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docs/mathematics/linear-algebra/tensors/tensor-formalism.md
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docs/mathematics/linear-algebra/tensors/tensor-formalism.md
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# Tensor formalism
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We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$ and a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}.$ In the following sections we make use of the Einstein summation convention introduced in [vector analysis](/en/physics/mathematical-physics/vector-analysis/curvilinear-coordinates/) and $\mathbb{K} = \mathbb{R} \lor\mathbb{K} = \mathbb{C}.$
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## Definition
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> *Definition 1*: a **tensor** is a multilinear mapping of the type
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>
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> $$
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> \mathbf{T}: \underbrace{V^* \times \dots \times V^*}_p \times \underbrace{V \times \dots \times V}_q \to \mathbb{K},
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> $$
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>
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> with $p, q \in \mathbb{N}$. Tensors are collectively denoted as
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>
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> $$
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> \mathbf{T} \in \underbrace{V \otimes \dots \otimes V}_p \otimes \underbrace{V^* \otimes \dots \otimes V^*}_q = \mathscr{T}_q^p(V),
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> $$
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>
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> with $\mathscr{T}_0^0(V) = \mathbb{K}$.
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We refer to $\mathbf{T} \in \mathscr{T}_q^p(V)$ as a $(p, q)$-tensor; a mixed tensor of **contravariant rank** $p$ and **covariant rank** $q.$ It may be observed that we have $\dim \mathscr{T}_q^p (V) = n^{p+q}$ with $\dim V = n \in \mathbb{N}$.
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It follows from definition 1 and by virtue of the isomorphism between $V^{**}$ and $V$ that $\mathbf{T} \in \mathscr{T}_1^0(V) = V^*$ is a covector and $\mathbf{T} \in \mathscr{T}_0^1(V) = V$ is a vector.
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## Kronecker tensor
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> *Definition 2*: let $\mathbf{k} \in \mathscr{T}_1^1(V)$ be the **Kronecker tensor** be defined such that
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>
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> $$
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> \mathbf{k}(\mathbf{\hat e}^i, \mathbf{e}_j) = \delta^i_j,
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> $$
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>
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> with $\delta_j^i$ the Kronecker symbol.
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Let $\mathbf{\hat u} = u_i \mathbf{\hat e}^i \in V^*$ and $\mathbf{v} = v^j \mathbf{e}_j \in V$ then the tensor properties and the definition of the Kronecker tensor imply that
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$$
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\begin{align*}
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\mathbf{k}(\mathbf{\hat u}, \mathbf{v}) &= \mathbf{k}(u_i \mathbf{\hat e}^i, v^j \mathbf{e}_j), \\
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&= u_i v^j \mathbf{k}(\mathbf{\hat e}^i, \mathbf{e}_j), \\
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&= u_i v^j \delta^i_j, \\
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&= u_i v^i.
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\end{align*}
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$$
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## Outer product
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> *Definition 3*: the outer product $f \otimes g: X \times Y \to \mathbb{K}$ of two scalar functions $f: X \to \mathbb{K}$ and $g: Y \to \mathbb{K}$ is defined as
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>
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> $$
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> (f \otimes g)(x,y) = f(x) g(y),
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> $$
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>
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> for all $(x,y) \in X \times Y$.
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The outer product is associative and distributive with respect to addition and scalar multiplication, but not commutative.
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Note that although the same symbol is used for the outer product and the denotation of a tensor space, these are not equivalent.
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The following statements are given with $p=q=r=s=1$ without loss of generality.
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> *Definition 4*: the mixed $(p, q)$-tensor $\mathbf{e}_i \otimes \mathbf{\hat e}^j \in \mathscr{T}_q^p(V)$ is defined as
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>
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> $$
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> (\mathbf{e}_i \otimes \mathbf{\hat e}^j)(\mathbf{\hat u}, \mathbf{v}) = \mathbf{k}(\mathbf{\hat u}, \mathbf{e}_i) \mathbf{k}(\mathbf{\hat e}^j, \mathbf{v}),
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> $$
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>
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> for all $(\mathbf{\hat u}, \mathbf{v}) \in V^* \times V$.
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From this definition the subsequent theorem follows naturally.
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> *Theorem 1*: let $\mathbf{T} \in \mathscr{T}_q^p(V)$ be a tensor, then there exists **holors** $T_j^i \in \mathbb{K}$ such that
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>
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> $$
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> \mathbf{T} = T^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j,
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> $$
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>
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> with $T^i_j = \mathbf{T}(\mathbf{\hat e}^i, \mathbf{e}_j)$.
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??? note "*Proof*:"
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Let $\mathbf{T} \in \mathscr{T}_q^p(V)$ such that
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$$
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\begin{align*}
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\mathbf{T}(\mathbf{\hat e}^i, \mathbf{e}_j) &= T^k_l (\mathbf{e}_k \otimes \mathbf{\hat e}^l)(\mathbf{\hat e}^i, \mathbf{e}_j), \\
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&= T^k_l \mathbf{k}(\mathbf{\hat e}^i, \mathbf{e}_k) \mathbf{k}(\mathbf{\hat e}^l,\mathbf{e}_j), \\
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&= T^k_l \delta^i_k \delta^l_j, \\
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&= T^i_j.
