) = j(y)j(x)$ for any $x,y \in A$.
For an arbitrary algebra $A$ with involution $j$, the subspace $\sym^+(A,j)$
is closed with respect to the half of the anticommutator $x \circ y = \frac 12
(xy + yx)$, and thus forms a (commutative) algebra with respect to $\circ$. The
operation $\circ$ will be also frequently referred as the \emph{Jordan product},
despite that the ensuing algebras are, generally, not Jordan. Similarly, the
subspace $\sym^-(A,j)$ is closed with respect to the commutator
$[x,y] = xy - yx$, and thus forms an (anticommutative) algebra with respect to $\liebrack$.
We have the following obvious inclusions:
\begin{alignat}{1}\label{eq-inc}
&\sym^+(A,j) \circ \sym^+(A,j) \subseteq \sym^+(A,j) \notag \\
&\sym^+(A,j) \circ \sym^-(A,j) \subseteq \sym^-(A,j) \\
&\sym^-(A,j) \circ \sym^-(A,j) \subseteq \sym^+(A,j) \notag
\end{alignat}
and
\begin{alignat}{1}\label{eq-incbrack}
&[\sym^+(A,j), \sym^+(A,j)] \subseteq \sym^-(A,j) \notag \\
&[\sym^+(A,j), \sym^-(A,j)] \subseteq \sym^+(A,j) \\
&[\sym^-(A,j), \sym^-(A,j)] \subseteq \sym^-(A,j) . \notag
\end{alignat}
If $(A,j)$ and $(B,k)$ are two vector spaces, respectively algebras, with
involution, then their tensor product $(A \otimes B, j\otimes k)$, is a vector
space, respectively algebra, with involution. Here $j\otimes k$ acts on
$A \otimes B$ in an obvious way:
$$
(j \otimes k) (a \otimes b) = j(a) \otimes k(b)
$$
for any $a \in A$, $b \in B$.
\subsection{Matrix algebras}\label{ss-matrix}
$M_n(K)$ denotes the (associative) algebra of $n \times n$ matrices with entries
in $K$. The matrix transposition, denoted by ${}^\top$, is an involution on
$M_n(K)$. $\Tr(X)$ denotes the trace of a matrix $X$, and $E$ denotes the unit
matrix. We use the shorthand notation $M_n^+(K) = \sym^+(M_n(K),{}^\top)$ and
$M_n^-(K) = \sym^-(M_n(K),{}^\top)$ for the spaces of symmetric and
skew-symmetric $n \times n$ matrices, respectively.
The algebra $M_n^+(K)$ with respect to the Jordan product is a simple Jordan
algebra. The space $M_n^-(K)$ is an irreducible Jordan module over $M_n^+(K)$
(see, for example, \cite[Chapter VII, \S 3, Theorem 7]{jacobson}). In
particular, $M_n^+(K) \circ M_n^-(K) = M_n^-(K)$.
The algebra $M_n^-(K)$ with respect to the commutator is the orthogonal Lie
algebra, customarily denoted by $\mathfrak{so}_n(K)$. We have
$\mathfrak{so}_1(K) = 0$, and $\mathfrak{so}_2(K) \simeq K$, the one-dimensional
(abelian) Lie algebra. If $n = 3$ or $n \ge 5$, the Lie algebra
$\mathfrak{so}_n(K)$ is simple; if $n = 4$, $\mathfrak{so}_4(K)$ is isomorphic
to the direct sum of two copies of the $3$-dimensional simple Lie algebra with
the basis $\{e_1,e_2,e_3\}$ and the multiplication table $[e_1,e_2] = e_3$,
$[e_2,e_3] = e_1$, $[e_3,e_1] = e_2$, denoted by us as $\mathfrak{su}_2(K)$
(of course, isomorphic to $\mathfrak{sl}_2(K)$ if $K$ is algebraically closed).
If $n \ge 3$, the $\mathfrak{so}_n(K)$-module $M_n^+(K)$, being isomorphic to
the symmetric square of the tautological module, decomposes as the direct sum
$KE \oplus SM_n(K)$, where $KE$ is the trivial $1$-dimensional module spanned by
the unit matrix, and the vector space
$$
SM_n(K) = \set{X \in M_n^+(K)}{\Tr(X) = 0}
$$
forms the $\frac{n^2 + n - 2}{2}$-dimensional irreducible module. In the case
$n=4$, the latter $\mathfrak{su}_2(K) \oplus \mathfrak{su}_2(K)$-module is
isomorphic to the tensor product $\mathfrak{su}_2(K) \otimes \mathfrak{su}_2(K)$
of two irreducible adjoint modules over two copies of $\mathfrak{su}_2(K)$.
(See, for example, \cite[Lemma 3.1]{bbm}.) In particular,
$[M_n^-(K),M_n^+(K)] = SM_n(K)$.
\begin{lemma}\label{lemma-1}
If $x \in M_n^-(K)$ is such that $x \circ M_n^-(K) = 0$, then $x=0$.
\end{lemma}
\begin{proof}
For $n=1$ the statement is vacuous, so assume $n \ge 2$. Considering this on the
Lie algebra level, we have $xy + yx = 0$ for any $y \in \mathfrak{so}_n(K)$.
Taking the trace of the both sides of this equality, we have $\Tr(xy) = 0$. But
the trace form $(x,y) \mapsto \Tr(xy)$ is nondegenerate on $\mathfrak{so}_n(K)$ (this can be verified directly, or see, for example,
\cite[p.~66]{kap}), and, consequently, $x=0$.
\end{proof}
\begin{lemma}\label{lemma-circ}
If $m \in M_n^+(K)$ is such that $[m,M_n^-(K)] = 0$ or $[m,M_n^+(K)] = 0$, then
$m$ is a multiple of $E$.
\end{lemma}
\begin{proof}
\emph{Case of $[m,M_n^-(K)] = 0$} for $n=1,2$ is verified immediately, and for
$n \ge 3$ the proof follows from the above description of $M_n^+(K)$ as an
$\mathfrak{so}_n(K)$-module.
\emph{Case of $[m,M_n^+(K)] = 0$}.
It is easy to check that this condition implies
$$
(m,s,t) = (s,m,t) = (s,t,m) = 0
$$
for any $s,t \in M_n^+(K)$, where
$(x,y,z) = (x \circ y) \circ z - x \circ (y \circ z)$ is the Jordan associator,
i.e., $m$ lies in the center of the simple Jordan algebra $(M_n^+(K), \circ)$,
which coincides with $KE$.
\end{proof}
\subsection{Octonion algebras}\label{ss-octonion}
Octonion algebras over an arbitrary field $K$ form the $3$-parametric family
$\mathbb O_\mu(K)$, where $\mu = (\mu_1,\mu_2,\mu_3)$ is a triple of nonzero
elements of $K$. Let us recall its multiplication table in the standard basis
$\{1, e_1, \dots, e_7\}$ (by abuse of notation, the basis element $1$ is the
unit of the algebra):
\vskip 14pt
\begin{center}
{\small
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
& $e_1$ & $e_2$ & $e_3$ & $e_4$ & $e_5$ & $e_6$ & $e_7$ \\ \hline
$e_1$ & $\mu_1 1$ & $-e_3$ & $-\mu_1 e_2$ & $-e_5$ & $-\mu_1 e_4$ & $e_7$ & $\mu_1 e_6$ \\ \hline
$e_2$ & $e_3$ & $\mu_2 1$ & $\mu_2 e_1$ & $-e_6$ & $-e_7$ & $-\mu_2 e_4$ & $-\mu_2 e_5$ \\ \hline
$e_3$ & $\mu_1 e_2$ & $-\mu_2 e_1$ & $-\mu_1 \mu_2 1$ & $-e_7$ & $-\mu_1 e_6$ & $\mu_2 e_5$ & $\mu_1 \mu_2 e_4$ \\ \hline
$e_4$ & $e_5$ & $e_6$ & $e_7$ & $\mu_3 1$ & $\mu_3 e_1$ & $\mu_3 e_2$ & $\mu_3 e_3$ \\ \hline
$e_5$ & $\mu_1 e_4$ & $e_7$ & $\mu_1 e_6$ & $-\mu_3 e_1$ & $-\mu_1 \mu_3 1$ & $-\mu_3 e_3$ & $-\mu_1 \mu_3 e_2$ \\ \hline
$e_6$ & $-e_7$ & $\mu_2 e_4$ & $-\mu_2 e_5$ & $-\mu_3 e_2$ & $\mu_3 e_3$ & $-\mu_2 \mu_3 1$ & $\mu_2 \mu_3 e_1$ \\ \hline
$e_7$ & $-\mu_1 e_6$ & $\mu_2 e_5$ & $-\mu_1 \mu_2 e_4$ & $-\mu_3 e_3$ & $\mu_1 \mu_3 e_2$ & $-\mu_2 \mu_3 e_1$ & $\mu_1 \mu_2 \mu_3 1$ \\ \hline
\end{tabular}}
\end{center}
\vskip 14pt
\noindent
(the table, up to obvious notational changes, is reproduced from
\cite[p.~5]{schafer}). Over some fields, there are isomorphisms within this
family; for example, if the field is algebraically closed or finite, all
octonion algebras are isomorphic to each other. As explained below, in the
proofs of our main results we may assume the ground field to be algebraically
closed, so we are free to choose any form of an octonion algebra we wish. The
two most natural candidates would be $\mathbb O_{(-1,-1,-1)}(K)$ (for example,
over $\mathbb R$ this is the single octonion division algebra), or the split
octonion algebra $\mathbb O_{(-1,-1,1)}(K)$.
We have decided that for our calculations the most convenient will be the
algebra $\mathbb O_{(-1,-1,-1)}(K)$, denoted just by $\mathbb O(K)$ in the
sequel\footnote{
Of course, it is also possible to perform all our calculations in the case of
generic $3$-parametric octonion algebra $\mathbb O_\mu(K)$, but then they will
be somewhat more cumbersome.
