PhyslibAlpha.AlgebraicFramework.JordanOrderUnit.Operator

For an element $a$ in a Jordan algebra $E$, the $n$-th Jordan power $a^{[n]}$ is defined recursively by $a^{[0]} = 1$ and $a^{[n+1]} = a \circ a^{[n]}$ for all $n \in \mathbb{N}$, where $\circ$ denotes the Jordan product.

The notation $a^{[n]}$ denotes the $n$-th power of an element $a$ in a Jordan algebra, which is defined by the function $\text{jpow}(a, n)$. For $n \in \mathbb{N}$, the power is defined recursively by $a^{[0]} = 1$ and $a^{[n+1]} = a \circ a^{[n]}$, where $\circ$ represents the Jordan product.

For any element $a$ in a Jordan algebra $E$, the $0$-th Jordan power $a^{[0]}$ is equal to the identity element $1$.

For an element $a$ in a Jordan algebra $E$ and any natural number $n$, the $(n+1)$-th Jordan power of $a$ is the Jordan product of $a$ and the $n$-th Jordan power of $a$, expressed as $a^{[n+1]} = a \circ a^{[n]}$.

Let $E$ be a Jordan algebra. For any element $a \in E$, the first Jordan power of $a$ is equal to itself, i.e., $a^{[1]} = a$.

For any element $a$ in a Jordan algebra $E$, the second Jordan power $a^{[2]}$ is equal to the Jordan product of $a$ with itself, denoted $a \circ a$.

For an element $a$ in a Jordan algebra $E$, the left multiplication operator $L_a: E \to E$ is the $\mathbb{R}$-linear map defined by $L_a(b) = a \circ b$ for all $b \in E$, where $\circ$ denotes the Jordan product.

The notation $L$ represents the left multiplication operator in a Jordan algebra. For an element $a$, $L_a$ is the linear map defined by $L_a(b) = a \circ b$, where $\circ$ denotes the Jordan product.

For any elements $a$ and $b$ in a Jordan algebra $E$, the application of the left multiplication operator $L_a$ to $b$ is equal to the Jordan product $a \circ b$.

For any element $a$ in a Jordan algebra $E$ with identity element $1$, the left multiplication operator $L_a$ applied to $1$ is equal to $a$, satisfying $L_a(1) = a$.

For any element $a$ in a Jordan algebra $E$ with identity element $1$, the left multiplication operator $L_1$ applied to $a$ is equal to $a$, i.e., $L_1(a) = a$.

For an element $a$ in a Jordan algebra $E$, the quadratic representation $U_a$ is the $\mathbb{R}$-linear map from $E$ to $E$ defined by the formula $U_a = 2 L_a^2 - L_{a^{[2]}}$, where $L_a$ is the left multiplication operator $L_a(b) = a \circ b$ and $a^{[2]} = a \circ a$ is the second Jordan power of $a$. This operator serves as the Jordan-algebraic analogue of the two-sided conjugation map $b \mapsto aba$ in associative algebras.

The notation $U$ denotes the quadratic representation operator in a Jordan algebra. For an element $a$, the operator $U_a$ is a linear map defined by $U_a = 2 L_a^2 - L_{a^2}$, where $L_a$ is the left multiplication operator defined as $L_a(b) = a \circ b$. In the case of an associative algebra, this corresponds to the map $U_a(b) = aba$.

In a Jordan algebra $E$, for any elements $a, b \in E$, the quadratic representation $U_a$ applied to $b$ is given by the identity $$U_a(b) = 2(a \circ (a \circ b)) - a^{[2]} \circ b,$$ where $\circ$ denotes the Jordan product and $a^{[2]} = a \circ a$ is the second Jordan power of $a$.

In a Jordan algebra $E$, for any element $b \in E$, the quadratic representation of the zero element $0 \in E$ applied to $b$ is zero, denoted as $U_0(b) = 0$.

For any elements $a, b, c$ in a Jordan algebra $E$, the quadratic representation $U_a$ satisfies the identity $U_a(b + c) = U_a(b) + U_a(c)$. Here, $U_a$ is the linear operator defined as $U_a = 2 L_a^2 - L_{a^2}$, where $L_a$ is the left multiplication operator.

