PhyslibAlpha.ProbabilisticTheory.JordanOrderUnit.Examples.SpinFactor

Given a module $V$ over a commutative ring $R$ and a bilinear form $B: V \times V \to R$, the spin factor is the type $V \times R$. This structure is designed to be equipped with the Jordan product defined by $(x, a) \circ (y, b) = (ax + by, B(x, y) + ab)$, where $x, y \in V$ and $a, b \in R$. The definition is implemented as a type synonym to ensure that Jordan algebra structures associated with different bilinear forms remain distinct.

The spin factor $V \times R$, associated with a module $V$ over a commutative ring $R$ and a bilinear form $B: V \times V \to R$, is an additive abelian group.

Given a commutative ring $R$, a module $V$ over $R$, and a bilinear form $B: V \times V \to R$, the spin factor $V \times R$ (associated with $B$) is a module over $R$. This structure inherits the standard component-wise scalar multiplication from the product $V \times R$.

Let $V$ be a module over a commutative ring $R$, and let $B$ be a bilinear form on $V$. For any two elements $z, w$ in the spin factor $V \times R$ associated with $B$, if their first components (in $V$) are equal and their second components (in $R$) are equal, then $z = w$.

Let $V$ be a module over a commutative ring $R$, and let $B$ be a bilinear form on $V$. For any two elements $z, w$ in the spin factor $V \times R$ associated with $B$, the first component of their sum is the sum of their first components, i.e., $(z + w).1 = z.1 + w.1$.

In the spin factor $V \times R$ defined by a bilinear form $B$ on a module $V$ over a commutative ring $R$, for any two elements $z$ and $w$, the second component of their sum is equal to the sum of their individual second components, i.e., $(z + w)_2 = z_2 + w_2$.

In the spin factor $V \times R$ defined by a bilinear form $B$ on a module $V$ over a commutative ring $R$, the first component of the additive identity element $0$ is the zero vector $0 \in V$.

In the spin factor $V \times R$ defined by a bilinear form $B$ on a module $V$ over a commutative ring $R$, the second component of the additive identity element $0$ is the zero element $0 \in R$.

Let $R$ be a commutative ring, $V$ be an $R$-module, and $B$ be a bilinear form on $V$. For any element $z$ in the spin factor $V \times R$ associated with $B$, the first component of the additive inverse $-z$ is equal to the additive inverse of the first component of $z$. That is, $(-z)_1 = -z_1$, where $z_1 \in V$ denotes the first component of the pair $z$.

Let $R$ be a commutative ring, $V$ be an $R$-module, and $B$ be a bilinear form on $V$. For any element $z$ in the spin factor $V \times R$ associated with $B$, the second component of the additive inverse $-z$ is equal to the additive inverse of the second component of $z$. That is, $(-z)_2 = -z_2$, where the subscript $2$ denotes the second component of the pair.

Let $R$ be a commutative ring, $V$ be an $R$-module, and $B$ be a bilinear form on $V$. For any scalar $r \in R$ and any element $z$ in the spin factor $V \times R$ associated with $B$, the first component of the scalar multiplication $r \cdot z$ is equal to the scalar multiplication of $r$ with the first component of $z$. That is, $(r \cdot z)_1 = r \cdot z_1$, where $z_1 \in V$ denotes the first component of the pair $z$.

Let $V$ be a module over a commutative ring $R$ and $B$ be a bilinear form on $V$. For any scalar $r \in R$ and any element $z$ in the spin factor (defined as the product $V \times R$), the second component of the scalar multiplication $r \cdot z$ is equal to $r$ multiplied by the second component of $z$. That is, $(r \cdot z)_2 = r \cdot z_2$.

Given a module $V$ over a commutative ring $R$ and a bilinear form $B$ on $V$, the function maps a vector $x \in V$ and a scalar $a \in R$ to the corresponding element $(x, a)$ in the spin factor $V \times R$.

Let $V$ be a module over a commutative ring $R$ and $B$ be a bilinear form on $V$. For any vector $x \in V$ and scalar $a \in R$, the first component of the element $(x, a)$ in the spin factor $V \times R$ is equal to $x$. That is, $(x, a)_1 = x$.

Let $R$ be a commutative ring and $V$ be an $R$-module equipped with a bilinear form $B$. For any vector $x \in V$ and scalar $a \in R$, the second component of the element $(x, a)$ constructed in the spin factor $V \times R$ is equal to $a$.

Given a module $V$ over a commutative ring $R$ and a bilinear form $B$ on $V$, the multiplication operation on the spin factor $V \times R$ is defined such that for any two elements $z = (x, a)$ and $w = (y, b)$ (where $x, y \in V$ and $a, b \in R$), their product $z \circ w$ is given by $(ax + by, B(x, y) + ab)$.

Let $R$ be a commutative ring and $V$ be an $R$-module equipped with a bilinear form $B: V \times V \to R$. For any two elements $z$ and $w$ in the spin factor $V \times R$, let $z_1, w_1 \in V$ denote their vector components and $z_2, w_2 \in R$ denote their scalar components. Then the first component (the vector part) of their Jordan product $z \circ w$ is given by $z_2 \cdot w_1 + w_2 \cdot z_1$.