}.
A quick glance at the multiplication table reveals the following properties of
the basis elements we will need: $e_i^2 = -1$, $e_ie_j = -e_je_i$, and, denoting
by $B_i$ the $6$-dimensional linear span of all the basis elements except for
$1$ and $e_i$, we have $e_i B_i = B_i e_i = B_i$, for any $i=1,\dots,7$. By
$$
*: \{1,\dots,7\} \times \{1,\dots,7\} \to \{1,\dots,7\}
$$
we denote the partial binary operation such that
$e_ie_j = -e_je_i = \pm e_{i*j}$ for any $i \ne j$.
Extending the base field $K$ to its algebraic closure $\overline K$, we have an
isomorphism of $\overline K$-algebras
\begin{equation}\label{eq-oisom}
\mathbb O_\mu(K) \otimes_K \overline K \simeq \mathbb O(\overline K) .
\end{equation}
The standard conjugation in $\mathbb O_\mu(K)$, denoted by $\topbar$, and
defined by $\overline 1 = 1$, $\overline e_i = -e_i$, turns $\mathbb O_\mu(K)$
into an algebra with involution. We have
$\sym^+(\mathbb O_\mu(K),\topbar) = K1$, and $\sym^-(\mathbb O_\mu(K),\topbar)$
is the $7$-dimensional subspace of imaginary octonions, linearly spanned by
$e_1, \dots, e_7$. The latter subspace forms a $7$-dimensional simple Malcev
algebra with respect to the commutator. We will use the shorthand notation
$\mathbb O_\mu^-(K) = \sym^-(\mathbb O_\mu(K),\topbar)$ and
$\mathbb O^-(K) = \sym^-(\mathbb O(K),\topbar)$.
Since for any $a \in \mathbb O_\mu(K)$, the elements $a + \overline a$ and
$a \overline a$ belong to $K1$, we can define the linear map
$T: \mathbb O_\mu(K) \to K$ and the quadratic map $N: \mathbb O_\mu(K) \to K$
by $T(a) = a + \overline a$ and $N(a) = a\overline a$, called the \emph{trace}
and \emph{norm}, respectively. Any element $a \in \mathbb O_\mu(K)$ satisfies the quadratic equality
\begin{equation}\label{eq-r}
a^2 - T(a)a + N(a)1 = 0
\end{equation}
(see, for example, \cite[Chapter III, \S 4]{schafer} or
\cite[p.~233, Exercise 1]{jacobson}).
For any two elements $a,b\in \mathbb O_\mu^-(K)$, writing the equality
(\ref{eq-r}) for the element $a+b$, subtracting from it the same equalities for
$a$ and for $b$, and taking into account that $T(a) = T(b) = 0$, yields
\begin{equation}\label{eq-o}
ab + ba = -N(a,b) 1 ,
\end{equation}
where
$$
N(a,b) = N(a+b) - N(a) - N(b) .
$$
\subsection{Algebras of Hermitian and skew-Hermitian matrices over octonions}
Our main characters, the algebras of Hermitian and skew-Hermitian matrices over
octonions, are defined as $\sym^+(M_n(\mathbb O_\mu(K)),J)$ and
$\sym^-(M_n(\mathbb O_\mu(K)),J)$ respectively, where
$M_n(\mathbb O_\mu(K))$ is the algebra of $n \times n$ matrices with entries in
$\mathbb O_\mu(K)$. The involution on $M_n(\mathbb O_\mu(K))$ is defined as
$J: (a_{ij}) \mapsto (\overline{a_{ji}})$, i.e., the matrix is transposed and
each entry is conjugated, simultaneously.
The algebras $\sym^+(M_n(\mathbb O_\mu(K)),J)$ contain the unit matrix, so they
are unital. These algebras for small $n$'s are Jordan algebras, well-known from
the literature: for $n=1$, this is nothing but the ground field $K$; for $n=2$,
they are $10$-dimensional simple Jordan algebras of symmetric nondegenerate
bilinear form (see, for example, \cite[Chapter IX, Exercise 4]{involutions} and
\cite[\S 6]{ruhaak}); and for $n=3$, they are the famous $27$-dimensional
exceptional simple Jordan algebras. For $n \ge 4$, they are no longer Jordan
algebras.
Interestingly enough, the algebras $\sym^+(M_4(\mathbb O_\mu(K)),J)$ were
considered already in a little-known dissertation \cite{ruhaak} (for a more
accessible exposition, see \cite[\S 5]{phd-new}), under the direction of Hel
Braun and Pascual Jordan. More recently, the algebra
$\sym^+(M_4(\mathbb O(\mathbb R)),J)$ appeared in \cite[\S 4]{toppan} under the
name ``octonionic M-algebra'', where it was suggested as an alternative to the
standard M-algebra (a sort of generalization of the Poincar\'e algebra of
spacetime symmetries). This algebra features some M-theory numerology (lesser
number of real bosonic generators, equivalence between supermembrane and
super-five-brane sectors) which, as suggested in \cite{toppan}, could make this
algebra a better alternative.
The algebras $\sym^-(M_n(\mathbb O_\mu(K),J)$ are less prominent: for $n=1$
these are the $7$-dimensional simple Malcev algebras $\mathbb O_\mu^-(K)$; it
seems that the only place where they appeared in the literature in the case of
(small) $n>1$ is \cite{bremner-hentzel}, where identities of these algebras were
studied.
Due to the isomorphism of algebras
$$
M_n(\mathbb O_\mu(K)) \simeq M_n(K) \otimes \mathbb O_\mu(K) ,
$$
the algebra with involution $(M_n(\mathbb O_\mu(K)),J)$ can be represented as
the tensor product of two algebras with involution: $(M_n(K), {}^\top)$, the
associative algebra of $n \times n$ matrices over $K$ with involution defined by
the matrix transposition, and $(\mathbb O_\mu(K), \topbar)$.
Finally, due to isomorphism (\ref{eq-oisom}), we have an isomorphism of
$\overline K$-algebras:
\begin{equation}\label{eq-isom}
\sym^\pm(M_n(\mathbb O_\mu(K), J)) \otimes_K \overline K
\simeq \sym^\pm(M_n(\mathbb O(\overline K)), J) .
\end{equation}
\section{Simplicity}\label{sec-simp}
We start with rewriting our matrix algebras as the vector space direct sums of
certain tensor products, which appears to be more convenient for computations.
For this, we need the following simple lemma of linear algebra.
\begin{lemma}[\protect{\cite[Lemma 1.1]{low}}]\label{lemma-1.1}
Let $V, W$ be two vector spaces,
$\varphi, \varphi^\prime \in \Hom\,(V, \>\cdot\>)$,
$\psi, \psi^\prime \in \Hom\,(W, \>\cdot\>)$.
Then
\begin{multline}\notag
\Ker(\varphi \otimes \psi) \cap \Ker(\varphi^\prime \otimes \psi^\prime) \\
\simeq
(\Ker\varphi \cap \Ker \varphi^\prime) \otimes W +
\Ker\varphi \otimes \Ker \psi^\prime +
\Ker\varphi^\prime \otimes \Ker \psi +
V \otimes (\Ker\psi \cap \Ker\psi^\prime).
\end{multline}
\end{lemma}
\begin{proposition}\label{prop-1}
For any two vector spaces with involution $(V,j)$ and $(W,k)$, there are
isomorphisms of vector spaces
\begin{alignat*}{1}
&\sym^{+}(V \otimes W, j \otimes k) \simeq
\sym^{+}(V,j) \otimes \sym^{+}(W,k) \>\dotplus\>
\sym^{-}(V,j) \otimes \sym^{-}(W,k)
\\
&\sym^{-}(V \otimes W, j \otimes k) \simeq
\sym^{+}(V,j) \otimes \sym^{-}(W,k) \>\dotplus\>
\sym^{-}(V,j) \otimes \sym^{+}(W,k) .
\end{alignat*}
\end{proposition}
\begin{proof}
Let us prove the first isomorphism, the proof of the second one is completely
similar. By definition, an element $\sum_{i\in \mathbb I} v_i \otimes w_i$ of
$V \otimes W$, where $\mathbb I$ is a set of indices, belongs to
$\sym^{+}(V \otimes W, j \otimes k)$, if and only if
$$
\sum_{i\in \mathbb I} \Big(j(v_i) \otimes k(w_i) - v_i \otimes w_i\Big) = 0 .
$$
Applying to this equality the linear maps $(\id_V + j) \otimes \id_W$ and
$(\id_V - j) \otimes \id_W$, we get respectively:
$$
\sum_{i\in \mathbb I} (j(v_i) + v_i) \otimes (k(w_i) - w_i) = 0
$$
and
$$
\sum_{i\in \mathbb I} (j(v_i) - v_i) \otimes (k(w_i) + w_i) = 0 .
$$
Applying Lemma \ref{lemma-1.1} to the last two equalities, we can replace
$v_i$'s and $w_i$'s by their linear combinations in such a way that the index
set splits into the disjoint union
$\mathbb I =
\mathbb I_{11} \cup \mathbb I_{12} \cup \mathbb I_{21} \cup \mathbb I_{22}$,
where
\begin{alignat*}{5}
&v_i&\in& \sym^-(V,j), \>&v_i&\in& &\sym^+(V,j) \quad
&\text{for }& i\in \mathbb I_{11} ,
\\
&v_i&\in& \sym^-(V,j), \>&w_i&\in& &\sym^-(W,k) \quad
&\text{for }& i\in \mathbb I_{12} ,
\\
&v_i&\in& \sym^+(V,j), \>&w_i&\in& &\sym^+(W,k) \quad
&\text{for }& i\in \mathbb I_{21} ,
\\
&w_i&\in& \sym^+(W,k), \>&w_i&\in& &\sym^-(W,k) \quad
&\text{for }& i\in \mathbb I_{22} .