For any real scalar $r \in \mathbb{R}$ and elements $a, b$ in a Jordan algebra $E$, the quadratic representation $U_a$ satisfies $U_a(r \cdot b) = r \cdot U_a(b)$. Here $U_a$ is the $\mathbb{R}$-linear map defined by $U_a = 2 L_a^2 - L_{a^2}$, where $L_a$ is the left multiplication operator.

For elements $a, b$ in a Jordan algebra $E$, the polarized quadratic representation $U_{a, b} : E \to E$ is the $\mathbb{R}$-linear map defined by $$U_{a, b} = 2(L_a L_b + L_b L_a - L_{a \circ b})$$ where $L_x$ is the left multiplication operator $L_x(y) = x \circ y$ and $\circ$ denotes the Jordan product. This definition provides the bilinear polarization (or "cross-effect") of the quadratic representation $U_a = 2L_a^2 - L_{a^2}$. In the specific case of a special Jordan algebra, this operator acts on an element $x$ as $U_{a, b}(x) = axb + bxa$.

For elements $a$ and $b$ in a Jordan algebra $E$, the bilinear quadratic operator is the $\mathbb{R}$-linear map $E \to E$ defined by $$L_a L_b + L_b L_a - L_{a \circ b}$$ where $L_x$ is the left multiplication operator $L_x(y) = x \circ y$ and $\circ$ denotes the Jordan product. Explicitly, it maps an element $x \in E$ to $a \circ (b \circ x) + b \circ (a \circ x) - (a \circ b) \circ x$.

In a Jordan algebra $E$, for any elements $a, b, x \in E$, the bilinear quadratic operator applied to $x$ is defined by the following identity: $$\text{quadRepBilin}(a, b)(x) = a \circ (b \circ x) + b \circ (a \circ x) - (a \circ b) \circ x$$ where $\circ$ denotes the Jordan product.

In a Jordan algebra $E$, for any elements $a, b \in E$, the bilinear quadratic operator is symmetric in its two arguments: $$\text{quadRepBilin}(a, b) = \text{quadRepBilin}(b, a)$$ where $\text{quadRepBilin}(a, b)$ is the linear map defined by $L_a L_b + L_b L_a - L_{a \circ b}$, $L_x$ is the left multiplication operator, and $\circ$ denotes the Jordan product.

Let $E$ be a Jordan algebra. For any elements $a, b, c \in E$, the bilinear quadratic operator $\text{quadRepBilin}$ is additive in its first argument: $$\text{quadRepBilin}(a + b, c) = \text{quadRepBilin}(a, c) + \text{quadRepBilin}(b, c)$$ where $\text{quadRepBilin}(x, y)$ is the $\mathbb{R}$-linear map $E \to E$ given by $L_x L_y + L_y L_x - L_{x \circ y}$, with $L_x$ being the left multiplication operator and $\circ$ being the Jordan product. This identity expresses the additivity in the first variable of the standard quadratic polarization.

Let $E$ be a Jordan algebra. For any real scalar $r \in \mathbb{R}$ and any elements $a, b \in E$, the bilinear quadratic operator $\text{quadRepBilin}$ is homogeneous with respect to its first argument: $$\text{quadRepBilin}(r \cdot a, b) = r \cdot \text{quadRepBilin}(a, b)$$ where $\text{quadRepBilin}(a, b)$ is the $\mathbb{R}$-linear map $E \to E$ defined by $L_a L_b + L_b L_a - L_{a \circ b}$, with $L_x$ being the left multiplication operator $L_x(y) = x \circ y$ and $\circ$ being the Jordan product.

Let $E$ be a Jordan algebra. For any elements $a, b, c \in E$, the bilinear quadratic operator $\text{quadRepBilin}$ is additive in its second argument: $$\text{quadRepBilin}(a, b + c) = \text{quadRepBilin}(a, b) + \text{quadRepBilin}(a, c)$$ where $\text{quadRepBilin}(a, b)$ is the $\mathbb{R}$-linear map $E \to E$ given by $L_a L_b + L_b L_a - L_{a \circ b}$, with $L_x$ being the left multiplication operator and $\circ$ being the Jordan product.

