PhyslibAlpha.AlgebraicFramework.JordanOrderUnit.Quadratic.Fundamental

In a Jordan algebra $E$, for any elements $a, b, c, x \in E$, the following identity holds: $$(a \cdot ((b \cdot c) \cdot x) - (b \cdot c) \cdot (a \cdot x)) + (b \cdot ((c \cdot a) \cdot x) - (c \cdot a) \cdot (b \cdot x)) + (c \cdot ((a \cdot b) \cdot x) - (a \cdot b) \cdot (c \cdot x)) = 0$$ where $\cdot$ denotes the Jordan product. This expression is the fully polarized degree-four Jordan identity, expressing that the cyclic sum of commutators of the form $[L_a, L_{b \cdot c}]$ applied to an element $x$ vanishes.

In a Jordan algebra $E$, for any elements $a, b, c, x \in E$, the following identity holds: $$a \cdot ((b \cdot c) \cdot x) + b \cdot ((c \cdot a) \cdot x) + c \cdot ((a \cdot b) \cdot x) = (b \cdot c) \cdot (a \cdot x) + (c \cdot a) \cdot (b \cdot x) + (a \cdot b) \cdot (c \cdot x)$$ This expression is the fully polarized degree-four Jordan relation. It represents the cyclic commutator identity with its three commutators collected and rearranged, serving as a key identity in the operator normalization argument.

In a Jordan algebra $E$, for any elements $a, b, c \in E$, let $L_x$ denote the left multiplication operator defined by $L_x(y) = x \cdot y$, where $\cdot$ is the Jordan product. The multiplication operator associated with the element $a \cdot (b \cdot c)$ can be normalized to a combination of compositions of multiplication operators of single and double products via the following identity: $$L_{a \cdot (b \cdot c)} = (L_a L_{b \cdot c} + L_b L_{a \cdot c} + L_c L_{a \cdot b}) - (L_b L_a L_c + L_c L_a L_b)$$ where the composition of operators is denoted by juxtaposition.

In a Jordan algebra $E$ over the real numbers $\mathbb{R}$, let $L_x$ denote the left multiplication operator by an element $x \in E$, defined as $L_x(y) = x \cdot y$. For any element $a \in E$, the multiplication operator of the third Jordan power $a^3 = a \cdot (a \cdot a)$ can be expressed as a polynomial in the operators $L_a$ and $L_{a^2}$. Specifically, the following identity holds: $$L_{a^3} = 3 L_a L_{a^2} - 2 L_a^3$$ where $L_a L_{a^2}$ and $L_a^3$ denote the composition of operators. This identity is a specialization of the more general normalization identity for the operator $L_{a \cdot (b \cdot c)}$.

In a Jordan algebra $E$ over the real numbers $\mathbb{R}$, for any elements $a, x \in E$, let $a^2 = a \cdot a$ denote the Jordan product of $a$ with itself. The multiplication by the element $(a^2)^2$ is related to the multiplication operators $L_a$ and $L_{a^2}$ by the identity: $$((a^2)^2) \cdot x = a^2 \cdot (a^2 \cdot x) + 4 a \cdot (a \cdot (a^2 \cdot x)) - 4 a \cdot (a \cdot (a \cdot (a \cdot x)))$$ This formula constitutes a normalization step that reduces the operator $L_{(a^2)^2}$ to a linear combination of the compositions of the commuting operators $L_a$ and $L_{a^2}$.

In a Jordan algebra $E$ over the real numbers $\mathbb{R}$, let $U_a$ denote the quadratic representation of an element $a \in E$. For any element $a \in E$, the quadratic representation of its Jordan square $a^2 = a \cdot a$ is equal to the square (the composition with itself) of the quadratic representation of $a$. That is, $$U_{a^2} = U_a \circ U_a$$ where $\circ$ denotes the composition of linear operators.

In a Jordan algebra $E$ over the real numbers $\mathbb{R}$, for any elements $a, b \in E$, the quadratic representation of the square of their sum $(a+b)^2$ is equal to the composition of the operator $U_a + 2 U_{a,b} + U_b$ with itself: $$U_{(a+b)^2} = (U_a + 2 U_{a,b} + U_b) \circ (U_a + 2 U_{a,b} + U_b)$$ where $U_x$ is the quadratic representation $U_x = 2L_x^2 - L_{x^2}$, and $U_{a,b}$ is the bilinear quadratic operator $U_{a,b} = L_a L_b + L_b L_a - L_{a \circ b}$. Here, $L_x$ denotes the left multiplication operator $L_x(y) = x \circ y$ and $\circ$ represents the composition of linear maps.

