Order theoretical structures in atomic JBW-algebras: disjointness, bands, and centres

Abstract

Every atomic JBW-algebra is known to be a direct sum of JBW-algebra factors of type I. Extending Kadison’s anti-lattice theorem, we show that each of these factors is a disjointness free anti-lattice. We characterise disjointness, bands, and disjointness preserving bijections with disjointness preserving inverses in direct sums of disjointness free anti-lattices and, therefore, in atomic JBW-algebras. We show that in unital JB-algebras the algebraic centre and the order theoretical centre are isomorphic. Moreover, the order theoretical centre is a Riesz space of multiplication operators. A survey of JBW-algebra factors of type I is included.

Appendix

This appendix contains a survey of the factors of atomic JBW-algebras.

1.1 Quaternionic Hilbert spaces and their operators

By introducing the multiplication rules $i^2=j^2=k^2=ijk=-1$ on the symbols i, j, and k, we induce the structure of a unital associative algebra on the four dimensional real vector space

$$\begin{aligned} {\mathbb {H}}:=\bigl \{a1+bi+cj+dk:a,b,c,d\in {\mathbb {R}}\bigr \} \end{aligned}$$

where the general product distributes over the sum as usual, with unit 1. This algebra ${\mathbb {H}}$ is referred to as the quaternions, and i, j, and k are called the imaginary units. The multiplication on ${\mathbb {H}}$ is not commutative as $ij=-ji=k$, $ki=-ik=j$, and $jk=-kj=i$. The algebraic centre of ${\mathbb {H}}$ equals ${\mathbb {R}}1$, and every non-zero quaternion is invertible as

$$\begin{aligned} (a1+bi+cj+dk)(a1-bi-cj-dk)=(a^2+b^2+c^2+d^2)1. \end{aligned}$$

For $q=a1+bi+cj+dk$, the quaternionic conjugate is defined to be $q^*:=a1-bi-cj-dk$, which defines an involution ${}^*:{\mathbb {H}}\rightarrow {\mathbb {H}}$ that reverses the order of multiplication. That is, for $q,r\in {\mathbb {H}}$, we have $(qr)^*=r^*q^*$. The real part of q is denoted by ${\text {Re}}(q)$ and is given by ${\text {Re}}(q):=\frac{1}{2}(q+q^*)=a1$. Furthermore, this conjugation gives rise to the multiplicative norm $|q|:=\sqrt{q^*q}$ on ${\mathbb {H}}$. Note that ${\mathbb {R}}1$ has been identified with ${\mathbb {R}}$ here.

An abelian group $(V,+)$ that admits a right action $\cdot :V\times {\mathbb {H}}\rightarrow V$ is called a quaternionic vector space if the action distributes over $+$ in V and the sum of quaternions. That is, we have $(v + w)\cdot q=v\cdot q + w\cdot q$, $v\cdot (q+r)=v\cdot q + v\cdot r$, $(v \cdot r)\cdot q=v\cdot (qr)$, and $v\cdot 1 =v$ for all $q,r\in {\mathbb {H}}$ and all $v,w\in V$. The reason for choosing a right action on V is so that an $n\times n$ matrix A acting on the left as usual on a vector $x\in {\mathbb {H}}^n$, that is, $x\mapsto Ax$, is now ${\mathbb {H}}$-linear.

A quaternionic inner product on V is a ${\mathbb {H}}$-sesquilinear form $\langle \cdot ,\cdot \rangle :V\times V\rightarrow {\mathbb {H}}$, so $\langle \cdot , \cdot \rangle$ satisfies $\langle u, v\cdot q +w\rangle =\langle u,v\rangle q+\langle u,w\rangle$ and $\langle v,w\rangle =\langle w,v\rangle ^*$ for all $q\in {\mathbb {H}}$ and all $u,v,w\in V$, which in addition satisfies $\langle v,v\rangle \ge 0$ for all $v\in V$ and $\langle v,v \rangle =0$ if and only if $v=0$. It follows that $\Vert v\Vert :=\sqrt{\langle v,v \rangle }$ yields a norm on V, which we prove next by using the quaternionic version of the Cauchy–Schwarz inequality.

Lemma A.1

(Cauchy–Schwarz) Let V be a quaternionic vector space equipped with a quaternionic inner product. Then, for any $v,w\in V$, it follows that $|\langle v,w \rangle |\le \Vert v\Vert \Vert w\Vert$ with equality if and only if v and w are ${\mathbb {H}}$-linearly dependent.

Proof

If $w=0$, the statement clearly holds, so we may assume that $w\ne 0$. Let $q:=\langle w,v \rangle \Vert w\Vert ^{-2}$ and observe that

$$\begin{aligned} 0 \le \Vert v-w\cdot q\Vert ^2&= \Vert v\Vert ^2 - q^*\langle w,v \rangle - \langle v,w \rangle q + q^* \Vert w\Vert ^2 q \\&= \Vert v\Vert ^2 - \frac{|\langle v,w \rangle |^2}{\Vert w\Vert ^2} - \frac{|\langle v,w \rangle |^2}{\Vert w\Vert ^2} + \frac{|\langle v,w \rangle |^2}{\Vert w\Vert ^2}\\&= \Vert v\Vert ^2 - \frac{|\langle v,w \rangle |^2}{\Vert w\Vert ^2}, \end{aligned}$$

and hence, we have $|\langle v,w \rangle |\le \Vert v\Vert \Vert w\Vert$. Moreover, in case of an equality above, it follows that $v=w\cdot q$ and if $v=w\cdot r$ for some $r\in {\mathbb {H}}$, then $|\langle v,w \rangle |=|\langle w,w\rangle ||r|=|r|\Vert w\Vert \Vert w\Vert =\Vert w\cdot r\Vert \Vert w\Vert =\Vert v\Vert \Vert w\Vert$. $\square$

It now follows from Lemma A.1 that, for $v,w\in V$, we have

$$\begin{aligned} \Vert v+w\Vert ^2\le \Vert v\Vert ^2+2|\langle v,w \rangle |+\Vert w\Vert ^2\le (\Vert v\Vert +\Vert w\Vert )^2, \end{aligned}$$

showing that the triangle inequality is satisfied. If V is complete with respect to the norm $\Vert \cdot \Vert$, then V is a quaternionic Hilbert space. For more details, see [21]. Quaternionic Hilbert spaces will from now on be denoted by ${\mathcal {H}}_q$. Most of the theory for Hilbert spaces passes over analogously to quaternionic Hilbert spaces as will be shown. Two vectors $v,w\in {\mathcal {H}}_q$ are said to be orthogonal if $\langle v,w \rangle =0$, and similarly, the orthogonal complement $S^\perp$ of a set is defined. A subset $B\subseteq {\mathcal {H}}_q$ is orthonormal if the vectors in B have norm one and are pairwise orthogonal. The Pythagorean theorem also holds for the quaternionic inner product. That is, if $v_1,\ldots ,v_n\in {\mathcal {H}}_q$ are pairwise orthogonal, then

$$\begin{aligned} \Vert v_1 + \cdots + v_n\Vert ^2=\Vert v_1\Vert ^2 + \cdots + \Vert v_n\Vert ^2\qquad \qquad {\text {(Pythagorean identity)}}. \end{aligned}$$

The Pythagorean theorem can be used in turn to prove Bessel’s inequality, stating that, for an orthonormal set $\{b_n :n\in {\mathbb {N}}\}$ and any $v\in {\mathcal {H}}_q$, we have

$$\begin{aligned} \sum _{k=1}^\infty |\langle v,b_k \rangle |^2\le \Vert v\Vert ^2\qquad \qquad {\text {(Bessel's inequality)}}. \end{aligned}$$

We call a subset $B\subseteq {\mathcal {H}}_q$ an orthonormal basis for ${\mathcal {H}}_q$ if it is a maximal orthonormal set. An application of Zorn’s lemma tells us that every quaternionic Hilbert space ${\mathcal {H}}_q\ne \{0\}$ has an orthonormal basis. Indeed, the set

$$\begin{aligned} {\mathscr {B}}:=\left\{ B\subseteq {\mathcal {H}}_q:B\text { is an orthonormal set}\right\} \end{aligned}$$

is non-empty and partially ordered by set inclusion. Let $(B_i)_i$ be a chain in ${\mathscr {B}}$. Then $B_0:=\bigcup _{i}B_i$ is an orthonormal set and Zorn’s lemma implies that ${\mathscr {B}}$ contains a maximal element B. Hence, there is no non-zero vector $v\in {\mathcal {H}}_q$ such that $\langle v,b \rangle =0$ for all $b\in B$, showing that B is an orthonormal basis for ${\mathcal {H}}_q$. By Bessel’s inequality, for any $v\in {\mathcal {H}}_q$, there are at most countably many vectors in B such that $\langle v,b \rangle \ne 0$, and it follows that

$$\begin{aligned} v=\sum _{b\in B} b\cdot \langle v,b \rangle \qquad \text {and}\qquad \Vert v\Vert ^2=\sum _{b\in B} |\langle v,b \rangle |^2. \end{aligned}$$

(A.1)

For more details, see [8, Lemma I.4.12] and [8, Theorem I.4.13] (the arguments are the same for quaternionic Hilbert spaces).

