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##  Appendix A Direct Sums of EJAs

The first part of this Appendix collects some basic facts about direct sums of
EJAs that are used in the body of the paper. The second part contains a proof
that $C^{\ast}(A\oplus B)=C^{\ast}(A)\oplus C^{\ast}(B)$, and that the direct
sum of universally reversible EJAs is again UR.

Direct Summands and Central Projections The direct sum of EJAs $A$ and $B$ is
$A\oplus B:=A\times B$, equipped with the slotwise operations, so that the
canonical projections $\pi_{1}:A\times B\rightarrow A$ and $\pi_{2}:A\times
B\rightarrow B$ are unital Jordan homomorphisms. Identifying $A$ and $B$ with
$A\times\\{0\\}$ and $\\{0\\}\times B$, respectively, we write $a+b$ for
$(a,0)+(0,b)$. Note that $A$ and $B$ are then ideals in $A\oplus B$, and that
$B=A^{\perp}:=\\{z\in A\oplus B|\langle a,z\rangle=0\ \forall a\in A\\}$.
Conversely, it’s easy to check that if $E$ is an EJA and $A$ is an ideal in
$A$, then $A^{\perp}$ is also an ideal, and $ab=0$ for all $a\in A,b\in
A^{\perp}$; hence, $E\simeq A\oplus A^{\perp}$.

Suppose $E$ is an EJA and $A\leq E$ is an ideal: let $B=A^{\perp}$. Then for
all $z\in E$,

$\langle a,zb\rangle=\langle az,b\rangle=0$

since $az\in A$. Thus, $B$ is also an ideal, and $E=A\oplus B$ as a vector
space. Finally, if $a\in A$ and $b\in B$, then $ab\in A\cap B=\\{0\\}$. Hence,
if $a,x\in A$ and $b,y\in B$ then $(a+b)(x+y)=ax+by$, i.e., in the
representation of $A\oplus B$ as $A\times B$, operations are slotwise. What is
not entirely obvious is that $A$ contains a unit element.

######  Lemma A.1.

Let $A$ be an ideal in an EJA $E$. Then there exists a projection $p\in E$
such that $pa=a$ for every $a\in A$. Thus, $A=pA$, and $E=pA\oplus
p^{\prime}A$, where $p^{\prime}=1-p$.

For a proof in the more general setting of JBW algebras, see [3], Propositions
2.7, 2.39 and 2.41.

The center of an EJA $E$ is the set of elements operator-commuting with all
other elements. Denote this by $C(E)$ If $E=A\oplus B$, and $p$ is the unit of
$A$, so that $A=pA$, then it’s easy to check that $p\in C(E)$. Conversely, if
$p$ is a central projection, then $pA$ is an ideal, with unit element $p$. If
$p$ is a minimal central projection, then $pA$ is a minimal direct summand of
$E$. If $E$ is simple, then its only central projections are $0$ and $1$, and
conversely.

One can show that for every projection $e$ in an EJA $E$, there exists a
unique miniml projection $c(e)\in C(E)$, the central cover of $p$, such that
$e\leq c(e)$. Then $A:=c(p)E$ is an ideal of $E$, in which $c(p)$ is the unit.
If $A$ is a minimal ideal, then elements of $A$ are exactly those with central
cover $c(p)$. [[3], 2.37, 2.39]. More generally, two elements of $E$ having
the same central cover have nonzero components in exactly the same ideals of
$E$.

Recall that a symmetry of $A$ is an element $s\in A$ with $s^{2}=u$.
Projections $e,f$ in $A$ are exchanged by a symmetry $s$ iff $U_{s}(e)=f$. If
there exists a sequence of symmetries $s_{1},...,s_{n}$ with
$U_{s_{n}}\circ\cdots\circ U_{s_{1}}(e)=f$, then $e$ and $f$ are equivalent.

######  Lemma A.2 ([3], Lemma 3.9).

Equivalent projections have the same central cover.

The universal $C^{\ast}$-algebra of a direct sum Recall that a sequence of
vector spaces and linear maps, or of Jordan algebras and Jordan homomorphisms,
or of $C^{\ast}$ algebras and $\ast$-homomorphisms

$A\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}B\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}C$

is said to be exact at $B$ iff the image of $\alpha$ is the kernel of $\beta$.
A short exact sequence is one of the form

$0\longrightarrow
A\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}B\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}C\longrightarrow
0$

that is exact at $A$, $B$ and $C$ (with the maps on the ends being the only
possible ones). This means that $\alpha$ is injective (its kernel is $0$),
while $\beta$ is surjective (its image is the kernel of the zero map, i.e.,
all of $C$).

Let EJA and Cstar be the categories of EJAs and Jordan homomorphisms, and of
$C^{\ast}$-algebras and $\ast$-homomorphisms, respectively.

######  Theorem A.3 ([21]).

$A\mapsto C^{\ast}(A)$ is an exact functor from EJA to Cstar. In other words,
if
$A\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}B\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}C$
is an exact sequence in EJA, then $C^{\ast}(A)\stackrel{{\scriptstyle
C^{\ast}(\alpha)}}{{\longrightarrow}}B\stackrel{{\scriptstyle
C^{\ast}(\beta)}}{{\longrightarrow}}C^{\ast}(C)$ is an exact sequence in
Cstar.

We are going to use this to show that $C^{\ast}(A\oplus B)=C^{\ast}(A)\oplus
C^{\ast}(B)$. We need some preliminaries. The following is standard:

######  Lemma A.4.

Let

$0\longrightarrow
A\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}C\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}B\longrightarrow
0$

be a short exact sequence of vector spaces. Then the following are equivalent:

  * (a)

There is an isomorphism $\phi:A\oplus B\simeq C$ such that $\alpha$ and
$\beta$ are respectively the canonical injection and surjection given by

$\alpha(a)=\phi(a,0)\ \ \mbox{and}\ \ \beta(\phi(a,b))=b$

  * (b)

The sequence is split at $B$: there exists a linear mapping
$\gamma:B\rightarrow C$ such that $\beta\circ\gamma=\mbox{id}_{B}$.

The idea is that, given $\phi$, we can define $\gamma$ by
$\gamma(b)=\phi(0,b)$ and, given $\gamma$, we can define $\phi$ by
$\phi(a,b)=\alpha(a)+\gamma(b)$.

If $A$, $B$ and $C$ are Jordan algebras or $C^{\ast}$ algebras, the
implication from (a) to (b) is obviously valid, but the converse requires
additional assumptions.

