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 $S_{i}$ on the $i$ -fold direct product $M \times \dots \times M$ , given by permuting the factors:

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

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

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

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 $z \in T^{i} (M)$ is called:

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

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 $i$ -tensors forms a submodule of $T^{i} (M)$ , as does the collection of alternating $i$ -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 there is also an induced action on the quotient modules $S^{i} (M)$ and $\overset{i}{⋀} (M)$ . Moreover, we have:

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

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

Symmetrization and skew-symmetrization

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

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

It is straightforward to verify these $i$ -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 $Sym : T^{i} (M) \to T^{i} (M)$ and $Alt : T^{i} (M) \to T^{i} (M)$ whose images lie in the submodule of symmetric and alternating tensors, respectively.

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

Sym (z) = i! \cdot z .

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

Alt (z) = i! \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. For each nonnegative integer $i$ , if $i!$ is a unit in $R$ then there is an $R$ -module isomorphism between $S^{i} (M)$ (respectively, $\overset{i}{⋀} (M)$ ) and the submodule of $T^{i} (M)$ consisting of all symmetric $i$ -tensors (respectively, alternating $i$ -tensors).

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

Suggested next note

Noetherian modules

An action of the symmetric group on Tk(M)

Symmetrization and skew-symmetrization

Suggested next note

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