An ${\mathbb {H}}$-linear operator $T :{\mathcal {H}}_q \rightarrow {\mathcal {H}}_q$ is bounded in the same way an operator between Hilbert spaces is bounded, so T is bounded if and only if $\sup \{\Vert Tv\Vert :\Vert v\Vert \le 1\}<\infty$, with this supremum denoted by $\Vert T\Vert$. Since multiplication in ${\mathbb {H}}$ is not commutative, the bounded operators on ${\mathcal {H}}_q$ can only be a equipped with the structure of a real vector space. For $r\in {\mathbb {R}}$, the ${\mathbb {H}}$-linear operator rT is defined by $rTx:=Tx\cdot r$, but linearity fails if r is replaced by a general quaternion $q\in {\mathbb {H}}$. To illustrate this with an example, for an operator T on ${\mathcal {H}}_q$, it follows that, for $S:=jT$, we have $S(v\cdot i)\ne Sv\cdot i$. The real vector space of bounded ${\mathbb {H}}$-linear operators is denoted by $\textrm{B}({\mathcal {H}}_q)$, and becomes a Banach space when equipped with the norm $T\mapsto \Vert T\Vert$. In order to define the quaternionic adjoint of an operator in $\textrm{B}({\mathcal {H}}_q)$, we need the quaternionic version of the Riesz representation theorem. Similarly, a ${\mathbb {H}}$-linear functional $\varphi :{\mathcal {H}}_q \rightarrow {\mathbb {H}}$ is bounded if $\sup \{|\varphi (v)| :\Vert v\Vert \le 1\}<\infty$.

Lemma A.2

(Riesz representation theorem) If $\varphi :{\mathcal {H}}_q \rightarrow {\mathbb {H}}$ is a bounded ${\mathbb {H}}$-linear functional, then there exists a unique $v \in {\mathcal {H}}_q$ such that $\varphi (w)=\langle v,w \rangle$ for all $w \in {\mathcal {H}}_q$.

Proof

Since $\varphi$ is continuous, it follows that $\ker \varphi$ is closed in ${\mathcal {H}}_q$. Clearly, if $\varphi =0$, then we can take $v=0$ to represent $\varphi$. So, suppose that $\varphi \ne 0$. By choosing an orthonormal basis for $\ker \varphi$, it follows that this basis is not maximal in ${\mathcal {H}}_q$, so there is a $v \in \ker \varphi ^\perp$ such that $\varphi (v)=1$. For $w \in {\mathcal {H}}_q$, we have $\varphi (w-v\cdot \varphi (w))=0$, so $w-v\cdot \varphi (w) \in \ker \varphi$. Hence, $\langle v,w \rangle - \Vert v\Vert ^2\varphi (w) = \langle v,w - v\cdot \varphi (w) \rangle = 0$ and so $\varphi (w) = \langle \Vert v\Vert ^{-2}v,w \rangle$ for all $w\in {\mathcal {H}}_q$. Note that if $u \in {\mathcal {H}}_q$ is such that $\langle v,w \rangle = \langle u,w \rangle$ for all $w \in {\mathcal {H}}_q$, then, by choosing $w=u-v$, we get $\langle u-v, u-v \rangle = 0$, so $u=v$. $\square$

Given a bounded operator $T\in \textrm{B}({\mathcal {H}}_q)$, we have a well-defined linear operator $S :{\mathcal {H}}_q \rightarrow {\mathcal {H}}_q$ given by the relation $\langle w,Tv \rangle = \langle Sw,v \rangle$ as a consequence of Lemma A.2, since, for any $v \in {\mathcal {H}}_q$, the map $w \mapsto \langle v,Tw \rangle$ is ${\mathbb {H}}$-linear and bounded by Lemma A.1. The properties of a quaternionic inner product also show that this operator S is unique, and bounded with $\Vert S\Vert =\Vert T\Vert$. We say that S is the quaternionic adjoint of T, and will be denoted by $T^*$. It is again similar to the case of dealing with operators on a Hilbert space to find that $T^{**}=T$ and $\Vert T^*T\Vert =\Vert T\Vert ^2$. An operator $T\in \textrm{B}({\mathcal {H}}_q)$ is self-adjoint if $T^*=T$ and the subspace of self-adjoint operators will be denoted by $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. Since $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is closed in $\textrm{B}({\mathcal {H}}_q)$, it is a Banach space as well. Note that $\langle Tv,v \rangle \in {\mathbb {R}}1$ for all $T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. An orthogonal projection in $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is an idempotent operator, and orthogonal projections are in bijection with the closed subspaces of ${\mathcal {H}}_q$. Indeed, for a closed subspace ${\mathcal {K}}\subseteq {\mathcal {H}}_q$, we have that ${\mathcal {K}}=\{0\}$ is precisely the range of the projection $P=0$. Otherwise, let B be an orthonormal basis for ${\mathcal {K}}$, and, for any finite set $F\subseteq B$, define $P_F :{\mathcal {H}}_q \rightarrow {\mathcal {H}}_q$ by

$$\begin{aligned} P_Fv:=\sum _{b\in F} b\cdot \langle b,v \rangle . \end{aligned}$$

Then, it follows that $P_F$ is linear, idempotent, and $\Vert P_F\Vert =1$. Furthermore, note that, for $v,w\in {\mathcal {H}}_q$, we have

$$\begin{aligned} \langle P_Fv,w \rangle= & {} \sum _{b\in F} \langle b\cdot \langle b,v \rangle ,w \rangle = \sum _{b\in F} (\langle w,b \rangle \langle b, v\rangle )^* = \sum _{b\in F} \langle v,b \rangle \langle b, w\rangle \\= & {} \sum _{b\in F} \langle v,b\cdot \langle b,w \rangle \rangle = \langle v,P_Fw \rangle , \end{aligned}$$

hence, $P_F$ is self-adjoint. Since

$$\begin{aligned} Pv:=\sum _{b\in B} \langle v,b \rangle \cdot b \end{aligned}$$

is the limit of the net $\{P_F v :F\subseteq B \text { finite}\}$, it follows that P is an idempotent linear operator with $\Vert P\Vert =1$. Moreover, for $v,w\in {\mathcal {H}}_q$, we have

$$\begin{aligned} \bigl | \langle Pv,w \rangle - \langle v,Pw \rangle \bigr |= & {} \bigl | \langle (P-P_F)v,w \rangle - \langle v,(P-P_F)w \rangle \bigr | \le \Vert (P-P_F)v\Vert \Vert w\Vert \\{} & {} +\Vert (P-P_F)w\Vert \Vert v\Vert \end{aligned}$$

and hence, we find that P is self-adjoint making it an orthogonal projection. By (A.1), the range of P equals ${\mathcal {K}}$. The uniqueness of P follows from the fact that any orthogonal projection Q with range ${\mathcal {K}}$ must agree with P as $\textrm{ran}\, P = \textrm{ran}\, Q = {\mathcal {K}}$ and $\ker P = \ker Q = {\mathcal {K}}^\perp$. Conversely, any orthogonal projection P yields a closed subset $\textrm{ran}\, P \subseteq {\mathcal {H}}_q$.