######  Lemma A.5.

Let

$0\longrightarrow
A\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}C\stackrel{{\scriptstyle\beta}}{{\longrightarrow}}B\longrightarrow
0.$

be a short exact sequence of $\ast$-algebras and $\ast$-homomorphisms, split
by a $\ast$-homomorphism $\gamma:B\rightarrow C$ with
$\beta\circ\gamma=\mbox{id}_{B}$. Let $\phi:A\oplus B\rightarrow C$ be as
defined above, i.e, $\phi(a,b)=\alpha(a)+\gamma(b)$ for $a\in A$, $b\in B$.
Then the following are equivalent:

  * (a)

$\gamma(B)$ is a (2-sided) $\ast$-ideal in $C$;

  * (b)

$\phi$ is multiplicative, and thus a $\ast$-isomorphism;

  * (c)

There exists a $\ast$-homomorphism $\delta:C\rightarrow A$ with

$0\longleftarrow
A\stackrel{{\scriptstyle\delta}}{{\longleftarrow}}C\stackrel{{\scriptstyle\gamma}}{{\longleftarrow}}B\longleftarrow
0$

exact.

Proof: (a) $\Rightarrow$ (b). It is easy to see that a $C^{\ast}$-algebra $C$
is the direct sum of two $\ast$-ideals $A,B\leq C$ iff $A+B=C$ and $A\cap
C=\\{0\\}$, i.e., iff $C$ is the vector-space direct sum of $A$ and $B$. We
already know that $\alpha(A)+\beta(B)=C$ (since $\phi$ is a linear
isomorphism). It therefore suffices to show that $\alpha(A)$ and $\gamma(B)$
are $\ast$-ideals with zero intersection. We are assuming that $\gamma(B)$ is
a $\ast$ ideal. As it’s the kernel of a $\ast$-homomorphism, $\alpha(A)$ is
automatically a $\ast$-ideal. To see that $\alpha(A)\cap\gamma(C)=\\{0\\}$,
let $c\in C$ with $c=\alpha(a)=\gamma(b)$ for some $a\in A$ and $b\in B$. Then
we have

$b=\beta(\gamma(b))=\beta(\alpha(a))=0$

whence, $c=\gamma(0)=0$.

(b) $\Rightarrow$ (c). If $\phi$ is a $\ast$-isomorphism, then let
$\delta=\pi_{A}\circ\phi^{-1}$ where $\pi_{A}:A\oplus B\rightarrow A$ is the
projection $\pi_{A}(a,b)=a$. Note that $\delta$ is the composition of two
$\ast$-homomorphisms, and thus, a $\ast$-homomorphism. To verify exactness,
note that as $\phi(a,b)=\alpha(a)+\gamma(b)$, we have $\phi(0,b)=\gamma(b)$,
whence, $\phi^{-1}(\gamma(b))=(0,b)$. Thus,
$\delta(\gamma(b)){=\pi_{A}(0,b)}=0$.

(c) $\Rightarrow$ (a). By exactness, $\gamma(C)$ is the kernel of the
$\ast$-homomorphism $\delta$, and hence, a $\ast$-ideal. $\Box$

Now let $E=A\oplus B$. Then we have a short exact sequence

$0\longrightarrow A\stackrel{{\scriptstyle j}}{{\longrightarrow}}A\oplus
B\stackrel{{\scriptstyle p}}{{\longrightarrow}}B\longrightarrow 0.$

where $j(a)=(a,0)$ and $p(a,b)=b$. This is split by the homomorphism
$k:A\rightarrow A\oplus B$ given by $k(b)=(0,b)$. Hanche-Olsen’s exactness
theorem (2.1) gives us a short exact sequence

$0\longrightarrow C^{\ast}(A)\stackrel{{\scriptstyle
C^{\ast}(j)}}{{\longrightarrow}}C^{\ast}(A\oplus B)\stackrel{{\scriptstyle
C^{\ast}(p)}}{{\longrightarrow}}C^{\ast}(B)\longrightarrow 0.$

By functoriality, $C^{\ast}(p)\circ C^{\ast}(j)=\mbox{id}_{C^{\ast}(B)}$, so
this is again split. Thus, regarded as a vector space, $C^{\ast}(A\oplus B)$
is canonically isomorphic to $C^{\ast}(A)\oplus C^{\ast}(B)$. On the other
hand, we also have an exact sequence

$0\longleftarrow A\stackrel{{\scriptstyle q}}{{\longleftarrow}}A\oplus
B\stackrel{{\scriptstyle k}}{{\longleftarrow}}B\longleftarrow 0$

where $q(a,b)=a$. By the same argument, then, we have a short exact sequence

$0\longleftarrow C^{\ast}(A)\stackrel{{\scriptstyle
C^{\ast}(q)}}{{\longleftarrow}}C^{\ast}(A\oplus B)\stackrel{{\scriptstyle
C^{\ast}(k)}}{{\longleftarrow}}C^{\ast}(B)\longleftarrow 0.$

Applying the preceding Lemma, we have

######  Proposition A.6.

If $A$ and $B$ are EJAs, then

$C^{\ast}(A\oplus B)\simeq C^{\ast}(A)\oplus C^{\ast}(B).$

Notice that if $\Phi_{A}$ and $\Phi_{B}$ are, respectively, the canonical
involutions on $C^{\ast}(A)$ and $C^{\ast}(B)$ fixing points of $A$ and $B$,
then $\Phi_{A}\oplus\Phi_{B}$ is a $\ast$-involution on $C^{\ast}(A)\oplus
C^{\ast}(B)$ fixing points of $A\oplus B$. But there is only one such
$\ast$-involution on $C^{\ast}(A\oplus B)$, the canonical one. In other words,
in identifying $C^{\ast}(A\oplus B)$ with $C^{\ast}(A)\oplus C^{\ast}(B)$, we
also identify $\Phi_{A\oplus B}$ with $\Phi_{A}\oplus\Phi_{B}$.

Recalling now the fact ([21], Lemma 4.2) that an EJA $A$ is universally
reversible (UR) iff $A$ coincides with the set of self-adjoint fixed points in
$C^{\ast}(A)$ of the canonical $\ast$-involution $\Phi_{A}$, we have the

######  Corollary A.7.

If $A$ and $B$ are UR, then so is $A\oplus B$.

