Euclidean Jordan Algebra (1) - Definitions and Properties

A Euclidean Jordan algebra is a triple $(V, \circ, \left< \cdot,\, \cdot \right>)$, where $(V, \left< \cdot,\, \cdot \right>)$ is a finite dimensional inner product space over $\mathbb{R}$ and $(x,y) \mapsto x \circ y : V \times V \rightarrow V$ is a bilinear mapping satisfying the following conditions:

• $x \circ y = y \circ x$ for all $x,y \in V$.
• $x^2 \circ (x \circ y) = x \circ (x^2 \circ y)$ for all $x,y \in V$ where $x^2=x \circ x$. (Jordan identity)
• $\left< x \circ y,\, z \right> = \left< x,\, y \circ z \right>$ for all $x,y,z \in V$.

and there exists a (unique) unit element $e \in V$ such that $x \circ e = x$ for all $x \in V$. Henceforth, we simply say that $V$ is a Euclidean Jordan algebra and $x \circ y$ is called the Jordan product of $x$ and $y$. 

Example 1. Typical examples of Euclidean Jordan algebras are:

1. Euclidean Jordan algebra of $n$-dimensional vectors:
   $$ V = \mathbb{R}^n, \quad \left< x,\, y \right> = \sum_{i=1}^{n}x_i y_i, \quad x \circ y = x \ast y,$$ where $x \ast y$ denotes the component-wise product of vectors $x$ and $y$. Here, the unit element is $e = (1,\ldots,1) \in \mathbb{R}^n$. We simply use $\mathbb{R}^n$ to denote this algebra.
2. Euclidean Jordan algebra of $n \times n$ symmetric matrices:
   $$ V = \mathcal{S}^n, \quad \left< X,\, Y \right> = \operatorname{tr}(XY), \quad X \circ Y = \frac{1}{2}(XY + YX).$$ Here, $\operatorname{tr}$ denotes the trace of a matrix. The identity matrix $I \in \mathcal{S}^n$ is the unit element of this algebra. This algebra is denoted by $\mathcal{S}^n$ in short.
3. Euclidean Jordan algebra of quadratic forms (Jordan spin algebra):
   $$ V = \mathbb{R}^n, \quad \left< x,\, y \right> = \sum_{i=1}^{n}x_i y_i, \quad x \circ y = (x_1, \bar{x}) \circ (y_1, \bar{y}) = (\left< x,\, y \right>, x_1 \bar{y} + y_1 \bar{x}), $$ where $\bar{x}, \bar{y} \in \mathbb{R}^{n-1}$. In this algebra, the unit element is $e = (1,0,\ldots,0) \in \mathbb{R}^n$. This algebra is denoted by $\mathcal{L}^n$.

A Euclidean Jordan algebra $V$ is simple if it does not contain any non-trivial ideal (that is, ideals of $V$ are $\{e\}$ and $V$ only). The classification theorem says that there are, up to isomorphism, only five simple Euclidean Jordan algebras. In the above example, the items 2, 3 are such simple Euclidean Jordan algebras. Other three examples are 

     4.  the algebra $\mathcal{H}^n$ of $n \times n$ complex Hermitian matrices,

     5.  the algebra $\mathcal{Q}^n$ of $n \times n$ quaternion Hermitian matrices,

     6.  the algebra $\mathcal{O}^3$ of $3 \times 3$ octonian Hermitian matrices.

However, the item 1 is not simple. That is simply because $\mathbb{R} = \mathcal{S}^1 \times \mathcal{S}^1 \times \cdots \times \mathcal{S}^1$.

Theorem 1. Any Euclidean Jordan is, in a unique way, a direct sum of simple Euclidean Jordan algebras. Moreover, the symmetric cone in a given Euclidean Jordan algebra is, in a unique way, a direct sum of symmetric cones in the constituent simple Euclidean Jordan algebras.