The commutative bilinear product $\circ :\textrm{B}({\mathcal {H}}_q)_\textrm{sa}\times \textrm{B}({\mathcal {H}}_q)_\textrm{sa}\rightarrow \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ defined by $T\circ S:=\frac{1}{2}(TS+ST)$ turns $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ into a real Jordan algebra and the norm $T\mapsto \Vert T\Vert$ satisfies $\Vert S\circ T\Vert \le \Vert S\Vert \Vert T\Vert$, and also $\Vert T^2\Vert =\Vert T^* T\Vert =\Vert T\Vert ^2$ for all $S,T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. Furthermore, for $S,T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ and $v\in {\mathcal {H}}_q$ with $\Vert v\Vert \le 1$, it follows that

$$\begin{aligned} \Vert Tv\Vert ^2= & {} \langle Tv,Tv\rangle \le \langle Tv,Tv\rangle \!+\! \langle Sv,Sv\rangle = \langle T^2v,v\rangle \!+\! \langle S^2v,v\rangle = \langle (T^2\!+\!S^2)v,v\rangle \\\le & {} \Vert T^2+S^2\Vert \end{aligned}$$

by Lemma A.1, so $\Vert T^2\Vert \le \Vert T^2+S^2\Vert$, and we conclude that $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is a JB-algebra with identity I. The spectrum of an operator T in $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is denoted by $\sigma (T)$ and is defined to be

$$\begin{aligned} \sigma (T):=\{\lambda \in {\mathbb {R}}:T-\lambda I \text { is not invertible in }\textrm{B}({\mathcal {H}}_q)_\textrm{sa}\}. \end{aligned}$$

Note that real scalar multiples of I are considered here as $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is a real vector space. Note that by the functional calculus [3, Corollary 1.19], the spectrum of an operator is never empty. The numerical range of an operator $T \in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$, which is defined to be $N(T):=\{\langle Tv,v \rangle :\Vert v\Vert =1\}$, is related to the spectrum of T as follows.

Lemma A.3

For $T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$, we have $\sigma (T)\subseteq \overline{N(T)}$.

Proof

If $\lambda$ is such that $T-\lambda I$ is not injective, then there is a normalised $v\in {\mathcal {H}}_q$ such that $Tv=\lambda v$, so $\langle Tv,v\rangle = \lambda$, and thus $\lambda \in N(T)$. If $T-\lambda I$ is not surjective, then there are two cases to distinguish, the range of $T-\lambda I$ is not dense in ${\mathcal {H}}_q$, and the range of $T-\lambda I$ is not closed but dense in ${\mathcal {H}}_q$. Firstly, suppose there is a normalised vector v in the orthogonal complement of $\overline{\textrm{ran}(T-\lambda I)}$. Then we find that $\langle Tv,v \rangle - \lambda = \langle (T-\lambda I)v,v \rangle = 0$, so $\lambda \in N(T)$. Secondly, if the range of $T-\lambda I$ is not closed but dense in ${\mathcal {H}}_q$, then there is no $\mu >0$ such that $\Vert (T-\lambda I)v\Vert \ge \mu \Vert v\Vert$ for all $v \in {\mathcal {H}}_q$ as the range is not closed, so there is a sequence of normalised vectors $(v_n)_{n\ge 1}$ in ${\mathcal {H}}_q$ such that $(T-\lambda I)v_n \rightarrow 0$. It follows that $\langle Tv_n,v_n \rangle - \lambda = \langle (T-\lambda I)v_n,v_n \rangle \rightarrow 0$, so $\lambda \in \overline{N(T)}$. $\square$

The partial ordering on $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ can be formulated via the following equivalent properties.

Lemma A.4

For an operator $T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$, the following statements are equivalent.

• (i) $\langle Tv,v \rangle \ge 0$ for all $v\in {\mathcal {H}}_q$.

• (ii) $T=S^2$ for some $S \in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$.

• (iii) $\sigma (T)\subseteq [0,\infty )$.

Proof

$(i)\Longrightarrow (iii)$: If $\lambda < 0$ and $v \in {\mathcal {H}}_q$, then

$$\begin{aligned} \Vert (T-\lambda I)v\Vert ^2 = \Vert Tv\Vert ^2 -2\lambda \langle Tv,v \rangle + \lambda ^2\Vert v\Vert ^2 \ge \lambda ^2\Vert v\Vert ^2 \end{aligned}$$

which implies that $T-\lambda I$ is injective. The same inequality shows that the range of T is closed in ${\mathcal {H}}_q$, so we can define a left inverse $S \in \textrm{B}({\mathcal {H}}_q)$ of T by $S:= T'\oplus I_{\textrm{ ran }\, T^\perp }$, where $T'$ is the inverse of T restricted to the range of T. The fact that $T'$ is bounded follows from the inverse mapping theorem for bounded operators on Banach spaces [8, Theorem III.12.5] where the argument also works for operators on ${\mathcal {H}}_q$. Since T is self-adjoint, it follows that T also has a right inverse. Note that $T^{-1}$ needs to be self-adjoint, as, for any $v \in {\mathcal {H}}_q$, we have

$$\begin{aligned} \langle T^{-1}v,v \rangle = \langle T^{-1}Tw,Tw \rangle = \langle w,Tw \rangle = \langle Tw,w \rangle = \langle v,T^{-1}v \rangle . \end{aligned}$$

We conclude that $\lambda \notin \sigma (T)$. Hence, $\sigma (T)\subseteq [0,\infty )$.

$(iii)\Longrightarrow (ii)$: The existence of an operator S such that $T=S^2$ in this case follows from the functional calculus [3, Corollary 1.19].

$(ii)\Longrightarrow (i)$: If S is such that $T=S^2$, then it follows directly that $\langle Tv,v \rangle = \langle S^2v,v \rangle = \langle Sv,Sv \rangle \ge 0$ for all $v\in {\mathcal {H}}_q$. $\square$

Lemma A.5

The JB-algebra $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is monotone complete.

Proof

Without loss of generality, we may assume that $(T_i)_i$ is an increasing net such that $0\le T_i \le I$. By the functional calculus [3, Proposition 1.12], we have $S^2\le S$ for all $0\le S\le I$, so that, for any $v\in {\mathcal {H}}_q$ and $i\le j$, we have

$$\begin{aligned} \Vert (T_j-T_i)v\Vert ^2=\langle (T_j -T_i)v,(T_j-T_i)v \rangle = \langle (T_j-T_i)^2 v,v \rangle \le \langle (T_j-T_i) v,v \rangle \end{aligned}$$

(A.2)

by Lemma A.4. Furthermore, as $\langle T_iv,v \rangle \le \Vert v\Vert ^2$ for all $v\in {\mathcal {H}}_q$, the increasing net $(\langle T_iv,v \rangle )_i$ is Cauchy in ${\mathbb {R}}1$, which implies that $(T_iv)_i$ is a Cauchy net in ${\mathcal {H}}_q$ for every $v\in {\mathcal {H}}_q$ by (A.2). Hence, we can define a linear operator T via the pointwise norm limits $Tv:=\lim _i T_iv$. It follows that $\Vert T\Vert \le 1$ and since we also have that $\langle T_iv,v \rangle \rightarrow \langle Tv,v \rangle$ for all $v\in {\mathcal {H}}_q$, we have $T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. Suppose that $S \in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is such that $T_i\le S$ for all i. Then

$$\begin{aligned} \langle (S-T)v,v \rangle = \langle (S-T_i)v,v \rangle + \langle (T_i-T)v,v \rangle \ge \langle (T_i-T)v,v \rangle \rightarrow 0, \end{aligned}$$

so $T\le S$ by Lemma A.4 and T is the supremum of $(T_i)_i$. $\square$

Let $v \in {\mathcal {H}}_q$ be such that $\Vert v\Vert =1$. Then $\varphi _v:\textrm{B}({\mathcal {H}}_q)_\textrm{sa}\rightarrow {\mathbb {R}}$ defined by $\varphi _v(T):=\langle Tv,v \rangle$ is a positive linear functional with $\varphi _v(I)=1$, so it is a state on $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. Furthermore, if $(T_i)_i$ is an increasing net with supremum T in $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$, then we saw in the proof of Lemma A.5 that $\varphi _v(T_i)=\langle T_iv,v \rangle \rightarrow \langle Tv,v \rangle = \varphi _v(T)$. Hence, we have that $\varphi _v$ is a normal state. These states are referred to as vector states.

Lemma A.6

The vector states $\varphi _v$ on $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ are pure states.