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\end{align*}
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$$
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For $\mathbf{T} \in \mathscr{T}^0_q(V)$ it follows that there exists holors $T_i \in \mathbb{K}$ such that $\mathbf{T} = T_i \mathbf{\hat e}^i$ with $T_i = \mathbf{T}(\mathbf{e}_i)$, are referred to as the **covariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$.
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For $\mathbf{T} \in \mathscr{T}^p_0(V)$ it follows that there exists holors $T^i \in \mathbb{K}$ such that $\mathbf{T} = T^i \mathbf{e}_i$ with $T^i = \mathbf{T}(\mathbf{\hat e}^i)$, are referred to as the **contravariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$.
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If $\mathbf{T} \in \mathscr{T}^p_q(V)$, it follows that there exists holors $T^i_j \in \mathbb{K}$ are coined the **mixed components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$.
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By definition tensors are basis independent. Holors are basis dependent.
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> *Theorem 2*: let $\mathbf{S} \in \mathscr{T}^p_q(V)$ and $\mathbf{T} \in \mathscr{T}^r_s(V)$ be tensors with
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>
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> $$
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> \mathbf{S} = S^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j \quad \land \quad \mathbf{T} = T^r_s \mathbf{e}_r \otimes \mathbf{\hat e}^s,
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> $$
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>
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> then the outer product of $\mathbf{S}$ and $\mathbf{T}$ is given by
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>
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> $$
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> \mathbf{S} \otimes \mathbf{T} = S^i_j T^k_l \mathbf{e}_i \otimes \mathbf{e}_k \otimes \mathbf{\hat e}^j \otimes \mathbf{\hat e}^l,
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> $$
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>
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> with $\mathbf{S} \otimes \mathbf{T} \in \mathscr{T}^{p+r}_{q+s}(V)$.
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??? note "*Proof*:"
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Let $\mathbf{S} \in \mathscr{T}^p_q(V)$ and $\mathbf{T} \in \mathscr{T}^r_s(V)$ with
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$$
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\mathbf{S} = S^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j \quad \land \quad \mathbf{T} = T^r_s \mathbf{e}_r \otimes \mathbf{\hat e}^s,
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$$
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then
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$$
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\begin{align*}
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\mathbf{S} \otimes \mathbf{T} &= S^i_j (\mathbf{e}_i \otimes \mathbf{\hat e}^j) \otimes T^r_s (\mathbf{e}_r \otimes \mathbf{\hat e}^s), \\
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&= S^i_j T^r_s \mathbf{e}_i \otimes \mathbf{e}_r \otimes \mathbf{\hat e}^j \otimes \mathbf{\hat e}^s.
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\end{align*}
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$$
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Which maps two vectors and two covectors, therefore $\mathbf{S} \otimes \mathbf{T} \in \mathscr{T}^{p+r}_{q+s}(V)$.
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We have from theorem 2 that the outer product of two tensors yields another tensor, with ranks adding up.
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## Inner product
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> *Definition 5*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to \mathbb{K}$ which satisfies
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>
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> 1. for all $\mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \overline{\bm{g}}(\mathbf{v}, \mathbf{u}),$
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> 2. for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $\lambda, \mu \in \mathbb{K}: \;\bm{g}(\mathbf{u}, \lambda \mathbf{v} + \mu \mathbf{w}) = \lambda \bm{g}(\mathbf{u}, \mathbf{v}) + \mu \bm{g}(\mathbf{u}, \mathbf{w}),$
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> 3. for all $\mathbf{u} \in V\backslash \{\mathbf{0}\}: \bm{g}(\mathbf{u},\mathbf{u}) > 0,$
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> 4. for $\mathbf{u} = \mathbf{0} \iff \bm{g}(\mathbf{u},\mathbf{u}) = 0.$
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It may be observed that $\bm{g} \in \mathscr{T}_2^0$. Unlike the Kronecker tensor, the existence of an inner product is never implied.
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> *Definition 6*: let $G$ be the Gram matrix with its components $G \overset{\text{def}}= (g_{ij})$ defined as
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>
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> $$
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> g_{ij} = \bm{g}(\mathbf{e}_i, \mathbf{e}_j).
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> $$
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For $\mathbf{u} = u^i \mathbf{e}_i, \mathbf{v} = v^j \mathbf{e}_j \in V$ we then have
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$$
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\begin{align*}
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\bm{g}(\mathbf{u}, \mathbf{v}) &= \bm{g}(u^i \mathbf{e}_i, v^j \mathbf{e}_j), \\
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&= u^i v^j \bm{g}(\mathbf{e}_i, \mathbf{e}_j), \\
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&\overset{\text{def}}= u^i v^j g_{ij}.
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\end{align*}
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$$
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> *Proposition 1*: the Gram matrix $G$ is symmetric and nonsingular such that
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>
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> $$
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> g^{ik} g_{kj} = \delta^i_j,
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> $$
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>
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> with $G^{-1} \overset{\text{def}}= (g^{ij})$.
