PhyslibAlpha.ProbabilisticTheory.JordanOrderUnit.Quadratic.Fundamental

In a Jordan algebra $E$, for any elements $a, b, c, x \in E$, the following identity holds: $$(a \cdot ((b \cdot c) \cdot x) - (b \cdot c) \cdot (a \cdot x)) + (b \cdot ((c \cdot a) \cdot x) - (c \cdot a) \cdot (b \cdot x)) + (c \cdot ((a \cdot b) \cdot x) - (a \cdot b) \cdot (c \cdot x)) = 0$$ This formula represents the polarized Jordan identity evaluated at a fourth element $x$. In terms of the left multiplication operator $L_a(x) = a \cdot x$ and the commutator $[L_a, L_b] = L_a L_b - L_b L_a$, this can be written as the cyclic sum $\sum_{cyc} [L_a, L_{b \cdot c}]x = 0$.

Let $E$ be a Jordan algebra. For any elements $a, b, c, x \in E$, the following identity involving cyclic sums holds: $$a \cdot ((b \cdot c) \cdot x) + b \cdot ((c \cdot a) \cdot x) + c \cdot ((a \cdot b) \cdot x) = (b \cdot c) \cdot (a \cdot x) + (c \cdot a) \cdot (b \cdot x) + (a \cdot b) \cdot (c \cdot x)$$ where $\cdot$ denotes the Jordan product. This identity represents a polarized form of the Jordan identity with collected commutators.

In a Jordan algebra $E$, for any elements $a, b, c \in E$, the left multiplication operator $L_{a \circ (b \circ c)}$ corresponding to the triple product can be expressed in terms of the multiplication operators of single and double products as: $$L_{a \circ (b \circ c)} = L_a L_{b \circ c} + L_b L_{a \circ c} + L_c L_{a \circ b} - L_b L_a L_c - L_c L_a L_b$$ where $\circ$ denotes the Jordan product, $L_x$ is the left multiplication operator defined by $L_x(y) = x \circ y$, and the juxtaposition of operators (e.g., $L_b L_a L_c$) denotes operator composition.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any element $a \in E$, let $a^2 = a \cdot a$ and $a^3 = a \cdot (a \cdot a)$ denote the second and third Jordan powers of $a$. The left multiplication operator $L_{a^3}$ (defined by $L_{a^3}(x) = a^3 \cdot x$) can be expressed as a polynomial in the operators $L_a$ and $L_{a^2}$ according to the identity: $$L_{a^3} = 3 L_a \circ L_{a^2} - 2 L_a^3$$ where $\circ$ denotes the composition of linear maps and $L_a^3 = L_a \circ L_a \circ L_a$.

In a Jordan algebra $E$, for any elements $a, x \in E$, the left multiplication by $(a^2)^2$ can be normalized and expanded in terms of the left multiplication operators $L_a$ and $L_{a^2}$. Specifically, the following identity holds: $$((a \cdot a) \cdot (a \cdot a)) \cdot x = (a \cdot a) \cdot ((a \cdot a) \cdot x) + 4 a \cdot (a \cdot ((a \cdot a) \cdot x)) - 4 a \cdot (a \cdot (a \cdot (a \cdot x)))$$ where $\cdot$ denotes the Jordan product and $4$ denotes scalar multiplication by the real number 4.

In a Jordan algebra $E$ over $\mathbb{R}$, for any element $a \in E$, the quadratic representation of the square $a^2$ is equal to the square of the quadratic representation of $a$. Specifically, if $U_a$ denotes the quadratic representation map, then $$U_{a^2} = U_a \circ U_a$$ where $\circ$ denotes the composition of linear maps.