Proof

Let v be a normalised vector and $\varphi _v$ be the corresponding vector state. Suppose that, for some $0<t<1$ and states $\psi$, $\eta$, we have $\varphi _v = t\psi +(1-t)\eta$. For the projection $P_vw:=v\cdot \langle v,w \rangle$, it follows that $\varphi _v(P_v)=1$ and as $0 \le \psi (P_v),\eta (P_v)\le 1$, since $0\le P_v\le I$, it follows that $\psi (P_v)=\eta (P_v)=1$. The symmetric bilinear form $(S,T) \mapsto \psi (S\circ T)$ is positive semi-definite, and so

$$\begin{aligned} |\psi (T\circ (I-P_v))|^2\le |\psi (T)||\psi (I-P_v)|=0 \end{aligned}$$

by the generalised Cauchy–Schwarz inequality, [9, 5.5.3]. Hence $\psi (T)=\psi (T\circ P_v)$ for all $T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. Using that $(P_v\circ T)\circ P_v = \frac{1}{2}P_vTP_v+\frac{1}{2}T\circ P_v$ and the fact that $P_vT P_v$ is self-adjoint, it follows that

$$\begin{aligned} \psi (T)=\psi ((P_v\circ T)\circ P_v)={\textstyle \frac{1}{2}}\psi (P_vT P_v)+ {\textstyle \frac{1}{2}}\psi (T\circ P_v) = {\textstyle \frac{1}{2}}\psi (P_vT P_v)+ {\textstyle \frac{1}{2}}\psi (T), \end{aligned}$$

so that $\psi (T)=\psi (P_v T P_v)$. Since $(P_v T P_v)w=v\cdot \langle Tv,v \rangle \langle v,w \rangle = \varphi _v(T)P_v w$ for all $w\in {\mathcal {H}}_q$, we conclude that $\psi (T)=\psi (P_v T P_v)=\varphi _v(T)\psi (P_v)=\varphi _v(T)$ for all $T \in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$, so $\psi =\varphi _v$. Hence, $\varphi _v = \psi = \eta$ and $\varphi _v$ is a pure state. $\square$

Note that if $T,S \in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ are such that $\langle (S-T)v,v \rangle = 0$ for all $v \in {\mathcal {H}}_q$, then $S\le T$ and $T\le S$ by Lemma A.4 and so $T=S$. We find that the vector states separate the points of $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ and hence, it follows that $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is a JBW-algebra.

We will show that $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is an atomic JBW-algebra. Indeed, let P be a non-zero orthogonal projection in $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. Then the range of P is a closed subspace of ${\mathcal {H}}_q$, so we may choose an orthonormal basis for it. Let v be an element of this orthonormal basis and note that the orthogonal projection $P_vw:=v\cdot \langle v,w \rangle$ satisfies $P_v=PP_v=P_vP$, so $P-P_v$ is idempotent and self-adjoint, so $\langle (P-P_v)w,w \rangle \ge 0$ for all $w\in {\mathcal {H}}_q$ and therefore $P_v\le P$. The following lemma shows that $P_v$ is an atom from which we can conclude that $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is atomic.

Lemma A.7

For $v \in {\mathcal {H}}_q$ with $\Vert v\Vert =1$, the projection $P_v$ is an atom, and every atom in $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is of this form.

Proof

Suppose that $P_v=Q+R$ for some orthogonal projections Q and R. If Qv and Rv are non-zero, then they are ${\mathbb {H}}$-linearly independent and so the dimension over ${\mathbb {H}}$ of the range of $Q+R$ is at least two, which contradicts the fact that the range of $P_v$ equals ${\mathbb {H}}v:=\{v\cdot q:\,q\in {\mathbb {H}}\}$. Hence, we may assume without loss of generality that $Qv = v$ and $Rv=0$. But $RQR=0$ by [3, Proposition 2.18] and as

$$\begin{aligned} \Vert RQR\Vert =\Vert RQ^2R\Vert =\Vert (QR)^*QR\Vert =\Vert QR\Vert ^2, \end{aligned}$$

it follows that $QR=0$. Hence, $P_vw = Q(v\cdot \langle v,w \rangle )= Qw+QRw = Qw$, so $Q=P_v$ and $P_v$ is an atom.

Let P be an atom in $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. Then the range of P must be of the form ${\mathbb {H}}v$ for some $v\in {\mathcal {H}}_q$ with $\Vert v\Vert =1$. If $w\in {\mathcal {H}}_q$ and $Pw=v\cdot q$, then it follows from $w=Pw+(I-P)w$ that $q=\langle v,w \rangle$ by taking the inner product with v as $v\in \ker P^\perp$. Hence, $P=P_v$. $\square$

Lemma A.8

The JBW-algebra $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ is a factor.

Proof

Suppose $T\in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ operator commutes with all $S \in \textrm{B}({\mathcal {H}}_q)_\textrm{sa}$. For any normalised $v\in {\mathcal {H}}_q$, the atom $P_v$ yields

$$\begin{aligned} {\textstyle \frac{1}{2}}(Tv+v\cdot \langle Tv,v\rangle )=T\circ (P_v\circ P_v)v= P_v\circ (T\circ P_v)v=v\cdot {\textstyle \frac{3}{4}}\langle Tv,v \rangle +{\textstyle \frac{1}{4}}Tv, \end{aligned}$$

hence, $Tv=v\cdot \langle Tv,v \rangle$. If v and w are two linearly independent normalised vectors in ${\mathcal {H}}_q$, then the linearity of T implies that

$$\begin{aligned} (v+w)\cdot \langle T(v+w),v+w \rangle \Vert v+w\Vert ^{-2}= & {} T(v+w)=Tv+Tw\\= & {} \langle Tv,v \rangle v+\langle Tw,w \rangle w \end{aligned}$$

and so $\langle Tv,v\rangle = \langle Tw,w\rangle$. Hence, it follows that $T=\lambda I$ for some $\lambda \in {\mathbb {R}}$ showing that the algebraic centre of $\textrm{B}({\mathcal {H}}_q)_\textrm{sa}$ equals ${\mathbb {R}}I$. $\square$

1.2 Spin factors

Let H be a real Hilbert space of dimension at least two. If we equip the direct sum $H\oplus {\mathbb {R}}$ with the product

$$\begin{aligned} (x,\lambda )\circ (y,\mu ):=(\mu x+\lambda y,\langle x,y \rangle +\lambda \mu ), \end{aligned}$$

(A.3)

then $H\oplus {\mathbb {R}}$ becomes a Jordan algebra with unit (0, 1). The Cauchy–Schwarz inequality implies that we can define the norm $\Vert (x,\lambda )\Vert :=\sqrt{\langle x,x \rangle }+|\lambda |$ on $H\oplus {\mathbb {R}}$ and it follows that this defines a JB-algebra norm.

Lemma A.9

The norm on $H \oplus {\mathbb {R}}$ is a JB-algebra norm.

Proof

Let $(x,\lambda )$ and $(y,\mu )$ be in $H \oplus {\mathbb {R}}$. Then

$$\begin{aligned}&\Vert (x,\lambda )\circ (y,\mu )\Vert \\&\quad = \Vert (\mu x+\lambda y,\langle x,y \rangle \!+\! \lambda \mu )\Vert \!=\! \sqrt{ \mu ^2\langle x,x \rangle \!+\! 2\lambda \mu \langle x,y \rangle \!+\! \lambda ^2\langle y,y \rangle } \!+\! |\langle x,y \rangle \!+\! \lambda \mu | \\&\quad \le \sqrt{\mu ^2\langle x,x \rangle + 2|\lambda ||\mu |\sqrt{\langle x,x \rangle } \sqrt{\langle y,y \rangle } + \lambda ^2\langle y,y \rangle } + \sqrt{\langle x,x \rangle }\sqrt{\langle y,y\rangle } +|\lambda ||\mu | \\&\quad = |\mu |\sqrt{\langle x,x \rangle } + |\lambda |\sqrt{\langle y,y \rangle } + \sqrt{\langle x,x \rangle }\sqrt{\langle y,y\rangle } +|\lambda ||\mu | \\&\quad = (\sqrt{\langle x,x \rangle } + |\lambda |)(\sqrt{\langle y,y \rangle } + |\mu |) \\&\quad = \Vert (x,\lambda )\Vert \Vert (y,\mu )\Vert , \end{aligned}$$

by the Cauchy–Schwarz inequality, and we can explicitly check the identity

$$\begin{aligned} \Vert (x,\lambda )^2\Vert= & {} \Vert (2\lambda x,\langle x,x\rangle +\lambda ^2)\Vert = 2|\lambda |\sqrt{\langle x,x \rangle }+\langle x,x \rangle + \lambda ^2 \\= & {} (\sqrt{\langle x,x \rangle }+|\lambda |)^2 = \Vert (x,\lambda )\Vert ^2. \end{aligned}$$