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??? note "*Proof*:"
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Let $G$ be the Gram matrix, symmetry of $G$ follows from defintion 5. Suppose that $G$ is singular, then there exists $\mathbf{u} = u^i \mathbf{e}_i \in V \backslash \{\mathbf{0}\}$ such that $G \mathbf{u} = \mathbf{0} \implies u^i g_{ij} = 0$, as a result we find that
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$$
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\forall \mathbf{v} = v^j \mathbf{e}_j \in V: 0 = u^i g_{ij} v^j = u^i \bm{g}(\mathbf{e}_i, \mathbf{e}_j) v^j = \bm{g}(u^i \mathbf{e}_i, v^j \mathbf{e}_j) = \bm{g}(\mathbf{u}, \mathbf{v}),
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$$
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which contradicts the non-degeneracy of the pseudo inner product in definition 5.
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> *Theorem 3*: there exists a bijective linear map $\mathbf{g}: V \to V^*$ with inverse $\mathbf{g}^{-1}$ such that
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>
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> 1. $\forall \mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \mathbf{k}(\mathbf{g}(\mathbf{u}), \mathbf{v})$,
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> 2. $\forall \mathbf{\hat u} \in V^*, \mathbf{v} \in V: \; \bm{g}(\mathbf{g}^{-1}(\mathbf{\hat u}), \mathbf{v}) = \mathbf{k}(\mathbf{\hat u}, \mathbf{v})$,
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>
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> with $\mathbf{g}(\mathbf{v}) = G \mathbf{v}$ for all $\mathbf{v} \in V$.
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??? note "*Proof*:"
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Let $\mathbf{u} \in V$ and let $\mathbf{\hat u} \in V^*$, suppose $\mathbf{\hat u}: \mathbf{v} \mapsto \bm{g}(\mathbf{u}, \mathbf{v})$ then we may define $\mathbf{g}: V \to V^*: \mathbf{u} \mapsto \mathbf{g}(\mathbf{u}) \overset{\text{def}} = \mathbf{\hat u}$.
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Let $\mathbf{v} \in V \backslash \{\mathbf{0}\}: \mathbf{g}(\mathbf{v}) = \mathbf{0}$, then
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$$
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0 = \mathbf{k}(\mathbf{g}(\mathbf{v}), \mathbf{w}) \overset{\text{def}} = \bm{g}(\mathbf{v}, \mathbf{w}),
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$$
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for all $\mathbf{w} \in V$, which contradicts the non-degeneracy of the pseude inner product in definition 5. Hence $\mathbf{g}$ is injective, since $\dim V$ is finite $\mathbf{g}$ is also bijective.
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Let $\mathbf{u} = u^i \mathbf{e}_i, \mathbf{v} = v^j \mathbf{e}_j \in V$ and define $\mathbf{g}(\mathbf{e}_i) = \text{g}_{ij} \mathbf{\hat e}^j$ such that
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$$
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\mathbf{k}(\mathbf{g}(\mathbf{u}), \mathbf{v}) \overset{\text{def}} = \bm{g}(\mathbf{u}, \mathbf{v}) = g_{ij} u^i v^j,
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$$
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but also
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$$
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\mathbf{k}(\mathbf{g}(\mathbf{u}), \mathbf{v}) = \text{g}_{ij} u^i v^k\mathbf{k}(\mathbf{\hat e}^j, \mathbf{e}_k) = \text{g}_{ij} u^i v^k \delta^j_k = \text{g}_{ij} u^i v^j.
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$$
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Since $u^i, v^j \in \mathbb{K}$ are arbitrary it follows that $\text{g}_{ij} = g_{ij}$.
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Consequently, the inverse $\mathbf{g}^{-1}: V^* \to V$ has the property $\mathbf{g}^{-1}(\mathbf{\hat u}) = G^{-1} \mathbf{\hat u}$ for all $\mathbf{\hat u} \in V^*$. The bijective linear map $\mathbf{g}$ is commonly known as the **metric** and $\mathbf{g}^{-1}$ as the **dual metric**.
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It follows from theorem 3 that for $\mathbf{u} = u^i \mathbf{e}_i \in V$ and $\mathbf{\hat u} = u_i \mathbf{\hat e}^i \in V^*$ we have
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$$
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\mathbf{g}(\mathbf{u}) = g_{ij} u^i \mathbf{\hat e}^j = u_j \mathbf{\hat e}^j = \mathbf{\hat u},
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$$
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with $u_j = g_{ij} u^i$ and
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$$
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\mathbf{g}^{-1}(\mathbf{\hat u}) = g^{ij} u_i \mathbf{e}_j = u^j \mathbf{e}_j = \mathbf{u},
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$$
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with $u^j = g^{ij} u_i$.
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> *Definition 7*: the basis $\{\mathbf{e}_i\}$ of $V$ induces a **reciprocal basis** $\{\mathbf{g}^{-1}(\mathbf{\hat e}^i)\}$ of $V$ given by
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>
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> $$
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> \mathbf{g}^{-1}(\mathbf{\hat e}^i) = g^{ij} \mathbf{e}_j.
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> $$
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>
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> Likewise, the basis $\{\mathbf{\hat e}^i\}$ of $V^*$ induces a **reciprocal dual basis** $\{\mathbf{g}(\mathbf{e}_i)\}$ of $V^*$ given by
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>
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> $$
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> \mathbf{g}(\mathbf{e}^i) = g_{ij} \mathbf{\hat e}^j.
