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Let $V$ be a vector space of dimension $n$ over $\mathbb{C}$.

Q.1 When thinking of $V\otimes V\otimes V $, if $V$ has basis $\{x_1,x_2,\cdots,x_n\}$ then can we think $V\otimes V\otimes V$ as $\mathbb{C}$-vector space of polynomials $\sum_{i,j} a_{ij}{x_ix_jx_k}$ where $x_i$ and $x_j$ are non-commuting for $i\neq j$.

If this is true, then consider the natural action of $S_3$ on $V\otimes V\otimes V$. This action makes $V\otimes V\otimes V$ an $\mathbb{C}[S_n]$-module. Consider the element of the group algebra $$\alpha=(1)+(12)+(13)+(23)+(123)+(132).$$ The action of $\alpha$ on $V\otimes V\otimes V$ leaves the subspace $S{\rm ym}^3(V)$ invariant. Now I want to prove that converse is also true:

if an element $\beta\in \mathbb{C}[S_n]$ leaves ${\rm Sym}^3(V)$ invariant, then $\beta$ should some scalar multiple of $\alpha$. How should I proceed for it?

I was thinking elements of $V\otimes V\otimes V$ as in Q.1, but I couldn't proceed for the question on $\beta$

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    $\begingroup$ This is wrong, as @anon noticed. Perhaps you rather want to say "if an element $\beta \in \mathbb{C}\left[S_n\right]$ sends $V \otimes V \otimes V$ to $\operatorname{Sym}^3 V$, then $\beta$ is a scalar multiple of $\alpha$". This seems more likely to be true to me (but I am not sure and don't have the time to check). $\endgroup$ – darij grinberg Sep 1 '18 at 23:58
  • $\begingroup$ @darijgrinberg That follows from the fact every element of the group algebra is expressible as a $\Bbb C[G]$-linear combination of the isotypical projectors $e_V$ (of which the normalized $\alpha$ is one). Probably a more elementary explanation for $S_3$ though. $\endgroup$ – anon Sep 2 '18 at 0:01
  • $\begingroup$ Actually, it's almost correct. It's false if $\dim V < 2$, but it's true if $\dim V \geq 3$. $\endgroup$ – darij grinberg Sep 2 '18 at 0:05
  • $\begingroup$ @darij: Yes, this is actually the thing I was missing from my understanding. But thanks for the clear statement and repairing my doubt. $\endgroup$ – Beginner Sep 3 '18 at 3:45
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${\rm Sym}^n(V)$ is already a subrepresentation of $V^{\otimes n}$; it is a subspace stabilized by every element of the group algebra. As for elements $\beta$ which fix each element $x$ of ${\rm Sym}^n(V)$, observe

$$ \beta x=\left(\sum_g c(g)g\right)x=\sum_g c(g)gx=\sum_g c(g)x=\left(\sum_g c(g)\right)x. $$

That is, $\beta$ fixes nonzero symmetric tensors $x$ if and only if the sum of its coefficients is $1$. In that case, the sum of the coefficients of $\beta-\alpha$ is $0$, and so $\beta$ fixes them if and only if it is in the coset $\alpha+A$, where $A$ is the augmentation ideal of $\Bbb C[G]$ and $\alpha=|G|^{-1}\sum_g g$ is the normalized sum of the group elements.

Note this argument works on any $\Bbb C[G]$-module, where $V^{\otimes n}$ is replaced with $V$, $S_n$ is replaced with $G$, and ${\rm Sym}^n(V)$ is replaced by the invariant subspace $V^G$ (assuming it's nontrivial).

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  • $\begingroup$ Careful: You want $V^G \neq 0$; otherwise, any element of $\mathbb{C}\left[G\right]$ leaves any element of $V^G$ invariant. $\endgroup$ – darij grinberg Sep 1 '18 at 23:57
  • $\begingroup$ Yeah, that's true. $\endgroup$ – anon Sep 2 '18 at 0:00

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