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Updated theorem in tensor symmetries.

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Luc Bijl 2024-05-14 18:04:56 +02:00
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docs/en/mathematics/linear-algebra/tensors/tensor-symmetries.md
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@ -170,21 +170,21 @@ $$
An interesting result of the definition of the (anti)symmetric product is given in the theorem below.
> *Theorem 2*: let $\mathbf{T} \in \mathscr{T}^0_q(V)$ and $\mathbf{S} \in \mathscr{T}^0_s(V)$ be tensors with $q,s \in \mathbb{N}$, the symmetric product of $\mathbf{T}$ and $\mathbf{S}$ may be given by
> *Theorem 2*: let $\mathbf{\hat u}_{1,2} \in V^*$ be covectors, the symmetric product of $\mathbf{\hat u}_1$ and $\mathbf{\hat u}_2$ may be given by
>
> $$
> (\mathbf{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)),
> (\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),
> $$
>
> for all $(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) \in V^{q+s}$ with $(j,k)$ denoting the entry of the matrix over which the permanent is taken.
> for all $(\mathbf{v}_1, \mathbf{v}_2) \in V \times V$ with $(i,j)$ denoting the entry of the matrix over which the permanent is taken.
>
> The antisymmetric product of $\mathbf{T}$ and $\mathbf{S}$ may be given by
> The antisymmetric product of $\mathbf{\hat u}_1$ and $\mathbf{\hat u}_2$ may be given by
>
> $$
> (\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)),
> (\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 $(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) \in V^{q+s}$ with $(j,k)$ denoting the entry of the matrix over which the determinant is taken.
> for all $(\mathbf{v}_1, \mathbf{v}_2) \in V \times V$ with $(j,k)$ denoting the entry of the matrix over which the determinant is taken.
??? note "*Proof*:"