\end{alignat*}
All elements with indices from $\mathbb I_{11}$ and $\mathbb I_{22}$ vanish, and
we are done.
\end{proof}
In the particular case $(V,j) = (M_n(K), {}^\top)$ and
$(W,k) = (\mathbb O_\mu(K), \topbar)$, denoting
$J = {}^\top \otimes \topbar$, and taking into account that
$\sym^+(\mathbb O_\mu(K),\topbar) = K1$, we get:
\begin{equation}\label{eq-dec}
\sym^+(M_n(\mathbb O_\mu(K)),J) \simeq
M_n^+(K) \otimes 1 \>\dotplus\> M_n^-(K) \otimes \mathbb O_\mu^-(K) .
\end{equation}
(In the case where $n=3$ and $K$ is algebraically closed and of characteristic
zero, and so $\sym^+(M_3(\mathbb O_\mu(K),J))$ is the $27$-dimensional
exceptional simple Jordan algebra, this decomposition was noted in \cite[\S 3.3]{draper}.)
In particular,
$$
\dim \sym^+(M_n(\mathbb O_\mu(K),J)) =
\frac{n(n+1)}{2} + 7 \cdot \frac{n(n-1)}{2} = 4n^2 - 3n .
$$
For any $m,s\in M_n^+(K)$, we have
$$
(m \otimes 1) \circ (s \otimes 1) = (m \circ s) \otimes 1 ,
$$
what implies that $M_n^+(K) \otimes 1$ is a (Jordan) subalgebra of
$\sym^+(M_n(\mathbb O_\mu(K),J))$. Moreover, for any $x,y \in M_n^-(K)$, and
$a \in \mathbb O_\mu^-(K)$, we have:
\begin{alignat*}{6}
&(m \otimes 1) \>&\circ&\> (x \otimes a) \>&=& &(m \circ x) &\otimes a ,
\\
&(x \otimes a) \>&\circ&\> (y \otimes a) \>&=& -N(a) &(x \circ y) &\otimes 1 .
\end{alignat*}
It follows that $M_n^+(K) \otimes 1 \dotplus M_n^-(K) \otimes a$ is a subalgebra
of $\sym^+(M_n(\mathbb O_\mu(K),J))$; let us denote this subalgebra by
$\mathscr L^+(a)$. If $N(a) \ne 0$, we have an isomorphism of Jordan algebras
$\mathscr L^+(a) \otimes_K K^q \simeq M_n(K^q)$; the isomorphism is provided by
sending $m \otimes 1$ to $m$ for $m \in M_n^+(K^q)$, and $x \otimes a$ to
$\sqrt{-N(a)}\,x$ for $x \in M_n^-(K^q)$.
Further,
$$
(M_n^+(K) \otimes 1) \circ (M_n^-(K) \otimes \mathbb O_\mu^-(K)) =
M_n^-(K) \otimes \mathbb O_\mu^-(K) .
$$
On the other hand, the subspace $M_n^-(K) \otimes \mathbb O_\mu^-(K)$ is not a
subalgebra. The formula for multiplication in this subspace in terms of the
decomposition (\ref{eq-dec}) is obtained using (\ref{eq-o}): for any
$x,y \in M_n^-(K)$ and $a, b\in \mathbb O_\mu^-(K)$, we have
\begin{multline*}
(x \otimes a) \circ (y \otimes b) = \frac 12 (xy \otimes ab + yx \otimes ba) =
\frac 14 (xy+yx) \otimes (ab+ba) + \frac 14 (xy - yx) \otimes (ab - ba) \\ =
-\frac{N(a,b)}{2} \> (x \circ y) \otimes 1 + \frac 14 [x,y] \otimes [a,b] .
\end{multline*}
Similarly, we have
\begin{equation}\label{eq-decomp-minus}
\sym^-(M_n(\mathbb O_\mu(K)),J) \simeq M_n^-(K) \otimes 1 \>\dotplus\>
M_n^+(K) \otimes \mathbb O_\mu^-(K) ,
\end{equation}
and
$$
\dim \sym^-(M_n(\mathbb O_\mu(K)),J) =
\frac{n(n-1)}{2} + 7 \cdot \frac{n(n+1)}{2} = 4n^2 + 3n.
$$
For any $x,y\in M_n^-(K)$, $m,s\in M_n^+(K)$, and $a \in \mathbb O_\mu^-(K)$, we
have:
\begin{alignat*}{6}
&[x \otimes 1, y \otimes 1] \>&=&\> &[&x,y] \>&\otimes&\> 1 \\
&[x \otimes 1, m \otimes a] \>&=&\> &[&x,m] \>&\otimes&\> a \\
&[m \otimes a, s \otimes a] \>&=&\> N(a) &[&s,m] \>&\otimes&\> 1 .
\end{alignat*}
It follows that both $M_n^-(K) \otimes 1$ and
$$
\mathscr L^-(a) = M_n^-(K) \otimes 1 \dotplus M_n^+(K) \otimes a
$$
are Lie subalgebras of $\sym^-(M_n(\mathbb O_\mu(K)),J)$, isomorphic to
$\mathfrak{so}_n(K)$, and, provided $N(a) \ne 0$, to a form of
$\mathfrak{gl}_n(K^q)$, respectively; the isomorphisms are defined by sending
$x \otimes 1$ to $x$ for $x \in M_n^-(K)$, and $m \otimes a$ to
$\sqrt{-N(a)}\,m$ for $m\in M_n^+(K^q)$.
Moreover,
$$
[M_n^-(K) \otimes 1, M_n^+(K) \otimes \mathbb O_\mu^-(K)] =
SM_n(K) \otimes \mathbb O_\mu^-(K) \subset M_n^+(K) \otimes \mathbb O_\mu^-(K) .
$$
The subspace $M_n^+(K) \otimes \mathbb O_\mu^-(K)$ is not a subalgebra: for any
$m,s \in M_n^+(K)$, $a,b \in \mathbb O_\mu^-(K)$, we have
\begin{multline}\label{eq-mult}
[m \otimes a, s \otimes b] \\ =
\frac 12 (ms - sm) \otimes (ab+ba) + \frac 12 (ms + sm) \otimes (ab - ba) =
-\frac{N(a,b)}{2} [m,s] \otimes 1 + (m \circ s) \otimes [a,b] .
\end{multline}
\begin{theorem}\label{th-1}
The algebras $\sym^+(M_n(\mathbb O_\mu(K)), J)$ and
$\sym^-(M_n(\mathbb O_\mu(K)), J)$ are simple for any $n \ge 1$.
\end{theorem}
Before we plunge into the proof, a few remarks are in order:
\begin{enumerate}[\upshape(i)]
\item
The cases of $\sym^+(M_n(\mathbb O_\mu(K)), J)$ for $n=1,2,3$, and of
$\sym^-(M_n(\mathbb O_\mu(K)), J)$ for $n=1$ are well-known, due to the known
structure of the algebras in question in these cases (see \S \ref{sec-1}); however, our proofs, uniform for all $n$, appear to
be new. The case of $\sym^+(M_4(\mathbb O_\mu(K)), J)$ is stated
without proof in \cite[Satz 8.1]{ruhaak}.
\item
In \cite{stewart} it is proved that ideals of the tensor product $A \otimes B$
of two algebras $A$ and $B$, where $A$ is central (i.e., its centroid coincides
with the ground field) and simple, and $B$ satisfies some other conditions (like
having a unit), are of the form $A \otimes I$, where $I$ is an ideal of $B$. In
particular, the tensor product of two central simple algebras, for example,
$M_n(K) \otimes \mathbb O_\mu(K)$, is simple. Our method of
proof of Theorem~\ref{th-1}, based on application of the (version of) Jacobson
density theorem, resembles those in \cite{stewart}.
\item
Another related result about simplicity of nonassociative algebras is
established in \cite[Satz 5.1]{ruhaak}: the matrix algebra over a composition
algebra with respect to the Jordan product $\circ$, is simple; a particular case
is the algebra $(M_n(\mathbb O_\mu(K)), \circ)$.
\end{enumerate}
We will need the following version of the Jacobson density theorem.
\begin{proposition}\label{prop-dens}\footnote{
Added March 27, 2022: As stated, the statement of the proposition is wrong. Like
in the classical Jacobson density theorem, one needs to require that the module
elements are independent over the ring of $R$-module endomorphisms of $M$, and
not just over the ground field. Alternatively, one may require that the ground
field is algebraically closed, as it is enough for our purposes. Thanks to are due
Alberto Elduque for spotting this mistake.
}
Let $R$ be an associative algebra with unit, and $M_1, \dots, M_n$ pairwise non
isomorphic right irreducible $R$-modules. Then for any linearly independent
elements
$x_1^{(i)}, \dots, x_{k_i}^{(i)} \in M_i$, and any elements
$y_1^{(i)}, \dots, y_{k_i}^{(i)} \in M_i$, $i=1,\dots,n$, there is an element
$a\in R$ such that $x_j^{(i)} \bullet a = y_j^{(i)}$ for any $i=1,\dots,n$,
$j=1,\dots,k_i$.
\end{proposition}
(Here $\bullet$ denotes the right action of $A$ on its modules).
\begin{proof}
This is, essentially, the Jacobson density theorem formulated for a completely
reducible module $M = M_1 \oplus \dots \oplus M_n$. Perhaps, the easiest way to
derive it in our formulation is the following. First, apply the classical
Jacobson density theorem to each irreducible $R$-module $M_i$ to get elements
$a_i \in R$ such that $x_j^{(i)} \bullet a_i = y_j^{(i)}$ for any $i=1,\dots,n$,
$j=1,\dots,k_i$. By \cite[Chapter XVII, Theorem 3.7]{lang} (which is a
consequence of the Jacobson density theorem for semisimple modules formulated in
terms of bicommutants of modules, see \cite[Chapter XVII, Theorem 3.2]{lang}),
there are elements $e_i \in R$ such that $e_i$ acts as the identity on $M_i$,
and $M_j \bullet e_i = 0$ for $j \ne i$. Then $a = e_1 a_1 + \dots + e_n a_n$ is
the required element.
