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Luc Bijl 2025-09-20 21:44:37 +02:00
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docs/mathematics/linear-algebra/tensors/tensor-symmetries.md
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@ -1,6 +1,6 @@
# 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 corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}$ and a pseudo inner product $\bm{g}$ on $V.$
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 $\{\mathbf{\hat e}^i\}$, a field F and a pseudo inner product $\bm{g}$ on $V$.
## Symmetric tensors
@ -53,7 +53,7 @@ Alternatively one may write $\bigwedge_q(V) = V^* \otimes_a \cdots \otimes_a V^*
It follows from the definitions of symmetric and antisymmetric tensors that for $0$-tensors we have
$$
{\bigvee}_0(V) = {\bigvee}^0(V) = {\bigwedge}_0(V) = {\bigwedge}^0(V) = \mathbb{K}.
{\bigvee}_0(V) = {\bigvee}^0(V) = {\bigwedge}_0(V) = {\bigwedge}^0(V) = F.
$$
Furthermore, for $1$-tensors we have
@ -188,6 +188,42 @@ An interesting result of the definition of the symmetric and antisymmetric produ
??? note "*Proof*:"
Will be added later.
We have for the symmetric case
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.
$$
\begin{align*}
\mathbf{\hat u}_1 \vee \mathbf{\hat u}_2 &= 2 \mathscr{S}(\mathbf{\hat u}_1 \otimes \mathbf{\hat u}_2),\\
&= u^1_{(i} u^2_{j)} \mathbf{\hat e}_i \otimes \mathbf{\hat e}^j,\\
&= u^1_i u^2_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j + u^2_i u^1_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j,
\end{align*}
$$
such that
$$
\begin{align*}
\mathbf{\hat u}_1 \vee \mathbf{\hat u}_2 (\mathbf{v}_1, \mathbf{v}_2) &= u^1_i u^2_j v_1^i v_2^j + u^2_i u^1_j v_1^i v_2^j,\\
&= \mathrm{perm}\big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j)\big).
\end{align*}
$$
And we have for the antisymmetric case
$$
\begin{align*}
\mathbf{\hat u}_1 \wedge \mathbf{\hat u}_2 &= 2 \mathscr{A}(\mathbf{\hat u}_1 \otimes \mathbf{\hat u}_2),\\
&= u^1_{[i} u^2_{j]} \mathbf{\hat e}_i \otimes \mathbf{\hat e}^j,\\
&= u^1_i u^2_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j - u^2_i u^1_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j,
\end{align*}
$$
such that
$$
\begin{align*}
\mathbf{\hat u}_1 \wedge \mathbf{\hat u}_2 (\mathbf{v}_1, \mathbf{v}_2) &= u^1_i u^2_j v_1^i v_2^j - u^2_i u^1_j v_1^i v_2^j,\\
&= \det\big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j)\big).
\end{align*}
$$
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.