We note that the ‘direct sum’ in the theorem refers to the orthogonal as well as the Jordan product direct sum. Thus given a Euclidean Jordan algebra $V$ and the corresponding symmetric cone $V_+$, we may write

$$ V = V^{(1)} \times V^{(2)} \times \cdots \times V^{(k)} \qquad \text{and} \qquad V_+ = V_+^{(1)} \times V_+^{(2)} \times \cdots \times V_+^{(k)}, $$ where each $V^{(j)}$ is simple with the corresponding symmetric cone $V_+^{(j)}$. Moreover, for $x = (x^{(1)}, \ldots, x^{(j)})$ and $y = (y^{(1)}, \ldots, y^{(j)})$ in $V$ with $x^{(j)}, y^{(j)} \in V^{(j)}$, we have

$$ x \circ y = (x^{(1)} \circ y^{(1)}, \ldots, x^{(k)} \circ y^{(k)}), \qquad \left< x,\, y \right> = \sum_{j=1}^{k} \left< x^{(j)},\, y^{(j)} \right>, \qquad \Vert x \Vert^2 = \sum_{j=1}^{k} \Vert x^{(j)} \Vert^2. $$

An element $c \in V$ is an idempotent if $c^2 = c$; it is a primitive idempotent if it is nonzero and cannot be written as a sum of two nonzero idempotents. We say a finite set $\{ e_{1}, e_{2}, \ldots , e_{r} \}$ of primitive idempotents in $V$ is a Jordan frame if

$$ e_{i} \circ e_{j} = 0 \ \text{if} \ i \neq j \quad \text{and} \quad \sum_{i=1}^{r}{e_{i}}=e. $$ Note that $ \left< e_{i} , e_{j} \right> = \left< e_{i} \circ e_{j}, e \right> = 0 $ whenever $i \neq j$. 

When $V$ is not simple (that is, when $k > 1$ in the previous setting), it can be verified that every primitive idempotent element $c$ of $V$ has necessarily the form $c = (0, 0, \ldots, c^{(j)}, 0, \ldots, 0)$ for some primitive idempotent element $c^{(j)}$ in $V^{(j)}$.

The rank of $V$ is defined as $r = \max \{ \operatorname{deg}(x) : x \in V \}$, where $\operatorname{deg}(x)$ is the degree of $x \in V$ given by $\operatorname{deg}(x) = \min \{k>0 : \{e, x, x^2, \ldots, x^k \}$ is linearly dependent$\}$. 

In $V$, it is known that the set of squares $V_+ := \{ x^2 : x \in V \}$ is a symmetric cone. This means that $V_+$ is a closed convex cone which is self-dual (proper cone and $(V_+)^* = V_+$) and homogeneous (for any two elements $x,\, y \in V_+$, there exists an invertible linear transformation $\Gamma : V \rightarrow V$ such that $\Gamma(V_+) = V_+$ and $\Gamma(x) = y$.)

Example 2.

1. In $\mathbb{R}^n$, the symmetric cone is $\mathbb{R}_+^n$, the nonnegative orthant of $\mathbb{R}^n$. The rank of $\mathbb{R}^n$ is $n$.

2. The symmetric cone of $\mathcal{S}^n$ is given by $\mathcal{S}_+^n$, the set of all $n \times n$ positive semidefinite matrices. It is known that $\operatorname{rank}(\mathcal{S}^n) = n$, whereas the dimension is $\dim(\mathcal{S}^n) = n(n+1)/2$.

3. For Jordan spin algebra $\mathcal{L}^n$, we have

   $$ \mathcal{L}_+^n = \{ x = (x_1, \bar{x}) \in \mathbb{R} \times \mathbb{R}^{n-1} : x_1 \geq \Vert \bar{x} \Vert \}, $$ where $\Vert \cdot \Vert$ denotes the Euclidean norm on $\mathbb{R}^{n-1}$. The symmetric cone $\mathcal{L}_+^n$ is called the Lorentz cone (or the second-order cone). Moreover, we have  $\operatorname{rank}(\mathcal{L}^n) = 2$.

We end this subsection by recalling some basic properties.

• $x \in V_+$ if and only if $\left< x,\, y \right> \geq 0$ holds for all $y \in V_+$. Moreover, $x \in \operatorname{int}(V_+)$ if and only if $\left< x,\, y \right> > 0$ holds for all $y \in V_+ \backslash \{0\}$.

• For $x,y \in V_+$, orthogonality condition $\left< x,\, y \right> = 0$ is equivalent to $x \circ y = 0$. In this case, $x$ and $y$ operator commute, that is, $x \circ (y \circ z) = y \circ (x \circ z)$ for all $z \in V$.