Let $E$ be a Jordan algebra over $\mathbb{R}$. For any element $a \in E$, let $U_a$ denote the quadratic representation $U_a = 2 L_a^2 - L_{a^2}$, where $L_a$ is the left multiplication operator and $a^2$ is the Jordan square. For any $a, b \in E$, let $U_{a,b}$ denote the bilinear quadratic operator $L_a L_b + L_b L_a - L_{a \circ b}$. Then the positive polarization of the square identity for the quadratic representation 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 $U \circ V$ (or simply $UV$) denotes the composition of linear maps, $U^2$ denotes $U \circ U$, and $a \circ b$ denotes the Jordan product.

Let $E$ be a Jordan algebra. For any elements $a, b \in E$, let $U_a$ denote the quadratic representation of $a$ (defined as $U_a = 2L_a^2 - L_{a^2}$) and let $U_{x, y}$ denote the bilinear quadratic operator associated with $x$ and $y$ (defined as $U_{x,y} = L_x L_y + L_y L_x - L_{x \cdot y}$). The following operator identity holds: $$U_a^2 + 4 U_{a \cdot b} + U_b^2 - 4 U_{a^2, a \cdot b} + 2 U_{a^2, b^2} - 4 U_{a \cdot b, b^2} = U_a^2 + 4 U_{a,b}^2 + U_b^2 + U_a U_b + U_b U_a - 2(U_a U_{a,b} + U_{a,b} U_a + U_b U_{a,b} + U_{a,b} U_b)$$ where $\cdot$ denotes the Jordan product and operator juxtaposition (or $X^2$) denotes the composition of linear maps. This identity serves as the signed companion to the additive polarization identity; subtracting the two cancels cubic terms and isolates specific mixed quadratic coefficients.

Let $E$ be a Jordan algebra. For any elements $a, b \in E$, let $U_x$ denote the quadratic representation $U_x = 2L_x^2 - L_{x^2}$ and $U_{x,y}$ denote the bilinear quadratic operator $U_{x,y} = L_x L_y + L_y L_x - L_{x \circ y}$, where $L_x$ is the left multiplication operator and $\circ$ (or juxtaposition) is the Jordan product. Then the following identity holds: $$4 (U_{a,b} \circ U_{a,b}) = 4 U_{a \circ b} + 2 U_{a^2, b^2} - (U_a \circ U_b + U_b \circ U_a)$$ where $\circ$ between operators denotes their composition.

In a Jordan algebra $E$, for any elements $a, b, y \in E$, the quadratic representation $U_{a+b}$ applied to $y$ satisfies the expansion: $$U_{a+b}(y) = U_a(y) + 2 U_{a,b}(y) + U_b(y)$$ where $U_x$ is the quadratic representation $U_x = 2L_x^2 - L_{x^{[2]}}$ and $U_{a,b}$ is the bilinear quadratic operator $L_a L_b + L_b L_a - L_{a \circ b}$. Here $L_x$ denotes the left multiplication operator $L_x(y) = x \circ y$.

In a Jordan algebra $E$, for any elements $a, b, y \in E$, the quadratic representation $U_{a-b}$ applied to $y$ satisfies the identity $$U_{a-b}(y) = U_a(y) - 2 U_{a,b}(y) + U_b(y)$$ where $U_x$ is the quadratic representation of $x$ (defined as $U_x = 2 L_x^2 - L_{x^2}$) and $U_{a,b}$ is the bilinear quadratic operator defined by $U_{a,b} = L_a L_b + L_b L_a - L_{a \circ b}$.

In a Jordan algebra $E$, for any elements $a, b, x \in E$, the following $(2,1)$-bidegree polarization of the cubic commutation law 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 $U_a(y) = 2 a \circ (a \circ y) - (a \circ a) \circ y$, and $U_{a,b}$ is the bilinear quadratic operator $U_{a,b}(y) = a \circ (b \circ y) + b \circ (a \circ y) - (a \circ b) \circ y$.

Let $E$ be a Jordan algebra. 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 (written as `*` in the formal statement). - $U_b$ is the quadratic representation of $b$, defined by $U_b = 2L_b^2 - L_{b^2}$, where $L_b$ is the left multiplication operator $L_b(y) = b \circ y$. - $U_{a,b}$ is the bilinear quadratic operator defined by $U_{a,b} = L_a L_b + L_b L_a - L_{a \circ b}$. This identity represents the mixed bidegree $(1, 2)$ component of the polarized cubic commutation law for the quadratic representation.

