From cc9c62307ae68e9c0cf342aad6825abea6688f0e Mon Sep 17 00:00:00 2001 From: Luc Date: Tue, 14 May 2024 17:51:53 +0200 Subject: [PATCH] Finished tensor symmetries. --- .../tensors/tensor-formalism.md | 12 +- .../tensors/tensor-symmetries.md | 175 +++++++++++++++++- 2 files changed, 174 insertions(+), 13 deletions(-) diff --git a/docs/en/mathematics/linear-algebra/tensors/tensor-formalism.md b/docs/en/mathematics/linear-algebra/tensors/tensor-formalism.md index 431a4d1..170ddf7 100644 --- a/docs/en/mathematics/linear-algebra/tensors/tensor-formalism.md +++ b/docs/en/mathematics/linear-algebra/tensors/tensor-formalism.md @@ -55,9 +55,9 @@ $$ The outer product is associative and distributive with respect to addition and scalar multiplication, but not commutative. -Note that although the same symbol is used for the outer product and the denotion of a tensor space, these are not equivalent. +Note that although the same symbol is used for the outer product and the denotation of a tensor space, these are not equivalent. -For the following statements we take $p=q=r=s=1$ without loss of generality. +The following statements are given with $p=q=r=s=1$ without loss of generality. > *Definition 4*: the mixed $(p, q)$-tensor $\mathbf{e}_i \otimes \mathbf{\hat e}^j \in \mathscr{T}_q^p(V)$ is defined as > @@ -141,7 +141,7 @@ We have from theorem 2 that the outer product of two tensors yields another tens > 2. for all $\mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \overline{\bm{g}}(\mathbf{v}, \mathbf{u})$, > 3. for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $\lambda, \mu \in \mathbb{K}: \;\bm{g}(\mathbf{u}, \lambda \mathbf{v} + \mu \mathbf{w}) = \lambda \bm{g}(\mathbf{u}, \mathbf{v}) + \mu \bm{g}(\mathbf{u}, \mathbf{w}).$ -It may be observed that $\bm{g} \in \mathscr{T}_2^0$. Unlike the Kronecker tensor, the existance of an inner product is never implied. +It may be observed that $\bm{g} \in \mathscr{T}_2^0$. Unlike the Kronecker tensor, the existence of an inner product is never implied. > *Definition 6*: let $G$ be the Gram matrix with its components $G \overset{\text{def}}= (g_{ij})$ defined as > @@ -210,7 +210,7 @@ $$ Since $u^i, v^j \in \mathbb{K}$ are arbitrary it follows that $\text{g}_{ij} = g_{ij}$. -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**. +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**. 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 @@ -232,10 +232,10 @@ with $u^j = g^{ij} u_i$. > \mathbf{g}^{-1}(\mathbf{\hat e}^i) = g^{ij} \mathbf{e}_j. > $$ > -> Likewise the basis $\{\mathbf{\hat e}^i\}$ of $V^*$ induces a **reciprocal dual basis** $\{\mathbf{g}(\mathbf{e}_i)\}$ of $V^*$ given by +> Likewise, the basis $\{\mathbf{\hat e}^i\}$ of $V^*$ induces a **reciprocal dual basis** $\{\mathbf{g}(\mathbf{e}_i)\}$ of $V^*$ given by > > $$ > \mathbf{g}(\mathbf{e}^i) = g_{ij} \mathbf{\hat e}^j. > $$ -Sofar, a vector space $V$ and its associated dual space $V^*$ have been introduced as a priori independent entities. An inner product provides us with an explicit mechanism to construct a bijective linear mapping associated with each vector by virtue of the metric. \ No newline at end of file +So far, a vector space $V$ and its associated dual space $V^*$ have been introduced as a priori independent entities. An inner product