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> $$
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So far, a vector space $V$ and its associated dual space $V^*$ have been introduced as a priori independent entities. An inner product provides us with an explicit mechanism to construct a bijective linear mapping associated with each vector by virtue of the metric.
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docs/mathematics/linear-algebra/tensors/tensor-symmetries.md
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# Tensor symmetries
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We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n,$ a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}$ and a pseudo inner product $\bm{g}$ on $V.$
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## Symmetric tensors
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> *Definition 1*: let $\pi = [\pi(1), \dots, \pi(k)]$ be any permutation of the set $\{1, \dots, k\}$, then $\mathbf{T} \in \mathscr{T}^0_q(V)$ is a **symmetric covariant** $q$-tensor if for all $\mathbf{v}_1, \dots, \mathbf{v}_q \in V$ we have
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>
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> $$
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> \mathbf{T}(\mathbf{v}_{\pi(1)}, \dots, \mathbf{v}_{\pi(q)}) = \mathbf{T}(\mathbf{v}_1, \dots, \mathbf{v}_q),
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> $$
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>
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> with $k = q \in \mathbb{N}$.
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>
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> Likewise, $\mathbf{T} \in \mathscr{T}^p_0(V)$ is a **symmetric contravariant** $p$-tensor if for all $\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p \in V^*$ we have
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>
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> $$
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> \mathbf{T}(\mathbf{\hat u}_{\pi(1)}, \dots, \mathbf{\hat u}_{\pi(p)}) = \mathbf{T}(\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p),
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> $$
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>
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> with $k = p \in \mathbb{N}$.
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This symmetry implies that the ordering of the (co)vector arguments in a tensor evaluation do not affect the outcome.
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> *Definition 2*: the vector space of symmetric covariant $q$-tensors is denoted by $\bigvee_q(V) \subset \mathscr{T}^0_q(V)$ and the vector space of symmetric contravariant $p$-tensors is denoted by $\bigvee^p(V) \subset \mathscr{T}^p_0(V).$
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Alternatively one may write $\bigvee_q(V) = V^* \otimes_s \cdots \otimes_s V^*$ and $\bigvee^p(V) = V \otimes_s \cdots \otimes_s V.$
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## Antisymmetric tensors
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> *Definition 3*: let $\pi = [\pi(1), \dots, \pi(k)]$ be any permutation of the set $\{1, \dots, k\}$, then $\mathbf{T} \in \mathscr{T}^0_q(V)$ is an **antisymmetric covariant** $q$-tensor if for all $\mathbf{v}_1, \dots, \mathbf{v}_q \in V$ we have
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>
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> $$
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> \mathbf{T}(\mathbf{v}_{\pi(1)}, \dots, \mathbf{v}_{\pi(q)}) = \mathrm{sign}(\pi) \mathbf{T}(\mathbf{v}_1, \dots, \mathbf{v}_q),
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> $$
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>
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> with $k = q \in \mathbb{N}$.
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>
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> Likewise, $\mathbf{T} \in \mathscr{T}^p_0(V)$ is an **antisymmetric contravariant** $p$-tensor if for all $\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p \in V^*$ we have
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>
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> $$
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> \mathbf{T}(\mathbf{\hat u}_{\pi(1)}, \dots, \mathbf{\hat u}_{\pi(p)}) = \mathrm{sign}(\pi)\mathbf{T}(\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p),
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> $$
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>
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> with $k = p \in \mathbb{N}$.
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This antisymmetry implies that the ordering of the (co)vector arguments in a tensor evaluation only change the sign of the outcome.
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> *Definition 4*: the vector space of antisymmetric covariant $q$-tensors is denoted by $\bigwedge_q(V) \subset \mathscr{T}^0_q(V)$ and the vector space of antisymmetric contravariant $p$-tensors is denoted by $\bigwedge^p(V) \subset \mathscr{T}^p_0(V).$
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Alternatively one may write $\bigwedge_q(V) = V^* \otimes_a \cdots \otimes_a V^*$ and $\bigwedge^p(V) = V \otimes_a \cdots \otimes_a V.$
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It follows from the definitions of symmetric and antisymmetric tensors that for $0$-tensors we have
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$$
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{\bigvee}_0(V) = {\bigvee}^0(V) = {\bigwedge}_0(V) = {\bigwedge}^0(V) = \mathbb{K}.
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$$
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Furthermore, for $1$-tensors we have
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$$
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{\bigvee}_1(V) = {\bigwedge}_1(V) = V^*,
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$$
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and
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$$
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{\bigvee}^1(V) = {\bigwedge}^1(V) = V.
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$$
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## Symmetrisation maps
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The following statements are given with the covariant $q$-tensor without loss of generality.
