Physlib.SpaceAndTime.Space.Slice

For a natural number $d$ and an index $i \in \{0, \dots, d\}$, the map $\text{slice}(i)$ is a continuous linear equivalence (a topological vector space isomorphism) between $\text{Space}(d+1)$ and $\mathbb{R} \times \text{Space}(d)$. Given an element $x \in \text{Space}(d+1)$, the map extracts the $i$-th coordinate as the first component of the product and projects the remaining coordinates onto $\text{Space}(d)$. Specifically, it is defined by $x \mapsto (x_i, \hat{x})$, where $\hat{x}_j = x_{\sigma_i(j)}$ and $\sigma_i$ is the function that skips the index $i$.

For any natural number $d$, an index $i \in \{0, \dots, d\}$, a scalar $r \in \mathbb{R}$, and a vector $x \in \text{Space } d$, the inverse of the continuous linear equivalence $\text{slice}(i) : \text{Space}(d+1) \cong \mathbb{R} \times \text{Space } d$ applied to the pair $(r, x)$ is the vector in $\text{Space}(d+1)$ obtained by inserting $r$ at the $i$-th coordinate and filling the remaining coordinates with the components of $x$.

For any natural number $d$, index $i \in \{0, \dots, d\}$, real number $r \in \mathbb{R}$, and vector $x \in \text{Space } d$, the $i$-th coordinate of the vector obtained by applying the inverse of the slicing map $\text{slice}(i)$ to the pair $(r, x)$ is equal to $r$. Here, $(\text{slice } i)^{-1} : \mathbb{R} \times \text{Space } d \to \text{Space}(d+1)$ is the inverse of the continuous linear equivalence that extracts the $i$-th coordinate.

For any natural number $d$, index $i \in \{0, \dots, d\}$, scalar $r \in \mathbb{R}$, and vector $x \in \text{Space } d$, let $v = \text{slice}(i)^{-1}(r, x)$ be the vector in $\text{Space}(d+1)$ obtained by inserting $r$ at the $i$-th coordinate and filling the remaining positions with $x$. For any $j \in \{0, \dots, d-1\}$, the component of $v$ at the index $\text{succAbove}_i(j)$ is equal to the $j$-th component of $x$, where $\text{succAbove}_i(j)$ maps $j$ to the corresponding index in $\{0, \dots, d\} \setminus \{i\}$.

For any natural number $d$ and index $i \in \{0, \dots, d\}$, the inverse of the slicing map $(\text{slice } i)^{-1} : \mathbb{R} \times \text{Space } d \to \text{Space}(d+1)$ is a measurable embedding. The slicing map $\text{slice}(i)$ is the continuous linear equivalence that identifies $\text{Space}(d+1)$ with $\mathbb{R} \times \text{Space } d$ by extracting the $i$-th coordinate; its inverse $(\text{slice } i)^{-1}$ reconstructs a vector in $\text{Space}(d+1)$ by inserting a real value at the $i$-th position and filling the remaining coordinates with the components of a vector in $\text{Space } d$.

For any natural number $d$, index $i \in \{0, \dots, d\}$, real number $r \in \mathbb{R}$, and vector $x \in \text{Space } d$, the Euclidean norm of the vector in $\text{Space}(d+1)$ obtained by applying the inverse of the slicing map $\text{slice}(i)$ to the pair $(r, x)$ is given by: $$\|(\text{slice } i)^{-1}(r, x)\| = \sqrt{\|r\|^2 + \|x\|^2}$$ where $(\text{slice } i)^{-1} : \mathbb{R} \times \text{Space } d \to \text{Space}(d+1)$ is the isomorphism that inserts $r$ as the $i$-th coordinate and uses the components of $x$ for the remaining coordinates.

For any natural number $d$, index $i \in \{0, \dots, d\}$, real number $r \in \mathbb{R}$, and vector $x \in \text{Space } d$, the absolute value of $r$ is less than or equal to the Euclidean norm of the vector in $\text{Space}(d+1)$ produced by applying the inverse slicing map $(\text{slice } i)^{-1}$ to the pair $(r, x)$: $$|r| \le \|(\text{slice } i)^{-1}(r, x)\|$$ where $(\text{slice } i)^{-1}$ is the isomorphism that constructs a vector by inserting $r$ as the $i$-th coordinate and using the components of $x$ for the remaining dimensions.

For any natural number $d$, index $i \in \{0, \dots, d\}$, real number $r \in \mathbb{R}$, and vector $x \in \text{Space } d$, the Euclidean norm of $x$ is less than or equal to the Euclidean norm of the vector in $\text{Space}(d+1)$ obtained by applying the inverse slicing map $(\text{slice } i)^{-1}$ to the pair $(r, x)$: $$\|x\| \le \|(\text{slice } i)^{-1}(r, x)\|$$ where $(\text{slice } i)^{-1} : \mathbb{R} \times \text{Space } d \to \text{Space}(d+1)$ is the isomorphism that inserts $r$ as the $i$-th coordinate and uses the components of $x$ for the remaining coordinates.

