Euclidean Jordan Algebra (2) - Decompositions

The following theorem gives us a spectral decomposition for the elements in an Euclidean Jordan algebra.

Theorem 2. (Spectral decomposition theorem)

Suppose $V$ is a Euclidean Jordan algebra with rank $r$. Then, for every $x \in V$, there exists a Jordan frame $\{e_1, \ldots, e_r\}$ and real numbers $\lambda_1(x), \ldots, \lambda_r(x)$ such that $x = \lambda_1(x)e_1 + \cdots \lambda_r(x) e_r$. The numbers $\lambda_i(x)$'s are uniquely determined and called the eigenvalues of $x$.

It is easy to show that $x \in V_+$ if and only if $\lambda_i(x) \geq 0$ for all $i$. Due to the uniqueness of the eigenvalues, we can define the trace and the determinant of $x$ as

$$ \operatorname{tr}(x) := \sum_{i=1}^{r} \lambda_i(x), \qquad \text{and} \qquad \operatorname{det}(x) := \prod_{i=1}^{r} \lambda_i(x). $$ Note that $\operatorname{tr}(c) = 1$ for every primitive idempotent $c$ in $V$.

We recall that in any simple Euclidean Jordan algebra $V$, there exists a $\theta > 0$ such that $\left< x,\, y \right> = \theta \operatorname{tr}(x \circ y)$. Hence, $\theta = \left< e,\, e \right> = \Vert c \Vert^2$ for every primitive idempotent $c$ in $V$. In particular, we have $\Vert e_j \Vert^2 = \theta$ for every element of a Jordan frame $\{ e_1 \ldots, e_r \}$. 

In the case that $\left< x,\, y \right> = \operatorname{tr}(x \circ y)$, we say that $\left< \cdot,\, \cdot \right>$ is a canonical inner product. Various concepts, results, and decompositions remain the same when the given inner product is replaced by the canonical inner product. In particular, for an element of $V$, the spectral decomposition, eigenvalues, and trace remain the same. We note that under the canonical inner product, the norm of any primitive element is one and $\operatorname{tr}(x) = \left< x,\, e \right>$.

A useful tool in the theory of Euclidean Jordan algebras is the Peirce decomposition theorem which is stated as follows.

Theorem 3. (Peirce decomposition theorem)

Let $V$ be a Euclidean Jordan algebra with rank $r$ and $\{e_1, \ldots, e_r\}$ be a Jordan frame of $V$. For $i,j \in \{1,\ldots,r\}$, we define the eigenspaces

$$ \begin{aligned} V_{ii} & := \{ x \in V : x \circ e_i = x \} = \mathbb{R} e_i, \\ V_{ij} & := \{ x \in V : x \circ e_i = \tfrac{1}{2}x = x \circ e_j \}. \end{aligned} $$ Then, the space $V$ is the orthogonal direct sum of subspaces $V_{ij}$ ($i \leq j$). Furthermore,

• $V_{ij} \circ V_{ij} \subseteq V_{ii} + V_{jj}$,

• $V_{ij} \circ V_{jk} \subseteq V_{ik}$ if $i \neq k$,

• $V_{ij} \circ V_{kl} = \{0\}$ if $\{i,j\} \cap \{k,l\} = \emptyset$.

Thus, given any Jordan frame $\{e_1, \ldots, e_r\}$, we can write any element $x \in V$ as

$$ x = \sum_{1 \leq i \leq j \leq r} x_{ij} = \sum_{i=1}^{r} x_i e_i + \sum_{1 \leq i < j \leq r} x_{ij}, $$ where $x_i \in \mathbb{R}$ and $x_{ij} \in V_{ij}$. This expression is called the Pierce decomposition of $x$ associated with $\{e_1, \ldots, e_r\}$.

Example 3.

1. Let $V = \mathbb{R}^n$ and $\{e_1, \ldots, e_r\}$ be the Euclidean basis of $\mathbb{R}^n$, that is, $e_i$ is a vector with $1$ in the $i$-entry and $0$'s elsewhere. It is easily seen that $\{e_1, \ldots, e_r\}$ is a Jordan frame in $\mathbb{R}^n$ and

   $$ V_{ii} = \{ ae_i : a \in \mathbb{R} \}, \ i=1, \ldots, n \qquad \text{and} \qquad V_{ij} = \{0\}, i \neq j. $$ Hence, any element $x \in \mathbb{R}^n$ can be written as $x = \sum_{i=1}^{n} x_i e_i$, which denotes its Peirce decomposition associated with $\{e_1, \ldots, e_r\}$.

2. Let $V = \mathcal{S}^n$ and consider the set $\{E_1, \ldots, E_n\}$, where $E_i$ is a diagonal matrix with $1$ in the $(i,i)$-entry and $0$'s elsewhere. It can be also verified that this set is a Jordan frame in $\mathcal{S}^n$. Associated with this Jordan frame, we have 

   $$ V_{ii} = \{ aE_i : a \in \mathbb{R} \}, \ i=1, \ldots, n \qquad \text{and} \qquad V_{ij} = \{ b E_{ij} : b \in \mathbb{R}\}, i \neq j, $$ where $E_{ij}$ is a matrix with $1$ in the $(i,j)$ and $(j,i)$-entries and $0$'s elsewhere. Thus, any $X \in \mathcal{S}^n$ has the Peirce decomposition associated with $\{E_1, \ldots, E_n\}$ by

   $$ x = \sum_{i=1}^{r} x_i E_i + \sum_{1 \leq i < j \leq r} x_{ij}E_{ij}. $$

3. Let $V = \mathcal{L}^n$ and $\{e_1, e_2\}$ defined by $e_1 = (\frac{1}{2}, \frac{1}{2}, 0_{n-2})$ and $e_1 = (\frac{1}{2}, -\frac{1}{2}, 0_{n-2})$, where $o_{n-2}$ is a vector of zeros in $\mathbb{R}^{n-2}$. Clearly, this set is a Jordan frame of $\mathcal{L}^n$. It is easy to show that

   $$ V_{ii} = \{ ae_i : a \in \mathbb{R} \}, \ i=1,2 \qquad \text{and} \qquad V_{12} = \{ x \in \mathbb{R}^n : x_1 = x_2 = 0\}. $$ Thus, given an $x \in \mathcal{L}^n$, we can write

   $$ x = (x_1 + x_2)e_1 + (x_1 - x_2)e_2 + (0, 0, x_3, \ldots, x_n), $$ which is the Peirce decomposition of $x$ associated with $\{e_1, e_2\}$.