Is a tensor with zero symmetrization (resp. antisymmetrization) necessarily antisymmetric (resp. symmetric)?

Question

Let $V$ be a finite-dimensional vector space (over some ground field $F$ of characteristic $0$), let $k$ be a nonnegative integer, and let $T^kV = V^{\otimes k}$ be the $k$-fold tensor product (over $F$) of $V$ with itself.

• Let $S_k$ denote the symmetric group on $k$ elements, and let $\epsilon_+,\epsilon_- : S_k \rightarrow \{+1,-1\}$ be the two homomorphisms given by $\epsilon_+(\sigma) = 1$ and $\epsilon_-(\sigma) = \mathrm{sgn}(\sigma)$ for $\sigma \in S_k$.
• Let $\rho_+,\rho_- : S_k \rightarrow \mathrm{GL}_{F}(T^k V)$ be the "symmetric" and "antisymmetric" actions, given on e.g. a homogeneous element $w:= v_1 \otimes \cdots \otimes v_k \in T^k V$ by $\rho_{\pm}(\sigma)(w) = \epsilon_{\pm}(\sigma)\, v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(k)}$.
• We define $P_{\pm} := \frac{1}{k!} \sum_{\sigma \in S_k} \rho_{\pm}(\sigma) \in \mathrm{End}_F(T^k V)$ and $I_{\pm} := \{ x \in T^k V : \forall \sigma \in S_k , \rho_{\pm}(\sigma)(x) = x\}$.

Intuitively we think of $P_+,P_-$ as "symmetrization and antisymmetrization" operators, and $I_+,I_-$ as the "symmetric and antisymmetric" (contravariant degree $k$) tensors on $V$.

So far, if I'm not mistaken, I've shown that $\mathrm{ker}(P_+) \supseteq I_-$ and $\mathrm{ker}(P_-) \supseteq I_+$. Also $I_{\pm}$ equals the fixed-point space (eigenspace for eigenvalue +1) of $P_{\pm}$.

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However, I am wondering whether $\mathrm{ker}(P_{+}) \subseteq I_{-}$ and $\mathrm{ker}(P_{-}) \subseteq I_{+}$ hold in general? I'm not sure how to approach solving this question; would anyone have a suggestion?

Answer 1

Let $\mathbb Z[S_k]$ denote the group algebra of $S_n$ over the integers $\mathbb Z$, so that, for any field $F$, there is a unique homomorphism of rings $s_F \colon \mathbb Z[S_k]\to F[S_k]$ satisfying $s_F(\sigma) = \sigma$, and hence any $S_k$-representation on an $F$-vector space is an $\mathbb Z[S_k]$-module.

Now let $a_+=\sum_{\sigma \in S_k} \sigma = \sum_{\sigma\in S_k} \epsilon_+(\sigma)\sigma$ and $a_- = \sum_{\sigma\in S_k}\epsilon_-(\sigma)\sigma$. It is easy to see that if $\tau \in S_k$ then $\tau.a_+ = a_+$ and $\tau.a_- = \epsilon_-(\tau)a_-$. In particular, $$ a_+.a_- = \sum_{\tau\in S_k} \tau.a_- = (\sum_{\sigma \in S_k} \epsilon_-(\sigma))a_- =0. $$ where the final equality holds since $S_n = (12)\mathrm{Alt}_k \sqcup \mathrm{Alt}_k$ where $\mathrm{Alt}_k$ denotes the subgroup $\ker(\epsilon_-)$ and $\epsilon_-((12)\tau) = -\epsilon(\tau)$ for all $\tau \in \mathrm{Alt}_k$. Similarly one can check $a_-a_+ =0$.

Now if $\rho_+\colon S_k \to \mathrm{GL}_F(T^k(V))$ is the natural action of $S_k$ the operators $P_{\pm}\colon T^k(V)\to T^k(V)$ are given by $\frac{1}{k!}\rho\circ s_F(a_+)$ and $\frac{1}{k!}\rho\circ s_F(a_-)$ respectively. (Note that this means $P_{\pm}$ are only defined when $\mathrm{char}(F)>k$).

Since $a_+a_- = a_-a_+=0$ it follows immediately that $\ker(P_\pm)\supseteq I_{\mp}$. However, in general, for $k \geq 3$, $I_- \subsetneq \ker(P_+)$. Indeed $\dim(P_+(T^3(V))= \binom{n+2}{2}$ while $\dim(T^3(V)) = n^3$, and hence $$ \dim(\ker(P_+))= n^3-\binom{n+2}{2}\simeq n^3 $$ while $\dim(P_-(T^3(V)))=\binom{n}{3} \simeq n^3/6$, thus the kernel of $P_+$ is much larger in general that $I_-$.

What is happening here is that, for $k\geq 3$, the representation of $S_k$ on $T^k(V)$ contains copies of irreducible representation of $S_k$ other than the trivial and sign representations, and all of these are contained in the kernel of both $P_+$ and $P_-$.

If $2<\mathrm{char}(F)<k$ then you cannot define the operators $P_+$ and $P_-$. However the operators $\pi_{\pm}:=\rho_+(a_{\pm})$ are well-defined, and $$ T^k(V)/\ker(\pi_{-}) = \bigwedge{^k(}V), \quad T^k(V)/\ker(\pi_{+}) = S^k(V), $$ are the $k$-th exterior and symmetric powers of $V$ respectively. In this case however, $I_-\subseteq \ker(\pi_-)$ and $I_+\subseteq \ker(\pi_+)$, since both $\pi_-$ and $\pi_+$ square to zero.

Finally, in the case $\mathrm{char}(F)=2$ and $k=2$, then $a_+=a_- = e+(12) \in \mathbb Z[S_2]$, and if we let $a = e+(12)$ and $\pi= \rho_+(a)$, then $a^2=0$ so that $F[S_2] \cong F[t]/(t^2)$ via $a\mapsto t$. If $V$ is an $F$-vector space, then $S^2(V) = T^2(V)/\ker(\pi)$ and $\bigwedge\!{^2(}V) = T^2(V)/\mathrm{im}(\pi)$.

Answer 2

If $n := \dim V > 0$, then

• the space of $k$-tensors on $V$ has dimension $$\dim (V^{\otimes k}) = n^k ,$$
• the space $I_-$ of symmetric $k$-tensors on $V$ has dimension $$\dim I_+ = {{n + k - 1}\choose k},$$ and
• the space $I_-$ of antisymmetric $k$-tenors on $V$ has dimension $$\dim I_- = {n \choose k} .$$

Now:

• If $k = 0$, then $I_+ = I_- = F$ and $\ker P_\pm = \{0\}$.
• If $k = 1$, then $I_+ = I_- = V$ and $\ker P_\pm = \{0\}$.
• If $k = 2$, then $I_+ \cap I_- = \{0\}$ and $$\dim I_+ + \dim I_- = \frac12 n (n + 1) + \frac12 n (n - 1) = \dim(V^{\otimes k}) = n^2,$$ so $$V^{\otimes 2} = I_+ \oplus I_-,$$ hence $\ker P_\pm = I_\mp$.
• If $k > 2$, then $I_+ \cap I_- = \{0\}$ and $$\dim I_- + \dim I_- < \dim V^{\otimes k} ,$$ and the containments $I_{\mp} \subset \ker P_\pm$ are proper. For example, for $k = 3$, $\ker P_+ \cap \ker P_-$ has dimension $\frac23 n (n^2 - 1)$ (and it splits as an $S_3$-representation into a direct sum of $2$ isomorphic subrepresentations of dimension $\frac13 n (n^2 - 1)$).