For any elements $a$ and $b$ in a Jordan algebra $E$, the left multiplication operator $L_{U_a b}$ of the element $U_a b$ 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 $L_x$ denotes the left multiplication operator defined by $L_x(y) = x \circ y$, $U_a b = 2 a \circ (a \circ b) - a^2 \circ b$ is the quadratic representation of $a$ acting on $b$, and operator composition is denoted by juxtaposition. This formula normalizes the multiplication operator of the compound element $U_a b$ into a polynomial of basic multiplication operators $L_a, L_b, L_{a^2}$, and $L_{a \circ b}$.

In a Jordan algebra $E$, for any elements $a, b \in E$, the quadratic representation 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: - $U_x$ denotes the quadratic representation of an element $x$, defined by $U_x(y) = 2x \circ (x \circ y) - x^2 \circ y$. - $U_{x, y}$ denotes the bilinear quadratic operator defined by $U_{x, y}(z) = x \circ (y \circ z) + y \circ (x \circ z) - (x \circ y) \circ z$. - $a^2$ is the second Jordan power $a \circ a$. - $\circ$ denotes the Jordan product.

In a Jordan algebra $E$, let $D_{a,b}$ denote the inner derivation defined by elements $a, b \in E$, which is the linear map given by the commutator of left multiplication operators $D_{a,b} = [L_a, L_b] = L_a \circ L_b - L_b \circ L_a$. For any elements $a, b, c \in E$, the inner derivation is additive in its first argument: $$D_{a+b, c} = D_{a, c} + D_{b, c}$$

Let $E$ be a Jordan algebra. For any scalar $r \in \mathbb{R}$ and any elements $a, b \in E$, the inner derivation $D_{r \cdot a, b}$ satisfies $D_{r \cdot a, b} = r D_{a, b}$. Here, $D_{x,y}$ denotes the inner derivation operator defined by the commutator of left multiplication operators, $D_{x,y} = [L_x, L_y] = L_x \circ L_y - L_y \circ L_x$.

In a Jordan algebra $E$, the inner derivation $D_{a,b}$ is additive in its second argument. Specifically, for any elements $a, b, c \in E$, the identity $$D_{a, b+c} = D_{a, b} + D_{a, c}$$ holds, where $D_{x,y}$ denotes the inner derivation operator defined by the commutator of left multiplication operators $[L_x, L_y]$.

Let $E$ be a Jordan algebra. For any real scalar $r \in \mathbb{R}$ and any elements $a, b \in E$, the inner derivation $D_{a,b}$ satisfies the identity $$D_{a, r \cdot b} = r D_{a, b}$$ where $D_{x,y}$ denotes the inner derivation operator defined by the commutator of left multiplication operators, $D_{x,y} = [L_x, L_y] = L_x \circ L_y - L_y \circ L_x$.

In a Jordan algebra $E$, for any elements $a, b, c \in E$, the inner derivation $D_{a,b}$ satisfies the identity $D_{a-b, c} = D_{a, c} - D_{b, c}$.

In a Jordan algebra $E$, for any elements $a, x \in E$, the Jordan triple product $\{a, a, x\}$, defined as $\{a, b, c\} = a(bc) + c(ba) - (ac)b$, satisfies the identity $\{a, a, x\} = a^{[2]} \circ x$, where $a^{[2]}$ denotes the second Jordan power of $a$ and the juxtaposition (or $\circ$) denotes the Jordan product.

In a Jordan algebra $E$, for any elements $a, b, x, y \in E$, the inner derivation $D_{a,b}$ satisfies the derivation identity for the Jordan product: $$D_{a,b}(x \circ y) = (D_{a,b}x) \circ y + x \circ (D_{a,b}y)$$ where $D_{a,b} = L_a L_b - L_b L_a$ is the commutator of the left multiplication operators and $\circ$ denotes the Jordan product.

In a Jordan algebra $E$, for any elements $a, b, x \in E$, the inner derivation $D_{a,b} = [L_a, L_b]$ and the left multiplication operator $L_x$ satisfy the operator identity: $$[D_{a,b}, L_x] = L_{D_{a,b} x}$$ where $L_y(z) = y \circ z$ denotes the Jordan left multiplication operator and $[A, B] = A \circ B - B \circ A$ denotes the commutator of linear operators.

Let $E$ be a Jordan algebra. For any elements $a, b, c, d \in E$, the commutator of the inner derivations $D_{a,b}$ and $D_{c,d}$ satisfies the identity: $$[D_{a,b}, D_{c,d}] = D_{D_{a,b}c, d} + D_{c, D_{a,b}d}$$ where $D_{x,y}$ is the inner derivation defined as the commutator of the left multiplication operators $[L_x, L_y]$, and the bracket on the left-hand side denotes the commutator of operators $[A, B] = A \circ B - B \circ A$.