provides us with an explicit mechanism to construct a bijective linear mapping associated with each vector by virtue of the metric. \ No newline at end of file diff --git a/docs/en/mathematics/linear-algebra/tensors/tensor-symmetries.md b/docs/en/mathematics/linear-algebra/tensors/tensor-symmetries.md index 25f4e94..75146c1 100644 --- a/docs/en/mathematics/linear-algebra/tensors/tensor-symmetries.md +++ b/docs/en/mathematics/linear-algebra/tensors/tensor-symmetries.md @@ -1,30 +1,191 @@ # Tensor symmetries -We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$, a pseudo inner product $\bm{g}$ on $V$ and a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}.$ +We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n,$ a pseudo inner product $\bm{g}$ on $V$ and a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}.$ ## Symmetric tensors -> *Definition 1*: let $\pi = [\pi(1), \dots, \pi(k)]$ be any permutation of labels $\{1, \dots, k\}$, then $\mathbf{T} \in \mathscr{T}^0_q(V)$ is a symmetric covariant tensor if for all $\mathbf{v}_1, \dots, \mathbf{v}_q \in V$ we have +> *Definition 1*: let $\pi = [\pi(1), \dots, \pi(k)]$ be any permutation of the set $\{1, \dots, k\}$, then $\mathbf{T} \in \mathscr{T}^0_q(V)$ is a **symmetric covariant** $q$-tensor if for all $\mathbf{v}_1, \dots, \mathbf{v}_q \in V$ we have > > $$ > \mathbf{T}(\mathbf{v}_{\pi(1)}, \dots, \mathbf{v}_{\pi(q)}) = \mathbf{T}(\mathbf{v}_1, \dots, \mathbf{v}_q), > $$ > -> with $k = q$. +> with $k = q \in \mathbb{N}$. > -> Likewise $\mathbf{T} \in \mathscr{T}^p_0(V)$ is called a symmetric contravariant tensor if for all $\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p \in V^*$ we have +> Likewise, $\mathbf{T} \in \mathscr{T}^p_0(V)$ is a **symmetric contravariant** $p$-tensor if for all $\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p \in V^*$ we have > > $$ > \mathbf{T}(\mathbf{\hat u}_{\pi(1)}, \dots, \mathbf{\hat u}_{\pi(p)}) = \mathbf{T}(\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p), > $$ > -> with $k = p$. +> with $k = p \in \mathbb{N}$. This symmetry implies that the ordering of the (co)vector arguments in a tensor evaluation do not affect the outcome. -> *Definition 2*: the vector space of symmetric covariant $q$-tensors is denoted by $\bigvee_q(V) \subset \mathscr{T}^0_q(V)$ and the vector space of symmetric contravariant $p$-tensors is denoted by $\bigwedge^p(V) \subset \mathscr{T}^p_0(V).$ +> *Definition 2*: the vector space of symmetric covariant $q$-tensors is denoted by $\bigvee_q(V) \subset \mathscr{T}^0_q(V)$ and the vector space of symmetric contravariant $p$-tensors is denoted by $\bigvee^p(V) \subset \mathscr{T}^p_0(V).$ -Alternatively one may write $\bigvee_q(V) = V^* \otimes_s \cdots \otimes_s V^*$ and $\bigwedge^p(V) = V \otimes_s \cdots \otimes_s V$. +Alternatively one may write $\bigvee_q(V) = V^* \otimes_s \cdots \otimes_s V^*$ and $\bigvee^p(V) = V \otimes_s \cdots \otimes_s V.