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Denote by $\mathfrak{S}_k$ the symmetric group on $k$ elements. Let $\lambda=(n^2\times n)=(n^2,\ldots,n^2)$ be a rectangular partition and $k=n^3$. Denote by $S_\lambda$ the Specht module corresponding to $\lambda$ and set $V:=S_\lambda^{\otimes 3}=S_\lambda\otimes S_\lambda\otimes S_\lambda$. There is a (unique, up to scalars) nonzero element $v\in V$ with $\sigma\cdot v=v$ for all $\sigma\in\mathfrak{S}_{n^3}$, this is something I can prove. However, I would like to know what that element looks like. Does anyone have any idea?

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This is untrue if $m=1$ and $n>1$. In that case, $S_{\lambda}$ is the sign rep and there are no invariants in the tensor cube. I also have a gut instinct that no simple statement like this should be true -- that Kronecker coefficients should only be $1$ in "trivial" cases, where this case doesn't feel "trivial" enough. – David Speyer Jun 5 '12 at 13:57
That is correct, I was too quick with the generalization. I corrected my question. – Jesko Hüttenhain Jun 5 '12 at 16:29
Your claim is that if $W$ is the Specht module corresponding to that specific partition, $V=W^{\otimes 3}$ has a single copy of the trivial representation. I assume you proved this using some form of the Littlewood-Richardson rule. Specht modules are complicated to describe (and have at least two dual descriptions), so $V$ is going to be complicated. However, you can get a grip on the problem by viewing $W^{\otimes 3}\cong W^{\otimes 2}\otimes W$ and figuring out which subrepresentation of $W^{\otimes 2}$ gives rise to the trivial representation. – Aaron Jun 6 '12 at 16:23
A second approach would be to look at the natural projection operator onto the trivial rep, namely $\displaystyle\operatorname{Pr}(v)=\frac{1}{|G|}\sum_{g\in G} gv$ (which works for any group, hence the more generic notation). Unfortunately, it will likely be nontrivial to show that for a particular $v$ that $\operatorname{Pr}(v)\neq 0$. – Aaron Jun 6 '12 at 16:28

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