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Let $\{e_1,\ldots, e_n\}$ be the standard basis of $\mathbb{C}^n$. Consider the $m$-multilinear form $$v=\sum_{i=1}^n e_i^{\otimes m}\in (\mathbb{C}^n)^{\otimes m}$$ and consider the action of $\text{GL}_n(\mathbb{C})$ on $(\mathbb{C}^n)^{\otimes m}$.

Question: What is the stabilizer of $v$ in $\text{GL}_n(\mathbb{C})$ where $m>2$? If $m=2$ this is just the orthogonal group. If $m>2$ then this must contain the subgroup $S_n\wr C_m$, where the $i$-th copy of $C_m$ acts nontrivialy only on $e_i$. Does the stabilizer equal to this wreath product, or can it be bigger?

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1 Answer 1

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It is equal.

Since the ground field is of characteristic zero and $v$ is symmetric one may as well compute the stabilizer of the polynomial $f:=\sum_{i=1}^nx_i^m$.

Assume $g\in\mathrm{GL}_n(\mathbb C)$ stabilizes $f$. Then it also stabilizes the Hessian $$\det\nolimits_{ij}(\partial_i\partial_jf)\in\mathbb C^*(x_1\cdots x_n)^{m-2}$$ up to a factor. Hence it permutes the irreducibel factors $x_1,\ldots,x_n$ up to scalars, i.e., $g\in\mathbb C^*\wr S_n$. Plugging this back into $f$ we conclude $g\in C_m\wr S_n$.

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