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> *Definition 5*: the linear **symmetrisation map** $\mathscr{S}: \mathscr{T}^0_q \to \bigvee_q(V)$ is given by
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>
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> $$
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> \mathscr{S}(\mathbf{T})(\mathbf{v}_1, \dots, \mathbf{v}_q) = \frac{1}{q!} \sum_\pi \mathbf{T}(\mathbf{v}_{\pi(1)}, \dots, \mathbf{v}_{\pi(q)}),
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> $$
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>
|
||||
> for all $\mathbf{T} \in \mathscr{T}^0_q(V)$ in which summation runs over all permutations $\pi$ of the set $\{1, \dots, q\}$.
|
||||
|
||||
Let $\mathbf{T} = T_{i_1 \cdots i_q} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q} \in \mathscr{T}^0_q(V)$, then we have $\mathscr{S}(\mathbf{T}) = T_{(i_1 \cdots i_q)} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q} \in \bigvee_q(V)$ with
|
||||
|
||||
$$
|
||||
T_{(i_1 \cdots i_q)} = \frac{1}{q!} \sum_\pi T_{i_{\pi(1)} \cdots i_{\pi(q)}}.
|
||||
$$
|
||||
|
||||
If $\mathbf{T} \in \bigvee_q(V)$ then $\mathbf{T} = \mathscr{S}(\mathbf{T})$. The symmetrisation map is idempotent such that $\mathscr{S} \circ \mathscr{S} = \mathscr{S}.$
|
||||
|
||||
> *Definition 6*: the linear **antisymmetrisation map** $\mathscr{A}: \mathscr{T}^0_q(V) \to \bigwedge_q(V)$ is given by
|
||||
>
|
||||
> $$
|
||||
> \mathscr{A}(\mathbf{T})(\mathbf{v}_1, \dots, \mathbf{v}_q) = \frac{1}{q!} \sum_\pi \mathrm{sign}(\pi) \mathbf{T}(\mathbf{v}_{\pi(1)}, \dots, \mathbf{v}_{\pi(q)}),
|
||||
> $$
|
||||
>
|
||||
> for all $\mathbf{T} \in \mathscr{T}^0_q(V)$ in which summation runs over all permutations $\pi$ of the set $\{1, \dots, q\}$.
|
||||
|
||||
Let $\mathbf{T} = T_{i_1 \cdots i_q} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q} \in \mathscr{T}^0_q(V)$, then we have $\mathscr{A}(\mathbf{T}) = T_{[i_1 \cdots i_q]} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q} \in \bigwedge_q(V)$ with
|
||||
|
||||
$$
|
||||
T_{[i_1 \cdots i_q]} = \frac{1}{q!} \sum_\pi \mathrm{sign}(\pi) T_{i_{\pi(1)} \cdots i_{\pi(q)}}.
|
||||
$$
|
||||
|
||||
If $\mathbf{T} \in \bigwedge_q(V)$ then $\mathbf{T} = \mathscr{A}(\mathbf{T})$. The antisymmetrisation map is idempotent such that $\mathscr{A} \circ \mathscr{A} = \mathscr{A}.$
|
||||
|
||||
## Symmetric product
|
||||
|
||||
The outer product does not preserve (anti)symmetry. For this reason alternative product operators are introduced which preserve (anti)symmetry. The following statements are given with covariant tensors without loss of generality.
|
||||
|
||||
> *Definition 7*: the **symmetric product** between two tensors is defined as
|
||||
>
|
||||
> $$
|
||||
> \mathbf{T} \vee \mathbf{S} = (q+s)! \cdot \mathscr{S}(\mathbf{T} \otimes \mathbf{S}),
|
||||
> $$
|
||||
>
|
||||
> for all $\mathbf{T} \in \mathscr{T}^0_q(V)$ and $\mathbf{S} \in \mathscr{T}^0_s(V)$ with $q,s \in \mathbb{N}$.
|
||||
|
||||
It follows from definition 7 that the symmetric product is associative, bilinear and symmetric. Subsequently, we may write a basis of $\bigvee_q(V)$ as
|
||||
|
||||
$$
|
||||
\mathscr{S}(\mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q}) = \frac{1}{q!} \mathbf{\hat e}^{i_1} \vee \cdots \vee \mathbf{\hat e}^{i_q},
|
||||
$$
|
||||
|
||||
with $\{1 \leq i_1 \leq \dots \leq i_q \leq n\}$.
|
||||
|
||||
Let $\mathbf{T} \in \bigvee_q(V)$ and $\mathbf{S} \in \bigvee_s(V)$ then it follows that
|
||||
|
||||
$$
|
||||
\mathbf{T} \vee \mathbf{S} = \mathbf{S} \vee \mathbf{T}.
|
||||
$$
|
||||
|
||||
> *Definition 8*: the **antisymmetric product** between two tensors is defined as
|
||||
>
|
||||
> $$
|
||||
> \mathbf{T} \wedge \mathbf{S} = (q+s)! \cdot \mathscr{A}(\mathbf{T} \otimes \mathbf{S}),
|
||||
> $$
|
||||
>
|
||||
> for all $\mathbf{T} \in \mathscr{T}^0_q(V)$ and $\mathbf{S} \in \mathscr{T}^0_s(V)$ with $q,s \in \mathbb{N}$.
|
||||
|
||||
It follows from definition 8 that the antisymmetric product is associative, bilinear and antisymmetric. Subsequently, we may write a basis of $\bigwedge_q(V)$ as
|
||||
|
||||
$$
|
||||
\mathscr{A}(\mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q}) = \frac{1}{q!} \mathbf{\hat e}^{i_1} \wedge \cdots \wedge \mathbf{\hat e}^{i_q},
|
||||
$$
|
||||
|
||||
with $\{1 \leq i_1 < \dots < i_q \leq n\}$.