For any natural number $d$, index $i \in \{0, \dots, d\}$, and point $p \in \mathbb{R} \times \text{Space } d$, the Fréchet derivative of the inverse slice map $(\text{slice}_i)^{-1} : \mathbb{R} \times \text{Space } d \to \text{Space}(d+1)$ at the point $p$ is equal to the linear map $(\text{slice}_i)^{-1}$ itself.

For any natural number $d$, index $i \in \{0, \dots, d\}$, and fixed vector $x \in \text{Space } d$, consider the function $g: \mathbb{R} \to \text{Space}(d+1)$ defined by $g(r) = (\text{slice}_i)^{-1}(r, x)$, where $(\text{slice}_i)^{-1}$ is the inverse of the continuous linear equivalence that extracts the $i$-th coordinate. For any $r_1, r_2 \in \mathbb{R}$, the Fréchet derivative of $g$ at the point $r_1$ applied to the direction $r_2$ is given by: $$Df(r_1)(r_2) = (\text{slice}_i)^{-1}(r_2, 0)$$

For any natural number $d$, index $i \in \{0, \dots, d\}$, fixed real number $r \in \mathbb{R}$, and points $x_1, x_2 \in \text{Space } d$, the Fréchet derivative of the map $g: \text{Space } d \to \text{Space}(d+1)$ defined by $g(x) = (\text{slice}_i)^{-1}(r, x)$ at the point $x_1$ applied to the vector $x_2$ is equal to $(\text{slice}_i)^{-1}(0, x_2)$, where $(\text{slice}_i)^{-1}$ is the inverse of the $i$-th coordinate slice map.

For any natural number $d$, index $i \in \{0, \dots, d\}$, and fixed real number $r \in \mathbb{R}$, let $f: \text{Space}(d+1) \to F$ be a function that is differentiable at $(\text{slice}_i)^{-1}(r, x_1)$ for some point $x_1 \in \text{Space } d$. The Fréchet derivative of the composition $x \mapsto f((\text{slice}_i)^{-1}(r, x))$ at the point $x_1$ applied to a vector $x_2 \in \text{Space } d$ is equal to the Fréchet derivative of $f$ at the point $(\text{slice}_i)^{-1}(r, x_1)$ applied to the vector $(\text{slice}_i)^{-1}(0, x_2)$, where $(\text{slice}_i)^{-1}$ denotes the inverse of the $i$-th coordinate slice map.

For any natural number $d$, index $i \in \{0, \dots, d\}$, real numbers $r_1, r_2 \in \mathbb{R}$, vector $x \in \text{Space } d$, and function $f: \text{Space}(d+1) \to F$, suppose $f$ is differentiable at the point $p = (\text{slice}_i)^{-1}(r_1, x)$, where $(\text{slice}_i)^{-1}$ is the inverse of the continuous linear equivalence $\text{slice}_i: \text{Space}(d+1) \cong \mathbb{R} \times \text{Space } d$ that extracts the $i$-th coordinate. The Fréchet derivative of the composition $r \mapsto f((\text{slice}_i)^{-1}(r, x))$ at the point $r_1$ applied to the increment $r_2$ is given by: $$D(f \circ (\text{slice}_i)^{-1}(\cdot, x))(r_1)(r_2) = Df(p)((\text{slice}_i)^{-1}(r_2, 0))$$

For any natural number $d$ and index $i \in \{0, \dots, d\}$, let $\text{basis}_i$ be the $i$-th vector of the standard orthonormal basis of $\text{Space}(d+1)$. This basis vector is equal to the image of the pair $(1, 0) \in \mathbb{R} \times \text{Space } d$ under the inverse of the slicing map $\text{slice}(i)^{-1}: \mathbb{R} \times \text{Space } d \to \text{Space}(d+1)$, where $0$ is the zero element of $\text{Space } d$.

For any natural number $d$, index $i \in \{0, \dots, d\}$, and index $j \in \{0, \dots, d-1\}$, let $\text{basis}^{(d+1)}$ denote the standard orthonormal basis of $\text{Space}(d+1)$ and $\text{basis}^{(d)}$ denote the standard orthonormal basis of $\text{Space } d$. The basis vector $\text{basis}^{(d+1)}_{\text{succAbove}_i(j)}$ is equal to the image of the pair $(0, \text{basis}^{(d)}_j)$ under the inverse of the slicing map $\text{slice}(i)^{-1} : \mathbb{R} \times \text{Space } d \to \text{Space}(d+1)$. Here, $\text{succAbove}_i(j)$ is the index in $\{0, \dots, d\}$ obtained by skipping the index $i$.