In a Jordan algebra $E$, for any scalar $r \in \mathbb{R}$ and any elements $a, b \in E$, the bilinear quadratic operator $\text{quadRepBilin}(a, b)$ is homogeneous with respect to its second argument: $$\text{quadRepBilin}(a, r \cdot b) = r \cdot \text{quadRepBilin}(a, b)$$ where $\text{quadRepBilin}(a, b)$ is the $\mathbb{R}$-linear map $E \to E$ defined by $L_a L_b + L_b L_a - L_{a \circ b}$, where $L_x$ is the left multiplication operator $L_x(y) = x \circ y$ and $\circ$ denotes the Jordan product.

In a Jordan algebra $E$, let $Q(a, b)$ denote the bilinear quadratic operator defined by $Q(a, b) = L_a L_b + L_b L_a - L_{a \circ b}$, where $L_x$ is the left multiplication operator $L_x(y) = x \circ y$ and $\circ$ is the Jordan product. For any elements $a, b \in E$, it holds that $Q(a, -b) = -Q(a, b)$.

In a Jordan algebra $E$, for any elements $a, b \in E$, the bilinear quadratic operator $\text{quadRepBilin}(a, b)$, which is the $\mathbb{R}$-linear map defined by $L_a L_b + L_b L_a - L_{a \circ b}$ (where $L_x$ is the left multiplication operator $L_x(y) = x \circ y$), satisfies the identity: $$\text{quadRepBilin}(-a, b) = -\text{quadRepBilin}(a, b)$$ This expresses the fact that the operator is odd with respect to its first variable.

Let $E$ be a Jordan algebra. For any elements $a, b, c \in E$, the bilinear quadratic operator satisfies the identity $$\text{quadRepBilin}(a, b - c) = \text{quadRepBilin}(a, b) - \text{quadRepBilin}(a, c)$$ where $\text{quadRepBilin}(a, b)$ is the linear operator defined by $L_a L_b + L_b L_a - L_{a \circ b}$, and $L_x$ denotes the left multiplication operator $L_x(y) = x \circ y$ in the Jordan algebra.

For any elements $a$ and $b$ in a Jordan algebra $E$, the polarized quadratic representation $U_{a, b}$ is equal to twice the bilinear quadratic operator, which is given by the expression $L_a L_b + L_b L_a - L_{a \circ b}$, where $L_x$ denotes the left multiplication operator $L_x(y) = x \circ y$.

For any element $a$ in a Jordan algebra $E$, the bilinear quadratic operator evaluated at $a$ and $a$ is equal to the quadratic representation $U_a$. That is, $$\text{quadRepBilin}(a, a) = U_a$$ where $U_a$ is the linear operator $2L_a^2 - L_{a^2}$ and $\text{quadRepBilin}(a, b)$ is defined as $L_a L_b + L_b L_a - L_{a \circ b}$, with $L_x$ being the left multiplication operator $L_x(y) = x \circ y$.

For any elements $a, b, x$ in a Jordan algebra $E$, the polarized quadratic representation $U_{a, b}$ applied to $x$ is given by $$U_{a, b}(x) = 2(a \circ (b \circ x) + b \circ (a \circ x) - (a \circ b) \circ x)$$ where $\circ$ denotes the Jordan product.

Let $E$ be a Jordan algebra. For any elements $a, b \in E$, the polarized quadratic representation $U_{a, b}$ is symmetric in its two arguments, that is, $$U_{a, b} = U_{b, a}$$ where $U_{a, b}$ is the $\mathbb{R}$-linear map $E \to E$ defined by $U_{a, b} = 2(L_a L_b + L_b L_a - L_{a \circ b})$.