\end{proof}
We now specialize this to our situation. Let $A$ be an algebra, and $M$ a right
$A$-module. By the \emph{multiplication algebra} $\mathfrak M(A,M)$ we mean the
unital subalgebra in the associative algebra of all linear transformations of
$M$, generated by actions of all elements of $A$ on $M$. If $A$ acts on itself
via right multiplications, i.e., $M = A$, then $\mathfrak M(A,A)$ is called the
\emph{multiplication algebra of $A$}.
%\pagebreak
\begin{lemma}\label{lemma-dens}\hfill
\begin{enumerate}[\upshape(i)]
\item\label{it-jord}
For any linearly independent elements $m_1, \dots, m_k \in M_n^+(K)$,
$x_1, \dots, x_\ell \in M_n^-(K)$, and any elements
$m_1^\prime, \dots, m_k^\prime \in M_n^+(K)$,
$x_1^\prime, \dots, x_\ell^\prime \in M_n^-(K)$, there is a map
$R \in \mathfrak M(M_n^+(K), M_n(K))$ such that $R(m_i) = m_i^\prime\>$ for
$i=1,\dots,k$, and $R(x_i) = x_i^\prime\>$ for $i=1,\dots,\ell$.
\item\label{it-lien}
Let $n \ne 4$. For any linearly independent elements
$m_1, \dots, m_k \in SM_n(K)$, $x_1, \dots, x_\ell \in M_n^-(K)$, and any
elements $m_1^\prime, \dots, m_k^\prime \in SM_n(K)$,
$x_1^\prime, \dots, x_\ell^\prime \in M_n^-(K)$, there is a map
$R \in \mathfrak M(\mathfrak{so}_n(K), M_n(K))$ such that
$R(m_i) = m_i^\prime\>$ for $i=1,\dots,k$, and $R(x_i) = x_i^\prime\>$ for
$i=1,\dots,\ell$.
\end{enumerate}
\end{lemma}
(Here the Jordan algebra $M_n^+(K)$, respectively the Lie algebra
$\mathfrak{so}_n(K)$, acts via Jordan multiplications, respectively commutators,
on its ambient algebra $M_n(K)$.)
\begin{proof}
(i)
As follows from \S \ref{ss-matrix}, $M_n(K)$ is decomposed, as an
$M_n^+(K)$-module, into the direct sum of two irreducible non isomorphic Jordan
modules: $M_n(K) = M_n^+(K) \oplus M_n^-(K)$. Apply Proposition \ref{prop-dens}
to $R = \mathfrak M(M_n^+(K), M_n(K))$, and $M_1 = M_n^+(K)$, $M_2 = M_n^-(K)$.
(ii)
The statement is vacuous for $n=1$, and easily verified directly for $n=2$, so
assume $n \ge 3$. As follows from \S \ref{ss-matrix}, $M_n(K)$ is decomposed, as
an $\mathfrak{so}_n(K)$-module, into the direct sum of three non-isomorphic
modules:
$$
M_n(K) = KE \oplus SM_n(K) \oplus M_n^-(K) .
$$
Apply Proposition \ref{prop-dens} to
$$
R = \mathfrak M\big(\mathfrak{so}_n(K), M_n(K)\big) =
\mathfrak M\big(\mathfrak{so}_n(K), SM_n(K) \oplus M_n^-(K)\big) ,
$$
and $M_1 = SM_n(K)$, $M_2 = M_n^-(K)$.
\end{proof}
Note that the restriction $n\ne 4$ in Lemma \ref{lemma-dens}(\ref{it-lien}) is
essential. As noted in \S \ref{ss-matrix}, the adjoint module of
$\mathfrak{so}_4(K)$ decomposes into the direct sum of two irreducible
isomorphic modules, so Proposition \ref{prop-dens} is not applicable as is. It
is possible to devise more sophisticated versions of Proposition \ref{prop-dens}
and Lemma \ref{lemma-dens} which are trying to take account of this, but we
found it easier to treat the case $n=4$ below in a different way, avoiding more
sophisticated versions of the Jacobson density theorem.
\begin{proof}[Proof of Theorem \ref{th-1}]
As a form of a simple algebra is simple, it is enough to prove the theorem when
the ground field $K$ is algebraically closed. In this case, due to isomorphism
(\ref{eq-isom}), we may assume $\mathbb O_\mu(K) = \mathbb O(K)$.
\emph{Case of $\sym^+(M_n(\mathbb O(K)),J)$}.
Let $I$ be an ideal of $\sym^+(M_n(\mathbb O(K)), J)$. We argue in terms of the
decomposition (\ref{eq-dec}). Assume first that
$I \subseteq M_n^-(K) \otimes \mathbb O^-(K)$. Consider an element
\begin{equation*}
\sum_{i=1}^7 x_i \otimes e_i \in I ,
\end{equation*}
where $x_i \in M_n^-(K)$, and $e_1,\dots,e_7$ are elements of the standard basis
of $\mathbb O(K)$, as described in \S \ref{ss-octonion}. For any
$y \in M_n^-(K)$, and any $k=1,\dots,7$, we have
$$
(y \otimes e_k) \circ (\sum_{i=1}^7 x_i \otimes e_i) = - (x_k \circ y) \otimes 1
+ \text{ terms lying in } M_n^-(K) \otimes \mathbb O^-(K) .
$$
Hence, $x_k \circ y = 0$ for any $y \in M_n^-(K)$, and by Lemma \ref{lemma-1},
$x_k = 0$. This shows that $I=0$, and we may assume
$I \nsubseteq M_n^-(K) \otimes \mathbb O^-(K)$.
Now take an element
$$
m \otimes 1 + \sum_{i \in \mathbb I} x_i \otimes a_i \in I ,
$$
where $m \in M_n^+(K)$, $m \ne 0$, $x_i \in M_n^-(K)$, $i\in \mathbb I$ are
linearly independent, and $a_i \in \mathbb O^-(K)$. By
Lemma \ref{lemma-dens}(\ref{it-jord}), for any $m^\prime \in M_n^+(K)$ there is
a linear map $R: M_n(K) \to M_n(K)$, represented as the sum of products of the
form $R_{s_1} \dots R_{s_\ell}$, where each $s_i$ belongs to $M_n^+(K)$, and
$R_s$ is the Jordan multiplication by the element $s$, such that $R(m) = m^\prime$ and
$R(x_i) = 0$ for any $i=1,\dots,7$. We form the corresponding map
$\widetilde R$ from the multiplication algebra of $\sym^+(M_n(\mathbb O(K)),J)$
by replacing each $R_{s_i}$ by $R_{s_i \otimes 1}$. Then
$\widetilde R(m \otimes 1) = m^\prime \otimes 1$ and
$\widetilde R(x_i \otimes a_i) = 0$. Consequently, $m^\prime \otimes 1 \in I$,
and $I$ contains $M_n^+(K) \otimes 1$. This, in its turn, implies
$$
M_n^-(K) \otimes \mathbb O^-(K) =
(M_n^+(K) \otimes 1) \circ (M_n^-(K) \otimes \mathbb O^-(K)) \subseteq I ,
$$
and hence $I$ coincides with the whole algebra $\sym^+(M_n(\mathbb O(K)),J)$.
\emph{Case of $\sym^-(M_n(\mathbb O(K)),J)$}.
The proof goes largely along the same route as in the previous case, but with
some complications and modifications, notably in the case $n=4$. If $n=1$, the
algebra in question is isomorphic to the $7$-dimensional Malcev algebra
$\mathbb O^{-}(K)$, whose simplicity is well known (and can be established by
an easy modification of some of the reasonings below), so assume $n \ge 2$.
Let $I$ be an ideal of $\sym^-(M_n(\mathbb O(K)),J)$. Assume first
$I \subseteq M_n^+(K) \otimes \mathbb O^-(K)$. Consider an element
\begin{equation*}
\sum_{i=1}^7 m_i \otimes e_i \in I ,
\end{equation*}
where $m_i \in M_n^+(K)$. For any $s \in M_n^+(K)$, and any $k=1,\dots,7$, we
have
$$
[s \otimes e_k, \sum_{i=1}^7 m_i \otimes e_i] = [m_k,s] \otimes 1 +
\text{terms lying in } M_n^+(K) \otimes \mathbb O^-(K) .
$$
Hence, $[m_k,s] = 0$ for any $s\in M_n^+(K)$, and by Lemma \ref{lemma-circ},
$m_k = \lambda_k E$ for some $\lambda_k \in K$. Therefore, any element of $I$
is of the form
$\sum_{i=1}^7 \lambda_i E \otimes e_k \in E \otimes \mathbb O^-(K)$, and
$I = E \otimes S$ for some subspace $S \subseteq \mathbb O^-(K)$. But then
$$
[M_n^+(K) \otimes \mathbb O^-(K), E \otimes S] =
M_n^+(K) \otimes [\mathbb O^-(K),S] \subseteq E \otimes S ,
$$
this can happen only if $[\mathbb O^-(K),S] = 0$, hence $S=0$ and $I=0$.
Therefore, we may assume $I \nsubseteq M_n^+(K) \otimes \mathbb O^-(K)$.
Consider an element
\begin{equation}\label{eq-x}
x \otimes 1 + \sum_{i\in \mathbb I} m_i \otimes a_i \in I ,
\end{equation}
where $x \in M_n^-(K)$ is non-zero, $m_i \in M_n^+(K)$ for $i\in \mathbb I$ are
linearly independent, and $a_i \in \mathbb O^-(K)$ are non-zero. Taking the
commutator of this element with an element $y \otimes 1$, where $y \in M_n^-(K)$
is such that $[x,y] \ne 0$, we may assume that $m_i \in SM_n(K)$.
