1 Jordan Algebra

What is a Jordan Algebra?

Motivation:

• Copenhagen models observables → Hermitian/adjoint operators in Hilbert space
  • In quantum mechanics (under the Copenhagen interpretation), physical quantities that can be measured—such as energy, position, momentum, and spin—are called “observables”. These are not represented by ordinary numbers, but by mathematical objects known as Hermitian operators or matrices.
  • We say matrix $A$ is Hermitian if $A^{\ast }=A$.
• What if all Hermitian matrices form an algebra?
  • Take $A,B$ as two Hermitian matrices.
  • Unfortunately, $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }\ne A^{\ast }{B}^{\ast }$ is not Hermitian.
  • Note that $A^2$ and $A+B$ are Hermitian.
  • Hence, $AB+BA=A^2+B^2-(A+B{)}^2$ is Hermitian.
  • Maybe we can define the multiplication as $A\circ B=AB+BA$.

Definition

Let $\mathcal{J}$ be an algebra. Jordan suggested the following axioms for $(J,\circ )$:

• $a\circ b=b\circ a$
• $a^2\circ (b\circ a)=(a^2\circ b)\circ a$ (Jordan identity)

Define $[a,b]=ab-ba$ and $(a,b,c)=(ab)c-a(bc)$, then the axioms above mean $[a,b]=0$ and $(a^2,b,a)=0$.

Such algebra $(J,\circ )$ is called Jordan algebra.

Special and Exceptional

Jordan wanted to describe simple (finite-dimensional) Jordan algebras.

In 1933, they were looking for algebras which do not have associative algebra “governing” it: If $A$ is an associative algebra, then $A^{+}=(A,\circ )$ with $a\circ b=ab+ba$ will be an Jordan algebra.

Definition

We call Jordan algebra $\mathcal{J}$ special if it can be embedded/ is a subalgebra in $A^{+}$ for some associative algebra $A$. Otherwise, it is called exceptional.

Examples.

• $M_n^{+}$ is a special Jordan algebra.
• Suppose $A$ is an associative algebra with involution $\ast$, then $H(A,\ast )=\{a\in A:a^{\ast }=a\}$ is a Jordan subalgebra in $A^{+}$, because $(a\circ b{)}^{\ast }=(ab+ba{)}^{\ast }=ba+ab=a\circ b$. Obviously it is special.

Classification of JNW (Jordan-Neumann-Wigner)

It is a classification of finite-dimensional, formally real Jordan algebras.

1. ${\mathbb{R}}^{+}$;

2. The three families $H_n(\mathbb{R}),H_n(\mathbb{C}),H_n(\mathbb{H})$.

   • Take associative algebra $M_n(\mathbb{R}),M_n(\mathbb{C}),M_n(\mathbb{H})$, where $\mathbb{H}$ is Hamiltonian algebra quaternions, and then define corresponding Jordan algebra as follows:
   • Define $(a_{ij}{)}^{\ast }=(\overline{a_{ij}}{)}^T$, then $H_n(\mathbb{R}),H_n(\mathbb{C}),H_n(\mathbb{H})$ with respect to this involution are special, which are denoted by ${\mathrm{Sym}}_n,M_n(\mathbb{R}{)}^{+}$.

3. Clifford Type Jordan Algebra or Spin Factor

   • For a vector space $V$, define $\mathcal{J}(V,f)=V\oplus \mathbb{R}\cdot 1$, where $f$ is a symmetric bilinear form. For example, $f$ is an inner product in ${\mathbb{R}}^n$ and
   $$(v+\alpha )(u+\beta )=(\beta v+\alpha u)+(⟨v,u⟩+\alpha \beta ).$$
   • Exercise. Prove that $\mathcal{J}(V,f)$ is a special Jordan algebra. (Hint: Show that $\mathcal{J}(V,f)\hookrightarrow {\mathrm{Clif}}^{+}(f)$, where $\mathrm{Clif}(f)$) is a Clifford algebra corresponding to $f$.

4. For the associative algebra $M_3(\mathbb{O})$, where $\mathbb{O}$ is Octonions = double quaternions. Define $\mathbb{O}=\{1,e_1,\cdots ,e_7\}$ with $e_i^2=-1$, $e_i{e}_{i+1}=e_{i+3}(\mathrm{mod}7)$ and $e_i{e}_j=-e_j{e}_i$. If $e_i{e}_j=e_k$, then $e_{\sigma (i)}{e}_{\sigma (j)}=\mathrm{sgn}(\sigma )e_{\sigma (k)}$. Then ${\mathbb{H}}_3(\mathbb{O})$ is a Jordan algebra.

   • $H_3(\mathbb{O})$, where $\{a_{ij}{\}}^{\ast }=\overline{a_{ji}}$, $(e_i{)}^{\ast }=-e_i$, $(1{)}^{\ast }=1$. If $M$ is an element of ${\mathbb{H}}_3(\mathbb{O})$, then

     $$M=\left[\begin{array}{ccc}\alpha & a& b\\ \overline{a}& \beta & c\\ \overline{b}& \overline{c}& \gamma \end{array}\right],$$

     where $\alpha ,\beta ,\gamma \in \mathbb{R}$ and $a,b,c\in \mathbb{O}$. Hence, we know $\dim {\mathbb{H}}_3(\mathbb{O})=27$.

   • In 1934, Albert proved that $H_3(\mathbb{O})$ is indeed exceptional. Furthermore, ${\mathbb{H}}_2(\mathbb{O})$ is special and ${\mathbb{H}}_n(\mathbb{O})$ with $n⩾4$ is not Jordan.

A Property

Proposition

Let $A$ be a simple associative algebra with $1\in A$. Then $A^{+}$ is a simple Jordan algebra. Recall that an algebra is simple if it does not contain non-trivial ideals.

\begin{proof} Let $I^{+}$ be an ideal of $A^{+}$. For any $x\in I^{+}$ and $a\in A$, one can check

$$(x\circ a)\circ x-x^2\circ a=(xa+ax)x+x(xa+ax)-x^{2}a-a{x}^2=2xax\in I^{+}$$

and so $xax\in I^{+}$. (We assume that $\mathrm{char}R\ne 2$.) It deduces that $x^{2}a=x\circ (xa)-xax\in I^{+}$ and $a{x}^2=(ax)\circ x-xax\in I^{+}$. Then the ideal of $A$ generated by $x^2$ is equal to $A$ or to $0$ as $A$ is simple.

If we have $x\in I^{+}$ such that $x^2\ne 0$, then $A=⟨x^2⟩\subseteq I^{+}$ and so $I^{+}=A$.

If $x^2=0$ for all $x\in I^{+}$, then $x\circ y=(x+y{)}^2-x^2-y^2=0$ for any $x,y\in I^{+}$. It deduces that $2xax=(x\circ a)\circ x-x^2\circ a=0$ for any $x\in I^{+}$ and $a\in A$. Hence $xAx=0$ and so $(⟨x⟩{)}^2=0$. Since $A$ is simple and $⟨x⟩$ is an ideal, either $⟨x⟩=A$ or $⟨x⟩=0$, where the former case is impossible. Therefore, $⟨x⟩=0$ and so $I=0$. \end{proof}

Exercise. (Herstein theorem) Let $A$ be an associative algebra with involution. If $(A,\ast )$ is $\ast$-simple, then $H(A,\ast )$ is simple Jordan algebra. See here.

Proposition

$(A,\ast )$ is $\ast$-simple if $A\ne 0$ and $A$ does not contain non-trivial ideals $I$ such that $I^{\ast }=I$.

Remark. This proposition is used for Classification of JNW (Jordan-Neumann-Wigner).