Proof: Let $\Phi=\Phi_{A}\oplus\Phi_{B}$ be the canonical involution on
$C^{\ast}(A\oplus B)=C^{\ast}(A)\oplus C^{\ast}(B)$. For $(a,b)\in
C^{\ast}(A)\oplus C^{\ast}(B)$, we have
$\Phi(a,b)=(\Phi_{A}(a),\Phi_{B}(b))=(a,b)$ iff $\Phi_{A}(a)=a$ and
$\Phi_{B}(b)=b$. Since $A$ and $B$ are UR, this holds iff $a\in A$ and $b\in
B$, i.e., iff $(a,b)\in A\oplus B$. Thus, $A\oplus B$ is exactly the set of
fixed-points of $\Phi$, and so, is UR. $\Box$

##  Appendix B The Quabit

In this appendix, we show that the standard tensor product $Q_{2}\odot Q_{2}$
of two quabits is $R_{16}$, the self-adjoint part of the real matrix algebra
$M_{1}6(\mathbb{R})$.

The symplectic representation A quaternion $q=a+bi+cj+dk$ can alternatively be
expressed in the form $(a+bi)+(c+di)j$, i.e., $z+wj$ where $z,w\in\mathbb{C}$,
and also has a standard representation as a $2\times 2$ complex matrix, namely

$[q]:=\left[\begin{array}[]{cc}z&w\\\
-\overline{w}&\overline{z}\end{array}\right].$

Treating $\mathbb{H}$ as a $\ast$-algebra over $\mathbb{C}$, the mapping
$q\mapsto[q]$ is a $\ast$-homomorphism from $\mathbb{H}$ into
$M_{2}(\mathbb{C})$. This gives us a natural representation — that is,
$\ast$-homomorphism — $\pi_{o}:M_{n}(\mathbb{H})\rightarrow
M_{n}(M_{2}(\mathbb{C}))\simeq M_{2n}(\mathbb{C})$, namely,

$\pi_{o}(a)_{p,q}=[a_{p,q}]$

for $a\in M_{n}(\mathbb{H})$ and $p,q=1,...,n$. An equivalent, but for our
purposes, more useful, representation is given by

$\pi(a)=F\pi_{o}(a)F$

where $F=\left[\begin{array}[]{cccc}1&0&0&0\\\ 0&0&1&0\\\ 0&1&0&0\\\
0&0&0&1\end{array}\right]$. This is called the symplectic representation of
$M_{n}(\mathbb{H})$.

If we express each entry of $a$ in the form $a_{p,q}=z_{p,q}+w_{p,q}j$, then
$a=\Gamma_{1}+\Gamma_{2}j$ where $(\Gamma_{1})_{p,q}=z_{p,q}$ and
$(\Gamma_{2})_{p,q}=w_{p,q}$. Computing, one finds that

$\pi(a)=\left[\begin{array}[]{cr}\Gamma_{1}&\Gamma_{2}\\\
-\overline{\Gamma}_{2}&\overline{\Gamma}_{1}\end{array}\right].$

Notice that $\pi(a)$ is self-adjoint iff $\Gamma_{1}$ is self-adjoint and
$\Gamma_{2}$ is anti-symmetric, and that this is the case iff $a$ is self-
adjoint in $M_{2}(\mathbb{H})$.

From now on, we identify $Q_{2}$ with $\pi(M_{2}(\mathbb{H})_{{sa}})\leq
M_{4}(\mathbb{C})_{{sa}}$. Regarded as an embedded EJA in this way, the
standard tensor product $Q_{2}\odot Q_{2}$ is defined to be the Jordan
subalgebra of $M_{4}(\mathbb{C})\otimes M_{4}(\mathbb{C})=M_{16}(\mathbb{C})$
generated by $Q_{2}\odot Q_{2}$. Our main goal in this Appendix is to prove

######  Proposition B.1.

$C^{\ast}(Q_{2}\odot Q_{2})\simeq M_{16}(\mathbb{C})$.

Since $Q_{2}\odot Q_{2}$ is UR, this will follow from ([21], Theorem 4.4) if
we can show that (i) $Q_{2}\odot Q_{2}$ generates $M_{16}(\mathbb{C})$ as a
$\ast$-algebra, and (ii) there is a $\ast$-involution on $M_{16}(\mathbb{C})$
fixing elements of $Q_{2}\odot Q_{2}$ pointwise.

Quaternionic Pauli Matrices In order to show that $Q_{2}\odot Q_{2}$ generates
$M_{16}(\mathbb{C})$, We begin by writing down some useful elements of
$Q_{2}$. The analogues of the Pauli matrices
$\sigma_{x},\sigma_{y},\sigma_{z}\in M_{2}(\mathbb{C})$ are the following
_quaternionic Pauli matrices_ :

$\begin{array}[]{ccccc}\sigma_{o}=\left[\begin{array}[]{cc}1&0\\\
0&1\end{array}\right]&&\sigma_{1}=\left[\begin{array}[]{cr}1&0\\\
0&-1\end{array}\right]&&\sigma_{2}=\left[\begin{array}[]{cc}0&1\\\
1&0\end{array}\right]\\\ &&&&\\\ \sigma_{3}=\left[\begin{array}[]{cr}0&-i\\\
i&0\end{array}\right]&&\sigma_{4}=\left[\begin{array}[]{cr}0&-j\\\
j&0\end{array}\right]&&\sigma_{5}=\left[\begin{array}[]{cr}0&-k\\\
k&0\end{array}\right]\end{array}$

Evidently, $\sigma_{1},\sigma_{2}$ and $\sigma_{3}$ are the standard Pauli
matrices $\sigma_{z}$, $\sigma_{x}$ and $\sigma_{y}$, respectively. Note that
these are all traceless and self-adjoint. Applying the representation $\pi$
gives us six elements of $Q_{2}$, $s_{o}=\pi(\sigma_{o})$,
$s_{1}=\pi(\sigma_{1})$, …, $s_{5}=\pi(\sigma_{5})$. Direct computation
reveals that