Let $E$ be a Jordan algebra. For any elements $a, b, c \in E$, the polarized quadratic representation $U_{a, c}$ is additive in its first variable: $$U_{a + b, c} = U_{a, c} + U_{b, c}$$ where $U_{x, y} : E \to E$ is the $\mathbb{R}$-linear map defined by $U_{x, y} = 2(L_x L_y + L_y L_x - L_{x \circ y})$, and $L_x$ denotes the left multiplication operator $L_x(y) = x \circ y$.

Let $E$ be a Jordan algebra. For any scalar $r \in \mathbb{R}$ and any elements $a, b \in E$, the polarized quadratic representation $U_{a, b}$ satisfies the homogeneity property in its first argument: $$U_{r \cdot a, b} = r \cdot U_{a, b}$$ where $U_{a, b}$ is the $\mathbb{R}$-linear map $E \to E$ defined by $U_{a, b} = 2(L_a L_b + L_b L_a - L_{a \circ b})$.

Let $E$ be a Jordan algebra. For any elements $a, b, c \in E$, the polarized quadratic representation $U_{a, b}$ is additive in its second argument: $$U_{a, b + c} = U_{a, b} + U_{a, c}$$ where $U_{x, y}$ is the $\mathbb{R}$-linear map $E \to E$ defined by $U_{x, y} = 2(L_x L_y + L_y L_x - L_{x \circ y})$, and $L_x$ denotes the left multiplication operator $L_x(y) = x \circ y$.

For any real number $r$ and elements $a, b$ in a Jordan algebra $E$, the polarized quadratic representation $U_{a, b}$ satisfies $U_{a, r \cdot b} = r \cdot U_{a, b}$.

For any element $a$ in a Jordan algebra $E$, the polarized quadratic representation $U_{a, a}$ is equal to twice the quadratic representation $U_a$. That is, $$U_{a, a} = 2 U_a$$ where $U_x$ denotes the quadratic representation of an element $x$.

For any elements $a, b$, and $x$ in a Jordan algebra $E$, the quadratic representation of the sum $a+b$ applied to $x$ satisfies the identity: $$U_{a+b}(x) = U_a(x) + U_{a, b}(x) + U_b(x)$$ where $U_a$ is the quadratic representation and $U_{a, b}$ is the polarized quadratic representation. This identity shows how the quadratic representation splits into its diagonal components $U_a, U_b$ and its polarized cross-effect $U_{a, b}$.

For any elements $a$ and $b$ in a Jordan algebra $E$, the quadratic representation of their sum satisfies the identity $$U_{a+b} = U_a + 2 U_{a,b} + U_b$$ where $U_x$ is the quadratic representation of $x$, and $U_{a,b}$ is the bilinear quadratic operator defined by $U_{a,b} = L_a L_b + L_b L_a - L_{a \circ b}$, where $L_x$ denotes the left multiplication operator.

For any elements $a$ and $b$ in a Jordan algebra $E$, the quadratic representation of their difference satisfies the identity $$U_{a-b} = U_a - 2 U_{a,b} + U_b$$ where $U_x$ is the quadratic representation of $x$ (defined as $2 L_x^2 - L_{x^2}$) and $U_{a,b}$ is the bilinear quadratic operator defined by $U_{a,b} = L_a L_b + L_b L_a - L_{a \circ b}$, with $L_x$ denoting the left multiplication operator. This identity provides the signed additive polarization of the quadratic representation.

For any elements $a, b, c$ in a Jordan algebra $E$, the quadratic representation of their sum is given by the expansion: $$U(a + b + c) = U(a) + U(b) + U(c) + 2 U(a, b) + 2 U(a, c) + 2 U(b, c)$$ where $U(x)$ is the quadratic representation $2 L_x^2 - L_{x^2}$ and $U(x, y)$ is the bilinear quadratic operator (the canonical symmetric bilinear cross-effect) defined as $L_x L_y + L_y L_x - L_{x \circ y}$, where $L_x$ denotes the left multiplication operator.

In a Jordan algebra $E$, for any elements $a, b \in E$, the expansion of the square of their sum is given by $$(a + b) \cdot (a + b) = a \cdot a + 2(a \cdot b) + b \cdot b$$ where $\cdot$ denotes the Jordan product and $2$ is a scalar in $\mathbb{R}$. This identity represents the exact square expansion used in positive polarization evaluations.