In a Jordan algebra $E$, for any elements $a, b \in E$, the quadratic representation of the square of their sum, $U_{(a+b)^2}$, is equal to the composition of the operator $U_a + 2U_{a,b} + U_b$ with itself. This identity evaluates the general square identity $U_{x^2} = U_x^2$ at the point $x = a + b$, expanding $U_{a+b}$ into its constituent quadratic and bilinear parts. Mathematically, it is expressed as: $$U_{(a+b)^2} = (U_a + 2U_{a,b} + U_b) \circ (U_a + 2U_{a,b} + U_b)$$ where $U_x$ is the quadratic representation of $x$, $U_{a,b}$ is the bilinear quadratic operator (the polarization of $U$), and $\circ$ denotes the composition of linear endomorphisms.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b \in E$, let $U_a$ denote the quadratic representation of $a$ and $U_{a, b}$ denote the bilinear quadratic operator. The following identity holds: $$U_a^2 + 4 U_{a \circ b} + U_b^2 + 4 U_{a^2, a \circ b} + 2 U_{a^2, b^2} + 4 U_{a \circ b, b^2} = U_a^2 + 4 U_{a, b}^2 + U_b^2 + U_a U_b + U_b U_a + 2 (U_a U_{a, b} + U_{a, b} U_a + U_b U_{a, b} + U_{a, b} U_b)$$ where $U_x^2$ denotes the composition $U_x \circ U_x$, juxtaposition of operators (e.g., $U_a U_b$) denotes their composition, and $a \circ b$ denotes the Jordan product. This equation represents the fully expanded positive polarization (substituting $a+b$ for $x$) of the identity $U_{x^2} = U_x^2$.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b \in E$, let $U_a$ denote the quadratic representation of $a$ and $U_{a, b}$ denote the bilinear quadratic operator. The following identity holds: $$U_a^2 + 4 U_{a \circ b} + U_b^2 - 4 U_{a^2, a \circ b} + 2 U_{a^2, b^2} - 4 U_{a \circ b, b^2} = U_a^2 + 4 U_{a, b}^2 + U_b^2 + U_a U_b + U_b U_a - 2 (U_a U_{a, b} + U_{a, b} U_a + U_b U_{a, b} + U_{a, b} U_b)$$ where $U_x^2$ denotes the composition $U_x \circ U_x$, juxtaposition of operators (e.g., $U_a U_b$) denotes their composition, and $a \circ b$ denotes the Jordan product. This equation represents the fully expanded negative polarization (substituting $a-b$ for $x$) of the fundamental Jordan identity $U_{x^2} = U_x^2$. Adding this identity to its positive counterpart (`quadRep_add_square_polarized`) cancels the cubic terms.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b \in E$, the bilinear quadratic operator $U_{a, b}$ and the quadratic representation $U$ satisfy the mixed polarization identity: $$4 (U_{a, b} \circ U_{a, b}) = 4 U_{a \circ b} + 2 U_{a^2, b^2} - U_a \circ U_b - U_b \circ U_a$$ where $x \circ y$ (or simply $xy$) denotes the Jordan product, $x^2 = x \circ x$, $U_x = 2L_x^2 - L_{x^2}$ is the quadratic representation, and $U_{x, y} = L_x L_y + L_y L_x - L_{x \circ y}$ is the bilinear quadratic operator.

In a Jordan algebra $E$ over $\mathbb{R}$, for any elements $a, b, y \in E$, the quadratic representation of the sum $a+b$ acting on $y$ satisfies the expansion: $$U_{a+b}(y) = U_a(y) + 2 U_{a,b}(y) + U_b(y)$$ where $U_a$ is the quadratic representation of $a$ and $U_{a,b}$ is the bilinear quadratic operator (the polarization of $U$).

In a Jordan algebra $E$ over the real numbers $\mathbb{R}$, for any elements $a, b, y \in E$, the quadratic representation $U_{a-b}$ applied to $y$ expands as: $$U_{a-b}(y) = U_a(y) - 2 U_{a,b}(y) + U_b(y)$$ where $U_x$ is the quadratic representation of $x$ (defined by $U_x = 2L_x^2 - L_{x^2}$) and $U_{a,b}$ is the bilinear quadratic operator (the polarization of $U$).