$\begin{array}[]{ccccc}s_{o}=\left[\begin{array}[]{cc}\mathbf{1}&0\\\
0&\mathbf{1}\end{array}\right]=\sigma_{o}\otimes\sigma_{o}&&s_{1}=\left[\begin{array}[]{cr}\sigma_{z}&0\\\
0&\sigma_{z}\end{array}\right]=\sigma_{o}\otimes\sigma_{z}&&s_{2}=\left[\begin{array}[]{cc}\sigma_{x}&0\\\
0&\sigma_{x}\end{array}\right]=\sigma_{o}\otimes\sigma_{x}\\\ &&&&\\\
s_{3}=\left[\begin{array}[]{cc}\sigma_{y}&0\\\
0&-\sigma_{y}\end{array}\right]=\sigma_{z}\otimes\sigma_{y}&&s_{4}=\left[\begin{array}[]{cr}0&-i\sigma_{y}\\\
i\sigma_{y}&0\end{array}\right]=\sigma_{y}\otimes\sigma_{y}&&s_{5}=\left[\begin{array}[]{cc}0&\sigma_{y}\\\
\sigma_{y}&0\end{array}\right]=\sigma_{x}\otimes\sigma_{x}\\\ \end{array}$

These again obey the Pauli-like identities above. Using these, we can compute
(associative) products of these matrices, e.g.,

$s_{3}s_{4}=(\sigma_{z}\otimes\sigma_{y})(\sigma_{y}\otimes\sigma_{y})=\sigma_{z}\sigma_{y}\otimes\sigma_{y}\sigma_{y}=-i\sigma_{x}\otimes\sigma_{o}.$

######  Lemma B.2.

$Q_{2}\otimes Q_{2}$ generates $M_{16}(\mathbb{C})$ as a $\ast$-algebra.

Proof: Begin by noting that the elements

$\sigma_{a}\otimes\sigma_{b}\otimes\sigma_{c}\otimes\sigma_{c},$

where $a,b,c,d\in\\{0,x,y,z\\}$, are a basis for $M_{16}(\mathbb{C})$. For
each $a\in\\{0,x,y,z\\}$, let

$x_{1}(a)=\sigma_{a}\otimes\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{o}$
$x_{2}(a)=\sigma_{o}\otimes\sigma_{a}\otimes\sigma_{o}\otimes\sigma_{o}$
$x_{3}(a)=\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{a}\otimes\sigma_{o}$
$x_{4}(a)=\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{a}$

Then
$x_{1}(a)x_{2}(b)x_{3}(c)x_{4}(d)=\sigma_{a}\otimes\sigma_{b}\otimes\sigma_{c}\otimes\sigma_{d}$.
These last form a basis for $M_{16}(\mathbb{C})$, so it will suffice to show
that, for $a\in\\{x,y,z\\}$, the elements $x_{p}(a)$ can be manufactured from
elements of $Q_{2}\otimes Q_{2}$ by forming (associative) products.

For a start, notice that

$s_{1}\otimes
s_{o}=\sigma_{o}\otimes\sigma_{z}\otimes\sigma_{o}\otimes\sigma_{o}=x_{2}(z)\
\ \mbox{and}\ \ s_{o}\otimes
s_{1}=\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{z}=x_{4}(z).$

Similarly, with $s_{2}$ in place of $s_{1}$, we have $x_{2}(x)$ and $x_{4}(x)$
in $Q_{2}\odot Q_{2}$.

As noted above, $s_{3}s_{4}=-i\sigma_{x}\otimes\sigma_{o}$. Similarly,

$s_{3}s_{5}=i\sigma_{y}\otimes\sigma_{o}\ \mbox{and}\
s_{4}s_{5}=-i\sigma_{z}\otimes\sigma_{o}.$

Hence,

$(s_{3}\otimes s_{4})(s_{4}\otimes s_{4})=s_{3}s_{4}\otimes
s_{4}s_{4}=-i\sigma_{y}\otimes\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{o}=-ix_{1}(y)$

and similarly

$((s_{3}\otimes s_{5})(s_{5}\otimes s_{5})=ix_{1}(y)\ \ \mbox{and}\ \
(s_{4}\otimes s_{5})(s_{5}\otimes s_{5})=-ix_{1}(z).$

Thus, we have $x_{1}(a)$ for all $a$. In an entirely similar way, we find that
considering

$(s_{3}\otimes s_{3})(s_{3}\otimes s_{4})=-ix_{3}(y),\ (s_{3}\otimes
s_{3})(s_{3}\otimes s_{5})=ix_{3}(x),\ (s_{4}\otimes s_{4})(s_{4}\otimes
s_{5})=-ix_{3}(z).$

So we have $x_{3}(a)\in(Q_{2}\odot Q_{2})^{2}$ for all $a$.

It remains to obtain $x_{2}(y)$ and $x_{4}(y)$. But now we have

$(s_{3}\otimes
s_{o})x_{1}(z)=(\sigma_{z}\otimes\sigma_{y}\sigma_{o}\otimes\sigma_{o})(\sigma_{z}\otimes\sigma_{o}\otimes\sigma_{o}\otimes\sigma_{o})=x_{2}(y)$

and similarly

$x_{3}(z)(s_{o}\otimes s_{3})=x_{3}(y).$

Hence, $x_{2}(y)$ and $x_{4}(y)$ also belong to $(Q_{2}\odot Q_{2})^{2}$,
completing the proof. $\Box$

An Involution We now wish to find an involution on $M_{16}(\mathbb{C})$ fixing
$Q_{2}\odot Q_{2}$ pointwise. If $\phi:\mathbf{M}\rightarrow\mathbf{M}$ is a
$\ast$-involution on a complex $\ast$-algebra $\mathbf{M}$, recall that we
write $\mathbf{M}^{\phi}_{{sa}}$ for the set of self-adjoint fixed-points of
$\phi$. In finite dimensions, this is always an EJA. Suppose $\phi$ and $\psi$
are $\ast$-involutions on complex $\ast$-algebras $\mathbf{M}$ and
$\mathbf{N}$, respectively: if $A$ is a Jordan subalgebra of
$\mathbf{M}^{\phi}_{{sa}}$ and $B$ is a Jordan subalgebra of
$\mathbf{N}^{\psi}_{{sa}}$, then it’s easy to see that $A\odot B$ is a Jordan
subalgebra of $(\mathbf{M}\otimes\mathbf{N})_{{sa}}^{\phi\otimes\psi}$.