|
||||
|
||||
Let $\mathbf{T} \in \bigwedge_q(V)$ and $\mathbf{S} \in \bigwedge_s(V)$ then it follows that
|
||||
|
||||
$$
|
||||
\mathbf{T} \wedge \mathbf{S} = (-1)^{qs} \mathbf{S} \wedge \mathbf{T}.
|
||||
$$
|
||||
|
||||
> *Theorem 1*: the dimension of the vector space of symmetric covariant $q$-tensors is given by
|
||||
>
|
||||
> $$
|
||||
> \dim \Big({\bigvee}_q(V) \Big) = \binom{n+q-1}{q},
|
||||
> $$
|
||||
>
|
||||
> and for antisymmetric covariant $q$-tensors the dimension is given by
|
||||
>
|
||||
> $$
|
||||
> \dim \Big({\bigwedge}_q(V) \Big) = \binom{n}{q}.
|
||||
> $$
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
An interesting result of the definition of the symmetric and antisymmetric product is given in the theorem below.
|
||||
|
||||
> *Theorem 2*: let $\mathbf{\hat u}_{1,2} \in V^*$ be covectors, the symmetric product of $\mathbf{\hat u}_1$ and $\mathbf{\hat u}_2$ may be given by
|
||||
>
|
||||
> $$
|
||||
> (\mathbf{\hat u}_1 \vee \mathbf{\hat u}_2)(\mathbf{v}_1, \mathbf{v}_2) = \mathrm{perm}\big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j)\big),
|
||||
> $$
|
||||
>
|
||||
> for all $(\mathbf{v}_1, \mathbf{v}_2) \in V \times V$ with $(i,j)$ denoting the entry of the matrix over which the permanent is taken.
|
||||
>
|
||||
> The antisymmetric product of $\mathbf{\hat u}_1$ and $\mathbf{\hat u}_2$ may be given by
|
||||
>
|
||||
> $$
|
||||
> (\mathbf{\hat u}_1 \wedge \mathbf{\hat u}_2)(\mathbf{v}_1, \mathbf{v}_2) = \det \big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j) \big),
|
||||
> $$
|
||||
>
|
||||
> for all $(\mathbf{v}_1, \mathbf{v}_2) \in V \times V$ with $(i,j)$ denoting the entry of the matrix over which the determinant is taken.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
In some literature theorem 2 is used as definition for the symmetric and antisymmetric product from which the relation with the symmetrisation maps can be proven. Either method is valid, however it has been chosen that defining the products in terms of the symmetrisation maps is more general.
|
||||
|
|
@ -0,0 +1,97 @@
|
|||
# Tensor transformations
|
||||
|
||||
We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n,$ a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}_{i=1}^n$ and a pseudo inner product $\bm{g}$ on $V.$
|
||||
|
||||
Let us introduce a different basis $\{\mathbf{f}_i\}_{i=1}^n$ of $V$ with a corresponding dual basis $\{\mathbf{\hat f}^i\}_{i=1}^n$ of $V^*$ which are related to the former basis $\{\mathbf{e}_i\}_{i=1}^n$ by
|
||||
|
||||
$$
|
||||
\mathbf{f}_j = A^i_j \mathbf{e}_i,
|
||||
$$
|
||||
|
||||
so that $\mathbf{\hat e}^i = A^i_j \mathbf{\hat f}^j$.
|
||||
|
||||
## Transformation of tensors
|
||||
|
||||
Recall from the section of [tensor-formalism]() that a holor depends on the chosen basis, but the corresponding tensor itself does not. This implies that holors transform in a particular way under a change of basis, which is characteristic for tensors.
|
||||
|
||||
> *Theorem 1*: let $\mathbf{T} \in \mathscr{T}^p_q(V)$ be a tensor with $p=q=1$ without loss of generality and $B = A^{-1}$. Then $\mathbf{T}$ may be decomposed into
|
||||
>
|
||||
> $$
|
||||
> \begin{align*}
|
||||
> \mathbf{T} &= T^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j, \\
|
||||
> &= \overline T^i_j \mathbf{f}_i \otimes \mathbf{\hat f}^j,
|
||||
> \end{align*}
|
||||
> $$
|
||||
>
|
||||
> with the holors related by
|
||||
>
|
||||
> $$
|
||||
> \overline T^i_j = B^i_k A^j_l T^k_l.
|
||||
> $$
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
The homogeneous nature of the tensor transformation implies that a holor equation of the form $T^i_j = 0$ holds relative to any basis if it holds relative to a particular one.
|
||||
|
||||
## Transformation of volume forms
|
||||
|
||||
> *Lemma 1*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form with
|
||||
>
|
||||
> $$
|
||||
> \begin{align*}
|
||||
> \bm{\mu} &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\
|
||||
> &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n},
|
||||
> \end{align*}
|
||||
> $$
|
||||
>
|
||||
> then we have
|
||||
>
|
||||
> $$
|
||||
> \overline \mu_{j_1 \dots j_n} = A^{i_1}_{j_1} \cdots A^{i_n}_{j_n} \mu_{i_1 \dots i_n} = \mu_{j_1 \dots j_n} \det (A).
