Finished tensor symmetries.

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
 
@@ -232,7 +232,7 @@ 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.

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.