Let $J=\left[\begin{array}[]{cc}\mathbf{0}&-\mathbf{1}\\\
\mathbf{1}&\mathbf{0}\end{array}\right]$. Then

$\phi(a):=-Ja^{T}J=-(JaJ)^{T}$

is a $\ast$-involution (a $\ast$-antiautomorphism of period two) on
$M_{n}(\mathbb{C})$. This fixes $Q_{2}$ pointwise, i.e., $\phi(\pi(a))=\pi(a)$
for every $a\in M_{2}(\mathbb{H})_{{sa}}$. Identifying $M_{16}(\mathbb{C})$
with $M_{4}(M_{4}(\mathbb{C}))=M_{4}(\mathbb{C})\otimes M_{4}(\mathbb{C})$, we
then have an involution

$\Phi=\phi\otimes\phi:M_{16}(\mathbb{C})\rightarrow M_{16}(\mathbb{C}).$

By the comments above, we have

######  Lemma B.3.

$\Phi=\phi\otimes\phi$ fixes every element of $Q_{2}\odot Q_{2}$.

This completes the proof of Proposition 1. Moreover, as a consequence of ([21]
Theorem 4.4), we have

######  Corollary B.4.

$Q_{2}\odot Q_{2}=M_{16}(\mathbb{C})^{\Phi}$.

Since $C^{\ast}(A)\simeq\mathbf{M}\simeq C^{\ast}(B)$ (as $\ast$-algebras with
involution) implies $A\simeq\mathbf{M}^{\phi}_{{sa}}\simeq B$, and since
$M_{16}(\mathbb{C})\simeq C^{\ast}(M_{16}(\mathbb{R})_{{sa}})$, we have

######  Corollary B.5.

$Q_{2}\odot Q_{2}=M_{16}(\mathbb{R})_{{sa}}=:R_{16}$.

##  Appendix C Spin factors

The Jordan-von Neumann-Wigner Classification Theorem [28] singles out exactly
three classes of finite dimensional simple euclidean Jordan algebras: the
matrix algebras $R_{n},C_{n},Q_{n}$, the exceptional Jordan algebra
$M_{3}(\mathbb{O})_{\text{sa}}$, and one further type. The remaining type were
dubbed Spin Factors by Topping in [37]. For each finite $\mathbb{N}\ni n>1$,
there exists a unique spin factor of dimension $1+n$ (up to Jordan isomorphism
[22]) denoted by $V_{n}$. Abstractly, $V_{n}$ is generated as a Jordan algebra
by a spin system of cardinality $n$: a collection of $2\leq n\in\mathbf{N}$
symmetries (i.e. self-adjoint unitaries) $s_{p}$ in a unital JB algebra
$\mathcal{A}$, with $s_{p}\neq\pm u_{\mathcal{A}}$ such that
$s_{p}\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}s_{q}=u_{\mathcal{A}}\delta_{p,q}$.
It follows that $V_{n}\cong\mathbb{R}\oplus\mathbb{R}^{n}$ as a real inner
product space; moreover as a euclidean Jordan algebra with

$\big{(}\lambda_{0}\oplus\vec{\lambda}\big{)}\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}\big{(}\mu_{0}\oplus\vec{\mu}\big{)}=\lambda_{0}\mu_{0}+\langle\vec{\lambda},\vec{\mu}\rangle\oplus\lambda_{0}\vec{\mu}+\mu_{0}\vec{\lambda}\text{.}$
(1)

Concretely, the usual complex Pauli matrices can be used to define the spin
factors. We recall the usual complex Pauli matrices as follows

$u_{\mathcal{C}_{2}}=\sigma_{0}=\begin{pmatrix}1&\hskip 5.69046pt0\\\ 0&\hskip
5.69046pt1\end{pmatrix}\qquad\sigma_{1}=\begin{pmatrix}1&\hskip 5.69046pt0\\\
0&-1\end{pmatrix}\qquad\sigma_{2}=\begin{pmatrix}0&\hskip 5.69046pt1\\\
1&\hskip 5.69046pt0\end{pmatrix}\qquad\sigma_{3}=\begin{pmatrix}0&-i\\\
i&\hskip 5.69046pt0\end{pmatrix}\text{.}$ (2)

Following [22], we define for each finite $\mathbb{N}\ni n>1$ and $\forall
1\leq p\leq n$, with $\left\lfloor{\cdot}\right\rfloor$ and
$\left\lceil{\cdot}\right\rceil$ the usual floor and ceiling functions

$\displaystyle
t_{p}=\begin{cases}\begin{cases}\sigma_{3}^{\otimes^{\left\lceil{\frac{p}{2}}\right\rceil-1}}\otimes\sigma_{1}\otimes\sigma_{0}^{\otimes^{\left\lfloor{\frac{n}{2}}\right\rfloor-\left\lceil{\frac{p}{2}}\right\rceil}}&p\text{
odd}\\\
\sigma_{3}^{\otimes^{\left\lceil{\frac{p}{2}}\right\rceil-1}}\otimes\sigma_{2}\otimes\sigma_{0}^{\otimes^{\left\lfloor{\frac{n}{2}}\right\rfloor-\left\lceil{\frac{p}{2}}\right\rceil}}&p\text{
even}\end{cases}&n\text{ even}\\\\[28.45274pt]
\begin{cases}\sigma_{3}^{\otimes^{\left\lceil{\frac{p}{2}}\right\rceil-1}}\otimes\sigma_{1}\otimes\sigma_{0}^{\otimes^{\left\lceil{\frac{n}{2}}\right\rceil-\left\lceil{\frac{p}{2}}\right\rceil}}&p\text{
odd}\\\
\sigma_{3}^{\otimes^{\left\lceil{\frac{p}{2}}\right\rceil-1}}\otimes\sigma_{2}\otimes\sigma_{0}^{\otimes^{\left\lceil{\frac{n}{2}}\right\rceil-\left\lceil{\frac{p}{2}}\right\rceil}}&p\text{
even}\end{cases}&n\text{ odd}\end{cases}$ (3)

where our notation is such that $x^{\otimes^{0}}=1\in\mathbb{R}$, $x\otimes
1=x=x^{\otimes^{1}}$, $x^{\otimes^{2}}=x\otimes x$, and so on. One can easily
check that for each $n>1$, $\\{t_{1},\dots,t_{n}\\}$ generates a spin factor
of dimension $1+n$ with
$t_{p}\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}t_{q}=(t_{p}t_{q}+t_{q}t_{p})/2$.
It turns out [21], with finite $k\in\mathbb{N}$, that the maps

