Symmetric and alternating tensors

This is a topic in undergraduate algebra that comes up a lot whenever I deal with some sort of differential forms and vector bundles. I am just summarizing it here to prevent my own confusion. We follow Chapter 11 of Dummit and Foote.

Fix a commutative ring $R.$ Given any $n \geq 0,$ we have an endomorphism functor on $\mathrm{Mod}_{R}$ given by $M \mapsto M^{\otimes n}.$ Note that $M^{\otimes 0} = R$ and given an $R$-module map $\varphi : M \rightarrow N,$ the map $M^{\otimes n} \rightarrow N^{\otimes n}$ is determined by $$x_{1} \otimes \cdots \otimes x_{n} \mapsto \varphi(x_{1}) \otimes \cdots \otimes \varphi(x_{n}).$$ We write $$\mathrm{Sym}^{n}(M) := \frac{M^{\otimes n}}{(x_{1} \otimes \cdots \otimes x_{n} - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})_{\sigma \in S_{n}}},$$ and note that we also get an endomorphism functor on $\mathrm{Mod}_{R}$ given by $M \mapsto \mathrm{Sym}^{n}(M).$ We write $$\bigwedge^{n}M := \frac{M^{\otimes n}}{(x_{1} \otimes \cdots \otimes x_{n} : x_{i} = x_{j} \text{ for some } i \neq j)},$$ which also gives a functor. If (the image of) $2$ is invertible in $R,$ then the ideal in the definition the $n$-the exterior power can be replaced by $$(x_{1} \otimes \cdots \otimes x_{n} - \mathrm{sgn}(\sigma)x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})_{\sigma \in S_{n}}.$$ Note that there are evident universal properties associated to constructions of $M^{\otimes n}, \mathrm{Sym}^{n}(M),$ and $\bigwedge^{n}M.$

Symmetric and alternating tensors. Given a module $M$ over $R,$ the $n$-tensor is an element in $M^{\otimes n}.$ Note that $S_{n}$ acts on $M^{\otimes n}$ by $$\sigma \cdot (x_{1} \otimes \cdots \otimes x_{n}) := x_{\sigma^{-1}(1)} \otimes \cdots \otimes x_{\sigma^{-1}(n)}$$ for each $\sigma \in S_{n}.$ An $n$-tensor $t \in M^{\otimes n}$ is

• symmetric if $\sigma t = t$ for all $\sigma \in S_{n}$;
• alternating if $\sigma t = \mathrm{sgn}(\sigma)t$ for all $\sigma \in S_{n}.$

Given an $n$-tensor $t,$ we define $$S(t) := \sum_{\sigma \in S_{n}}\sigma t,$$ and $$A(t) := \sum_{\sigma \in S_{n}}\mathrm{sgn}(\sigma)\sigma t.$$ It is not difficult to check that $S(t) = n!t$ if $t$ is an $n$-symmetric tensor and $A(t) = n!t$ if $t$ is an $n$-alternating tensor. Denote by $S(M^{\otimes n})$ the submodule of symmetric $n$-tensors and $A(M^{\otimes n})$ the submodule of alternating $n$-tensors in $M^{\otimes n}.$

Symmetric powers are symmetric tensors. Let $n!$ be invertible in $R.$ Then the image of the map $(n!)^{-1}S : M^{\otimes n} \rightarrow M^{\otimes n}$ is $S(M^{\otimes n})$, and its kernel is $(x_{1} \otimes \cdots \otimes x_{n} - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})_{\sigma \in S_{n}}.$ Thus, this map induces an isomorphism $$\mathrm{Sym}^{n}(M) \simeq S(M^{\otimes n})$$ such that $x_{1} \otimes \cdots \otimes x_{n} \mapsto x_{1} \otimes \cdots \otimes x_{n}.$

Proof. Since $n!$ is a unit in $R,$ we have $(n!)^{-1}S(t) = t$ for all $t \in S(M^{\otimes n}),$ so the image of $(n!)^{-1}S$ is $S(M^{\otimes n}).$ It is clear that $x_{1} \otimes \cdots \otimes x_{n} - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)}$ is in the kernel for any $\sigma \in S_{n},$ so we may define the desired map $$\mathrm{Sym}^{n}(M) \rightarrow S(M^{\otimes n}).$$ To check it is an isomorphism, let $t$ be any $n$-tensor such that $(n!)^{-1}S(t) = (n!)^{-1}\sum_{\sigma \in S_{n}}\sigma t = 0.$ We have $$t = t - \frac{1}{n!}\sum_{\sigma \in S_{n}}\sigma t = \frac{1}{n!}\sum_{\sigma \in S_{n}}\left(t - \sigma t \right),$$ but the image of $t - \sigma t$ is $0$ in $\mathrm{Sym}^{n}(M),$ so the image of $t$ must be $0$ as well. $\Box$