Assume $n \ne 4$. By Lemma \ref{lemma-dens}(\ref{it-lien}), for any
$x^\prime \in M_n^-(K)$ there is a linear map $R: M_n(K) \to M_n(K)$ of the form
\begin{equation}\label{eq-rr}
R = \lambda \id + R^\prime ,
\end{equation}
where $\lambda \in K$, and $R^\prime$ is the sum of products of the form
$\ad y_1 \dots \ad y_\ell$, where each $y_i$ belongs to $M_n^-(K)$, and $\ad y$
denotes the commutator with $y$, such that $R(x) = x^\prime$, and $R(m_i) = 0$
for each $i=1,\dots,7$. (Note that the term $\lambda\id$ in (\ref{eq-rr}) occurs
from the necessity to adjoin the unit to the multiplication algebra generated by
commutators with elements of $\mathfrak{so}_n(K)$; this term does not occur in
the previous case, where the multiplication algebra was formed by Jordan
multiplications by elements of $M_n^+(K)$, as the
latter already contains the unit: the Jordan product with the unit matrix.)
We have $R^\prime(x) = x^\prime - \lambda x$, and
$R^\prime(m_i) = -\lambda m_i$. Replacing in $R^\prime$ each $\ad y_i$ by
$\ad (y_i \otimes 1)$, we get the map $\widetilde R$ in the multiplication
algebra of $\sym^-(M_n(\mathbb O(K)),J)$ such that
$\widetilde R(x \otimes 1) = (x^\prime - \lambda x) \otimes 1$ and
$\widetilde R(m_i \otimes a_i) = -\lambda m_i \otimes a_i$,
and thus
$$
\widetilde R\>(x \otimes 1 + \sum_{i=1}^7 m_i \otimes a_i) =
(x^\prime - \lambda x) \otimes 1 - \lambda \sum_{i=1}^7 m_i \otimes a_i \in I .
$$
Adding to this element the element (\ref{eq-x}) multiplied by $\lambda$,
we get $x^\prime \otimes 1 \in I$ for any $x^\prime \in M_n^-(K)$, i.e., $I$
contains $M_n^-(K) \otimes 1$. Hence,
$$
SM_n(K) \otimes \mathbb O^-(K) =
[M_n^-(K) \otimes 1, M_n^+(K) \otimes \mathbb O^-(K)] \subseteq I .
$$
The formula (\ref{eq-mult}), in its turn, implies
$$
[m \otimes e_i, s \otimes e_j] = \pm 2 (m \circ s) \otimes e_{i*j} ,
$$
for any $m,s \in SM_n(K)$, and $i,j=1,\dots,7$. Since
$SM_n(K) \circ SM_n(K) = M_n^+(K)$, and $i*j$ runs through all the range
$1,\dots,7$, we conclude that $I$ contains $M_n^+(K) \otimes \mathbb O^-(K)$,
and hence coincides with the whole algebra $\sym^-(M_n(\mathbb O(K)),J)$.
Now consider the case $n=4$. Consider an element of $I$ of the form
(\ref{eq-x}), where $m_i \in SM_4(K)$ for any $i\in \mathbb I$. By the
(classical) Jacobson density theorem for the case of an irreducible module (or,
equivalently, by Lemma \ref{lemma-dens}(\ref{it-lien}) in the case $n=4$ where
the ``$M_n^-(K)$ part'' is ignored), for any $m \in SM_4(K)$, and any
$k \in \mathbb I$, there is a map of the form (\ref{eq-rr}), where $R^\prime$ is
formed by the commutators with elements of $M_4^-(K)$, such that $R(m_k) = m$,
and $R(m_i) = 0$, $i \ne k$. Deriving from this the map $\widetilde R$ in the
multiplication algebra of $\sym^-(M_4(\mathbb O(K)),J)$ as above, we get:
\begin{gather*}
\widetilde R(M_4^-(K) \otimes 1) \subseteq M_4^-(K) \otimes 1 \\
\widetilde R(m_k \otimes a_k) = (m - \lambda m_k) \otimes a_k \\
\widetilde R(m_i \otimes a_i) = -\lambda m_i \otimes a_i \>,\> i \ne k .
\end{gather*}
Consequently, $\widetilde R$, being applied to the element (\ref{eq-x}),
produces the element
$$
x^\prime \otimes 1 + (m - \lambda m_k) \otimes a_k
- \lambda \sum_{i \in \mathbb I \backslash \{k\}} m_i \otimes a_i \in I ,
$$
where $x^\prime \in M_4^-(K)$. Adding to this element the element (\ref{eq-x})
multiplied by $\lambda$, we get the element
$$
x^{\prime\prime} \otimes 1 + m \otimes a_k \in I ,
$$
where $x^{\prime\prime} \in M_4^-(K)$.
To summarize: for any $a \in \mathbb O^-(K)$ which appears as one of $a_i$'s in
the decomposition (\ref{eq-x}) of some nonzero element of $I$, and any
$m \in SM_4(K)$, there is an element $x \in M_4^-(K)$ such that
$x \otimes 1 + m \otimes a \in I$. Fixing here $a$ and varying $m$, we also vary
$x$, but since
$$
\dim SM_4(K) = 9 > \dim M_4^-(K) = 6 ,
$$
we will get nonzero elements with vanishing $x$, i.e., of the form
$m \otimes a$. Now taking commutators of such an element with elements from
$M_4^-(K) \otimes 1$, we get the whole $SM_4(K) \otimes a \subseteq I$.
This means that the ideal $I$ is homogeneous with respect to the decomposition
(\ref{eq-decomp-minus}), i.e., is of the form
$$
I =
T \otimes 1 \dotplus S_1 \otimes e_1 \dotplus \dots \dotplus S_7 \otimes e_7 ,
$$
where $T$ is a nonzero linear subspace of $M_4^-(K)$, and each of the linear
subspaces $S_i \subseteq M_4^+(K)$ is either zero, or contains $SM_4(K)$. Taking
commutators of elements from $T \otimes 1$ with elements from
$M_4^+(K) \otimes e_i$, we see that each $S_i$ is nonzero. Now,
$$
[SM_4^+(K) \otimes e_i, SM_4^+(K) \otimes e_i] =
[SM_4^+(K), SM_4^+(K)] \otimes 1 = M_4^-(K) \otimes 1 ;
$$
thus, $T = M_4^-(K)$. Finally, according to (\ref{eq-mult}), for any $i \ne j$,
we have
$$
[SM_4(K) \otimes e_i, SM_4(K) \otimes e_j] =
(SM_4(K) \circ SM_4(K)) \otimes e_{i*j} =
M_4^+(K) \otimes e_{i*j} ;
$$
thus, $S_i = M_4^+(K)$ for each $i$, and $I$ coincides with the whole algebra
$\sym^-(M_4(\mathbb O(K)),J)$.
\end{proof}
\section{$\delta$-derivations}\label{sec-der}
In \cite{petyt}, derivations of the algebras $\sym^+(M_n(\mathbb O_\mu(K)),J)$
and $\sym^-(M_n(\mathbb O_\mu(K)),J)$ were computed. Here we extend this result
by computing $\delta$-derivations of these algebras. Recall that a
\emph{$\delta$-derivation} of an algebra $A$ is a linear map $D: A \to A$ such
that
\begin{equation}\label{eq-delta}
D(xy) = \delta D(x)y + \delta xD(y)
\end{equation}
for any $x,y \in A$ and some fixed $\delta \in K$. This notion generalizes
simultaneously the notions of derivation and of centroid (any element of the
centroid is, obviously, a $\frac 12$-derivation).
The set of $\delta$-derivations of an algebra $A$, denoted by $\Der_\delta(A)$,
is a vector space. Moreover, as noted, for example, in \cite[\S 1]{filippov},
$$
[\Der_\delta(A), \Der_{\delta^\prime}(A)] \subseteq
\Der_{\delta\delta^\prime}(A) ,
$$
so the vector space $\Delta(A)$ linearly spanned by all $\delta$-derivations,
for all possible values of $\delta$, is a Lie algebra, an extension of the
Lie algebra $\Der(A)$ of (the ordinary) derivations of $A$.
\begin{theorem}\label{th-delta}
Let $D$ be a nonzero $\delta$-derivation of the algebra
$\sym^+(M_n(\mathbb O_\mu(K)),J)$ or $\sym^-(M_n(\mathbb O_\mu(K)),J)$. Then
either $\delta = 1$ (i.e., $D$ is a derivation), or $\delta = \frac 12$ and $D$
is a multiple of the identity map.
\end{theorem}
Note that $\delta$-derivations do not change under field extensions. Namely, an
obvious argument, the same as in the case of ordinary derivations, cocycles, or
any other ``linear'' structures, shows that
\begin{equation*}%\label{eq-dk}
\Der_\delta(A) \otimes_K \overline K \simeq
\Der_\delta(A \otimes_K \overline K)
\end{equation*}
for any $K$-algebra $A$, and $\delta \in K$. In view of this, it is enough to
prove the theorem in the case when $K$ is algebraically closed, and
$\mathbb O_\mu(K) = \mathbb O(K)$.
The case of $\sym^+(M_n(\mathbb O(K)),J)$ is easier, as the algebra contains
a unit, and $\delta$-derivations of algebras with unit are tackled by the simple
\begin{lemma}\label{lemma-a}
Let $D$ be a $\delta$-derivation of a commutative algebra $A$ with unit. Then
either $\delta=1$ (i.e., $D$ is a derivation), or $\delta = \frac 12$ and
$D = R_a$ for some $a \in A$ such that
\begin{equation}\label{eq-a}
2(xy)a - (xa)y - (ya)x = 0
\end{equation}
for any pair of elements $x,y \in A$.
\end{lemma}
\begin{proof}
This is, essentially, \cite[Theorem 2.1]{kayg-first} with a bit more (trivial)
details. Repeatedly substituting the unit $1$ in the equality (\ref{eq-delta})
gives that either $\delta = 1$ and $D(1) = 0$, or $\delta = \frac 12$ and
$D(x) = xD(1)$ for any $x\in A$. In the latter case, denoting $D(1) = a$, the
condition (\ref{eq-delta}) is equivalent to (\ref{eq-a}).
