PhyslibAlpha.ProbabilisticTheory.JordanOrderUnit.Quadratic.Projection

In a commutative Jordan algebra $E$, let $p$ be a Jordan projection, meaning it is idempotent ($p^2 = p$). For any element $x \in E$, the operator $L_p$ (defined by $L_p(x) = p \circ x$, where $\circ$ is the Jordan product) satisfies the Peirce polynomial $2 L_p^3 - 3 L_p^2 + L_p = 0$. Specifically, the following identity holds: $$2 (p \circ (p \circ (p \circ x))) - 3 (p \circ (p \circ x)) + p \circ x = 0$$ This identity is a specialization of the polarized Jordan identity and ensures that the quadratic representation $U_p$ acts as an idempotent compression.

Given an element $p$ in a Jordan algebra $E$ over $\mathbb{R}$, the Peirce-1 subspace $\text{peirceOne}(p)$ is the $\mathbb{R}$-submodule of $E$ consisting of all elements $x \in E$ that are fixed under Jordan multiplication by $p$, such that $p \circ x = x$.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $p, x \in E$, $x$ belongs to the Peirce-1 subspace of $p$ (denoted $\text{peirceOne}(p)$) if and only if the Jordan product of $p$ and $x$ satisfies $p \circ x = x$. This characterizes membership in the subspace as satisfying the eigenvalue-one equation for the operator $L_p$.

Given a Jordan projection $p$ in a Jordan algebra $E$ over $\mathbb{R}$, the Peirce-0 subspace (or eigenspace) is the $\mathbb{R}$-submodule of $E$ consisting of all elements $x \in E$ such that the Jordan product $p \circ x = 0$.

Let $E$ be a real Jordan algebra and $p \in E$ be a Jordan projection. An element $x \in E$ belongs to the Peirce-0 subspace of $p$ if and only if $p \circ x = 0$, where $\circ$ denotes the Jordan product.

For a Jordan projection $p$ in a real Jordan algebra $E$, `peirceHalf p` is the $\mathbb{R}$-submodule of $E$ corresponding to the $\frac{1}{2}$-eigenspace of the Jordan multiplication operator $L_p$. It consists of all elements $x \in E$ satisfying the condition $2(p \circ x) = x$, where $\circ$ denotes the Jordan product.

Let $E$ be a real Jordan algebra and $p \in E$ be a Jordan projection. An element $x \in E$ belongs to the Peirce-$1/2$ subspace of $p$ if and only if $2(p \circ x) = x$, where $\circ$ denotes the Jordan product.

For a Jordan projection $p$ in a real Jordan algebra $E$, the Peirce-$0$ subspace (the $0$-eigenspace of the Jordan multiplication operator $L_p$) and the Peirce-$1/2$ subspace (the $\frac{1}{2}$-eigenspace of $L_p$) are disjoint. That is, their intersection is $\{0\}$.

For any Jordan projection $p$ in a Jordan algebra $E$ over $\mathbb{R}$, the Peirce-0 subspace $\{x \in E \mid p \circ x = 0\}$ and the Peirce-1 subspace $\{x \in E \mid p \circ x = x\}$ are disjoint.

Let $E$ be a real Jordan algebra and $p \in E$ be a Jordan projection. The Peirce-$1/2$ subspace of $p$, consisting of all $x \in E$ such that $p \circ x = \frac{1}{2}x$, and the Peirce-$1$ subspace of $p$, consisting of all $x \in E$ such that $p \circ x = x$, are disjoint. That is, their intersection contains only the zero element: $$\text{peirceHalf}(p) \cap \text{peirceOne}(p) = \{0\}$$

In a Jordan algebra $E$, for a Jordan projection $p$ and an element $x$, the Peirce-0 component of $x$ relative to $p$ is the projection of $x$ onto the eigenspace of the left multiplication operator $L_p$ (where $L_p(x) = p \cdot x$) corresponding to the eigenvalue $0$. It is calculated using the formula $x - 3(p \cdot x) + 2(p \cdot (p \cdot x))$.

Given a Jordan projection $p$ and an element $x$ in a Jordan algebra $E$, this function computes the Peirce $1/2$-component of $x$ relative to $p$. This component corresponds to the projection of $x$ onto the eigenspace of the multiplication operator $L_p(y) = p \cdot y$ associated with the eigenvalue $1/2$.

In a commutative Jordan algebra $E$, let $p \in E$ be a Jordan projection (an idempotent element satisfying $p^2 = p$). For any element $x \in E$, let $x_0$ denote the Peirce-0 component of $x$ relative to $p$, which is the projection of $x$ onto the eigenspace of the multiplication operator $L_p$ corresponding to the eigenvalue $0$. Then, the Jordan product of $p$ and $x_0$ is zero: $$p \cdot x_0 = 0$$

In a commutative Jordan algebra $E$, let $p \in E$ be a Jordan projection (an idempotent element satisfying $p \cdot p = p$). For any element $x \in E$, let $x_0$ denote the Peirce $0$-component of $x$ relative to $p$. Then $x_0$ belongs to the Peirce $0$-subspace of $p$, which is the submodule of $E$ consisting of all elements $y$ such that $p \cdot y = 0$.