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b, x \in E$, the following identity holds: $$U_a(b \circ x) + 2 U_{a, b}(a \circ x) = b \circ U_a(x) + 2 (a \circ U_{a, b}(x))$$ where: - $\circ$ denotes the Jordan product. - $U_a$ is the quadratic representation defined by $U_a = 2L_a^2 - L_{a^2}$. - $U_{a, b}$ is the bilinear quadratic operator (the polarization of $U$) defined by $U_{a, b} = L_a L_b + L_b L_a - L_{a \circ b}$. - $L_a$ is the Jordan multiplication operator $L_a(y) = a \circ y$. This identity represents the $(2, 1)$-component of the commutation relation between the multiplication operator $L_{a+b}$ and the quadratic representation $U_{a+b}$.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b, x \in E$, the following identity holds: $$U_b(a \circ x) + 2 U_{a, b}(b \circ x) = a \circ U_b(x) + 2 (b \circ U_{a, b}(x))$$ where $\circ$ denotes the Jordan product, $U_b$ is the quadratic representation defined by $U_b = 2L_b^2 - L_{b^2}$, and $U_{a, b}$ is the bilinear quadratic operator defined by $U_{a, b} = L_a L_b + L_b L_a - L_{a \circ b}$.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b \in E$, the Jordan multiplication operator $L_{U_a b}$ corresponding to the quadratic image $U_a b$ is given by the identity: $$L_{U_a b} = 2 L_a L_{a \circ b} + L_b L_{a^2} - 2 L_a^2 L_b - 2 L_b L_a^2 + 2 L_a L_b L_a$$ where: - $L_x: E \to E$ is the Jordan multiplication operator defined by $L_x(y) = x \circ y$. - $U_a: E \to E$ is the quadratic representation defined by $U_a = 2L_a^2 - L_{a^2}$. - $a \circ b$ denotes the Jordan product, and $a^2 = a \circ a$. - Juice juxtaposition of operators denotes functional composition (e.g., $L_a L_b = L_a \circ L_b$).

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b \in E$, the quadratic representation $U$ of the element $U_a b$ satisfies the following identity: $$U_{U_a b} = 4 U_{a \circ (a \circ b)} - 4 U_{a \circ (a \circ b), a^2 \circ b} + U_{a^2 \circ b}$$ where: - $\circ$ denotes the Jordan product and $a^2 = a \circ a$. - $U_x$ is the quadratic representation operator defined by $U_x y = 2x \circ (x \circ y) - x^2 \circ y$. - $U_{x, y}$ is the bilinear quadratic operator defined by $U_{x, y} z = x \circ (y \circ z) + y \circ (x \circ z) - (x \circ y) \circ z$.

In a Jordan algebra $E$, for any elements $a, b, c \in E$, the inner derivation $D_{a,b}$, defined as the commutator of the Jordan multiplication operators $[L_a, L_b]$, is additive in its first argument: $D_{a+b, c} = D_{a, c} + D_{b, c}$.

Let $E$ be a Jordan algebra over the real numbers $\mathbb{R}$. For any scalar $r \in \mathbb{R}$ and elements $a, b \in E$, the inner derivation $D_{a,b}$ (defined as the commutator of the Jordan multiplication operators $L_a$ and $L_b$) satisfies the scalar homogeneity property in its first argument: $D_{r \cdot a, b} = r \cdot D_{a, b}$.

In a Jordan algebra $E$, the inner derivation $D_{a,b} = [L_a, L_b]$ is additive in its second argument. That is, for any $a, b, c \in E$, the following identity holds: $$D_{a, b+c} = D_{a,b} + D_{a,c}$$ where $D_{x,y}$ is the $\mathbb{R}$-linear map defined as the commutator of the Jordan multiplication operators $L_x$ and $L_y$, such that $D_{x,y}(z) = x \circ (y \circ z) - y \circ (x \circ z)$.

