Physlib.Mathematics.KroneckerDelta

For any type $\alpha$ and elements $i, j \in \alpha$, the Kronecker delta $\delta_{ij}$ is defined as $1$ if $i = j$ and $0$ if $i \neq j$.

The notation $\delta[i, j]$ represents the Kronecker delta function applied to the indices $i$ and $j$.

Let $M$ be an additive monoid and $\alpha$ be a type. For any elements $i, j \in \alpha$ and any function $f : \alpha \to \alpha \to M$, if $f(i, i) = 0$, then $\delta_{ij} \cdot f(i, j) = 0$, where $\delta_{ij}$ is the Kronecker delta, defined as $1$ if $i = j$ and $0$ if $i \neq j$.

Let $M$ be an additive monoid and $\alpha$ be a set. For any elements $i, j \in \alpha$ and any function $f : \alpha \to \alpha \to M$, the scalar multiplication of the Kronecker delta $\delta_{ij}$ and $f(i, j)$ is zero, i.e., $\delta_{ij} \cdot f(i, j) = 0$, if and only if $i \neq j$ or $f(i, i) = 0$. Here, $\delta_{ij}$ is defined to be $1$ if $i = j$ and $0$ if $i \neq j$.

Let $M$ be an additive monoid and $\alpha$ be a type. For any element $i \in \alpha$ and any function $f : \alpha \to \alpha \to M$, the condition $\delta_{ij} \cdot f(i, j) = 0$ holds for all $j \in \alpha$ if and only if $f(i, i) = 0$. Here, $\delta_{ij}$ denotes the Kronecker delta, defined as $1$ if $i = j$ and $0$ if $i \neq j$.

Let $M$ be an additive monoid and $\alpha$ be a type. For any function $f : \alpha \to \alpha \to M$, the condition $\delta_{ij} \cdot f(i, j) = 0$ holds for all $i, j \in \alpha$ if and only if $f(i, i) = 0$ for all $i \in \alpha$. Here, $\delta_{ij}$ denotes the Kronecker delta, defined as $1$ if $i = j$ and $0$ if $i \neq j$, and $\cdot$ denotes scalar multiplication of a natural number on an element of the additive monoid.

For any elements $i$ and $j$ of a type $\alpha$, the Kronecker delta $\delta_{ij}$ is symmetric in its indices, satisfying the identity $\delta_{ij} = \delta_{ji}$. Here, $\delta_{ij}$ is defined to be $1$ if $i = j$ and $0$ if $i \neq j$.

For any type $\alpha$ and any additive monoid $M$, let $f: \alpha \times \alpha \to M$ be a function. For any elements $i, j \in \alpha$, the following equality holds: $$\delta_{ij} \cdot f(j, i) = \delta_{ij} \cdot f(i, j)$$ where $\delta_{ij}$ is the Kronecker delta, defined as $1$ if $i = j$ and $0$ if $i \neq j$, and $\cdot$ denotes the scalar multiplication of a natural number on an element of the additive monoid.

For any additive monoid $M$, any function $f : \alpha \to \alpha \to M$, and any elements $i, j \in \alpha$, the following identity holds: $$\delta_{ij} \cdot (f(i, j) + f(j, i)) = (2\delta_{ij}) \cdot f(i, j)$$ where $\delta_{ij}$ is the Kronecker delta, which equals $1$ if $i = j$ and $0$ otherwise, and the multiplication denotes the scalar action of natural numbers on the monoid.

Let $M$ be an additive commutative monoid and $K$ be a semifield of characteristic zero such that $M$ is a $K$-module. For any $i, j \in \alpha$ and any function $f : \alpha \to \alpha \to M$, the following identity holds: $$\delta_{ij} \cdot \frac{1}{2}(f(i, j) + f(j, i)) = \delta_{ij} f(i, j)$$ where $\delta_{ij}$ is the Kronecker delta, which is $1$ if $i = j$ and $0$ otherwise.

For any additive group $M$, type $\alpha$, elements $i, j \in \alpha$, and function $f : \alpha \to \alpha \to M$, the product of the Kronecker delta $\delta_{ij}$ and the difference $f(i, j) - f(j, i)$ is zero, i.e., $\delta_{ij} \cdot (f(i, j) - f(j, i)) = 0$.

For any finite type $\alpha$ and elements $i, j \in \alpha$, the sum over all $k \in \alpha$ of the product of Kronecker deltas $\delta_{ik}$ and $\delta_{kj}$ is equal to $\delta_{ij}$, expressed as: $$\sum_{k \in \alpha} \delta_{ik} \delta_{kj} = \delta_{ij}$$ where $\delta_{ab}$ is the Kronecker delta, which is $1$ if $a = b$ and $0$ otherwise.

Let $\alpha$ be a finite set. For any element $i \in \alpha$ and any function $f : \alpha \to M$ mapping into a module $M$, the sum over all $j \in \alpha$ of the Kronecker delta $\delta_{ij}$ acting on $f(j)$ is equal to $f(i)$. That is, $$\sum_{j \in \alpha} \delta_{ij} \cdot f(j) = f(i)$$ where $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ if $i \neq j$.

Let $\alpha$ be a finite set and $f: \alpha \to \alpha \to M$ be a function into a module $M$. If $f(i, i) = 0$ for all $i \in \alpha$, then the double sum over $\alpha$ of the scalar multiplication of the Kronecker delta $\delta_{ij}$ and $f(i, j)$ is zero: $$\sum_{i \in \alpha} \sum_{j \in \alpha} \delta_{ij} \cdot f(i, j) = 0$$ where $\delta_{ij}$ is the Kronecker delta, defined as $1$ if $i = j$ and $0$ if $i \neq j$.

Let $s$ be a finite set of elements of type $\alpha$, $i$ be an element of $\alpha$, and $f: \alpha \to M$ be a function into an additive group $M$. The sum of the scalar multiplication of the Kronecker delta $\delta_{ij}$ and $f(j)$ over all $j \in s$ is given by: $$\sum_{j \in s} \delta_{ij} f(j) = \begin{cases} f(i) & \text{if } i \in s \\ 0 & \text{if } i \notin s \end{cases}$$ where $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ if $i \neq j$.

Let $s$ and $s'$ be finite sets of elements of type $\alpha$, and let $f: \alpha \to \alpha \to M$ be a function into an additive group $M$. If $f(i, i) = 0$ for all $i \in s \cap s'$, then the double sum of the scalar multiplication of the Kronecker delta $\delta_{ij}$ and $f(i, j)$ over $i \in s$ and $j \in s'$ is zero: $$\sum_{i \in s} \sum_{j \in s'} \delta_{ij} f(i, j) = 0$$ where $\delta_{ij}$ is the Kronecker delta, which is $1$ if $i = j$ and $0$ if $i \neq j$.