Let $E$ be a commutative Jordan algebra and $p \in E$ be a Jordan projection (i.e., an idempotent satisfying $p \cdot p = p$). For any element $x \in E$, let $x_{1/2}$ denote the Peirce $1/2$-component of $x$ relative to $p$. This component is an eigenvector of the multiplication operator $L_p(y) = p \cdot y$ with eigenvalue $1/2$, satisfying the identity: $$2(p \cdot x_{1/2}) = x_{1/2}$$ where $\cdot$ denotes the Jordan product.

Let $E$ be a commutative Jordan algebra and $p \in E$ be a Jordan projection (i.e., $p \circ p = p$). For any element $x \in E$, the Peirce $1/2$-component of $x$ relative to $p$ (denoted $x_{1/2}$) belongs to the Peirce $1/2$-subspace of $p$, which is defined as the eigenspace of the Jordan multiplication operator $L_p(y) = p \circ y$ with eigenvalue $1/2$.

Let $E$ be a Jordan algebra and $p \in E$ be a Jordan projection (an idempotent satisfying $p \circ p = p$). For any element $x \in E$, $x$ can be decomposed into its Peirce components relative to $p$: $$x_0 + x_{1/2} + U_p x = x$$ where $x_0$ is the Peirce $0$-component, $x_{1/2}$ is the Peirce $1/2$-component, and $U_p x$ is the quadratic representation of $p$ acting on $x$ (which constitutes the Peirce $1$-component). These summands correspond to the eigenspaces of the Jordan multiplication operator $L_p$ with eigenvalues $0, 1/2$, and $1$ respectively.

Let $E$ be a Jordan algebra and $p \in E$ be a Jordan projection (satisfying $p \circ p = p$). For any element $x \in E$, the Jordan product of $p$ and the quadratic representation $U_p x$ satisfies: $$p \circ U_p(x) = U_p(x)$$ Here, $U_p$ is the quadratic representation defined by $U_p = 2L_p^2 - L_{p^2}$, where $L_p$ is the Jordan multiplication operator $L_p(y) = p \circ y$. This identity states that the image of the quadratic representation of a projection $p$ lies in the Peirce-1 eigenspace of $p$.

Let $E$ be a commutative Jordan algebra over $\mathbb{R}$. If $p \in E$ is a Jordan projection (satisfying $p \circ p = p$), then for any $x \in E$, the quadratic representation $U_p(x)$ belongs to the Peirce-1 subspace of $p$. The Peirce-1 subspace is the $\mathbb{R}$-submodule of $E$ consisting of all elements $y$ such that $p \circ y = y$.

Let $E$ be a Jordan algebra over $\mathbb{R}$. If $p \in E$ is a Jordan projection (satisfying $p \circ p = p$) and $x \in E$ is a Peirce-1 element with respect to $p$—meaning it satisfies the condition $p \circ x = x$—then the quadratic representation $U_p$ fixes $x$, such that $U_p x = x$. Here, the quadratic representation is defined as $U_p = 2L_p^2 - L_{p^2}$, where $L_p$ is the Jordan multiplication operator.

Let $E$ be a commutative Jordan algebra over $\mathbb{R}$ and let $p \in E$ be a Jordan projection (i.e., $p \circ p = p$). Let $U_p$ be the quadratic representation of $p$, defined as $U_p = 2L_p^2 - L_{p^2}$, where $L_p$ is the Jordan multiplication operator. Then the range of $U_p$ is equal to the Peirce-1 subspace of $p$, which is the set of all elements $x \in E$ such that $p \circ x = x$.

Let $E$ be a commutative Jordan algebra over $\mathbb{R}$. If $p \in E$ is a Jordan projection (satisfying $p \circ p = p$), then the quadratic representation $U_p: E \to E$ is idempotent. That is, for any $x \in E$, $U_p(U_p(x)) = U_p(x)$, or equivalently, $U_p \circ U_p = U_p$.

In a commutative Jordan algebra $E$, let $p$ be a Jordan projection (an element satisfying $p \circ p = p$). Let $U_p: E \to E$ be the quadratic representation of $p$. For any element $x \in E$ that lies in the range of $U_p$ (that is, $x = U_p(y)$ for some $y \in E$), it holds that $U_p(x) = x$.

Let $E$ be a commutative Jordan algebra over $\mathbb{R}$. If $p \in E$ is a Jordan projection (satisfying $p \circ p = p$), then for any $x \in E$, the residual element $x - U_p x$ lies in the kernel of the quadratic representation $U_p$. That is, $U_p(x - U_p x) = 0$, where $U_p$ is the quadratic representation defined by $U_p = 2L_p^2 - L_{p^2}$.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $p, x \in E$, it holds that $U_p x + (x - U_p x) = x$, where $U_p$ is the quadratic representation of $p$ defined as $U_p = 2L_p^2 - L_{p^2}$.