In a Jordan algebra $E$ over the real numbers $\mathbb{R}$, for any elements $a, b \in E$ and any scalar $r \in \mathbb{R}$, the inner derivation $D_{a,b}$ satisfies the identity: $$D_{a, r \cdot b} = r \cdot D_{a,b}$$ where $D_{a,b} = [L_a, L_b]$ is the commutator of the Jordan multiplication operators $L_a$ and $L_b$. This identity demonstrates that the inner derivation is homogeneous with respect to scalar multiplication in its second argument.

In a Jordan algebra $E$, the inner derivation $D_{a,b} = [L_a, L_b]$ is distributive over subtraction in its first argument. For any elements $a, b, c \in E$, the following identity holds: $$D_{a-b, c} = D_{a,c} - D_{b,c}$$ where $D_{x,y}$ is the $\mathbb{R}$-linear map defined by $D_{x,y}(z) = x \circ (y \circ z) - y \circ (x \circ z)$, and $\circ$ denotes the Jordan product.

For any elements $a$ and $x$ in a Jordan algebra $E$, the Jordan triple product $\{a, a, x\}$ satisfies the identity $$\{a, a, x\} = a^2 \circ x$$ where $a^2$ is the second Jordan power of $a$ and $\circ$ denotes the Jordan product. This identity shows that the triple product collapses to the product of the square $a^2$ and $x$ when the first two arguments are identical.

In a Jordan algebra $E$ over $\mathbb{R}$, for any elements $a, b, x, y \in E$, the inner derivation $D_{a,b}$ (defined as the commutator of the multiplication operators $[L_a, L_b]$) satisfies the Leibniz rule with respect to the Jordan product $\circ$: $$D_{a,b}(x \circ y) = (D_{a,b}x) \circ y + x \circ (D_{a,b}y)$$

For any elements $a, b, x$ in a Jordan algebra $E$ over $\mathbb{R}$, the inner derivation $D_{a,b}$ (defined as the commutator of multiplication operators $[L_a, L_b]$) and the Jordan multiplication operator $L_x$ satisfy the identity $$[D_{a,b}, L_x] = L_{D_{a,b} x}$$ where $[D_{a,b}, L_x] = D_{a,b} \circ L_x - L_x \circ D_{a,b}$ is the commutator of the linear operators.

In a Jordan algebra $E$ over $\mathbb{R}$, let $D_{a,b}$ denote the inner derivation for elements $a, b \in E$, defined as $D_{a,b} = [L_a, L_b] = L_a L_b - L_b L_a$. For any elements $a, b, c, d \in E$, the commutator of two inner derivations satisfies the identity $$[D_{a,b}, D_{c,d}] = D_{D_{a,b}(c), d} + D_{c, D_{a,b}(d)}$$ where $[D_{a,b}, D_{c,d}] = D_{a,b} \circ D_{c,d} - D_{c,d} \circ D_{a,b}$ is the commutator of the linear maps.

For any elements $a$ and $b$ in a Jordan algebra $E$ over the real numbers $\mathbb{R}$, the Jordan triple operator $V_{a,b}: E \to E$ is the $\mathbb{R}$-linear map defined as $V_{a,b} = L_{a \circ b} + D_{a,b}$, where $L_{a \circ b}$ is the Jordan multiplication operator by $a \circ b$ and $D_{a,b} = [L_a, L_b]$ is the inner derivation. This operator serves as a bundled linear map representing the Jordan triple product, such that for any $x \in E$, $V_{a,b}(x) = \{a, b, x\}$.

The notation $V$ is defined to represent the Jordan triple operator $V_{a,b}$. For any elements $a$ and $b$ in a Jordan algebra $E$, $V_{a,b}$ is the linear map from $E$ to $E$ given by $V_{a,b} = L_{a \circ b} + [L_a, L_b]$, where $L_x$ denotes the left multiplication operator $y \mapsto x \circ y$ and $[L_a, L_b] = L_a L_b - L_b L_a$ is the commutator of the multiplication operators.