Let $V$ be a module over a commutative ring $R$ and $B$ be a bilinear form on $V$. For any two elements $z, w$ in the spin factor $V \times R$, the second component of their Jordan product $z * w$ is given by $B(z_1, w_1) + z_2 w_2$, where $z_1, w_1 \in V$ and $z_2, w_2 \in R$ are the first and second components of $z$ and $w$, respectively.

Given a module $V$ over a commutative ring $R$ and a bilinear form $B: V \times V \to R$, the identity element $1$ in the spin factor $V \times R$ is defined as the element $(0, 1)$, where $0$ is the zero vector in $V$ and $1$ is the multiplicative identity in $R$. This element acts as the unit with respect to the Jordan product $(x, a) \circ (y, b) = (ax + by, B(x, y) + ab)$.

Let $V$ be a module over a commutative ring $R$, and let $B$ be a bilinear form on $V$. In the spin factor $V \times R$, the first component of the identity element $1$ is the zero vector $0 \in V$.

Let $V$ be a module over a commutative ring $R$, and let $B$ be a bilinear form on $V$. In the spin factor $V \times R$, the second component of the identity element $1$ is $1 \in R$.

Given a module $V$ over a commutative ring $R$ and a bilinear form $B: V \times V \to R$, the spin factor $V \times R$ is a non-associative ring. The multiplication is defined by the Jordan product $(x, a) \circ (y, b) = (ax + by, B(x, y) + ab)$, and the identity element is $(0, 1)$. This structure satisfies the left and right distributive laws of multiplication over addition, as well as the identity laws $1 \circ z = z$ and $z \circ 1 = z$.

Let $R$ be a commutative ring and $V$ be a module over $R$. Let $B: V \times V \to R$ be a symmetric bilinear form. For any elements $z, w$ in the spin factor associated with $B$, the Jordan product is commutative, i.e., $z \circ w = w \circ z$.

Let $V$ be a module over a commutative ring $R$, and let $B$ be a bilinear form on $V$. The spin factor associated with $B$ is the set $V \times R$ equipped with the Jordan product $\circ$ defined by $(x, a) \circ (y, b) = (ay + bx, B(x, y) + ab)$. For any scalar $r \in R$ and elements $z, w \in V \times R$, the scalar multiplication and Jordan product satisfy the compatibility condition $(r \cdot z) \circ w = r \cdot (z \circ w)$.

Let $R$ be a commutative ring, $V$ be an $R$-module, and $B: V \times V \to R$ be a bilinear form. In the spin factor $S = V \times R$ equipped with the Jordan product $\circ$, the scalar multiplication by $r \in R$ and the Jordan product commute. That is, for any $r \in R$ and any elements $z, w \in S$, it holds that $z \circ (r \cdot w) = r \cdot (z \circ w)$.

Let $R$ be a commutative ring and $V$ be a module over $R$. Given a symmetric bilinear form $B: V \times V \to R$, the spin factor $V \times R$ forms a non-associative commutative ring. The multiplication (the Jordan product) is defined by $(x, a) \circ (y, b) = (ax + by, B(x, y) + ab)$ for $x, y \in V$ and $a, b \in R$, and the multiplicative identity is $(0, 1)$.

Let $R$ be a commutative ring and $V$ be a module over $R$. If $B: V \times V \to R$ is a symmetric bilinear form, then the spin factor $V \times R$, equipped with the Jordan product $(x, a) \circ (y, b) = (ax + by, B(x, y) + ab)$, is a commutative Jordan algebra. This implies that the product is commutative and satisfies the Jordan identity $(z \circ z) \circ (w \circ z) = ((z \circ z) \circ w) \circ z$ for all elements $z, w$ in the spin factor.

For a spin factor associated with a module $V$ over a commutative ring $R$ and a bilinear form $B: V \times V \to R$, the determinant is a quadratic map from the spin factor $V \times R$ to $R$. For any element $z = (x, a)$, where $x \in V$ is the vector component and $a \in R$ is the scalar component, the determinant is defined as $a^2 - B(x, x)$.

Let $R$ be a commutative ring, $V$ a module over $R$, and $B : V \times V \to R$ a bilinear form. For any element $z = (x, a)$ in the spin factor $V \times R$ associated with $B$, where $x \in V$ is the vector component and $a \in R$ is the scalar component, the determinant of $z$ is given by $\text{det}(z) = a^2 - B(x, x)$.

Let $R$ be a commutative ring, $V$ a module over $R$, and $B: V \times V \to R$ a symmetric bilinear form. Let $S = V \times R$ be the spin factor associated with $B$, equipped with the identity element $\mathbf{1} = (0, 1)$ and the multiplication $z \circ w$. For any element $z = (x, a) \in S$ (where $x \in V$ and $a \in R$), the following quadratic Cayley–Hamilton identity holds: $$z^2 - 2a z + \det(z) \mathbf{1} = 0$$ where $z^2 = z \circ z$ and $\det(z) = a^2 - B(x, x)$.