$\displaystyle\psi_{2k}:V_{2k}\longrightarrow
M_{2^{k}}(\mathbb{C})_{\text{sa}}::s_{p}\longmapsto t_{p}$ (4)
$\displaystyle\psi_{2k+1}:V_{2k+1}\longrightarrow
M_{2^{k}}(\mathbb{C})_{\text{sa}}\oplus
M_{2^{k}}(\mathbb{C})_{\text{sa}}::s_{p}\longmapsto t_{p}$ (5)

are precisely the canonical injections of $V_{n}$ into their universal
C∗-algebras (i.e. their universal representations). Our standard
representation $\pi_{n}$ of $V_{n}$ differs from the universal representation
when $n$ is odd. Specifically, when $n$ is even we define $v_{p}=t_{p}$, and
when $n$ is odd we define $\forall 1\leq p<n$

$\displaystyle v_{p}$ $\displaystyle=$
$\displaystyle\begin{cases}\sigma_{3}^{\otimes^{\left\lceil{\frac{p}{2}}\right\rceil-1}}\otimes\sigma_{1}\otimes\sigma_{0}^{\otimes^{\left\lfloor{\frac{n}{2}}\right\rfloor-\left\lceil{\frac{p}{2}}\right\rceil}}&p\text{
odd}\\\
\sigma_{3}^{\otimes^{\left\lceil{\frac{p}{2}}\right\rceil-1}}\otimes\sigma_{2}\otimes\sigma_{0}^{\otimes^{\left\lfloor{\frac{n}{2}}\right\rfloor-\left\lceil{\frac{p}{2}}\right\rceil}}&p\text{
even}\end{cases}$ (6) $\displaystyle v_{n}$ $\displaystyle=$
$\displaystyle\sigma_{3}^{\otimes^{\left\lfloor{\frac{n}{2}}\right\rfloor}}$
(7)

and we embed $V_{n}$ into its standard C∗-algebra via the following Jordan
monomorphism with
$v_{p}\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}v_{q}=(v_{p}v_{q}+v_{q}v_{p})/2$

$\pi_{n}:V_{n}\longrightarrow
M_{2}(\mathbb{C})_{\text{sa}}^{\otimes^{\left\lfloor{\frac{n}{2}}\right\rfloor}}::s_{p}\longmapsto
v_{p}\text{.}$ (8)

For example, the universal and standard representations for the qubit — i.e.
$V_{3}$ — differ as follows

$\displaystyle t_{1}=\sigma_{1}\otimes\sigma_{0}\qquad v_{1}=\sigma_{1}$ (9)
$\displaystyle t_{2}=\sigma_{2}\otimes\sigma_{0}\qquad v_{2}=\sigma_{2}$ (10)
$\displaystyle t_{3}=\sigma_{3}\otimes\sigma_{1}\qquad v_{3}=\sigma_{3}$ (11)

hence the name standard representation. Incidentally, note that
$V_{3}\cong{C}_{2}$ as a euclidean Jordan algebra. Furthermore,
$V_{2}\cong{R}_{2}$, $V_{5}\cong{Q}_{2}$, and $V_{9}\cong
M_{2}(\mathbb{O})_{\text{sa}}$; the ambient spaces for the real, quaternionic,
and octonionic quantum bits.

##  Appendix D The inner product on a composite

In this appendix, we prove Proposition 4.3, which, for convenience, we
restate:

######  Proposition D.1.

Let $AB$ be a composite of EJAs $A$ and $B$. Then for all $a,x\in A$ and all
$b,y\in B$,

$\langle a\odot x,b\odot y\rangle=\langle a,x\rangle\langle b,y\rangle.$

Proof: We begin with the case in which $a$ and $b$ are minimal projections.
Since $a\odot b$ is then also a projection, the discussion in Section 3.2
tells us that

$\widehat{a\odot b}=\|a\odot b\|^{-2}(a\odot b)$

defines a state. Now define a positive bilinear form $\omega:A\times
B\rightarrow\mathbb{R}$ by setting

$\omega(x,y)=\langle\widehat{a\odot b},x\odot y\rangle=\langle a\odot b,x\odot
y\rangle$

for all $x\in A$, $y\in B$. Note that $\omega$ is normalized, i.e., a state in
the maximal tensor product $A\otimes B$. Now evaluate the first marginal of
$\omega$ at $a$:

$\displaystyle\omega_{1}(a)=\omega(a\otimes u_{B})$ $\displaystyle=$
$\displaystyle\langle\widehat{a\odot b},a\otimes u_{B}\rangle$
$\displaystyle=$ $\displaystyle\langle\widehat{a\odot b},a\odot b+a\odot
b^{\prime}\rangle$ $\displaystyle=$ $\displaystyle(\langle\widehat{a\odot
b},a\odot b\rangle+\langle\widehat{a\odot b},a\odot b^{\prime}\rangle.$

Since $\langle\widehat{a\odot b},a\odot b\rangle=1$, the secondond summand at
the end is $0$, and we have $\omega_{1}(a)=1$. Since $a$ is minimal, there is
only one such state: $\omega_{1}(a)=\langle\widehat{a}|$, i.e., the functional
$b\mapsto\langle a,b\rangle$. Moreover, this is a pure state. The same
argument shows that $\omega_{2}(b)=1$, so that
$\omega_{2}=\langle\widehat{b}|$. As is well known, if a non-signaling state
has pure marginals, then it’s the product of these marginals [9]. Thus,
$\omega=\langle\widehat{a}|\otimes\langle\widehat{b}|$. This gives us

$\langle a\odot b,x\odot y\rangle=c\langle a,x\rangle\langle b,y\rangle.$

where $c:=\frac{\|a\odot b\|^{2}}{\|a\|^{2}\|b\|^{2}}$.

We want to show that $c=1$. As a first step, note that $c$ is independent of
the choice of minimal projections $a$ and $b$ (Argue by symmetry: since $A$
and $B$ are simple, there are symmetries $\phi$ and $\psi$ taking any given
pair $a,b$ to any other such pair $a^{\prime},b^{\prime}$; hence, there’s a
symmetry taking $a\odot b$ to $a^{\prime}\odot b^{\prime}$ 4.8. This is the
only use we make of condition (c) above.) Extend $a$ and $b$ to orthogonal
decompositions of $u_{A}$ and $u_{B}$ as sums of projections:
$u_{A}=\sum_{a_{i}}a_{i}$ with $a=a_{1}$, and $u_{B}=\sum_{j}b_{j}$ with
$b=b_{1}$ Then we have $u_{AB}=\sum_{i,j}a_{i}\odot b_{j}$, so

$\sum_{i,j,k,l}\langle a_{i}\odot b_{j},a_{k}\odot b_{l}\rangle=\langle
u_{AB},u_{AB}\rangle=1$

but also, noting that all $a_{i}$ and $b_{j}$ here are minimal projections, so
that the constant $c$ above is the same for all choices of $a_{i}\odot b_{j}$,

$\sum_{i,j,k,l}\langle a_{i}\odot b_{j},a_{i}\odot
b_{l}\rangle=\sum_{i,j,k,l}c\langle a_{i}\odot b_{j},a_{k}\odot
b_{l}\rangle=c.$

Hence, $c=1$.