Lastly, by using the Cauchy–Schwarz inequality once more, it follows that

$$\begin{aligned}&\Vert (x,\lambda )^2 + (y,\mu )^2\Vert \\&\quad = \sqrt{ 4\lambda ^2 \langle x,x \rangle +8\lambda \mu \langle x,y \rangle + 4\mu ^2\langle y,y \rangle } + \langle x,x \rangle + \langle y,y \rangle + \lambda ^2+ \mu ^2 \\&\quad \ge \sqrt{ 4\lambda ^2 \langle x,x \rangle -8|\lambda ||\mu |\sqrt{\langle x,x \rangle }\sqrt{\langle y,y \rangle } + 4\mu ^2\langle y,y \rangle } + \langle x,x \rangle + \langle y,y \rangle + \lambda ^2+ \mu ^2 \\&\quad = 2|\lambda |\sqrt{\langle x,x \rangle } - 2|\mu |\sqrt{\langle y,y \rangle } + \langle x,x \rangle + \langle y,y \rangle + \lambda ^2+ \mu ^2 \\ {}&= (\sqrt{\langle x,x \rangle }+|\lambda |)^2 + (\sqrt{\langle y,y \rangle } - |\mu |)^2 \\&\quad \ge \Vert (x,\lambda )^2\Vert . \end{aligned}$$

Hence this norm satisfies the properties of a JB-algebra norm. $\square$

Furthermore, note that the Hilbert space direct sum norm $\Vert (x,\lambda )\Vert _2:=\sqrt{\langle x,x \rangle + \lambda ^2}$ on $H\oplus {\mathbb {R}}$ is equivalent to $\left\| \cdot \right\|$. Indeed, we have

$$\begin{aligned} \Vert (x,\lambda )\Vert _2 \le \sqrt{\langle x,x \rangle } + |\lambda | = \Vert (x,\lambda )\Vert , \end{aligned}$$

and by the concavity of the square root function, we find that

$$\begin{aligned} \Vert (x,\lambda )\Vert = \sqrt{\langle x,x \rangle } + |\lambda | \le \sqrt{2}\sqrt{\langle x,x \rangle + \lambda ^2} = \sqrt{2}\Vert (x,\lambda )\Vert _2. \end{aligned}$$

Hence, $H\oplus {\mathbb {R}}$ is reflexive by [20, Proposition 1.11.8]. It follows that $H\oplus {\mathbb {R}}$ is a JB-algebra that is a dual space, so by [3, Theorem 2.55] it is a JBW-algebra. These JBW-algebras are called a spin factor.

Lemma A.10

The projections in $H\oplus {\mathbb {R}}$ are precisely (0, 0), (0, 1), and $(x,\frac{1}{2})$ such that $\sqrt{\langle x,x \rangle } = \frac{1}{2}$. Moreover, the latter are precisely the atoms.

Proof

The equation $(x,\lambda )^2=(2\lambda x,\langle x,x \rangle +\lambda ^2)$ yields $(2\lambda -1)x=0$ and $\lambda ^2-\lambda + \langle x,x \rangle = 0$. So, if $x=0$, then $\lambda = 0$ or $\lambda = 1$ gives the idempotents (0, 0) and (0, 1), and if $x\ne 0$, then $\lambda = \frac{1}{2}$ and $\langle x,x \rangle = \frac{1}{4}$ as required. If x is such that $\sqrt{\langle x,x \rangle } = \frac{1}{2}$, then $(x,\frac{1}{2}) + (-x,\frac{1}{2}) = (0,1)$, and it is clear that $(x,\frac{1}{2})$ can not be written as the sum of two projections of the form $(y,\frac{1}{2})$ or (0, 1). Hence, the atoms are precisely of the form $(x,\frac{1}{2})$. $\square$

By [3, Lemma 1.10], spin factors are partially ordered by the cone of squares.

Lemma A.11

The cone of squares C in a spin factor $H \oplus {\mathbb {R}}$ equals $\Lambda :=\{(x,\lambda ):\sqrt{\langle x,x \rangle } \le \lambda \}$.

Proof

Observe that $(0,\lambda )$ is a square if and only if $\lambda \ge 0$. Suppose that $(x,\lambda )$ is not a multiple of (0, 1). Then $(x,\lambda )^2=(2\lambda x,\langle x,x \rangle +\lambda ^2)$ satisfies $\sqrt{\langle 2\lambda x,2\lambda x \rangle } = 2|\lambda |\sqrt{\langle x,x \rangle } \le \langle x,x \rangle +\lambda ^2$ as $(\sqrt{\langle x,x \rangle }-|\lambda |)^2\ge 0$ and it follows that $C\subseteq \Lambda$. Conversely, note that the two atoms $(x,\frac{1}{2})$ and $(-x,\frac{1}{2})$ are orthogonal since $(x,\frac{1}{2})\circ (-x,\frac{1}{2})=(0,0)$. So, if $(x,\lambda )$ is an element of $\Lambda$, then $(x,\lambda )$ is the square of

$$\begin{aligned} \sigma _1 \left( \frac{x}{2\sqrt{\langle x,x \rangle }},\frac{1}{2}\right) + \sigma _2\left( -\frac{x}{2\sqrt{\langle x,x \rangle }},\frac{1}{2}\right) \end{aligned}$$

where $\sigma _k:=\sqrt{\lambda + (-1)^{k+1}\sqrt{\langle x,x \rangle }}$ for $k=1,2$. Hence $\Lambda \subseteq C$. $\square$

To see that the spin factor $H\oplus {\mathbb {R}}$ is an atomic JBW-algebra, by Lemma A.10, it remains to show that there is an atom below (0, 1). Actually, for every atom $(x,\frac{1}{2})$, we have that $(x,\frac{1}{2}) \le (0,1)$ by Lemma A.11. Thus, $H\oplus {\mathbb {R}}$ is an atomic JBW-algebra.

Next, we consider the spectrum of elements of $H\oplus {\mathbb {R}}$. Suppose that $(x,\lambda )$ is not a multiple of (0, 1). Then we can write

$$\begin{aligned} (x,\lambda ) = \lambda (0,1) + \sqrt{\langle x,x \rangle }\left( \frac{x}{\sqrt{\langle x,x \rangle }},0\right) \end{aligned}$$

where $\langle x,x \rangle ^{-1/2}(x,0)$ squares to (0, 1). By the functional calculus [3, Corollary 1.19], the spectrum of the element $\langle x,x \rangle ^{-1/2}(x,0)$ must be $\{\pm 1\}$, so the spectrum of the element $(x,\lambda )$ must, therefore, equal $\{\lambda + \sqrt{\langle x,x \rangle },\lambda - \sqrt{\langle x,x \rangle }\}$. On the other hand, any multiple of (0, 1) has a spectrum containing at most two numbers, so any element of $H\oplus {\mathbb {R}}$ has a spectrum consisting of at most two numbers. Furthermore,

$$\begin{aligned} (x,\lambda ) = (\lambda + \sqrt{\langle x,x \rangle })\left( \frac{x}{2\sqrt{\langle x,x \rangle }},\frac{1}{2}\right) + (\lambda - \sqrt{\langle x,x \rangle })\left( -\frac{x}{2\sqrt{\langle x,x \rangle }},\frac{1}{2}\right) \end{aligned}$$

is the spectral decomposition of $(x,\lambda )$. Note that the spectrum of $(x,\lambda )$ is positive if and only if $(x,\lambda )$ is an element of C.

The cone $\Lambda :=\{(x,\lambda ):\sqrt{\langle x,x \rangle } \le \lambda \}$ in the vector space $H \oplus {\mathbb {R}}$ is called the Lorentz cone. Clearly, $(H \oplus {\mathbb {R}}, \Lambda )$ is an Archimedean partially ordered vector space and (0, 1) is an order unit. In particular, for every $n\ge 3$, the vector space ${\mathbb {R}}^n$ can be endowed with a Lorentz cone, by viewing ${\mathbb {R}}^n$ as ${\mathbb {R}}\times {\mathbb {R}}^{n-1}$ and considering the Euclidean inner product on ${\mathbb {R}}^{n-1}$.

Lemma A.12

A spin factor $H\oplus {\mathbb {R}}$ is in fact a factor.