\end{proof}
\begin{proof}[Proof of Theorem \ref{th-delta} in the case of
$\sym^+(M_n(\mathbb O(K)),J)$]
Due to Lemma \ref{lemma-a}, it amounts to description of the algebra elements
satisfying the condition (\ref{eq-a}). Let
$$
a = m \otimes 1 + \sum_{i=1}^7 x_i \otimes e_i
$$
be such an element, where $m \in M_n^+(K)$, $x_i \in M_n^-(K)$. Writing the
condition (\ref{eq-a}) for the pair of elements $s \otimes 1$, $t \otimes 1$
where $s,t \in M_n^+(K)$, and collecting terms lying in $M_n^+(K) \otimes 1$, we
get
$$
2(s \circ t) \circ m - (s \circ m) \circ t - (t \circ m) \circ s = 0
$$
for any $s,t \in M_n^+(K)$. The latter condition means that $R_m$ is a
$\frac 12$-derivation of the Jordan algebra $M_n^+(K)$, and by
\cite[Theorem 2.5]{kayg-first}, $m = \lambda E$ for some $\lambda \in K$. Since
the set of elements satisfying the condition (\ref{eq-a}) forms a vector space
(as, generally, the set of $\frac 12$-derivations does), by subtracting from $a$
the element $\lambda E \otimes 1$, we get an element still satisfying the
condition (\ref{eq-a}), so we may assume $\lambda = 0$.
Now writing the condition (\ref{eq-a}) for $a = \sum_{i=1}^7 x_i \otimes e_i$,
and the pair $x \otimes e_k$, $y \otimes e_\ell$, where $x,y \in M_n^-(K)$ and
$k,\ell = 1,\dots,7$, $k \ne \ell$, and again collecting terms lying in
$M_n^+(K) \otimes 1$, we get $[x,y] \circ x_{k*\ell} = 0$. Since
$[M_n^-(K),M_n^-(K)] = M_n^-(K)$, and the values of $k*\ell$ run over all
$1,\dots,7$, we see that $M_n^-(K) \circ x_i = 0$ for any $i=1,\dots,7$. By
Lemma \ref{lemma-1}, $x_i = 0$, which shows that any element
$a \in \sym^+(M_n(\mathbb O(K),J))$ satisfying (\ref{eq-a}), is a multiple of
the unit.
\end{proof}
Before turning to the proof of the $\sym^-(M_n(\mathbb O(K)),J)$ case, we need a
couple of auxiliary lemmas.
\begin{lemma}\label{lemma-gl}
Let $n>2$.
\begin{enumerate}[\upshape(i)]
\item
If $\delta \ne 1,\frac 12$, then the vector space
$\Der_{\delta}(\mathfrak{gl}_n(K))$ is $1$-dimensional, and each
$\delta$-derivation is a multiple of the map $\xi$ vanishing on
$\mathfrak{sl}_n(K)$, and sending $E$ to itself.
\item
The vector space $\Der_{\frac 12}(\mathfrak{gl}_n(K))$ is $2$-dimensional, with
a basis consisting of the two maps: the map $\xi$ from part {\rm (i)}, and the
map coinciding with the identity map on $\mathfrak{sl}_n(K)$, and vanishing on
$E$.
\end{enumerate}
\end{lemma}
\begin{proof}
This follows immediately from the fact that $\mathfrak{gl}_n(K)$ is the split
central extension of $\mathfrak{sl}_n(K)$:
$\mathfrak{gl}_n(K) = \mathfrak{sl}_n(K) \oplus KE$, and the fact, established
in numerous places, that each nonzero $\delta$-derivation of
$\mathfrak{sl}_n(K)$, $n>2$, is either an ordinary derivation ($\delta = 1$), or
an element of the centroid ($\delta = \frac 12$) (see, for example,
\cite[Corollary 4.16]{leger-luks} or \cite{filippov}).
\end{proof}
\begin{lemma}\label{lemma-xdm-comm}
Let $D: M_n^+(K) \to M_n^+(K)$ be a linear map such that
\begin{equation}\label{eq-dxm}
D([x,m]) = \delta [x,D(m)]
\end{equation}
for any $x \in M_n^-(K)$, $m \in M_n^+(K)$, and some fixed $\delta \in K$,
$\delta \ne 0,1$. Then the image of $D$ lies in the one-dimensional linear space
spanned by $E$.
\end{lemma}
\begin{proof}
Replacing in the equality (\ref{eq-dxm}) $x$ by $[x,y]$, where
$x,y\in M_n^-(K)$, and using the Jacobi identity, we get:
$$
D([x,[y,m]]) - D([y,[x,m]]) = \delta [[x,y],D(m)] .
$$
Using the fact that $[x,m], [y,m] \in M_n^+(K)$, applying again (\ref{eq-dxm})
to each term at the left-hand side twice, and using the Jacobi identity, we get
$[[x,y],D(m)] = 0$. Since $[M_n^-(K),M_n^-(K)] = M_n^-(K)$, the latter equality
is equivalent to $[M_n^-(K),D(m)] = 0$. By Lemma \ref{lemma-circ}, $D(m)$ is a
multiple of $E$ for any $m \in M_n^+(K)$.
\end{proof}
When considering restrictions of $\delta$-derivations to subalgebras, we arrive
naturally at the necessity to consider a more general notion of
$\delta$-derivations with values in not necessary the algebra itself, but in a
module over the algebra. Generally, this require to consider bimodules, but as
we will need this generalization only in the case of anticommutative (in fact,
Lie) algebras, we confine ourselves here with the following definition. Let $A$
be an anticommutative algebra, and $M$ a left $A$-module, with the action of $A$
on $M$ denoted by $\bullet$. A $\delta$-derivation of $A$ with values in $M$ is
a linear map $D: A \to M$ such that
\begin{equation*}
D(xy) = - \delta y \bullet D(x) + \delta x \bullet D(y)
\end{equation*}
for any $x,y \in A$.
\begin{proof}[Proof of Theorem \ref{th-delta} in the case of
$\sym^-(M_n(\mathbb O(K)),J)$]
If $n=1$, the algebra in question is the $7$-di\-men\-si\-o\-nal simple Malcev
algebra $\mathbb O^-(K)$, and the result is covered by
\cite[Lemma 3]{filippov-ass}.
Let $n > 2$ and $\delta \ne 1$. We may write
\begin{alignat*}{3}
&D(x \otimes 1) \>&=&\> d(x) \otimes 1 &+& \sum_{i=1}^7 d_i(x) \otimes e_i
\\
&D(m \otimes e_k) \>&=&\> f_k(m) \otimes 1 &+& \sum_{i=1}^7 f_{ki}(m) \otimes e_i
\end{alignat*}
for any $x \in M_n^-(K)$, $m\in M_n^+(K)$, $k=1,\dots,7$, and some linear maps
$d: M_n^-(K) \to M_n^-(K)$, $d_i: M_n^-(K) \to M_n^+(K)$,
$f_k: M_n^+(K) \to M_n^-(K)$, and $f_{ki}: M_n^+(K) \to M_n^+(K)$.
For a fixed $k=1,\dots,7$, consider the Lie subalgebra
$$
\mathscr L^-(e_k) = M_n^-(K) \otimes 1 \>\dotplus\> M_n^+(K) \otimes e_k
$$
of $\sym^-(M_n(\mathbb O),J)$, isomorphic, as noted in \S \ref{sec-simp}, to
$\mathfrak{gl}_n(K)$ (remember that $K$ is algebraically, and, in particular,
quadratically, closed). According to decomposition (\ref{eq-decomp-minus}),
$\sym^-(M_n(\mathbb O(K)),J)$ is decomposed, as an $\mathscr L^-(e_k)$-module,
into the direct sum of the adjoint module $\mathscr L^-(e_k)$, and the module
$M_n^+(K) \otimes B_k$ (note, however, that the latter is not a Lie module).
This implies that the restriction of $D$ to $\mathscr L^-(e_k)$, being composed
with the canonical projection
$\sym^-(M_n(\mathbb O(K)),J) \to \mathscr L^-(e_k)$, i.e., the map
\begin{alignat*}{3}
&x \otimes 1 \>&\mapsto&\> d(x) \otimes 1 \>&+&\> d_k(x) \otimes e_k \\
&m \otimes e_k \>&\mapsto&\> f_k(m) \otimes 1 \>&+&\> f_{kk}(m) \otimes e_k ,
\end{alignat*}
is a $\delta$-derivation of $\mathscr L^-(e_k)$ (with values in the adjoint
module).
By Lemma \ref{lemma-gl}, either $\delta \ne \frac 12$, and each such map is of
the form
\begin{alignat*}{3}
&x \otimes 1 \>\>&\mapsto&\>\> 0 \\
&m \otimes e_k \>\>&\mapsto&\>\> 0, \quad m\in SM_n(K) \\
&E \otimes e_k \>\>&\mapsto&\>\> \mu_k E \otimes e_k
\end{alignat*}
for some $\mu_k \in K$; or $\delta = \frac 12$, and each such map is of the form
\begin{alignat*}{3}
&x \otimes 1 \>\>&\mapsto&\>\> \lambda_k x \>&\otimes&\> 1 \\
&m \otimes e_k \>\>&\mapsto&\>\> \lambda_k m \>&\otimes&\> e_k,
\quad m\in SM_n(K)
\\
&E \otimes e_k \>\>&\mapsto&\>\> \mu_k E \>&\otimes&\> e_k
\end{alignat*}
for some $\lambda_k, \mu_k \in K$. (Recall from \S \ref{ss-matrix}, that
$SM_n(K)$ denotes the space of matrices from $M_n^+(K)$ with trace zero.) Taking
into account that one of these alternatives holds uniformly for all values of
$k$, we arrive at the following two cases:
\emph{Case 1}. $\delta \ne 1, \frac 12$, and $D(M_n^-(K) \otimes 1) = 0$.