In a Jordan algebra $E$, for any elements $a, b, x \in E$, the application of the Jordan triple operator $V_{a,b}$ to $x$ is equal to the Jordan triple product $\{a, b, x\}$. That is, $V_{a,b}(x) = \{a, b, x\}$, where the triple product is defined as $\{a, b, x\} = a \circ (b \circ x) + x \circ (b \circ a) - (a \circ x) \circ b$ and $\circ$ denotes the Jordan product.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b, c \in E$, the inner derivation $D_{a,b}$ (defined as the commutator of the Jordan multiplication operators $[L_a, L_b]$) satisfies the identity: $$D_{a, b-c} = D_{a, b} - D_{a, c}$$ where $D_{x, y}$ is the linear map defined by $D_{x, y}(z) = x \circ (y \circ z) - y \circ (x \circ z)$ and $\circ$ denotes the Jordan product.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $a, b, q, e \in E$, the inner derivation $D$ satisfies the identity: $$D_{a \circ b, q}(e) = D_{a, b \circ q}(e) + D_{b, a \circ q}(e)$$ where $\circ$ denotes the Jordan product and $D_{x, y}$ is the inner derivation defined by $D_{x, y}(z) = x \circ (y \circ z) - y \circ (x \circ z)$.

Let $E$ be a Jordan algebra. For any elements $a, b, c, d \in E$, the Jordan triple operators $V_{x,y}$ and the Jordan triple product $\{x, y, z\}$ satisfy the fundamental identity: $$[V_{a,b}, V_{c,d}] = V_{\{a,b,c\},d} - V_{c,\{b,a,d\}}$$ where $[V_{a,b}, V_{c,d}] = V_{a,b} \circ V_{c,d} - V_{c,d} \circ V_{a,b}$ is the commutator of the operators, and the Jordan triple product is defined as $\{a, b, c\} = a \circ (b \circ c) + c \circ (b \circ a) - (a \circ c) \circ b$.

In a Jordan algebra $E$ over $\mathbb{R}$, for any elements $x, y, z \in E$, the Jordan triple operator $V$ and the quadratic representation $U$ satisfy the following identity: $$V_{x,y}(U_x z) = 2 V_{x,z}(U_x y) - U_x(\{y, x, z\})$$ where $V_{a,b}$ is the triple operator defined by $V_{a,b}(c) = \{a, b, c\}$, $U_x$ is the quadratic representation defined by $U_x = 2L_x^2 - L_{x^2}$, and $\{y, x, z\}$ is the Jordan triple product. This identity corresponds to the specialization of the triple-operator commutator law to the sequence of elements $(x, y, x, z)$ evaluated at the element $x$.

In a Jordan algebra $E$ over $\mathbb{R}$, for any elements $x, y, z \in E$, the Jordan triple operator $V_{x,y}$ and the quadratic representation $U_x$ satisfy the identity: $$V_{x,y}(U_x z) = U_x(\{y, x, z\})$$ where $\{y, x, z\}$ denotes the Jordan triple product of $y, x$, and $z$.

In a Jordan algebra $E$ over $\mathbb{R}$, for any elements $x, y, z \in E$, the quadratic representation $U$ satisfies the pointwise form of the fundamental formula: $$U_{U_x y} z = U_x(U_y(U_x z))$$ where $U_a$ is the quadratic representation of an element $a \in E$, defined as $U_a = 2L_a^2 - L_{a^2}$.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any elements $x, y \in E$, the quadratic representation $U$ satisfies the fundamental formula: $$U_{U_x y} = U_x \circ U_y \circ U_x$$ where $U_x$ is the linear operator on $E$ defined as $U_x = 2L_x^2 - L_{x^2}$, with $L_x$ being the Jordan multiplication operator $L_x(z) = x \circ z$.

In a Jordan algebra $E$, for any element $a \in E$ and natural numbers $m, n, k$, let $a^n$ denote the $n$-th Jordan power of $a$ and $U_x$ denote the quadratic representation of an element $x$. The following identity, known as the fundamental formula, holds for powers of the same element: $$U_{U_{a^m} a^n} a^k = U_{a^m} (U_{a^n} (U_{a^m} a^k))$$