Proof

Let $x,y\in H$ be orthogonal unit vectors. Then $(x,0)\circ (y,0)^2 = (x,0)\circ (0,1) = (x,0)$ and $(y,0)\circ \bigl ((x,0)\circ (y,0)\bigr ) = (0,0)$, so (x, 0) does not operator commute with (y, 0), thus (x, 0) can not be in the algebraic centre of $H\oplus {\mathbb {R}}$. This implies that if $(x,\lambda )=(x,0)+\lambda (0,1)$ is an element of the algebraic centre, then (x, 0) must be an element of the algebraic centre as $\lambda (0,1)$ is. We conclude that $x=0$ and, therefore, the algebraic centre equals ${\mathbb {R}}(0,1)$ and $H\oplus {\mathbb {R}}$ is a factor. $\square$

We show that a functional $\varphi$ is a state of $H\oplus {\mathbb {R}}$ if and only if there exists $y\in H$ with $\langle y,y\rangle =1$ such that $\varphi ((x,\lambda ))=\langle (x,\lambda ),(y,1)\rangle$ for every $(x,\lambda )\in H\oplus {\mathbb {R}}$. Indeed, let $\varphi$ be a state of $H\oplus {\mathbb {R}}$. By the Riesz representation theorem, it follows that there is a $(y,\mu )$ such that $\varphi ((x,\lambda ))=\langle (x,\lambda ),(y,\mu ) \rangle = \langle x,y\rangle + \lambda \mu$ for all $(x,\lambda ) \in H\oplus {\mathbb {R}}$. Since $\varphi (0,1) = 1$, we must have $\mu =1$. If $y\ne 0$, then

$$\begin{aligned} \varphi \left( \left( \frac{-y}{2\sqrt{\langle y,y \rangle }},\frac{1}{2}\right) \right) = \left\langle \left( \frac{-y}{2\sqrt{\langle y,y \rangle }},\frac{1}{2}\right) ,(y,1)\right\rangle = \frac{1}{2}\left( -\sqrt{\langle y,y \rangle } + 1\right) \ge 0 \end{aligned}$$

since $\varphi$ is positive, so $\sqrt{\langle y,y \rangle } \le 1$. On the other hand, if $y\in H$ is such that $\sqrt{\langle y,y \rangle }\le 1$ and we define the linear functional $\psi$ by $\psi ((x,\lambda )):=\langle x,y \rangle + \lambda$, then $\psi ((0,1))=1$ and

$$\begin{aligned} \psi ((x,\lambda )^2)&= 2\lambda \langle x,y \rangle + \langle x,x \rangle + \lambda ^2 \ge -2|\lambda | |\langle x,y \rangle | + \langle x,x \rangle + \lambda ^2 \\&\ge -2|\lambda |\sqrt{\langle x,x \rangle } + \langle x,x \rangle + \lambda ^2 = \left( \sqrt{\langle x,x \rangle } - |\lambda |\right) ^2 \ge 0 \end{aligned}$$

by the Cauchy–Schwarz inequality, so $\psi$ is positive. Hence, $\psi$ is a state.

Lemma A.13

Let (y, 1) represent the state $\varphi$ on $H\oplus {\mathbb {R}}$. Then $\varphi$ is a pure state if and only if $\sqrt{\langle y,y \rangle }=1$.

Proof

Suppose that $y \in H$ is such that $\sqrt{\langle y,y \rangle }<1$. If $y=0$, then we can write $(0,1) = \frac{1}{2}(x,1)+\frac{1}{2}(-x,1)$ for a unit vector $x\in H$, and the states represented by (x, 1) and $(-x,1)$ are distinct since

$$\begin{aligned} \langle (x,1),(-x,1) \rangle = -\langle x,x \rangle + 1 = 0\qquad \text {and}\qquad \langle (-x,1),(-x,1) \rangle = \langle x,x \rangle + 1 = 2. \end{aligned}$$

Hence, the state represented by (0, 1) is not a pure state. If $y\ne 0$, then we can write

$$\begin{aligned} (y,1) = t\left( \frac{y}{\sqrt{\langle y,y \rangle }},1\right) + (1-t)\left( \frac{-y}{\sqrt{\langle y,y \rangle }},1\right) \end{aligned}$$

for some $0<t<1$, and the states represented by $(\pm \ \langle y,y \rangle ^{-1/2}y,1)$ are again distinct since

$$\begin{aligned} \left\langle \left( \frac{y}{\sqrt{\langle y,y \rangle }},1\right) ,\left( \frac{-y}{\sqrt{\langle y,y \rangle }},1\right) \right\rangle = 0 \quad \text {and}\quad \left\langle \left( \frac{-y}{\sqrt{\langle y,y \rangle }},1\right) ,\left( \frac{-y}{\sqrt{\langle y,y \rangle }},1\right) \right\rangle = 2. \end{aligned}$$

We conclude that the state represented by (y, 1) can, therefore, not be a pure state. Conversely, suppose that y is a unit vector in H. If (x, 1) and (z, 1) represent states such that a non-trivial convex combination of them equal the state represented by (y, 1), then $(y,1)=t(x,1)+(1-t)(z,1)$ for some $0<t<1$ and so $y=tx+(1-t)z$. Since the unit sphere in H is strictly convex, it follows that $x=z=y$, so (y, 1) represents a pure state. $\square$

1.3 Matrices with octonionic entries

We introduce the multiplication rules on $\{e_1,\ldots , e_7\}$ as follows. Set $e_i^2=-1$ for all $1\le i\le 7$ and determine the product of any two $e_i$ and $e_j$ via the so called Fano plane below.

The elements $e_i$ and $e_j$ lie on a unique line consisting of three elements, including the circle. The product is defined by following the arrow and using cyclic permutations (which preserve the directions of the arrows). For example, the elements $e_1$ and $e_2$ lie on the line $(e_1,e_2,e_4)$, so $e_1 e_2=e_4$. Furthermore, we also have that $e_1$ and $e_4$ lie on the line $(e_1,e_2,e_4)$, which yields the same line $(e_4,e_1,e_2)$ by applying cyclic permutations, so $e_4e_1=e_2$. Transversing in the opposite direction of the indicated arrow yields a minus sign, that is, $e_1e_4=-e_2$. The 8-dimensional real vector space

$$\begin{aligned} {\mathbb {O}}:=\bigl \{a_01 + a_1 e_1 +\dots + a_7 e_7 :a_0,\dots ,a_7 \in {\mathbb {R}}\bigr \} \end{aligned}$$

equipped with the multiplication rules described above and unit 1, where a general product distributes over the sums, forms the so called octonions. The product on the octonions is not commutative as we have seen, and it also fails to be associative. Indeed, note that $(e_1e_2)e_3=e_4e_3=-e_6$ and $e_1(e_2e_3)=e_1e_5=e_6$. The octonions are alternative, meaning that the subalgebra generated by two elements in ${\mathbb {O}}$ is associative. The real multiples of the identity 1 commute with all octonions. For $x=a_01+\sum _{k=1}^7 a_ke_k$, the octonionic conjugate of x is defined to be $x^*:= a_01-\sum _{k=1}^7a_ke_k$. The octonionic conjugate is an involution on ${\mathbb {O}}$ that reverses the order of multiplication, that is, for $x,y\in {\mathbb {O}}$, we have $(xy)^*=y^*x^*$. The real part of x is denoted by ${\text {Re}}(x)$ and is given by ${\text {Re}}(x):=\frac{1}{2}(x+x^*)=a_01$. Note that every non-zero octonion is invertible since $x^*x=(a_0^2+\dots +a_7^2)1$. Furthermore, the octonionic conjugation induces a norm on ${\mathbb {O}}$ given by $\Vert x\Vert :=\sqrt{x^*x}$, where ${\mathbb {R}}1$ has been identified with ${\mathbb {R}}$. Similar to the norm on the quaternions, the norm is multiplicative, that is, $\Vert xy\Vert =\Vert x\Vert \Vert y\Vert$ for all $x,y\in {\mathbb {O}}$. This implies that

$$\begin{aligned} \Vert xy\Vert ^2=(xy)^*(xy)=(y^*x^*)(xy)=y^*(x^*x)y=\Vert x\Vert ^2\Vert y\Vert ^2. \end{aligned}$$

The octonions can be equipped with the real inner product $\langle x,y \rangle := {\text {Re}}(xy^*)= \frac{1}{2}(xy^*+yx^*)$, where again the real multiples of the identity 1 are identified with the real numbers. The inner product coincides with the standard inner product on ${\mathbb {R}}^8$, that is, for $x:=a_01+\sum _{k=1}^7a_ke_k$ and $y:=b_01+\sum _{k=1}^7b_ke_k$ it follows that $\langle x,y \rangle = \sum _{k=0}^7a_kb_k$. Furthermore, note that the norm relates to the inner product as usual, $\Vert x\Vert =\sqrt{\langle x,x \rangle }$. For the reader interested in studying properties of the octonions in more detail, we recommend the well written and extensive exposition on the subject [4].