|
||||
> $$
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
Then $\det(A)$ is the volume scaling factor of the transformation with $A$. So that if $\bm{\mu}(\mathbf{e}_1, \dots, \mathbf{e}_n) = 1$, then $\bm{\mu}(\mathbf{f}_1, \dots, \mathbf{f}_n) = \det(A).$
|
||||
|
||||
> *Theorem 2*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form with
|
||||
>
|
||||
> $$
|
||||
> \begin{align*}
|
||||
> \bm{\mu} &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\
|
||||
> &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n},
|
||||
> \end{align*}
|
||||
> $$
|
||||
>
|
||||
> and if we define
|
||||
>
|
||||
> $$
|
||||
> \overline \mu_{i_1 \dots i_n} \overset{\text{def}}{=} \frac{1}{\det (A)} A^{j_1}_{i_1} \cdots A^{j_n}_{i_n} \mu_{j_1 \dots j_n},
|
||||
> $$
|
||||
>
|
||||
> then $\mu_{i_1 \dots i_n} = \overline \mu_{i_1 \dots i_n} = [i_1, \dots, i_n]$ is an invariant holor.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
## Transformation of Levi-Civita form
|
||||
|
||||
> *Theorem 3*: let $\bm{\epsilon} \in \bigwedge_n(V)$ be the Levi-Civita tensor with
|
||||
>
|
||||
> $$
|
||||
> \begin{align*}
|
||||
> \bm{\epsilon} &= \epsilon_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\
|
||||
> &= \overline \epsilon_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n},
|
||||
> \end{align*}
|
||||
> $$
|
||||
>
|
||||
> then $\epsilon_{i_1 \dots i_n} = \overline \epsilon_{i_1 \dots i_n}$ is an invariant holor.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
131
docs/mathematics/linear-algebra/tensors/volume-forms.md
Normal file
131
docs/mathematics/linear-algebra/tensors/volume-forms.md
Normal file
|
|
@ -0,0 +1,131 @@
|
|||
# Volume forms
|
||||
|
||||
We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n,$ a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}_{i=1}^n$ and a pseudo inner product $\bm{g}$ on $V.$
|
||||
|
||||
## n-forms
|
||||
|
||||
> *Definition 1*: let $\bm{\mu} \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$, if
|
||||
>
|
||||
> $$
|
||||
> \bm{\mu}(\mathbf{e}_1, \dots, \mathbf{e}_n) = 1,
|
||||
> $$
|
||||
>
|
||||
> then $\bm{\mu}$ is the **unit volume form** with respect to the basis $\{\mathbf{e}_i\}$.
|
||||
|
||||
Note that $\dim \bigwedge_n(V) = 1$ and consequently if $\bm{\mu}_1, \bm{\mu}_2 \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$, then $\bm{\mu}_1 = \lambda \bm{\mu}_2$ with $\lambda \in \mathbb{K}$.
|
||||
|
||||
> *Proposition 1*: the unit volume form $\bm{\mu} \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$ may be given by
|
||||
>
|
||||
> $$
|
||||
> \begin{align*}
|
||||
> \bm{\mu} &= \mathbf{\hat e}^1 \wedge \dots \wedge \mathbf{\hat e}^n, \\
|
||||
> &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \dots \otimes \mathbf{\hat e}^{i_n},
|
||||
> \end{align*}
|
||||
> $$
|
||||
>
|
||||
> with $\mu_{i_1 \dots i_n} = [i_1, \dots, i_n]$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
The normalisation of the unit volume form $\bm{\mu}$ requires a basis. Consequently, the identification $\mu_{i_1 \dots i_n} = [i_1, \dots, i_n]$ holds only relative to the basis.
|
||||
|
||||
> *Definition 2*: let $(V, \bm{\mu})$ denote the vector space $V$ endowed with an **oriented volume form** $\bm{\mu}$. For $\bm{\mu} > 0$ we have a positive orientation of $(V, \bm{\mu})$ and for $\bm{\mu} < 0$ we have a negative orientation of $(V, \bm{\mu})$.
|
||||
|
||||
For a vector space with an oriented volume $(V, \bm{\mu})$ we may write
|
||||
|
||||
$$
|
||||
\bm{\mu} = \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n},
|
||||
$$
|
||||
|
||||
or, equivalently
|
||||
|
||||
$$
|
||||
\bm{\mu} = \mu_{|i_1 \dots i_n|} \mathbf{\hat e}^{i_1} \wedge \cdots \wedge \mathbf{\hat e}^{i_n},
|
||||
$$
|
||||
|
||||
by convention, to resolve ambiguity with respect to the meaning of $\mu_{i_1 \dots i_n}$ without using another symbol or extra accents.
|
||||
|
||||
Using theorem 2 in the section of [tensor symmetries]() we may state the following.
|
||||
|
||||
> *Proposition 2*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form, then we have
|
||||
>
|
||||
> $$
|
||||
> \bm{\mu}(\mathbf{v}_1, \dots, \mathbf{v}_n) = \det \big(\mathbf{k}(\mathbf{\hat e}^i, \mathbf{v}_j) \big),
|
||||
> $$
|
||||
>
|
||||
> for all $\mathbf{v}_1, \dots, \mathbf{v}_n \in V$ with $(i,j)$ denoting the entry of the matrix over which the determinant is taken.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
Which reveals the role of the Kronecker tensor and thus the role of the dual space in the definition of $\bm{\mu}$. We may also conclude that an oriented volume $\bm{\mu} \in \bigwedge_n(V)$ on a vector space $V$ does not require an inner product.