\emph{Case 2}. $\delta = \frac 12$, and $D(x \otimes 1) = \lambda x \otimes 1$
for any $x \in M_n^-(K)$ and some fixed $\lambda \in K$.
Moreover, in both cases
$$
D\big(M_n^+(K) \otimes \mathbb O^-(K)\big) \subseteq
M_n^+(K) \otimes \mathbb O^-(K) .
$$
We will handle these two cases together, keeping in mind that $\lambda = 0$ if
$\delta \ne \frac 12$.
Consider now the restriction of $D$ to $M_n^+(K) \otimes \mathbb O^-(K)$. Since
\begin{equation*}
\Hom\big(M_n^+(K) \otimes \mathbb O^-(K), M_n^+(K) \otimes \mathbb O^-(K)\big)
\simeq
\Hom\big(M_n^+(K),M_n^+(K)\big) \otimes
\Hom\big(\mathbb O^-(K),\mathbb O^-(K)\big) ,
\end{equation*}
we may write
$$
D(m \otimes a) = \sum_{i \in \mathbb I} d_i(m) \otimes \alpha_i(a)
$$
for any $m \in M_n^+(K)$, $a \in \mathbb O^-(K)$, some index set $\mathbb I$,
and linear maps $d_i: M_n^+(K) \to M_n^+(K)$,
$\alpha_i: \mathbb O^-(K) \to \mathbb O^-(K)$, $i\in \mathbb I$. Writing the
condition of $\delta$-derivation (\ref{eq-delta}) for the pair $x \otimes 1$,
$m \otimes a$, where $x \in M_n^-(K)$, $m \in M_n^+(K)$, $a \in \mathbb O^-(K)$,
we get
\begin{equation}\label{eq-22}
\sum_{i\in \mathbb I}
\Big(d_i([x,m]) - \delta [x,d_i(m)]\Big) \otimes \alpha_i(a)
= \delta\lambda [x,m] \otimes a .
\end{equation}
In Case 1 the right-hand side of (\ref{eq-22}) vanishes, and hence we may assume
$d_i([x,m]) = \delta [x,d_i(m)]$ for any $x\in M_n^-(K)$, $m\in M_n^+(K)$, and
any $i\in \mathbb I$. By Lemma \ref{lemma-xdm-comm}, each $d_i(m)$ is a multiple
of $E$, and hence
$D(M_n^+(K) \otimes \mathbb O^-(K)) \subseteq E \otimes \mathbb O^-(K)$. But
then writing (\ref{eq-delta}) for the pair $m\otimes a$, $s\otimes b$, where
$m,s\in M_n^+(K)$, $a,b\in \mathbb O^-(K)$, and taking into account
(\ref{eq-mult}), we get $D((m \circ s) \otimes [a,b]) = 0$. Since
$(M_n(K),\circ)$ and $(\mathbb O^-(K), \liebrack)$ are perfect (in fact, simple)
algebras, the latter equality implies vanishing of $D$ on the whole
$M_n^+(K) \otimes \mathbb O^-(K)$, and thus on the whole
$\sym^-(M_n(\mathbb O(K)),J)$, a contradiction.
Hence, we are in Case 2, and $\delta = \frac 12$. Setting in this case
$d_\star = -\lambda \id_{M_n^+(K)}$, and $\alpha_\star = \id_{\mathbb O^-(K)}$,
the equality (\ref{eq-22}) can be rewritten as
$$
\sum_{i\in \mathbb I \cup \{\star\}}
\Big(d_i([x,m]) - \frac 12 [x,d_i(m)]\Big) \otimes \alpha_i(a) = 0 .
$$
As in the previous case, this means that there are new linear maps
$\widetilde d_i$, $\widetilde \alpha_i$ which are linear combinations of $d_i$
and $\alpha_i$, respectively, and such that
\begin{equation}\label{eq-tilde}
\sum_{i \in \mathbb I \cup \{\star\}}
\widetilde d_i \otimes \widetilde \alpha_i =
\sum_{i \in \mathbb I \cup \{\star\}} d_i \otimes \alpha_i ,
\end{equation}
and $\widetilde d_i([x,m]) = \frac 12 [x, \widetilde d_i(m)]$.
Lemma~\ref{lemma-xdm-comm} tells us, as previously, that each
$\widetilde d_i(m)$ is a multiple of $E$, and hence the image of the map in the
left-hand side of (\ref{eq-tilde}) lies in $E \otimes \mathbb O^-(K)$. Since the
right-hand side of (\ref{eq-tilde}) is equal to
$D + d_\star \otimes \alpha_\star$, we have
$$
D(m \otimes a) = \lambda m \otimes a + E \otimes \beta(m,a)
$$
for any $m \in M_n^+(K)$, $a \in \mathbb O^-(K)$, and some bilinear map
$\beta: M_n^+(K) \times \mathbb O^-(K) \to \mathbb O^-(K)$. Replacing $D$ by the
$\frac 12$-derivation $D - \lambda \id$, we arrive at the situation as in the
previous case: a $\delta$-derivation (with $\delta = \frac 12$) vanishing on
$M_n^-(K) \otimes 1$, and taking values in $E \otimes \mathbb O^-(K)$ on
$M_n^+(K) \otimes \mathbb O^-(K)$. Hence, $D - \lambda \id$ vanishes on the
whole $\sym^-(M_n(\mathbb O(K)),J)$, and $D = \lambda \id$, as claimed.
Finally, consider the case $n=2$. In this case Lemma \ref{lemma-gl} is not
applicable: in addition to the cases described in Lemma, there is the
$5$-dimensional space of $(-1)$-derivations of $\mathfrak{sl}_2(K)$, and thus
the corresponding $6$-dimensional space of $(-1)$-derivations of
$\mathfrak{gl}_2(K)$ (see \cite[Example 1.5]{hopkins} or
\cite[Example in \S 3]{filippov-5}). In view of this, to proceed like in the
proof of the case $n>2$, considering $\delta$-derivations of the Lie subalgebras
$\mathscr L^-(e_k)$, would be too cumbersome, and we are taking a somewhat
alternative route.
Denote by $H = \left(\begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix}\right)$ the
basis element of the $1$-dimensional space $M_2^-(K)$. Consider the subalgebra
$E \otimes \mathbb O^-(K)$ of $\sym^+(M_2(\mathbb O(K)),J)$, isomorphic to the
$7$-dimensional simple Malcev algebra $\mathbb O^-(K)$. As an
$E \otimes \mathbb O^-(K)$-module, $\sym^+(M_2(\mathbb O(K)),J)$ decomposes into
the direct sum of the trivial $1$-dimensional module $KH \otimes 1$, and the
module $M_2^+(K) \otimes \mathbb O^-(K)$ which is isomorphic to the direct sum
of $3$ copies of the adjoint module ($\mathbb O^-(K)$ acting on itself). Thus
$D$, being restricted to $E \otimes \mathbb O^-(K)$, is equal to the sum of a
$\delta$-derivation with values in the trivial module, which is obviously zero,
and $3$ $\delta$-derivations of $\mathbb O^-(K)$. By the result mentioned at the
beginning of this proof, the latter $\delta$-derivations are zero if
$\delta \ne 1,\frac 12$, and are multiples of the identity map if
$\delta = \frac 12$. Consequently, $D(E \otimes a) = m_0 \otimes a$ for any
$a \in \mathbb O^-(K)$, and some fixed $m_0 \in M_2^+(K)$.
Now write
$$
D(H \otimes 1) = \lambda H \otimes 1 + \sum_{i=1}^7 m_i \otimes e_i
$$
for some $\lambda \in K$, and $m_i \in M_2^+(K)$. Writing the condition of
$\delta$-derivation (\ref{eq-delta}) for the pair $H \otimes 1$,
$E \otimes e_k$, where $k=1,\dots,7$, we get
$$
2\sum_{1\le i \le 7, i\ne k} (\pm m_i \otimes e_{i*k}) +
[H,m_0] \otimes e_k = 0 .
$$
It follows that $m_i = 0$ for each $i=1,\dots,7$, and
$D(H \otimes 1) = \lambda H\otimes 1$.
Now let
$$
D(m \otimes a) = \beta(m,a) H \otimes 1 +
\text{terms lying in } M_2^+(K) \otimes \mathbb O^-(K)
$$
for any $m\in M_2^+(K)$, $a\in \mathbb O^-(K)$, and some bilinear map
$\beta: M_2^+(K) \otimes \mathbb O^-(K) \to K$. Writing the condition of
$\delta$-derivation for the pair $H \otimes 1$, $m \otimes a$, and collecting
terms which are multiples of $H \otimes 1$, we see that
$\beta(m,a) H \otimes 1 = 0$. Thus,
$$
D\big(M_2^+(K) \otimes \mathbb O^-(K)\big) \subseteq
M_2^+(K) \otimes \mathbb O^-(K) ,
$$
and we may proceed as in the generic case $n>2$ above.
\end{proof}
Note that it is also possible to pursue the case $\delta = 1$ along the same
lines; this would give us an alternative proof of the results of \cite{petyt},
as well as of the classical result that derivation algebra of the
$27$-dimensional exceptional simple Jordan algebra is isomorphic to the simple
Lie algebra of type $F_4$.
There is a vast literature devoted to $\delta$-derivations of algebras and
related notions (for a small, but representative sample, see \cite{hopkins},
\cite{filippov-5}--\cite{filippov-ass}, \cite{kayg-first}, \cite{leger-luks}).
