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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 @@ $$
An interesting result of the definition of the (anti)symmetric product is given in the theorem below. An interesting result of the definition of the (anti)symmetric product is given in the theorem below.
> *Theorem 2*: let $\mathbf{T} \in \mathscr{T}^0_q(V)$ and $\mathbf{S} \in \mathscr{T}^0_s(V)$ be tensors with $q,s \in \mathbb{N}$, the symmetric product of $\mathbf{T}$ and $\mathbf{S}$ may be given by > *Theorem 2*: let $\mathbf{\hat u}_{1,2} \in V^*$ be covectors, the symmetric product of $\mathbf{\hat u}_1$ and $\mathbf{\hat u}_2$ may be given by
> >
> $$ > $$
> (\mathbf{T} \vee \mathbf{S})(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) = T_{i_1 \cdots i_q} S_{i_{q+1} \cdots i_{q+s}} \mathrm{perm}(\mathbf{k}(\mathbf{\hat e}^{i_j}, \mathbf{v}_k)), > (\mathbf{\hat u}_1 \vee \mathbf{\hat u}_2)(\mathbf{v}_1, \mathbf{v}_2) = \mathrm{perm}\big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j)\big),
> $$ > $$
> >
> for all $(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) \in V^{q+s}$ with $(j,k)$ denoting the entry of the matrix over which the permanent is taken. > for all $(\mathbf{v}_1, \mathbf{v}_2) \in V \times V$ with $(i,j)$ denoting the entry of the matrix over which the permanent is taken.
> >
> The antisymmetric product of $\mathbf{T}$ and $\mathbf{S}$ may be given by > The antisymmetric product of $\mathbf{\hat u}_1$ and $\mathbf{\hat u}_2$ may be given by
> >
> $$ > $$
> (\mathbf{T} \vee \mathbf{S})(\mathbf{v}_1, \dots, \mathbf{v}_{q+s}) = T_{i_1 \cdots i_q} S_{i_{q+1} \cdots i_{q+s}} \det(\mathbf{k}(\mathbf{\hat e}^{i_j}, \mathbf{v}_k)), > (\mathbf{\hat u}_1 \wedge \mathbf{\hat u}_2)(\mathbf{v}_1, \mathbf{v}_2) = \det \big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j) \big),
> $$ > $$
> >
> 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*:" ??? note "*Proof*:"