Physlib.Relativity.Tensors.RealTensor.Vector.Pre.Contraction

For a natural number $d$, this defines the $\mathbb{R}$-bilinear map from the space of contravariant Lorentz vectors $\text{ContrMod } d$ and the space of covariant Lorentz vectors $\text{CoMod } d$ to the real numbers $\mathbb{R}$. Given a contravariant vector $\psi \in \text{ContrMod } d$ and a covariant vector $\phi \in \text{CoMod } d$, the map computes their contraction as the dot product of their coordinate representations: $$(\psi, \phi) \mapsto \psi \cdot \phi = \sum_{i \in \text{Fin } 1 \oplus \text{Fin } d} \psi^i \phi_i$$ where $\psi^i$ and $\phi_i$ are the real-valued components of the vectors in the $1+d$ dimensional spacetime.

For a given dimension $d$, the map defines a bilinear form $\text{CoMod } d \times \text{ContrMod } d \to \mathbb{R}$ that represents the contraction of a covariant Lorentz vector $\phi$ with a contravariant Lorentz vector $\psi$. Specifically, if $\phi \in \text{CoMod } d$ and $\psi \in \text{ContrMod } d$, the result is the dot product of their coordinate representations in $\mathbb{R}^{1+d}$, given by $\sum_{\mu=0}^d \phi_\mu \psi^\mu$.

For a $(1+d)$-dimensional spacetime, let $\text{Contr } d$ and $\text{Co } d$ be the representations of the Lorentz group corresponding to contravariant and covariant vectors, respectively. This definition represents the Lorentz-invariant contraction map from the tensor product representation $\text{Contr } d \otimes \text{Co } d$ to the trivial representation $\mathbb{R}$. Given a contravariant vector $\psi \in \text{Contr } d$ and a covariant vector $\phi \in \text{Co } d$, the map is defined by the sum over their components in the standard basis: $$\psi \otimes \phi \mapsto \sum_{i=0}^d \psi^i \phi_i$$ In index notation, this corresponds to the contraction $\psi^i \phi_i$. As a morphism in the category of representations, this map is intertwining, meaning it is invariant under the action of the Lorentz group.

For a contravariant Lorentz vector $\psi$ and a covariant Lorentz vector $\phi$ in $(1+d)$-dimensional spacetime, the Lorentz-invariant contraction $\langle \psi, \phi \rangle_m$ (representing the evaluation of the contraction morphism on the tensor product $\psi \otimes \phi$) is equal to the standard Euclidean dot product of their coordinate representations. Specifically, $\langle \psi, \phi \rangle_m = \sum_{i=0}^d \psi^i \phi_i$, where $\psi^i$ and $\phi_i$ are the components of $\psi$ and $\phi$ in the standard basis $\text{Fin } 1 \oplus \text{Fin } d$.

For a given dimension $d$, the morphism `Lorentz.coContrContract` is an equivariant linear map from the tensor product of the covariant Lorentz representation $\text{Co } d$ and the contravariant Lorentz representation $\text{Contr } d$ to the trivial representation $\mathbb{R}$. This map is defined by the contraction of a covariant vector $\phi$ and a contravariant vector $\psi$, expressed in index notation as $\phi_\mu \psi^\mu$, which corresponds to the dot product of their components in the standard basis.

For any spatial dimension $d$, let $\phi \in \text{Co } d$ be a covariant Lorentz vector and $\psi \in \text{Contr } d$ be a contravariant Lorentz vector. The contraction $\langle \phi, \psi \rangle_m$ is equal to the dot product of their coordinate representations in the standard basis of $\mathbb{R}^{1+d}$, denoted by $\phi.\text{toFin1d}\mathbb{R} \cdot \psi.\text{toFin1d}\mathbb{R}$.