Now suppose $a$ and $b$ are arbitrary elements of $A$ and $B$, respectively.
Spectrally decomposing $a$ and $b$ as

$a\ =\ \sum_{i}t_{i}a_{i}\ \mbox{and}\ b\ =\ \sum_{j}s_{j}b_{j}$

where $a_{i}$ and $b_{j}$ are pairwise orthogonal families of minimal
projections, we have

$\langle a\odot b|=\langle\sum_{i,j}t_{i}s_{j}a_{i}\odot
b_{j}|=\sum_{i,j}t_{i}s_{j}\langle a_{i}\odot b_{j}|.$

Hence, for all $x,y$,

$\displaystyle\langle a\odot b,x\odot y\rangle$ $\displaystyle=$
$\displaystyle\sum_{i,j}t_{i}s_{j}\langle a_{i}\odot b_{j},x\odot y\rangle$
$\displaystyle=$ $\displaystyle\sum_{i,j}t_{i}s_{j}\langle
a_{i}|x\rangle\langle b_{j},y\rangle$ $\displaystyle=$ $\displaystyle\langle
a,x\rangle\langle b,y\rangle$

as advertised. $\Box$

##  Appendix E Weak Composites

By a weak composite of EJAs $A$ and $B$, we mean an EJA $AB$ and a bilinear
mapping $\pi:A\otimes B\rightarrow AB$ satisfying parts (a) and (b), but not
necessarily part (c), of Definition 4.1. That is, we suspend the requirement
that $AB$ be generated, as a Jordan algebra, by $\pi(A\otimes B)$. We are
going to show that the Jordan subalgebra of $AB$ generated by the image of
$A\otimes B$ in $AB$ also satisfies these conditions, and hence, is a
composite in the strict sense.

Observe, first, that we made no use of Condition (c) in proving any of the
results leading up to, and including, Corollary 4.7. So all of these are also
satisfied by weak composites.

######  Proposition E.1.

Let $AB$ be a weak composite of EJAs $A$ and $B$. Then the Jordan subalgebra
of $AB$ generated by $A\otimes B$ is also a composite.

Proof: Identifying $A\otimes B$ with its image, $\pi(A\otimes B)$, in $AB$,
let $A\odot B=J(A\otimes B)$ denote the Jordan subalgebra of $AB$ generated by
$A\otimes B$. That the co-restriction $\pi:A\otimes B\rightarrow A\odot B$
satisfies the conditions for a composite (Definition 2.1) is straightfoward.
It is also straightforward that $A\odot B$ will satisfy the conditions for a
composite of Jordan models (Definition 4.1, provided that it satisfies the
conditions for a dynamical composite (Definition 2.3. To see that it does, let
$\phi\in G(A)$ and $\psi\in G(B)$. We need to show that $\phi\otimes\psi$
preserves $A\odot B$. We can expand $\phi$ as $U_{a}\circ g$ and $\psi$ as
$U_{b}\circ h$ for some interior elements $a\in A_{+}$, $b\in B_{+}$, and
Jordan homomorphisms $g\in G(A)$ and $h\in G(B)$ ([18], III.5.1). Now

$\phi\otimes\psi=(U_{a}\circ g)\otimes(U_{b}\circ h)=(U_{a}\otimes
U_{b})\circ(g\otimes h).$

Since $(g\otimes h)(u_{AB})=g(u_{A})\otimes h(u_{B})=u_{A}\otimes
u_{B}=u_{AB}$, $g\otimes h$ is a symmetry, hence, a Jordan automorphism of
$AB$ ([3], Theorem 2.8). As it maps $A\otimes B$ to itself, it also preserves
$J(A\otimes B)$, i.e., $A\odot B$.

It remains to show that $U_{a}\otimes U_{b}$ also preserves $A\odot B$. Begin
with the observation that if $A$ is an EJA and $X\subseteq A$, then for all
$x\in X$, $U_{x}(J(X))\subseteq J(X)$. This is evident from the fact that, as
$x\in J(X)$ and the latter is a Jordan subalgebra of $A$,
$U_{x}(y)=2x\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}(x\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}y)-(x^{2})\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}y$
for all $y\in J(X)$.

Now consider that the proof of the identities $(a\otimes
u)\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}(x\otimes
y)=a\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}x\otimes y$ and $(u\otimes
b)\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}(x\otimes y)=x\otimes
b\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}y$ relies only on (a) and (b),
and so, holds in our context. By Corollary 4.7, $U_{a\otimes
u}=U_{a}\otimes\mbox{id}$ and $U_{u\otimes b}=\mbox{id}_{A}\otimes U_{b}$. If
$a,b$ are interior elements, then $U_{a}\in G(A)$ and $U_{b}\in G(B)$, so by
(b), we have

$U_{a}\otimes U_{b}=(U_{a}\otimes\mbox{id}_{B})\circ(\mbox{id}_{A}\otimes
U_{b})=U_{a\otimes u_{B}}\circ U_{u_{A}\otimes b}.$

Since $U_{a\otimes u_{B}}$ and $U_{u_{A}\otimes b}$ preserve $J(A\otimes
B)=A\odot B$, by the remark above, so does $U_{a}\otimes U_{b}$, and the proof
is finished. $\Box$

##  Appendix F Extending order automorphisms

Let $A$ be an EJC-algebra, that is, $A$ a Jordan subagebra of
$\mathbf{M}_{{sa}}$ for a finite-dimensional complex $\ast$-algebra
$\mathbf{M}=\mathbf{M}_{A}$. Recall that $G(A)$ is the connected component of
the identity in the group of order-automorphisms of $A$.

######  Lemma F.1.

Let $A$ be reversible. Then any one-parameter group of order automorphisms of
$A$ extends to a one-parameter group of order-automorphisms of $\mathbf{M}$.

