How do I show that $\text{Sym}^k(V)\oplus\bigwedge^k(V)=V^{\otimes k}$?

Question

I have defined both $\text{Sym}^k(V)$ and $\bigwedge^k(V)$ as quotient space of vector power $V^{\otimes k}$. So how do I understand that $\text{Sym}^k(V)\oplus\bigwedge^k(V)=V^{\otimes k}$, since elements of symmetric and exterior power are equivalence classes?

Answer 1

I will be working in characteristic zero algebraically closed base field $K$ exclusively.

Symmetric and exterior powers are understood as quotients of tensor powers, specifically by the ideals generated by the relations $\sigma a=a$ and $\sigma a={\rm sgn}(\sigma)a$ for all $\sigma\in S_n$ respectively. We do have $V^{\otimes 2}=S(V)\oplus T(V)$ though, where $S^2(V)$ and $T^2(V)$ respectively denote the subspaces of symmetric and antisymmetric tensors. We also have isomorphisms

$${\rm Sym}^2(V)\cong S^2(V):vw\mapsto\frac{v\otimes w+w\otimes v}{2},$$ $${\rm Alt}^2(V)\cong T^2(V): v\wedge w\mapsto\frac{v\otimes w-w\otimes v}{2}$$

One can see these (anti-)symmetrization maps generalize to $n$th powers, not just $2$nd ones, and may verify these are indeed isomorphisms by computing dimensions.

The dimension formulas are given by

$$\dim V^{\otimes n}=(\dim V)^n,\quad \dim {\rm Sym}^n(V)=\binom{\dim V+n-1}{n},\quad \dim{\rm Alt}^n(V)=\binom{\dim V}{n}.$$

It is easy to see that $d^n\ne\binom{d+n-1}{n}+\binom{d}{n}$ as polynomials in $d$ unless $n=2$ by checking the leading coefficients of both sides ($1$ vs. $2/n!$), hence this decomposition doesn't work for $V^{\otimes n}$.

But, reviewing the case $n=2$ again, the isomorphism is hardly just one of vector spaces: in fact it is an isomorphism of ${\rm GL}(V)$-$S_n$ bimodules. The actions of ${\rm GL}(V)$ and $S_n$ on $V^{\otimes n}$ are given by

$$\begin{cases} A(v_1\otimes\cdots \otimes v_n)=Av_1\otimes\cdots\otimes Av_n & A\in{\rm GL}(V) \\ (v_1\otimes\cdots\otimes v_n)\sigma=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)} & \sigma\in S_n\end{cases}$$

Seeing that the $S_n$-action is in fact a right action can be tricky.

The decomposition $V^{\otimes 2}\cong{\rm Sym}^2(V)\oplus{\rm Alt}^2(V)$ generalizes to arbitrary $n$ through a cool piece of work called Schur-Weyl duality, a representation-theoretic theorem. I will recount the highlights.

The irreducible representations of $S_n$ can be constructed explicitly using idempotents called Young symmetrizers $c_\lambda\in K[S_n]$. Both they and the irreps are indexed by integer partitions $\lambda\vdash n$. The left and right Specht modules are given by $S_\lambda^{\rm L}\cong K[S_n]c_\lambda$ and $S_\lambda^{\rm R}\cong c_\lambda K[S_n]$; these are the irreducible representations of $S_n$. The group algebra decomposes as $K[S_n]\cong\bigoplus_\lambda S_\lambda^{\rm L}\otimes_K S_\lambda^{\rm R}$.

(Group algebras decompose this way in general; this is content of the Peter-Weyl theorem.) So

$$\large\begin{array}{ll} V^{\otimes n} & \cong V^{\otimes n}\otimes_{K[S_n\,]}K[S_n] \\ & \cong V^{\otimes n}\otimes_{K[S_n]}\left(\bigoplus_\lambda S_\lambda^{\rm L}\otimes_K S_\lambda^{\rm R}\right) \\ & \cong \bigoplus_\lambda V^{\otimes n}\otimes_{K[S_n]}\left(S_\lambda^{\rm L}\otimes_K S_\lambda^{\rm R}\right) \\ & \cong \bigoplus_\lambda \left(V^{\otimes n}\otimes_{K[S_n]} S_\lambda^{\rm L}\right)\otimes_K S_\lambda^{\rm R} \\ & \cong \bigoplus_\lambda {\Bbb S}_\lambda(V)\otimes_K S_\lambda^{\rm R} \end{array}$$

where ${\Bbb S}_\lambda$ are the Schur functors, given by ${\Bbb S}_\lambda(V)\cong V^{\otimes n}\otimes_{K[S_n]}S_\lambda^{\rm R}\cong V^{\otimes n}c_\lambda$ (one may verify that this construction is in fact functorial). Not only does this ${\rm GL}(V)$-$S_n$ bimodule decomposition hold, the ${\Bbb S}_\lambda(V)$s are the irreducible representations of ${\rm GL}(V)$, and the $K$-spans of ${\rm GL}(V)$ and $S_n$ inside ${\rm End}_K(V^{\otimes n})$ are each other's full mutual centralizers.

The symmetric, exterior powers correspond to ${\Bbb S}_\lambda$ with $\lambda=(n)$, $\lambda=(1~\cdots~1)$ respectively, with corresponding Specht modules isomorphic to $K$ with trivial and sign representation respectively.

Further reading.

• Representation Theory: A first course (Fulton & Harris). Presumably chapters 4 & 6.
• Young Tableaux and the Representations of the Symmetric Group (Zhao) Link.
• Lie Groups: An Approach through Invariants and Representations (Procesi). Specifically chapter 9 on tensor symmetry. Draft available online here.