For any contravariant Lorentz vector $\phi \in \text{Contr } d$ and any covariant Lorentz vector $\psi \in \text{Co } d$ in a $(1+d)$-dimensional spacetime, the Lorentz-invariant contraction $\langle \phi, \psi \rangle_m$ is equal to the contraction $\langle \psi, \phi \rangle_m$.

For any spatial dimension $d$, let $\phi \in \text{Co } d$ be a covariant Lorentz vector and $\psi \in \text{Contr } d$ be a contravariant Lorentz vector in a $(1+d)$-dimensional spacetime. The Lorentz-invariant contraction $\langle \phi, \psi \rangle_m$ is equal to the contraction $\langle \psi, \phi \rangle_m$.

For a $(1+d)$-dimensional spacetime, let $\text{Contr } d$ be the representation of the Lorentz group corresponding to contravariant vectors. The definition represents the Lorentz-invariant contraction morphism from the tensor product representation $\text{Contr } d \otimes \text{Contr } d$ to the trivial representation $\mathbb{R}$. Given two contravariant vectors $\psi, \phi \in \text{Contr } d$, the map is defined by first applying the isomorphism $\text{Contr.toCo } d$ to the second vector (effectively lowering its index using the Minkowski metric $\eta_{\mu\nu}$) and then applying the standard contravariant-covariant contraction $\text{contrCoContract}$. In components, this corresponds to the Minkowski inner product: $$\psi \otimes \phi \mapsto \eta_{\mu\nu} \psi^\mu \phi^\nu = \psi^\mu \phi_\mu$$ where $\eta_{\mu\nu}$ is the Minkowski metric tensor.

Let $\text{Contr } d$ denote the vector space of contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime. The definition is an $\mathbb{R}$-linear map from the tensor product space $(\text{Contr } d) \otimes_{\mathbb{R}} (\text{Contr } d)$ to the real numbers $\mathbb{R}$. This map computes the Minkowski inner product of two contravariant vectors. Specifically, for any two contravariant vectors $u, v \in \text{Contr } d$, the map is defined by lowering the index of the second vector via the Minkowski metric $\eta_{\mu\nu}$ and performing the standard contraction: $$u \otimes v \mapsto \eta_{\mu\nu} u^\mu v^\nu$$ where $\eta_{\mu\nu} = \text{diag}(1, -1, \dots, -1)$.

For any two contravariant Lorentz vectors $\phi$ and $\psi$ in a $(1+d)$-dimensional spacetime, the Minkowski inner product (or Lorentz contraction) $\langle \phi, \psi \rangle_m$ is equal to the product $\mathbf{v}_\phi^\top \eta \mathbf{v}_\psi$. Here, $\mathbf{v}_\phi, \mathbf{v}_\psi \in \mathbb{R}^{1+d}$ are the coordinate representations of the vectors in the standard basis, $\eta = \text{diag}(1, -1, \dots, -1)$ is the Minkowski matrix, and the operation corresponds to the Euclidean dot product of $\mathbf{v}_\phi$ with $\eta \mathbf{v}_\psi$.

For a given dimension $d$, the morphism `Lorentz.coCoContract` is an equivariant linear map from the tensor product of two covariant Lorentz representations $\text{Co } d \otimes \text{Co } d$ to the trivial representation $\mathbb{R}$. This map is defined by first converting the second covariant vector in the tensor product into a contravariant vector via the isomorphism `Co.toContr` (which corresponds to raising the index using the inverse Minkowski metric $\eta^{\mu\nu}$) and then applying the standard contraction `coContrContract`. In index notation, for two covariant vectors $\alpha_\mu$ and $\beta_\nu$, this operation computes the scalar $\eta^{\mu\nu} \alpha_\mu \beta_\nu$.