Exterior powers are alternating tensors. Let $n!$ be invertible in $R.$ Then the image of the map $(n!)^{-1}A : M^{\otimes n} \rightarrow M^{\otimes n}$ is $A(M^{\otimes n})$, and its kernel is $(x_{1} \otimes \cdots \otimes x_{n} : x_{i} = x_{j} \text{ for some } i \neq j).$ Thus, this map induces an isomorphism $$\bigwedge^{n}M \simeq A(M^{\otimes n})$$ such that $x_{1} \wedge \cdots \wedge x_{n} \mapsto x_{1} \otimes \cdots \otimes x_{n}.$

Proof. Since $n!$ is a unit in $R,$ we have $(n!)^{-1}A(t) = t$ for all $t \in A(M^{\otimes n}),$ so the image of $(n!)^{-1}A$ is $A(M^{\otimes n}).$ It is clear that $x_{1} \otimes \cdots \otimes x_{n}$ is in the kernel whenever we have some $i \neq j$ with $x_{i} = x_{j},$ by using the transpose switching $i$ and $j.$ Hence, we may define the desired map $$\bigwedge^{n}M \rightarrow A(M^{\otimes n}).$$ To check it is an isomorphism, let $t$ be any $n$-tensor such that $(n!)^{-1}A(t) = (n!)^{-1}\sum_{\sigma \in S_{n}}\mathrm{sgn}(\sigma)\sigma t = 0.$ We have $$t = t - \frac{1}{n!}\sum_{\sigma \in S_{n}} \mathrm{sgn}(\sigma) \sigma t = \frac{1}{n!}\sum_{\sigma \in S_{n}}\left(t - \mathrm{sgn}(\sigma)\sigma t \right),$$ but the image of $t - \mathrm{sgn}(\sigma)\sigma t$ is $0$ in $\bigwedge^{n}M,$ because $$0 = (x + y) \wedge (x + y) = x \otimes y + y \otimes x$$ so that $x \otimes y = - y \otimes x.$ This implies that the image of $t$ is $0$ in $\bigwedge^{n}M.$ $\Box$

Geometric tensors. Now, let $M$ be a free $R$-module with finite rank $m.$ Fix a basis $e_{1}, \dots, e_{m} \in M.$ Then we have an $R$-module isomorphism $$M \overset{\sim}{\longrightarrow} M^{\vee} = \mathrm{Hom}_{R}(M, R)$$ given by $e_{j} \mapsto e_{j}^{\vee}.$ This gives us an $R$-module isomorphism $$M^{\otimes n} \overset{\sim}{\longrightarrow} (M^{\otimes n})^{\vee}$$ given by $e_{j_{1}} \otimes \cdots \otimes e_{j_{n}} \mapsto (e_{j_{1}} \otimes \cdots \otimes e_{j_{n}})^{\vee}.$ We have $$(M^{\otimes n})^{\vee} = \mathrm{Hom}_{R}(M^{\otimes n}, R) \simeq \mathrm{Mult}_{R}(M^{n}, R)$$ given by $(e_{j_{1}} \otimes \cdots \otimes e_{j_{n}})^{\vee} \mapsto (e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}).$ We call the elements in $\mathrm{Mult}_{R}(M^{n}, R)$ geometric $n$-tensors of $M.$

Remark. The terminology is non-standard.