For elements $a, b$ in a Jordan algebra $E$, the Jordan triple operator $V_{a,b}$ is the $\mathbb{R}$-linear map from $E$ to itself defined by the sum: $$V_{a,b} = L_{a \cdot b} + D_{a,b}$$ where $L_{a \cdot b}$ is the left multiplication operator by the Jordan product $a \cdot b$, and $D_{a,b} = [L_a, L_b]$ is the inner derivation. The action of this operator on an element $x \in E$ corresponds to the Jordan triple product $\{a, b, x\}$.

The notation $V$ denotes the Jordan triple operator $V_{a, b}$, which is a linear map from a Jordan algebra $E$ to itself, parameterized by two elements $a, b \in E$. In the context of Jordan algebras, it is defined as $V_{a, b} = L_{a \cdot b} + [L_a, L_b]$, where $L_x$ is the left multiplication operator and $[L_a, L_b]$ is the commutator of the multiplication operators (also known as the inner derivation $D_{a,b}$).

In a Jordan algebra $E$, for any elements $a, b, x \in E$, the Jordan triple operator $V_{a,b}$ applied to $x$ is equal to the Jordan triple product $\{a, b, x\}$. That is, $$V_{a,b}(x) = \{a, b, x\}$$ where the Jordan triple product is defined as $\{a, b, x\} = a(bx) + x(ba) - (ax)b$.

Let $E$ be a Jordan algebra. For any elements $a, b, c \in E$, the inner derivation $D_{a, b-c}$ satisfies the identity: $$D_{a, b-c} = D_{a, b} - D_{a, c}$$ where $D_{x,y}$ denotes the inner derivation operator defined by the commutator of left multiplication operators, $D_{x,y} = [L_x, L_y]$.

In a Jordan algebra $E$, let $a, b, q, e$ be elements of $E$. The inner derivation $D_{x,y}$ (defined as the commutator of left multiplication operators $[L_x, L_y]$) satisfies a product rule for its first argument: $$D_{a*b, q}(e) = D_{a, b*q}(e) + D_{b, a*q}(e)$$ where $*$ denotes the Jordan product.

In a Jordan algebra $E$, for any elements $a, b, c, d \in E$, the commutator of the Jordan triple operators $V_{a,b}$ and $V_{c,d}$ satisfies the identity: $$[V_{a,b}, V_{c,d}] = V_{\{a,b,c\},d} - V_{c,\{b,a,d\}}$$ where $V_{x,y}$ is the Jordan triple operator and $\{x,y,z\}$ is the Jordan triple product $x(yz) + z(yx) - (xz)y$. This identity is the operator form of the fundamental identity for Jordan triple systems.

In a Jordan algebra $E$, 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 operator defined by the Jordan triple product $V_{a, b}(c) = \{a, b, c\}$, $U_a$ is the quadratic representation $U_a(b) = \{a, b, a\}$, and $\{y, x, z\}$ denotes the Jordan triple product. This identity serves as an intermediate step toward proving the fundamental formula $U_{U_x y} = U_x U_y U_x$.

In a Jordan algebra $E$, for any elements $x, y, z \in E$, the following identity holds: $$V_{x,y}(U_x z) = U_x(\{y, x, z\})$$ where $V_{x,y}$ is the Jordan triple operator (defined such that $V_{x,y} a = \{x, y, a\}$), $U_x$ is the quadratic representation of $x$, and $\{y, x, z\}$ is the Jordan triple product.

In a Jordan algebra $E$, for any elements $x, y, z \in E$, the quadratic representation $U$ satisfies the fundamental identity $$U_{U_x y} z = U_x(U_y(U_x z)),$$ where $U_x$ denotes the quadratic representation of an element $x$, defined as the linear map $U_x(b) = 2x \circ (x \circ b) - (x \circ x) \circ b$.

In a Jordan algebra $E$, for any elements $x, y \in E$, the quadratic representation $U$ satisfies the fundamental identity: $$U_{U_x y} = U_x \circ U_y \circ U_x$$ where $U_x$ is the quadratic representation of $x$ (defined as $U_x = 2L_x^2 - L_{x^2}$) and $\circ$ denotes the composition of linear operators.

Let $E$ be a Jordan algebra. For any element $a \in E$ and any natural numbers $m, n, k \in \mathbb{N}$, the fundamental identity for quadratic representations holds when restricted to the Jordan powers of $a$: $$U_{U_{a^{[m]}} a^{[n]}} a^{[k]} = U_{a^{[m]}} (U_{a^{[n]}} (U_{a^{[m]}} a^{[k]}))$$ where $a^{[p]}$ denotes the $p$-th Jordan power of $a$, and $U_x$ is the quadratic representation of $x \in E$.