For any dimension $d$, let $\phi, \psi \in \text{Co } d$ be two covariant Lorentz vectors. The contraction $\langle \phi, \psi \rangle_m$ is equal to the dot product of the coordinate representation of $\phi$ with the product of the Minkowski matrix $\eta$ and the coordinate representation of $\psi$. Mathematically, this is expressed as: $$\langle \phi, \psi \rangle_m = [\phi] \cdot (\eta [\psi])$$ where $[\phi], [\psi] \in \mathbb{R}^{1+d}$ are the representations of the vectors in the standard basis and $\eta = \text{diag}(1, -1, \dots, -1)$ is the Minkowski matrix.

In a spacetime with $d$ spatial dimensions, let $\langle \cdot, \cdot \rangle_m$ denote the Minkowski inner product (contraction) for contravariant vectors. For any Lorentz transformation $g$ in the Lorentz group and any contravariant vectors $x$ and $y$, the inner product is invariant under the group action $\rho(g)$. That is, $$\langle \rho(g)x, \rho(g)y \rangle_m = \langle x, y \rangle_m$$ where $\rho(g)$ is the representation of the Lorentz group acting on the space of contravariant vectors.

For any two contravariant Lorentz vectors $x$ and $y$ in a $(1+d)$-dimensional spacetime, their Minkowski inner product $\langle x, y \rangle_m$ is given by the difference between the product of their temporal components and the sum of the products of their spatial components: $$\langle x, y \rangle_m = x^0 y^0 - \sum_{i=1}^d x^i y^i$$ where $x^0$ and $y^0$ are the components indexed by the temporal dimension (represented by `Sum.inl 0`), and $x^i$ and $y^i$ are the components indexed by the $d$ spatial dimensions (represented by `Sum.inr i`).

For any two contravariant Lorentz vectors $x$ and $y$ in a $(1+d)$-dimensional spacetime, their Minkowski inner product $\langle x, y \rangle_m$ is equal to the product of their temporal components minus the Euclidean inner product of their spatial parts: $$\langle x, y \rangle_m = x^0 y^0 - \langle \vec{x}, \vec{y} \rangle$$ where $x^0$ and $y^0$ are the temporal components of $x$ and $y$ respectively (indexed by `Sum.inl 0`), and $\vec{x}, \vec{y} \in \mathbb{R}^d$ are the spatial vectors (obtained via `toSpace`) equipped with the standard Euclidean inner product $\langle \cdot, \cdot \rangle$.

In a $(1+d)$-dimensional spacetime, let $e_0$ be the standard basis vector for the temporal dimension (indexed by `Sum.inl 0`) in the space of contravariant Lorentz vectors. Its Minkowski inner product with itself is $1$: $$\langle e_0, e_0 \rangle_m = 1$$

For any two contravariant Lorentz vectors $x$ and $y$ in a $(1+d)$-dimensional spacetime, the Minkowski inner product $\langle \cdot, \cdot \rangle_m$ is symmetric: $$\langle x, y \rangle_m = \langle y, x \rangle_m$$

Let $x$ and $y$ be contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime. For any $(1+d) \times (1+d)$ real matrix $\Lambda$, let $\Lambda^*$ (denoted in the formal text as `dual Λ`) be its Minkowski dual matrix defined by $\Lambda^* = \eta \Lambda^\top \eta$, where $\eta = \text{diag}(1, -1, \dots, -1)$ is the Minkowski metric. The Minkowski inner product $\langle \cdot, \cdot \rangle_m$ satisfies: $$\langle x, \Lambda^* y \rangle_m = \langle \Lambda x, y \rangle_m$$

Let $x$ and $y$ be contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime. For any $(1+d) \times (1+d)$ real matrix $\Lambda$, let $\Lambda^*$ (the Minkowski dual, denoted as `dual Λ`) be defined by $\Lambda^* = \eta \Lambda^\top \eta$, where $\eta = \text{diag}(1, -1, \dots, -1)$ is the Minkowski metric. The Minkowski inner product $\langle \cdot, \cdot \rangle_m$ satisfies the adjoint property: $$\langle \Lambda^* x, y \rangle_m = \langle x, \Lambda y \rangle_m$$

Let $x$ and $y$ be contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime. Let $P$ denote the parity transformation (representing the action of the parity element of the Lorentz group), which leaves the temporal component unchanged and negates the spatial components. The Minkowski inner product of $x$ and the parity-transformed vector $Py$ is equal to the Euclidean inner product of their coordinate components: $$\langle x, Py \rangle_m = \sum_{i} x^i y^i$$ where the sum is taken over all temporal and spatial indices $i \in \{0, 1, \dots, d\}$.