For any elements $a$ and $b$ in a Jordan algebra $E$, the square of their difference satisfies the identity $(a - b) * (a - b) = a * a - 2 \cdot (a * b) + b * b$, where $*$ denotes the Jordan product and $\cdot$ represents scalar multiplication by the real number $2$.

In a Jordan algebra $E$, for any real scalar $r \in \mathbb{R}$ and any elements $a, b \in E$, the quadratic representation $U$ is homogeneous of degree two in its first argument, satisfying the identity: $$U_{r \cdot a}(b) = r^2 \cdot U_a(b)$$ where $U_a$ is the quadratic representation operator defined by $U_a = 2 L_a^2 - L_{a^2}$, and $L_a$ is the left multiplication operator $L_a(x) = a \circ x$.

For any real number $r \in \mathbb{R}$ and any element $a$ of a Jordan algebra $E$, the quadratic representation $U$ satisfies the homogeneity property $U_{r \cdot a} = r^2 U_a$. Here, $U_a$ is the linear operator defined by $U_a = 2 L_a^2 - L_{a^2}$, where $L_a$ denotes the left multiplication operator $L_a(b) = a \circ b$.

For any elements $a$ and $b$ in a Jordan algebra $E$, the quadratic representation $U$ satisfies the identity $U_{-a}(b) = U_a(b)$. Here, $U_a$ is the linear operator defined by $U_a = 2 L_a^2 - L_{a^2}$, where $L_a$ is the left multiplication operator.

For any element $a$ in a Jordan algebra $E$, the quadratic representation of the negative of $a$ is equal to the quadratic representation of $a$. That is, $U_{-a} = U_a$ as linear operators on $E$.

In a Jordan algebra $E$ with identity element $1$, the quadratic representation of the identity $U_1$ acts as the identity operator on $E$. That is, for any element $b \in E$, $U_1(b) = b$.

In a Jordan algebra $E$ with unit $1$, let $U_a$ denote the quadratic representation of an element $a \in E$. For any $a \in E$, the composition of the quadratic representation of the unit element $U_1$ with $U_a$ is equal to $U_a$, which is written as $U_1 \circ U_a = U_a$.

Let $E$ be a Jordan algebra with identity element $1$. For any element $a \in E$, the composition of the quadratic representation $U_a$ with the quadratic representation of the identity $U_1$ is equal to $U_a$: $$U_a \circ U_1 = U_a$$ where $U_x$ denotes the quadratic representation of an element $x$.

In a Jordan algebra $E$ with unit element $1$, for any element $a \in E$, the quadratic representation $U_a$ applied to the unit element $1$ is equal to the second Jordan power of $a$. That is, $U_a(1) = a^{[2]}$.

In a commutative Jordan algebra $E$, for any elements $a, x \in E$, the quadratic representation $U_a$ commutes with the left multiplication operator $L_a$. Specifically, $U_a(a \circ x) = a \circ (U_a x)$, where $a \circ x$ is the Jordan product and $U_a$ is defined as $2L_a^2 - L_{a^2}$.

In a commutative Jordan algebra $E$, for any elements $a, x \in E$, the quadratic representation $U_a$ commutes with left multiplication by the Jordan square $a^2$. Specifically, it holds that $$U_a(a^2 \circ x) = a^2 \circ U_a(x)$$ where $\circ$ denotes the Jordan product, $a^2 = a \circ a$ is the second Jordan power, and $U_a$ is the quadratic representation operator defined as $U_a = 2L_a^2 - L_{a^2}$. This identity indicates that $U_a$ preserves the associative algebra generated by $a$ at the operator level.

In a commutative Jordan algebra $E$, for any element $a \in E$, the quadratic representation $U_a$ and the left multiplication operator $L_a$ commute as linear operators. That is, $U_a \circ L_a = L_a \circ U_a$, where $L_a(x) = a \circ x$ and $U_a = 2L_a^2 - L_{a^2}$.