We have $M^{\otimes n} \simeq \mathrm{Mult}_{R}(M^{n}, R)$ given by $$(e_{j_{1}} \otimes \cdots \otimes e_{j_{n}})^{\vee} \mapsto (e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}).$$ We may define the product $$\mathrm{Mult}_{R}(M^{l}, R) \times \mathrm{Mult}_{R}(M^{n}, R) \rightarrow \mathrm{Mult}_{R}(M^{l+n}, R)$$ by concatenation $$(e_{i_{1}}^{\vee}, \dots, e_{i_{l}}^{\vee}) \cdot (e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}) = (e_{i_{1}}^{\vee}, \dots, e_{i_{l}}^{\vee}, e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}),$$ which is compatible with the product $M^{\otimes l} \times M^{\otimes n} \rightarrow M^{\otimes (l + n)}$ given by the tensor product. Note that given geometric $l$-tensor $f$ and $n$-tensor $g,$ we have $$(f \cdot g)(x_{1}, \dots, x_{l+n}) = f(x_{1}, \dots, x_{l}) g(x_{l+1}, \dots, x_{l+n}).$$ Considering the correspondence $e_{j_{1}} \otimes \cdots \otimes e_{j_{n}} \leftrightarrow (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee}),$ we may consider the action of $S_{n}$ on $\mathrm{Mult}_{R}(M^{n}, R)$ given by $$\sigma \cdot (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee}) := (e_{\sigma^{-1}(j_{1})}^{\vee}, \cdots, e_{\sigma^{-1}(j_{n})}^{\vee}).$$ Note that $$\begin{align*}(\sigma \cdot (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee}))(x_{1}, \dots, x_{n}) &= (e_{\sigma^{-1}(j_{1})}^{\vee}, \cdots, e_{\sigma^{-1}(j_{n})}^{\vee})(x_{1}, \dots, x_{n}) \\ &=  (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee})(x_{\sigma(1)}, \dots, x_{\sigma(n)}),\end{align*}$$ because if $x_{i} = e_{\sigma^{-1}(i)},$ then $x_{\sigma(i)} = e_{i}.$ Thus, for $f \in \mathrm{Mult}_{R}(M^{n}, R),$ we have $$(\sigma f)(x_{1}, \dots, x_{n}) = f(x_{\sigma(1)}, \dots, x_{\sigma(n)}).$$ We note that under the isomorphism $M^{\otimes n} \simeq \mathrm{Mult}_{R}(M, R),$ the geometric $n$-tensor $f$ correponds to

• a symmetric $n$-tensor if $\sigma f = f$ for all $\sigma \in S_{n}$;
• an alternating $n$-tensor if $\sigma f = \mathrm{sgn}(\sigma)f$ for all $\sigma \in S_{n}.$

If $f$ corresponds to a symmetric (resp. alternating) $n$-tensor, we say $f$ is symmetric (resp. alternating). The set of symmetric geometric $n$-tensors is denoted as $S_{R}(M^{n}, R),$ and the set of alternating geometric $n$-tensors is denoted as $A_{R}(M^{n}, R),$ both of which are $R$-submodules of $\mathrm{Mult}_{R}(M^{n}, R).$ The discussions above imply the following:

Theorem. If $n!$ is invertible in $R,$ then we have $$\mathrm{Sym}^{n}(M) \simeq S_{R}(M^{n}, R)$$ given by $$e_{i_{1}} \otimes \cdots \otimes e_{i_{n}} \mapsto e_{i_{1}}^{\vee} \otimes \cdots \otimes e_{i_{n}}^{\vee} = (e_{i_{1}}^{\vee}, \dots, e_{i_{n}}^{\vee}),$$ where for $(f, g) \in \mathrm{Mult}_{R}(M^{r}, R) \times \mathrm{Mult}_{R}(M^{s}, R),$ we used the notation $f \otimes g := f \cdot g,$ namely $$(f \otimes g)(x_{1}, \dots, x_{r+s}) := f(x_{1}, \dots, x_{r})g(x_{r+1}, \dots, x_{r+s}).$$ Moreover, we have $$\bigwedge^{n}M \simeq A_{R}(M^{n}, R),$$ given by $e_{i_{1}} \wedge \cdots \wedge e_{i_{n}} \mapsto e_{i_{1}}^{\vee} \wedge \cdots \wedge e_{i_{n}}^{\vee} = (e_{i_{1}}^{\vee}, \dots, e_{i_{n}}^{\vee}),$ where for $(f, g) \in \mathrm{Mult}_{R}(M^{r}, R) \times \mathrm{Mult}_{R}(M^{s}, R)$ (assuming $r!, s! \in R^{\times}$) we used the notation $$f \wedge g := \frac{A(f \otimes g)}{r!s!} := \frac{1}{r!s!} \sum_{\sigma \in S_{r + s}}\sigma (f \otimes g).$$ That is, we have $$(f \wedge g)(x_{1}, \dots, x_{r+s}) := \frac{1}{r!s!}\sum_{\sigma \in S_{r+s}}f(x_{\sigma(1)}, \dots, x_{\sigma(r)})g(x_{\sigma(r+1)}, \dots, x_{\sigma(r+s)}).$$ The whole posting was due to extract the following purpose of mine:

Corollary. Algebraic definitions of symmetric power and exterior power over characteristic $0$ match with those of definitions in Chapter 3 of Tu's book.