PhyslibAlpha.AlgebraicFramework.StarAlgebra.Jordan

For an algebra $A$ with a star-involution, the Jordan product (or anticommutator) of two self-adjoint elements $a, b \in A$ is defined as $a \circ b = ab + ba$. This operation maps a pair of self-adjoint elements to another self-adjoint element.

Let $A$ be an algebra with a star-involution. For any two self-adjoint elements $a, b \in \text{selfAdjoint}(A)$, the Jordan product $a \circ b$ is equal to the anticommutator $ab + ba$ when evaluated in the underlying algebra $A$.

For any self-adjoint elements $a$ and $b$ in a star-algebra $A$, the Jordan product (or anticommutator), defined as $a \circ b = ab + ba$, is commutative, such that $a \circ b = b \circ a$.

Let $A$ be an algebra with a star-involution. For any self-adjoint elements $a, b, c \in \text{selfAdjoint}(A)$, the Jordan product $x \circ y = xy + yx$ distributes over addition in its right argument: $$a \circ (b + c) = a \circ b + a \circ c$$

Let $A$ be an algebra with a star-involution. For any self-adjoint elements $a, b, c \in A$, the Jordan product (or anticommutator) defined by $x \circ y = xy + yx$ distributes over addition in its left argument: $$(a + b) \circ c = (a \circ c) + (b \circ c)$$

Let $A$ be an associative algebra equipped with a star-involution, and let $a, b \in A$ be self-adjoint elements. Let $a \circ b = ab + ba$ denote the Jordan product (or anticommutator) of two elements. Then the Jordan identity holds: $$(a \circ b) \circ (a \circ a) = a \circ (b \circ (a \circ a)).$$ This identity expresses a form of weak associativity, specifically that the operators of Jordan-multiplication by $a$ and by $a \circ a$ commute.

This instance defines the multiplication operator $*$ on the set of self-adjoint elements $\text{selfAdjoint}(A)$ of an algebra $A$. The product of two self-adjoint elements $a, b \in \text{selfAdjoint}(A)$ is given by the Jordan product (anticommutator), defined as $a * b = ab + ba$. This multiplication is provided as a scoped instance to avoid conflict with the standard associative product in cases where $A$ is commutative.

For any two self-adjoint elements $a$ and $b$ of an algebra $A$, the product $a * b$ (under the Jordan multiplication instance) is equal to the Jordan product (anticommutator) $a \circ b = ab + ba$.

This instance equips the set of self-adjoint elements $\text{selfAdjoint}(A)$ of a star-ring $A$ with the structure of a commutative magma. The multiplication operation used is the Jordan product (anticommutator), defined as $a \circ b = ab + ba$, which is shown to be commutative ($a \circ b = b \circ a$).

For a star-ring $A$, the set of self-adjoint elements $\text{selfAdjoint}(A) = \{a \in A \mid a^* = a\}$, equipped with the Jordan product (anticommutator) defined as $a \circ b = ab + ba$, satisfies the axioms of a commutative Jordan ring. This means the operation is commutative ($a \circ b = b \circ a$) and satisfies the Jordan identity $(a^2 \circ b) \circ a = a^2 \circ (b \circ a)$ for all $a, b \in \text{selfAdjoint}(A)$.

For a star-ring $A$, let $\text{Observable } A$ denote the type of self-adjoint elements (where $a^* = a$). The Jordan product of two observables $a, b \in \text{Observable } A$ is defined as the anticommutator $a \circ b = ab + ba$. This operation maps a pair of observables to another observable.

Let $A$ be a star-algebra over $\mathbb{R}$. For any self-adjoint elements $a$ and $b$ in $A$, the Jordan product $a * b$ (as defined in the `selfAdjoint` namespace) corresponds to the normalized anticommutator in the ambient algebra $A$: $$(a * b)_A = \frac{1}{2}(ab + ba)$$ where the products on the right-hand side are the standard associative multiplications in $A$.

Let $A$ be a star-algebra over $\mathbb{R}$. For any real scalar $c \in \mathbb{R}$ and any self-adjoint elements $a, b \in A$, the Jordan product (anticommutator) $a \circ b = ab + ba$ satisfies the property that the scalar can be pulled out of the left argument: $(c \cdot a) \circ b = c \cdot (a \circ b)$.

Let $A$ be an algebra over $\mathbb{R}$. For any self-adjoint elements $a, b$ in $A$ and any real scalar $c$, the Jordan product $a \circ b$ satisfies the property that the scalar can be factored out of the second argument: $a \circ (c \cdot b) = c \cdot (a \circ b)$.

Let $A$ be an algebra over $\mathbb{R}$ with a star-involution. For any self-adjoint elements $a, b, x \in A$, the right-nested Jordan product $a \circ (b \circ x)$ is related to the unnormalized anticommutator $\{u, v\} = uv + vu$ by the identity $$a \circ (b \circ x) = \frac{1}{4} \{a, \{b, x\}\}$$ where $u \circ v = \frac{1}{2}(uv + vu)$ denotes the normalized Jordan product.

For any self-adjoint element $a$ of an algebra $A$ with a star-involution, the Jordan product of $a$ with $0$ is $0$, denoted as $a \circ 0 = 0$. Here, the Jordan product is defined as the anticommutator $a \circ b = ab + ba$.

The set of self-adjoint elements $\text{selfAdjoint}(A)$ of a star-algebra $A$ carries the structure of a non-unital, non-associative, commutative ring. Under this structure, addition is the standard addition in $A$, and multiplication is defined by the Jordan product (anticommutator) $a \circ b = ab + ba$. This product is commutative and distributes over addition, though it is not generally associative.

The set of self-adjoint elements $\text{selfAdjoint}(A)$ of a star-algebra $A$ is equipped with the structure of a non-associative ring. In this structure, addition is the standard addition inherited from $A$, and multiplication is defined by the Jordan product (anticommutator) $a \circ b = ab + ba$. This ring structure includes an additive identity $0$, additive inverses, and a multiplicative identity $1$ such that $1 \circ a = a$ for all $a \in \text{selfAdjoint}(A)$, though the multiplication is not generally associative.

The set of self-adjoint elements $\text{selfAdjoint}(A)$ of a star-algebra $A$ is equipped with the structure of a unital, commutative, non-associative ring. Under this structure, the addition and the multiplicative identity $1$ are inherited from $A$, and the multiplication is defined by the Jordan product (anticommutator) $a \circ b = ab + ba$. This product is commutative and distributive over addition, but it is not generally associative.

For any real number $r \in \mathbb{R}$ and any two self-adjoint elements $a, b \in \text{selfAdjoint}(A)$, the scalar multiplication by $r$ commutes with the Jordan product $a \circ b = ab + ba$. Specifically, the identity $r \cdot (a \circ b) = a \circ (r \cdot b)$ holds.

Let $A$ be an algebra over $\mathbb{R}$. For any real scalar $r \in \mathbb{R}$ and any self-adjoint elements $a, b \in \text{selfAdjoint}(A)$, the real scalar multiplication satisfies the scalar tower property with respect to the Jordan product (the anticommutator $a \circ b = ab + ba$): $$(r \cdot a) \circ b = r \cdot (a \circ b).$$