Our strategy to prove Theorem \ref{th-delta} was to identify certain Lie
subalgebras of the algebra $\sym^-(M_n(\mathbb O(K)),J)$, and consider
$\delta$-derivations of those subalgebras with values in the whole
$\sym^-(M_n(\mathbb O(K)),J)$. Developing further the methods of the above-cited
papers, it is possible to prove that $\delta$-derivations of semisimple Lie
algebras of classical type with coefficients in finite-dimensional modules are
either (inner) derivations, or multiples of the identity map on irreducible
constituents of the module isomorphic to the adjoint module of the algebra, or,
in the case of the direct summands in the algebra isomorphic to
$\mathfrak{sl}_2(K)$, $(-1)$-derivations with values in the irreducible
constituents isomorphic to the adjoint $\mathfrak{sl}_2(K)$-modules. This
general fact would allow us to further simplify the proof of Theorem \ref{th-delta}, but establishing it would require
considerable (though pretty much straightforward) efforts, and would lead us far away from the topic of this paper. We hope to return to this elsewhere.
Since by \cite{petyt}, both $\Der(\sym^+(M_n(\mathbb O_\mu(K)),J))$ for
$n\ge 4$, and $\Der(\sym^-(M_n(\mathbb O_\mu(K)),J))$ for any $n$ are isomorphic
to the Lie algebra $G_2 \oplus \mathfrak{so}_n(K)$, then by
Theorem~\ref{th-delta}, both $\Delta(\sym^+(M_n(\mathbb O_\mu(K)),J))$ and
\linebreak
$\Delta(\sym^-(M_n(\mathbb O_\mu(K)),J))$ are isomorphic to the one-dimensional trivial central extension of
$G_2 \oplus \mathfrak{so}_n(K)$.
Finally, note an important
\begin{corollary}
The algebras $\sym^+(M_n(\mathbb O_\mu(K)),J)$ and
$\sym^-(M_n(\mathbb O_\mu(K)),J)$ are central simple.
\end{corollary}
\begin{proof}
By Theorem \ref{th-1}, these algebras are simple, and by Theorem \ref{th-delta}
their centroid coincides with the ground field.
\end{proof}
\section{Symmetric associative forms}\label{sec-form}
Let $A$ be an algebra. A bilinear symmetric form $\varphi: A \times A \to K$ is
called \emph{associative}, if
\begin{equation}\label{eq-f}
\varphi(xy,z) = \varphi(x,yz)
\end{equation}
for any $x,y,z \in A$. (In the context of Lie algebras, associative forms are
usually called \emph{invariant}, because in that case the condition (\ref{eq-f})
is equivalent to invariance of the form $\varphi$ with respect to the standard
action of the underlying Lie algebra on the space of symmetric bilinear forms.)
For a matrix $X = (a_{ij})$ from $M_n(\mathbb O_\mu(K))$, by $\overline X$ we
will understand the matrix $(\overline{a_{ij}})$, obtained by element-wise
application of conjugation in $\mathbb O_\mu(K)$.
\begin{theorem}\label{th-form}
Any bilinear symmetric associative form on $\sym^+(M_n(\mathbb O_\mu(K)),J)$, or
on $\sym^-(M_n(\mathbb O_\mu(K)),J)$, is a scalar multiple of the form
\begin{equation}\label{eq-form}
(X, Y) \mapsto \Tr\,(XY + \overline X \, \overline Y) .
\end{equation}
\end{theorem}
The form (\ref{eq-form}) is reminiscent of the Killing form on simple Lie
algebras of classical type, and of the generic trace form on simple Jordan
algebras (and \emph{is} such a form when restricted from the algebra
$\sym^+(M_n(\mathbb O_\mu(K)),J)$ to its Jordan subalgebra $M_n(K)$, and from
the algebra $\sym^-(M_n(\mathbb O_\mu(K)),J)$ to its Lie subalgebra
$\mathfrak{so}_n(K)$, see below).
\begin{proof}
According to Corollary in \S \ref{sec-der}, both algebras are central simple.
The standard linear algebra arguments show that any bilinear symmetric
associative form on a simple algebra is nondegenerate, and that any two
nondegenerate symmetric associative forms on a finite-dimensional central
algebra differ from each other by a scalar (see, e.g.,
\cite[pp.~30--31, Exercise 15(b)]{kap}). Thus, the vector space of bilinear
symmetric associative forms on a finite-dimensional central simple algebra is
either $0$- or $1$-dimensional.
Now it remains to observe that in both cases this space is $1$-dimensional by
verifying that the form (\ref{eq-form}) is indeed associative. The most
convenient way to do this is, perhaps, to rewrite the form in terms of
decompositions (\ref{eq-dec}) or (\ref{eq-decomp-minus}). On the algebra
$\sym^+(M_n(\mathbb O_\mu(K)),J)$ we obtain
\begin{alignat*}{3}
&(m \otimes 1, s \otimes 1) &\>\>\mapsto\>\>& 2\Tr(ms) \\
&(m \otimes 1, x \otimes a) &\>\>\mapsto\>\>& 0 \\
&(x \otimes a, y \otimes b) &\>\>\mapsto\>\>& (ab+ba) \Tr(xy) ,
\end{alignat*}
and on $\sym^-(M_n(\mathbb O_\mu(K)),J)$,
\begin{alignat*}{3}
&(x \otimes 1, y \otimes 1) &\>\>\mapsto\>\>& 2\Tr(xy) \\
&(x \otimes 1, m \otimes a) &\>\>\mapsto\>\>& 0 \\
&(m \otimes a, s \otimes b) &\>\>\mapsto\>\>& (ab+ba) \Tr(ms) .
\end{alignat*}
Here, as before in this paper, $x,y\in M_n^-(K)$, $m,s\in M_n^+(K)$, and
$a,b \in \mathbb O_\mu^-(K)$. (For the algebra
\linebreak
$\sym^+(M_n(\mathbb O_\mu(K)),J)$, the associativity follows also from
\cite[Satz 5.2]{ruhaak}, where it is proved that the form (\ref{eq-form}) is a
symmetric associative form on a larger algebra
$(M_n(\mathbb O_\mu(K)), \circ)$.)
\end{proof}
Note that it is possible to get an alternative, direct proof of
Theorem~\ref{th-form} without appealing to results of \S \ref{sec-der}, in the
linear algebra spirit of the proofs of Proposition \ref{prop-1} and
Theorem \ref{th-delta}.
\section{Further questions}\label{sec-q}
1)
Compute automorphism groups of the algebras $\sym^+(M_n(\mathbb O_\mu(K)),J)$
and $\sym^-(M_n(\mathbb O_\mu(K)),J)$. Are they isomorphic to $G_2 \times SO(n)$?
\smallskip
2)
For $n>3$, the algebras $\sym^+(M_n(\mathbb O_\mu(K)),J)$ are no longer Jordan.
How ``far'' they are from Jordan algebras? Which identities these algebras do
satisfy? (The last question was also asked in \cite{bremner-hentzel}, where it
is proved that $\sym^+(M_4(\mathbb O_{}(\mathbb Q)),J)$ does not satisfy
nontrivial identities of degree $\le 6$.) A starting point could be
investigation of (non-Jordan) representations of the Jordan subalgebras which
are forms of the full matrix Jordan algebra $M_n(K)$, mentioned in
\S~\ref{sec-simp}, in the whole $\sym^+(M_n(\mathbb O_\mu(K)),J)$.
\smallskip
3)
What can one say about subalgebras of the algebras in question? Say, what are
the maximal subalgebras? Maximal Jordan subalgebras of
$\sym^+(M_n(\mathbb O_\mu(K)),J)$? Some low-dimensional subalgebras of
$\sym^+(M_4(\mathbb O(\mathbb R)),J)$ were exhibited in
\cite[pp.~34--37]{jordan} (see also \cite[p.~37]{phd-new}). These subalgebras
belong to the class of so-called elementary algebras, defined by a certain identity of degree $5$. In that old
and seemingly forgotten paper, Jordan suggested to investigate which other
elementary subalgebras the octonionic matrix algebras may contain.
\smallskip
4)
Idempotents play an important role in Jordan algebras. Find idempotents in
$\sym^+(M_n(\mathbb O_\mu(K)),J)$. This amounts to solving a system of quadratic
equations in the Lie algebra $\mathfrak{so}_n(K)$.
\smallskip
5)
In \cite{sagle} it is proved that any anticommutative algebra with a bilinear
symmetric associative form is isomorphic to a ``minus'' algebra $A^{(-)}$ of a
noncommutative Jordan algebra $A$. In view of Theorem \ref{th-form}, which
noncommutative Jordan algebras arise in this way in connection with the algebras
$\sym^{-}(M_n(\mathbb O_\mu(K)),J)$?
\smallskip
6)
Investigate the case of characteristic $3$. Though this case is, perhaps, of
little interest for physics, in characteristic $3$ the $7$-dimensional algebra
$\mathbb O_\mu^-(K)$ is not merely a Malcev algebra, but isomorphic to a form of
the Lie algebra $\mathfrak{psl}_3(K)$ (see, for example,
\cite[Theorem 4.26]{elduque-kochetov}). This suggests that the algebras
$\sym^+(M_n(\mathbb O_\mu(K)),J)$ and $\sym^-(M_n(\mathbb O_\mu(K)),J)$ in this
characteristic may satisfy a different set of identities than in the generic
case, perhaps, more tractable and more closer to the classical identities (Jacobi, Jordan, etc.).
\smallskip
Note that, unlike the questions treated in this paper, some of these questions
are sensitive to the ground field, and are related to the subtle behavior of
quadratic forms, etc.
\section*{Acknowledgements}
Thanks are due to Francesco Toppan and Bernd Henschenmacher, who explained the
importance of algebras considered here, and pointed us to the relevant
literature (\cite{ruhaak}, \cite{jordan}, \cite{toppan}), and to Dimitry Leites
and the anonymous referee for significant improvements to the previous version
of the paper. GAP \cite{gap} was utilized to check some of the computations performed in the paper.
Arezoo Zohrabi was supported by grant SGS01/P\v{r}F/20-21 of the University of
Ostrava.
\begin{thebibliography}{KMRT}
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