Let $y$ be a contravariant Lorentz vector in a $(1+d)$-dimensional spacetime. Let $P$ denote the parity transformation, representing the action of the parity element of the Lorentz group. The Minkowski inner product of $y$ with its parity-transformed image $Py$ is zero if and only if $y$ is the zero vector: $$\langle y, Py \rangle_m = 0 \iff y = 0$$

Let $y$ be a contravariant Lorentz vector in a $(1+d)$-dimensional spacetime. The Minkowski metric is non-degenerate, meaning that the Minkowski inner product $\langle x, y \rangle_m$ is zero for all contravariant vectors $x$ if and only if $y$ is the zero vector: $$(\forall x \in \text{Contr } d, \langle x, y \rangle_m = 0) \iff y = 0$$

Let $x$ and $y$ be contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime, and let $\Lambda$ and $\Lambda'$ be $(1+d) \times (1+d)$ real matrices. Then the Minkowski inner product satisfies $\langle x, \Lambda y \rangle_m = \langle x, \Lambda' y \rangle_m$ if and only if $\langle x, (\Lambda - \Lambda') y \rangle_m = 0$.

Let $\text{Contr } d$ denote the vector space of contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime, and let $\langle \cdot, \cdot \rangle_m$ denote the Minkowski inner product. For any two real matrices $\Lambda$ and $\Lambda'$ of size $(1+d) \times (1+d)$, the linear transformations they induce are identical if and only if their associated bilinear forms under the Minkowski metric are identical for all vectors: $$(\forall v \in \text{Contr } d, \Lambda v = \Lambda' v) \iff (\forall v, w \in \text{Contr } d, \langle v, \Lambda w \rangle_m = \langle v, \Lambda' w \rangle_m)$$

Let $\text{Contr } d$ be the vector space of contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime, and let $\langle \cdot, \cdot \rangle_m$ denote the Minkowski inner product. For any two real matrices $\Lambda$ and $\Lambda'$ of size $(1+d) \times (1+d)$, the matrices are equal if and only if their associated bilinear forms under the Minkowski metric are identical for all vectors $v, w \in \text{Contr } d$: $$\Lambda = \Lambda' \iff \forall v, w \in \text{Contr } d, \langle v, \Lambda w \rangle_m = \langle v, \Lambda' w \rangle_m$$

Let $\text{Contr } d$ denote the vector space of contravariant Lorentz vectors in a $(1+d)$-dimensional spacetime, and let $\langle \cdot, \cdot \rangle_m$ denote the Minkowski inner product. For any real square matrix $\Lambda$ of size $(1+d) \times (1+d)$, $\Lambda$ is equal to the identity matrix $I$ if and only if for all vectors $v, w \in \text{Contr } d$, the Minkowski inner product of $v$ and $\Lambda w$ is equal to the Minkowski inner product of $v$ and $w$: $$\Lambda = I \iff \forall v, w \in \text{Contr } d, \langle v, \Lambda w \rangle_m = \langle v, w \rangle_m$$

A square matrix $\Lambda$ of size $(1+d) \times (1+d)$ is an element of the Lorentz group $O(1, d)$ if and only if it preserves the Minkowski inner product for all contravariant Lorentz vectors $v, w \in \text{Contr } d$. That is, $$\Lambda \in \text{LorentzGroup } d \iff \forall v, w \in \text{Contr } d, \langle \Lambda v, \Lambda w \rangle_m = \langle v, w \rangle_m$$ where $\langle \cdot, \cdot \rangle_m$ denotes the Minkowski inner product.