|
||||
|
||||
From proposition 2 it may also be observed that within a geometrical context the oriented volume form may represent the area of a parallelogram in $n=2$ or the volume of a parallelepiped in $n=3$, span by its basis.
|
||||
|
||||
## (n - k)-forms
|
||||
|
||||
> *Definition 3*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form and let $\mathbf{u}_1, \dots, \mathbf{u}_k \in V$ with $k \in \mathbb{N}[k < n]$. Let the $(n-k)$-form $\bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k \in \bigwedge_{n-k}(V)$ be defined as
|
||||
>
|
||||
> $$
|
||||
> \bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k(\mathbf{v}_{k+1}, \dots, \mathbf{v}_n) = \bm{\mu}(\mathbf{u}_1, \dots, \mathbf{u}_k, \mathbf{v}_{k+1}, \dots, \mathbf{v}_n),
|
||||
> $$
|
||||
>
|
||||
> for all $\mathbf{v}_{k+1}, \dots, \mathbf{v}_n \in V$ with $\lrcorner$ the insert operator.
|
||||
|
||||
It follows that $(n-k)$-form $\bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k \in \bigwedge_{n-k}(V)$ can be written as
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k &= u_1^{i_1} \cdots u_k^{i_k} (\bm{\mu} \lrcorner \mathbf{e}_{i_1} \lrcorner \dots \lrcorner \mathbf{e}_{i_k}), \\
|
||||
&= u_1^{i_1} \cdots u_k^{i_k} \mu_{i_1 \dots i_n} (\mathbf{\hat e}^{i_{k+1}} \wedge \cdots \wedge \mathbf{\hat e}^{i_{n}}),
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
for $\mathbf{u}_1, \dots, \mathbf{u}_k \in V$ with $k \in \mathbb{N}[k < n]$ and decomposition by $\mathbf{u}_q = u_q^{i_q} \mathbf{e}_{i_q}$ for $q \in \mathbb{N}[q \leq k]$.
|
||||
|
||||
If we have a unit volume form $\bm{\mu}$ with respect to $\{\mathbf{e}_i\}$ then
|
||||
|
||||
$$
|
||||
\bm{\mu}\lrcorner\mathbf{e}_1 \lrcorner \dots \lrcorner \mathbf{e}_k = \mathbf{\hat e}^{i_{k+1}} \wedge \cdots \wedge \mathbf{e}^{i_n},
|
||||
$$
|
||||
|
||||
for $k \in \mathbb{N}[k < n]$.
|
||||
|
||||
## Levi-Civita form
|
||||
|
||||
> *Definition 4*: let $(V, \bm{\mu})$ be a vector space with a unit volume form with invariant holor. Let $\bm{\epsilon} \in \bigwedge_n(V)$ be the **Levi-Civita tensor** which is the unique unit volume form of positive orientation defined as
|
||||
>
|
||||
> $$
|
||||
> \bm{\epsilon} = \sqrt{g} \bm{\mu},
|
||||
> $$
|
||||
>
|
||||
> with $g \overset{\text{def}}{=} \det (G)$, the determinant of the [Gram matrix]().
|
||||
|
||||
Therefore, if we decompose the Levi-Civita tensor by
|
||||
|
||||
$$
|
||||
\bm{\epsilon} = \epsilon_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \dots \otimes \mathbf{\hat e}^{i_n} = \epsilon_{|i_1 \dots i_n|} \mathbf{\hat e}^{i_1} \wedge \dots \wedge \mathbf{\hat e}^{i_n},
|
||||
$$
|
||||
|
||||
then we have $\epsilon_{i_1 \dots i_n} = \sqrt{g} \mu_{i_1 \dots i_n}$ and $\epsilon_{|i_1 \dots i_n|} = \sqrt{g}$.
|
||||
|
||||
> *Theorem 2*: let $(V, \bm{\mu})$ be a vector space with a unit volume form with invariant holor. Let $\mathbf{g}(\bm{\epsilon}) \in \bigwedge^n(V)$ be the **reciprocal Levi-Civita tensor** which is given by
|
||||
>
|
||||
> $$
|
||||
> \mathbf{g}(\bm{\epsilon}) = \frac{1}{\sqrt{g}} \bm{\mu}.
|
||||
> $$
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
We may decompose the reciprocal Levi-Civita tensor by
|
||||
|
||||
$$
|
||||
\mathbf{g}(\bm{\epsilon}) = \epsilon^{i_1 \dots i_n} \mathbf{e}_{i_1} \otimes \cdots \otimes \mathbf{e}_{i_n} = \epsilon^{|i_1 \dots i_n|} \mathbf{e}_{i_1} \wedge \cdots \wedge \mathbf{e}_{i_n},
|
||||
$$
|
||||
|
||||
then we have $\epsilon^{i_1 \dots i_n} = \frac{1}{\sqrt{g}} \mu^{i_1 \dots i_n}$ and $\epsilon^{|i_1 \dots i_n|} = \frac{1}{\sqrt{g}}$.
|
||||
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Add table
Add a link
Reference in a new issue