In view of the theory of JB-algebras, let $\textrm{M}_n({\mathbb {O}})$ denote the $n\times n$ matrices over the octonions which form a non-associative unital real algebra. Similar to the Hermitian adjoint, an involution can be defined on $\textrm{M}_n({\mathbb {O}})$ given by $(A^*)_{ij}:=(A_{ji})^*$. Since every JB-algebra is formally real, the subspace of self-adjoint matrices $\textrm{M}_n({\mathbb {O}})_\textrm{sa}$ are considered instead of $\textrm{M}_n({\mathbb {O}})$, equipped with the commutative product $A\circ B:=\frac{1}{2}(AB+BA)$ (note that squares coincide for both products). It was shown by Jordan, von Neumann, and Wigner in [15] that $\textrm{M}_n({\mathbb {O}})_\textrm{sa}$ is a Jordan algebra for $1\le n \le 3$ and not for $n\ge 4$, see also [13, Theorem 2.7.6, Theorem 2.7.8]. In particular, it turns out that $\textrm{M}_2({\mathbb {O}})_\textrm{sa}$ is a spin factor, see [4, p. 28].

Lemma A.14

$\textrm{M}_2({\mathbb {O}})_\textrm{sa}$ is a spin factor.

Proof

Define the map $f :\textrm{M}_2({\mathbb {O}})_\textrm{sa} \rightarrow {\mathbb {R}}^9 \oplus {\mathbb {R}}$ by

$$\begin{aligned} \begin{pmatrix} \alpha +\beta &{} x \\ x^* &{} \alpha -\beta \end{pmatrix} \mapsto ((x,\beta ),\alpha ). \end{aligned}$$

It is a straightforward verification that f is a linear bijection that maps the identity matrix $I_2$ to the unit (0, 1). Let

$$\begin{aligned} A:=\begin{pmatrix} \alpha _1+\beta _1 &{} x \\ x^* &{} \alpha _1-\beta _1 \end{pmatrix} \qquad \text {and}\qquad B:=\begin{pmatrix} \alpha _2+\beta _2 &{} y \\ y^* &{} \alpha _2-\beta _2 \end{pmatrix}. \end{aligned}$$

Using that $\langle x,y \rangle = \langle x^*,y^* \rangle$, it follows that

$$\begin{aligned} A\circ B = \begin{pmatrix} a+b &{} \alpha _2 x+\alpha _1 y \\ \alpha _2 x^*+\alpha _1 y^* &{} a-b \end{pmatrix} \end{aligned}$$

where $a=\alpha _1\alpha _2+\beta _1\beta _2+\langle x,y \rangle$ and $b=\alpha _1\beta _2+\alpha _2\beta _1$, so

$$\begin{aligned} f(A\circ B)&=((\alpha _2 x+\alpha _1 y,\alpha _1\beta _2+\alpha _2\beta _1),\alpha _1\alpha _2+\beta _1\beta _2+\langle x,y \rangle )\\&=((x,\beta _1),\alpha _1)\circ ((y,\beta _2),\alpha _2)=f(A)\circ f(B) \end{aligned}$$

showing that f is a Jordan homomorphism. Hence $\textrm{M}_2({\mathbb {O}})_\textrm{sa}$ is isomorphic to the spin factor ${\mathbb {R}}^9\oplus {\mathbb {R}}$. $\square$

Remark A.15

A similar argument proves that $\textrm{M}_2({\mathbb {H}})_\textrm{sa}$ is isomorphic to the spin factor ${\mathbb {R}}^5\oplus {\mathbb {R}}$, that $\textrm{M}_2({\mathbb {C}})_\textrm{sa}$ is isomorphic to the spin factor ${\mathbb {R}}^3\oplus {\mathbb {R}}$, and that $\textrm{M}_2({\mathbb {R}})_\textrm{sa}$ is isomorphic to the spin factor ${\mathbb {R}}^2\oplus {\mathbb {R}}$. Therefore, by Lemma A.10, all the minimal projections in $\textrm{M}_2({\mathbb {R}})_\textrm{sa}$ are of the form

$$\begin{aligned} \begin{pmatrix} \frac{1}{2}+x_2&{}x_1\\ x_1&{}\frac{1}{2}-x_2 \end{pmatrix} \end{aligned}$$

where $x_1^2+x_2^2=\frac{1}{4}$.

The self-adjoint $3\times 3$ matrices over the octonions is called the Albert algebra and is an exceptional Jordan algebra, as it is not Jordan isomorphic to a subalgebra of an associative real algebra A with the product $a\circ b:=\frac{1}{2}(ab+ba)$, see [3, Theorem 4.6] and [13, Corollary 2.8.5]. Furthermore, by [3, Theorem 3.32], the Albert algebra is a JBW-algebra and even a factor. Hence, it follows from [3, Lemma 1.10] that $\textrm{M}_3({\mathbb {O}})_\textrm{sa}$ is partially ordered by the cone of squares. The minimal projections (or atoms) in $\textrm{M}_3({\mathbb {O}})_\textrm{sa}$ can be characterised as follows.

Proposition A.16

The minimal projections P in $\textrm{M}_3({\mathbb {O}})_\textrm{sa}$ are of the form

$$\begin{aligned}P:= \begin{pmatrix} \Vert x_1\Vert ^2 &{} x_1x_2^* &{} x_1x_3^*\\ x_2x_1^* &{} \Vert x_2\Vert ^2 &{} x_2x_3^*\\ x_3x_1^* &{} x_3x_2^* &{} \Vert x_3\Vert ^2 \end{pmatrix} \end{aligned}$$

where $x_1,x_2,x_3\in {\mathbb {O}}$ associate, that is, $(x_1x_2)x_3=x_1(x_2x_3)$, and $\Vert x_1\Vert ^2+\Vert x_2\Vert ^2+\Vert x_3\Vert ^2=1$.

Proof

Let

$$\begin{aligned}A:= \begin{pmatrix} r_1 &{} y_1^* &{} y_2^*\\ y_1 &{} r_2 &{} y_3^*\\ y_2 &{} y_3 &{} r_3 \end{pmatrix} \end{aligned}$$

be so that $A^2=A$. Then, as

$$\begin{aligned} A^2= \begin{pmatrix} r_1^2+\Vert y_1\Vert ^2+\Vert y_2\Vert ^2 &{} (r_1+r_2)y_1^*+y_2^*y_3 &{} (r_1+r_3)y^*_2+y^*_1y_3^*\\ (r_1+r_2)y_1+y_3^*y_2 &{} r_2^2+\Vert y_1\Vert ^2+\Vert y_3\Vert ^2 &{} (r_2+r_3)y^*_3+y_1y_2^*\\ (r_1+r_3)y_2+y_3y_1 &{} (r_2+r_3)y_3+y_2y_1^* &{} r_3^2+\Vert y_2\Vert ^2+\Vert y_3\Vert ^2 \end{pmatrix}, \end{aligned}$$

(A.4)

it follows that $0\le r_1,r_2,r_3\le 1$ and not all the $r_i$ are zero, as otherwise $A=0$. Furthermore, from the system of equations

$$\begin{aligned} {\left\{ \begin{array}{ll} (1-r_1-r_2)y_1=y_3^*y_2\\ (1-r_1-r_3)y_2=y_3y_1\\ (1-r_2-r_3)y_3=y_2y_1^* \end{array}\right. } \end{aligned}$$

we see that $y_1,y_2$, and $y_3$ are in a subalgebra $N\subseteq {\mathbb {O}}$ generated by two elements (and 1). Since ${\mathbb {O}}$ is alternative, we must have that N is associative. Let $x\in N$ be non-zero. Since ${\mathbb {O}}$ has no zero divisors, the ${\mathbb {R}}$-linear map $L_x(y):=xy$ is injective on ${\mathbb {O}}$. Hence, the restriction of $L_x$ to N is injective as well and as N is finite-dimensional, it is a bijection. Let $z\in N$ be such that $xz=1$. It follows that $(zx)^2=(zx)(zx)=z(xz)x=zx$ as N is associative, and so $zx=1$, again since ${\mathbb {O}}$ has no zero divisors. This shows that x has an inverse z and, therefore, N is a real division algebra. By Hurwitz’s theorem [14], we have that N is isomorphic to ${\mathbb {R}}$, ${\mathbb {C}}$, or ${\mathbb {H}}$. It follows that the entries of A are elements of the algebra N which is isomorphic to ${\mathbb {H}}$. Under this isomorphism, the inner product on $N^3$ induced by the inner product of ${\mathbb {O}}^3$ coincides with the inner product of ${\mathbb {H}}^3$.