A square matrix $\Lambda$ of size $(1+d) \times (1+d)$ is an element of the Lorentz group $O(1, d)$ if and only if it preserves the Minkowski norm squared for every contravariant Lorentz vector $w \in \text{Contr } d$. That is, $$\Lambda \in \text{LorentzGroup } d \iff \forall w \in \text{Contr } d, \langle \Lambda w, \Lambda w \rangle_m = \langle w, w \rangle_m$$ where $\langle \cdot, \cdot \rangle_m$ denotes the Minkowski inner product.

For any contravariant Lorentz vector $v$ in a $(1+d)$-dimensional spacetime, let $v^0$ denote its temporal component (indexed by `Sum.inl 0`) and $v^i$ denote its spatial components (indexed by `Sum.inr i`). The square of the temporal component is equal to the sum of the Minkowski inner product of $v$ with itself and the sum of the squares of its spatial components: $$(v^0)^2 = \langle v, v \rangle_m + \sum_{i=1}^d (v^i)^2$$

For any contravariant Lorentz vector $v$ in a $(1+d)$-dimensional spacetime, the Minkowski inner product of $v$ with itself, denoted by $\langle v, v \rangle_m$, is less than or equal to the square of its temporal component $v^0$ (indexed by `Sum.inl 0`). That is, $$\langle v, v \rangle_m \le (v^0)^2$$

For any two contravariant Lorentz vectors $v$ and $w$ in a $(1+d)$-dimensional spacetime, the Minkowski inner product $\langle v, w \rangle_m$ satisfies the following inequality: $$v^0 w^0 - |\langle \vec{v}, \vec{w} \rangle| \le \langle v, w \rangle_m$$ where $v^0$ and $w^0$ are the temporal components of $v$ and $w$ (indexed by `Sum.inl 0`), $\vec{v}$ and $\vec{w}$ are the spatial components in $\mathbb{R}^d$ (obtained via `toSpace`), and $\langle \cdot, \cdot \rangle$ is the standard Euclidean inner product on $\mathbb{R}^d$.

For any two contravariant Lorentz vectors $v$ and $w$ in a $(1+d)$-dimensional spacetime, the Minkowski inner product $\langle v, w \rangle_m$ satisfies the following inequality: $$v^0 w^0 - \|\vec{v}\| \|\vec{w}\| \le \langle v, w \rangle_m$$ where $v^0$ and $w^0$ are the temporal components of $v$ and $w$ (indexed by `Sum.inl 0`), $\vec{v}$ and $\vec{w}$ are the spatial components in $\mathbb{R}^d$ (obtained via `toSpace`), and $\|\cdot\|$ denotes the standard Euclidean norm on $\mathbb{R}^d$.

For any contravariant Lorentz vector $v \in \text{Contr } d$ and any index $\mu \in \text{Fin } 1 \oplus \text{Fin } d$, the Minkowski inner product of the $\mu$-th standard basis vector $e_\mu$ and $v$ is equal to the product of the $\mu$-th diagonal component of the Minkowski metric $\eta$ and the $\mu$-th component of $v$: $$\langle e_\mu, v \rangle_m = \eta_{\mu\mu} v^\mu$$ where $e_\mu$ is the $\mu$-th element of the standard basis for $\text{ContrMod } d$, $\eta$ is the Minkowski matrix $\text{diag}(1, -1, \dots, -1)$, and $v^\mu$ is the component of $v$ at index $\mu$.