$ ## Antisymmetric tensors +> *Definition 3*: let $\pi = [\pi(1), \dots, \pi(k)]$ be any permutation of the set $\{1, \dots, k\}$, then $\mathbf{T} \in \mathscr{T}^0_q(V)$ is an **antisymmetric covariant** $q$-tensor if for all $\mathbf{v}_1, \dots, \mathbf{v}_q \in V$ we have +> +> $$ +> \mathbf{T}(\mathbf{v}_{\pi(1)}, \dots, \mathbf{v}_{\pi(q)}) = \mathrm{sign}(\pi) \mathbf{T}(\mathbf{v}_1, \dots, \mathbf{v}_q), +> $$ +> +> with $k = q \in \mathbb{N}$. +> +> Likewise, $\mathbf{T} \in \mathscr{T}^p_0(V)$ is an **antisymmetric contravariant** $p$-tensor if for all $\mathbf{\hat u}_1, \dots, \mathbf{\hat u}_p \in V^*$ we have +> +> $$ +> \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), +> $$ +> +> with $k = p \in \mathbb{N}$. + +This antisymmetry implies that the ordering of the (co)vector arguments in a tensor evaluation only change the sign of the outcome. + +> *Definition 4*: the vector space of antisymmetric covariant $q$-tensors is denoted by $\bigwedge_q(V) \subset \mathscr{T}^0_q(V)$ and the vector space of antisymmetric contravariant $p$-tensors is denoted by $\bigwedge^p(V) \subset \mathscr{T}^p_0(V).$ + +Alternatively one may write $\bigwedge_q(V) = V^* \otimes_a \cdots \otimes_a V^*$ and $\bigwedge^p(V) = V \otimes_a \cdots \otimes_a V.$ + +It follows from the definitions of symmetric and antisymmetric tensors that for $0$-tensors we have + +$$ + {\bigvee}_0(V) = {\bigvee}^0(V) = {\bigwedge}_0(V) = {\bigwedge}^0(V) = \mathbb{K}. +$$ + +Furthermore, for $1$-tensors we have + +$$ + {\bigvee}_1(V) = {\bigwedge}_1(V) = V^*, +$$ + +and + +$$ + {\bigvee}^1(V) = {\bigwedge}^1(V) = V. +$$ + +## Symmetrisation maps + +The following statements are given with the covariant $q$-tensor without loss of generality. + +> *Definition 5*: the linear **symmetrisation map** $\mathscr{S}: \mathscr{T}^0_q \to \bigvee_q(V)$ is given by +> +> $$ +> \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)}), +> $$ +> +> for all $\mathbf{T} \in \mathscr{T}^0_q(V)$ in which summation runs over all permutations $\pi$ of the set $\{1, \dots, q\}$. + +Let $\mathbf{T} = T_{i_1 \cdots i_q} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q} \in \mathscr{T}^0_q(V)$, then we have $\mathscr{S}(\mathbf{T}) = T_{(i_1 \cdots i_q)} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_q} \in \bigvee_q(V)$ with + +$$ + 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}. +> $$ + +??? note "*Proof*:" + + Will be added later. + +An interesting result of the definition of the (anti)symmetric product is given in the theorem below. + +> *Theorem 2*: let $\mathbf{T} \in \mathscr{T}^0_q(V)$ and $\mathbf{S} \in \mathscr{T}^0_s(V)$ be tensors with $q,s \in \mathbb{N}$, the symmetric product of $\mathbf{T}$ and $\mathbf{S}$ may be given by +> +> $$ +> (\mathbf{T} \vee \mathbf{S})(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) = T_{i_1 \cdots i_q} S_{i_{q+1} \cdots i_{q+s}} \mathrm{perm}(\mathbf{k}(\mathbf{\hat e}^{i_j}, \mathbf{v}_k)), +> $$ +> +> for all $(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) \in V^{q+s}$ with $(j,k)$ denoting the entry of the matrix over which the permanent is taken. +> +> The antisymmetric product of $\mathbf{T}$ and $\mathbf{S}$ may be given by +> +> $$ +> (\mathbf{T} \vee \mathbf{S})(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) = T_{i_1 \cdots i_q} S_{i_{q+1} \cdots i_{q+s}} \det(\mathbf{k}(\mathbf{\hat e}^{i_j}, \mathbf{v}_k)), +> $$ +> +> for all $(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) \in V^{q+s}$ with $(j,k)$ denoting the entry of the matrix over which the determinant is taken. + +??? note "*Proof*:" + + Will be added later. \ No newline at end of file