Proof: If $\\{\phi(t)\\}_{t\in\mathbb{R}}$ is a one-parameter group of order-
automorphisms of $A$, then $\phi(t)(a)=e^{tX}a$ where $X=\phi^{\prime}(0)$ is
a linear operator on $A$. By definition, $X$ is an order-derivation of $A$.
Now, order-derivations come in two basic types: those having the form
$X=L_{a}$ for some $a\in A$, and those having the property $Xu=0$ (which turns
out to be the same thing as being a Jordan derivation). The former are self-
adjoint with respect to the canonical inner product on $A$, by the definition
of a euclidean Jordan algebra, while the latter are skew-adjoint . Moreover,
every order-derivation has the form $\delta=L_{a}+\delta^{\prime}$ where
$\delta^{\prime}$ is skew ([3], Proposition 1.60).

Now, $L_{a}$ obviously extends from $A$ to $\mathbf{M}$, simply because
$a\in\mathbf{M}$ and the Jordan product on $A$ is the restriction of that on
$\mathbf{M}$. By ([38], Theorem 2.5), if $A$ is reversible in $\mathbf{M}$,
$\delta^{\prime}$ also extends to a Jordan derivation $\delta^{\prime\prime}$
on $\mathbf{M}$. Thus, we have an extension
$\widehat{\delta}=L_{a}+\delta^{\prime\prime}$ on $\mathbf{M}$. In particular,
$\delta^{\prime\prime}(A)\subseteq A$.

It follows that we have an order-automorphism
$\widehat{\phi}(t)=e^{t\widehat{\delta}}$ of $\mathbf{M}_{+}$. Note that this
preserves $A$, since

$\widehat{\phi}(t)x=\sum_{k=1}^{\infty}\frac{t^{k}}{k!}\widehat{\delta}^{k}x$

and
$\widehat{\delta}x=(L_{a}+\delta^{\prime\prime})x=a\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}x+\delta^{\prime\prime}(x)$,
which belongs to $A$ if $x$ does. $\Box$

######  Corollary F.2.

If $A$ is reversible, every element of $G(A)$ extends to an element of
$G((\mathbf{M}_{A})_{{sa}})$.

Now let $B$ also be a reversible EJC-algebra.

######  Lemma F.3.

Let $\delta$ be any $\ast$-derivation of $\mathbf{M}_{A}$ fixing $A$. Then
$\delta\otimes\mathbf{1}$ is a $\ast$-derivation of
$\mathbf{M}_{A}\otimes\mathbf{M}_{B}$ fixing $A\odot B$.

Proof: Let $\mathbf{M}$ and $\mathbf{N}$ be $\ast$-algebras, and let
$a,b\in\mathbf{M}$ and $x,y\in\mathbf{N}$. Then

$(a\otimes x)\bullet(b\otimes y)=\frac{1}{2}(ab\otimes xy+ba\otimes yx).$

If $\delta$ is a $\ast$-derivation of $\mathbf{M}$, then it is straighforward
to check that $\delta\otimes\mathbf{1}$ is a $\ast$-derivation of
$\mathbf{M}\otimes\mathbf{N}$, and that for all $a,b\in\mathbf{M}$ and
$x,y\in\mathbf{N}$,

$(\delta\otimes\mathbf{1})((a\otimes x)\bullet(b\otimes y))=(a\otimes
x)\bullet(\delta(b)\otimes y)+(\delta(a)\otimes x)\bullet(b\otimes y).$

In particular, if $A\subseteq\mathbf{M}$ and $\delta(A)\subseteq A$, it
follows that $(\delta\otimes\mathbf{1})(A\otimes B)\subseteq(A\otimes
B)\bullet(A\otimes B)$ for any $B\subseteq\mathbf{N}$. It follows easily that,
where $A$ and $B$ are EJCs and $\mathbf{M}=\mathbf{M}_{A}$ and
$\mathbf{N}=\mathbf{M}_{B}$, $\delta\otimes\mathbf{1}$ preserves $A\odot B$.
242424The details: let $\delta$ be a $\ast$-derivation on a $\ast$-algebra
$\mathbf{M}$, and let $X\subseteq M_{{sa}}$ with $\delta(X)\subseteq X$. Let
$Y=\\{a\in J(X)|\delta(a)\in J(X)\\}$. Evidently $X\subseteq Y$. Now if
$a,b\in Y$ and $t\in\mathbb{R}$, then $\delta(ta+b)=t\delta(a)+\delta(b)\in
J(X)$, so $J(X)$ is a subspace of $M$. If $a,b\in Y$ then
$\delta(a\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}b)=a\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}\delta(b)+\delta(a)\raisebox{-3.7889pt}{\scalebox{2.5}{$\cdot$}}b\in
J(X)$. Thus, $Y$ is a Jordan subalgebra of $\mathbf{M}_{{sa}}$, containing
$X$, and contained in $J(X)$. Ergo, $Y=J(X)$, and $\delta(J(X))\subseteq
J(X)$.

######  Proposition F.4.

If $\phi$ and $\psi$ are order-automorphisms of $A$ and $B$, respectively,
then $\phi\otimes\psi$ extends to an order-automorphism of $A\odot B$.

Proof: By Corollary F.2, we can assume that $\phi$ is an order-automorphism of
$M_{{sa}}$ fixing $A$. Then $\phi=e^{\delta}$ for an order-derivation $\delta$
of $M$, which (taking derivatives) must also fix $A$. It follows that

$\phi\otimes\mathbf{1}=e^{t\delta}\otimes\mathbf{1}=\sum_{n=0}^{\infty}\frac{t^{n}}{n!}\delta^{n}\otimes\mathbf{1}=\sum_{n=0}^{\infty}\frac{t^{n}}{n!}(\delta\otimes\mathbf{1})^{n}=e^{t(\delta\otimes\mathbf{1})}.$

By Lemma F.3, $\delta\otimes\mathbf{1}$ fixes $A\odot B$; thus, so does the
series at right, whence, so does $\phi\otimes\mathbf{1}$. It follows that if
$\phi$ is an order-automorphism of $M_{A}$ fixing $A$, then
$\phi\otimes\mathbf{1}$ is an order-automorphism of $A\otimes B$ fixing
$J(A\otimes B)=A\odot B$. Hence, if $\phi$ and $\psi$ are order-automorphisms
of $M_{A}$ and $M_{B}$, respectively fixing $A$ and $B$, then
$\phi\otimes\psi=(\phi\otimes\mathbf{1})\circ(\mathbf{1}\otimes\psi)$ fixes
$A\odot B$. $\Box$