For any indices $\mu, \nu \in \text{Fin } 1 \oplus \text{Fin } d$, let $e_\mu$ and $e_\nu$ be the corresponding vectors in the standard basis of the space of contravariant Lorentz vectors $\text{ContrMod } d$. For any real matrix $\Lambda$ of size $(1+d) \times (1+d)$, the Minkowski inner product of $e_\mu$ and the vector $\Lambda e_\nu$ (obtained by matrix-vector multiplication) is equal to the product of the $\mu$-th diagonal entry of the Minkowski metric $\eta$ and the $(\mu, \nu)$-th entry of the matrix $\Lambda$: $$\langle e_\mu, \Lambda e_\nu \rangle_m = \eta_{\mu\mu} \Lambda_{\mu\nu}$$ where $\eta = \text{diag}(1, -1, \dots, -1)$.

For any indices $\mu, \nu \in \text{Fin } 1 \oplus \text{Fin } d$, let $e_\mu$ and $e_\nu$ be the elements of the standard basis for the space of contravariant real Lorentz vectors $\text{ContrMod } d$. The Minkowski inner product of these two basis vectors is equal to the $(\mu, \nu)$-th entry of the Minkowski metric $\eta$: $$\langle e_\mu, e_\nu \rangle_m = \eta_{\mu\nu}$$ where $\eta$ is the $(1+d) \times (1+d)$ diagonal matrix $\text{diag}(1, -1, \dots, -1)$.

For any indices $\nu, \mu \in \text{Fin } 1 \oplus \text{Fin } d$, let $e_\nu$ and $e_\mu$ be the corresponding vectors in the standard basis of the space of contravariant Lorentz vectors $\text{ContrMod } d$. Let $\Lambda$ be a real matrix of size $(1+d) \times (1+d)$ and $\eta$ be the Minkowski metric $\text{diag}(1, -1, \dots, -1)$. The $(\nu, \mu)$-th entry of the matrix $\Lambda$ is given by the product of the $\nu$-th diagonal entry of the Minkowski metric $\eta$ and the Minkowski inner product of $e_\nu$ with the vector $\Lambda e_\mu$: $$\Lambda_{\nu\mu} = \eta_{\nu\nu} \langle e_\nu, \Lambda e_\mu \rangle_m$$

For any contravariant Lorentz vector $x \in \text{ContrMod } 3$ in $3+1$ dimensions, let $\langle x, x \rangle_m$ denote its Minkowski inner product with itself (the Minkowski norm squared). Let $M$ be the $2 \times 2$ self-adjoint complex matrix associated with $x$ via the linear isomorphism `toSelfAdjoint`, defined by the linear combination $M = x^0 \sigma_0 - x^1 \sigma_1 - x^2 \sigma_2 - x^3 \sigma_3$, where $\sigma_0$ is the identity matrix and $\sigma_1, \sigma_2, \sigma_3$ are the standard Pauli matrices. The Minkowski norm squared is equal to the determinant of this matrix: $$\langle x, x \rangle_m = \det(M)$$

In a $(1+d)$-dimensional spacetime, let $\{e^i\}$ be the standard basis for the contravariant representation $\text{Contr } d$ and $\{e_j\}$ be the standard basis for the covariant representation $\text{Co } d$, where indices $i, j$ belong to the index set $\text{Fin } 1 \oplus \text{Fin } d$. The evaluation of the Lorentz-invariant contraction morphism on the tensor product of these basis vectors satisfies: $$\text{contrCoContract}(e^i \otimes e_j) = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$$ This result represents the Kronecker delta $\delta_{ij}$ in the context of contravariant-covariant vector contraction.

For any dimension $d$, let $\{e^*_i\}$ be the standard basis of the covariant Lorentz representation $\text{Co } d$ and let $\{e^j\}$ be the standard basis of the contravariant Lorentz representation $\text{Contr } d$, where the indices $i, j$ range over the set $\text{Fin } 1 \oplus \text{Fin } d$. The contraction morphism $\text{coContrContract}$ applied to the tensor product of basis elements $e^*_i \otimes e^j$ is given by the Kronecker delta: $$\langle e^*_i, e^j \rangle_m = \delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$$