Symmetric and alternating tensors

An action of the symmetric group on $T^{k} (M)$

For any $R$ -module $M$ there is a left action of the symmetric group $S_{k}$ on the $k$ -fold direct product $M \times \dots \times M$ , given by permuting the factors:

σ (m_{1}, \dots, m_{k}) = (m_{σ^{- 1} (1)}, \dots, m_{σ^{- 1} (k)}) .

(The reason for $σ^{- 1}$ is to make this a left group action.) This balanced, multilinear map corresponds to a left group action of $S_{k}$ on $T^{k} (M)$ which is defined on simple tensors by

σ (m_{1} \otimes \dots \otimes m_{k}) = m_{σ^{- 1} (1)} \otimes \dots \otimes m_{σ^{- 1} (k)} .

For example,

(1, 3, 2) \cdot (m_{1} \otimes m_{2} \otimes m_{3}) = m_{2} \otimes m_{3} \otimes m_{1} .

Definition of symmetric and alternating tensors

Suppose $R$ is a commutative ring and $M$ is an $R$ -module. An element $z \in T^{k} (M)$ is called:

a symmetric $k$ -tensor if $σ z = z$ for all $σ \in S_{k}$
an alternating $k$ -tensor if $σ z = sign (σ) z$ for all $σ \in S_{k}$ .

For example, the elements $m \otimes m$ and $m_{1} \otimes m_{2} + m_{2} \otimes m_{1}$ are symmetric 2-tensors, while the element $m_{1} \otimes m_{2} - m_{2} \otimes m_{1}$ is an alternating 2-tensor. The 2-tensor $m_{1} \otimes m_{2}$ is neither symmetric nor alternating.

The collection of symmetric $k$ -tensors forms a submodule of $T^{k} (M)$ , as does the collection of alternating $k$ -tensors. These submodules are denoted ... nothing. At least as far as I can tell, Dummit & Foote doesn't use any notation for them.

One can prove that these submodules are stable under this action of $S_{k}$ , hence there is an induced action on the quotient modules $S^{k} (M)$ and $\overset{k}{⋀} (M)$ . Moreover, we have:

$σ w = w$ for every $w \in S^{k} (M)$
$σ w = sign (σ) w$ for every $w \in \overset{k}{⋀} (M)$

These actions seem identical to that of $S_{k}$ on the submodules of $T^{k} (M)$ consisting of the symmetric and alternating tensors, respectively. Let's investigate this a bit further.

Symmetrization and skew-symmetrization

For any $k$ -tensor $z \in T^{k} (M)$ , define

\begin{aligned} Sym (z) & = \sum_{σ \in S_{k}} σ z \\ Alt (z) & = \sum_{σ \in S_{k}} sign (σ) σ z . \end{aligned}

It is straightforward to verify these $k$ -tensors are symmetric and alternating, respectively. We call them the symmetrization and skew-symmetrization of $z$ .

One can verify that we have actually defined $R$ -module morphisms $Sym : T^{k} (M) \to T^{k} (M)$ and $Alt : T^{k} (M) \to T^{k} (M)$ whose images lie in the submodule of symmetric and alternating tensors, respectively.

Note that if $z$ is a symmetric $k$ -tensor, then

Sym (z) = k! \cdot z .

Similarly, if $z$ is an alternating $k$ -tensor, then

Alt (z) = k! \cdot z .

From these observations, it is not too difficult to prove the following:

Proposition

Suppose $R$ is a commutative ring and $M$ is an $R$ -module. If $k!$ is a unit in $R$ , then there is an $R$ -module isomorphism between $S^{k} (M)$ (respectively, $\overset{k}{⋀} (M)$ ) and the submodule of $T^{k} (M)$ consisting of all symmetric $k$ -tensors (respectively, alternating $k$ -tensors).

As an upshot, so long as $k!$ is a unit in $R$ , we can write $Sym (T^{k} (M))$ and $Alt (T^{k} (M))$ for the submodules of symmetric and alternating $k$ -tensors, respectively.

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Noetherian modules

An action of the symmetric group on Tk(M)

Symmetrization and skew-symmetrization

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An action of the symmetric group on $T^{k} (M)$