Hence, by Lemma A.7, there is a unital vector $x\in N^3$ such that $Ay=x \cdot \langle x,y \rangle$. It follows that $x:=(x_1,x_2,x_3)$ is a unital vector in ${\mathbb {O}}^3$ with $(x_1x_2)x_3=x_1(x_2x_3)$ and

$$\begin{aligned} A=\begin{pmatrix} \Vert x_1\Vert ^2 &{} x_1x_2^* &{} x_1x_3^*\\ x_2x_1^* &{} \Vert x_2\Vert ^2 &{} x_2x_3^*\\ x_3x_1^* &{} x_3x_2^* &{} \Vert x_3\Vert ^2 \end{pmatrix}. \end{aligned}$$

(A.5)

Conversely, suppose A is as in (A.5) for some unit vector $(x_1,x_2,x_3)\in {\mathbb {O}}^3$ such that $(x_1x_2)x_3=x_1(x_2x_3)$. Then it follows that $A^2=A$, and as the subalgebra $M\subseteq {\mathbb {O}}$ generated by $x_1$, $x_2$, and $x_3$ (and 1) is associative, it follows that M is isomorphic to ${\mathbb {R}}$, ${\mathbb {C}}$, or ${\mathbb {H}}$ by Hurwitz’s theorem once more. Therefore, the matrix A satisfies $Ay=x\cdot \langle x,y \rangle$ for $x=(x_1,x_2,x_3)$, so it is a minimal projection by Lemma A.7. $\square$

The trace of $A\in \textrm{M}_3({\mathbb {O}})_\textrm{sa}$ is defined as usual for matrices by $\textrm{trace}(A):=A_{11}+A_{22}+A_{33}$, where $A_{ii}$ are the diagonal entries of A. It follows that $\langle A,B \rangle := \textrm{trace}(A\circ B)$ is a real inner product on $\textrm{M}_3({\mathbb {O}})_\textrm{sa}$. Indeed, note that, by (A.4), it follows that $\langle A,A \rangle \ge 0$, and $\langle A,A \rangle = 0$ if and only if $A=0$. Furthermore, $\textrm{trace}(A\circ B)={\text {Re}}(\textrm{trace}(AB))$ and, by [10, Proposition V.2.2], we have that $\langle A\circ B, C \rangle = \langle A, B\circ C \rangle$ for all $A,B,C\in \textrm{M}_3({\mathbb {O}})_\textrm{sa}$. Hence, with this inner product $\textrm{M}_3({\mathbb {O}})_\textrm{sa}$ is a Euclidean Jordan algebra. For any $A\in \textrm{M}_3({\mathbb {O}})_\textrm{sa}$, there are unique $\lambda _1,\dots ,\lambda _m$ and unique pairwise orthogonal projections $P_1,\dots ,P_m$ such that $A=\lambda _1 P_1+\dots +\lambda _mP_m$ by [10, Theorem III.1.1]. This is the spectral decomposition of A. The spectrum of A, denoted by $\sigma (A)$, consists of the eigenvalues that occur in the spectral decomposition of A, that is, $\sigma (A)=\{\lambda _1,\ldots ,\lambda _m\}$.

Lemma A.17

Let $A\in \textrm{M}_3({\mathbb {O}})_\textrm{sa}$. Then the following statements are equivalent.

• (i) $A\ge 0$.

• (ii) $\sigma (A)\subseteq [0,\infty )$.

• (iii) $\langle A,B \rangle \ge 0$ for all $B\ge 0$.

Proof

$(i)\Longleftrightarrow (ii)$: If $A\ge 0$, then $A=B^2$ for some $B\in \textrm{M}_3({\mathbb {O}})_\textrm{sa}$, and the spectral decomposition of $B=\lambda _1P_1+\dots +\lambda _mP_m$ now yields $A=B^2=\lambda _1^2P_1+\dots +\lambda _m^2P_m$, so $\sigma (A)\subseteq [0,\infty )$. On the other hand, if $\sigma (A)\subseteq [0,\infty )$, then the spectral decomposition of $A=\mu _1Q_1+\dots +\mu _nQ_n$ yields $A=B^2$ for $B:=\sqrt{\mu _1}Q_1+\dots +\sqrt{\mu _n}Q_n$.

$(i)\Longleftrightarrow (iii)$: This equivalence follows from the fact that the cone of squares in a Euclidean Jordan algebra yields a symmetric cone by [10, Theorem III.2.1]. In particular, $\langle A,B\rangle \ge 0$ for all $B\ge 0$ if and only if $A\ge 0$. $\square$

By the Riesz representation theorem, for every functional $\varphi :\textrm{M}_3({\mathbb {O}})_\textrm{sa} \rightarrow {\mathbb {R}}$, there is a unique $B\in \textrm{M}_3({\mathbb {O}})_\textrm{sa}$ such that $\varphi (A)=\langle A, B\rangle$. Furthermore, it follows from Lemma A.17 that $\varphi :=\langle \cdot ,B\rangle$ is a state if and only if $B\ge 0$ and $\textrm{trace}(B)=1$.

Lemma A.18

A state $\langle \cdot ,B\rangle$ on $\textrm{M}_3({\mathbb {O}})_\textrm{sa}$ is pure if and only if B is a minimal projection, i.e., an atom.

Proof

Suppose that $\langle \cdot ,B\rangle$ is a pure state on $\textrm{M}_3({\mathbb {O}})_\textrm{sa}$. Let $B=\sum _{k=1}^m \lambda _k P_k$ be the spectral decomposition of B such that $\lambda _k\ne 0$ for all k. Since every $P_k$ can be written as the sum of minimal projections, we may assume that each $P_k$ is minimal. Suppose that there are two distinct minimal projections $P_i$ and $P_j$ in this decomposition. Then we can write

$$\begin{aligned} \langle \cdot ,B\rangle =\lambda _i\langle \cdot ,P_i\rangle +\textstyle {\sum _{k\ne j}\lambda _k\left( (\sum _{k\ne j}\lambda _k)^{-1}\left\langle \cdot ,\sum _{k\ne j}\lambda _kP_k\right\rangle \right) } \end{aligned}$$

(A.6)

and since $\textrm{trace}(P_k)=1$ for all k by Lemma A.16, it follows that (A.6) writes $\langle \cdot ,B\rangle$ as a non-trivial convex combination of two distinct states, which is impossible. Hence, we have that $\langle \cdot ,B\rangle = \lambda _k\langle \cdot ,P_k\rangle$ for some k and as $\textrm{trace}(B)=1$, we find that $B=P_k$.

On the other hand, let P be a minimal projection and suppose $\langle \cdot ,P\rangle =t\langle \cdot ,C\rangle +(1-t)\langle \cdot , D\rangle$ for $0<t<1$, $C,D\ge 0$ with $\textrm{trace}(C)=\textrm{trace}(D)=1$. Then we must have $P=tC+(1-t)D$ and so $C=\lambda P$ and $D=\mu P$ by [3, Proposition 2.15] and [3, Lemma 3.29] since P is a minimal projection. Because $\textrm{trace}(C)=\textrm{trace}(D)=1$, it follows that $\lambda =\mu =1$ and so $\langle \cdot ,P\rangle$ is a pure state. $\square$

1.4 The pre-duals of atomic JBW-algebra factors

We conclude this appendix by determining the pre-duals of all atomic JBW-algebra factors M. For this we need the notion of so called trace class elements. These are elements $x \in M$ that can be written as

$$\begin{aligned} x = \sum _{k=1}^\infty \lambda _k p_k, \end{aligned}$$

where $(p_k)_k$ is a sequence of pairwise orthogonal atoms in M and $(\lambda _k)_k \subseteq {\mathbb {R}}$ satisfies $\sum _{k=1}^\infty |\lambda _k| < \infty$. The set of trace class elements will be denoted by $M_{\textrm{tr}}$, and a trace can be define on $M_{\textrm{tr}}$ by

$$\begin{aligned} \textrm{tr}(x):= \sum _{k=1}^\infty \lambda _k. \end{aligned}$$

The trace does not depend on the representation of x, so it is well defined on $M_\textrm{tr}$; see [3, Definition 5.65] and the paragraph below for more details. Given $x \in M$, we consider the JB-subalgebra JB(x, e) of M generated by x and e, which is isomorphic to a space of continuous functions. Hence, in JB(x, e) the modulus |x| of x exists. The trace norm of $x \in M_{\textrm{tr}}$ is defined by

$$\begin{aligned} \Vert x\Vert _\textrm{tr}:=\textrm{tr}(|x|) = \sum _{k=1}^\infty |\lambda _k|. \end{aligned}$$

It follows from [3, Proposition 5.66] that $M_{\textrm{tr}}$ equipped with the trace norm is a Banach space. The pre-dual $M_*$ of M is isometrically isomorphic to $M_{\textrm{tr}}$. In particular, for all finite-dimensional factors and spin factors, the pre-dual is the same space equipped with the trace norm. For the self-adjoint bounded operators on a real, complex, or quaternionic Hilbert space, the pre-dual is identified with the three analogues of trace class operators.