Let $V_m$ be the $m$-dimensional complex vector space with basis $\{e_1, \dots, e_m\}$ and let $i\leq m$. Consider the element ${v}_0^i \in S^i(S^m(V_m))$, where ${v}_0$ is the element $e_1\dots e_m \in S^m(V_m)$. Then, is the $GL(V_m)$-submodule of $S^i(S^m(V_m))$ generated by the element ${v}_0^i$ is the whole of $S^i(S^m(V_m))$?
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2$\begingroup$ Welcome to MO, Shrawan ! $\endgroup$– Chandan Singh DalawatCommented Apr 27, 2012 at 21:33
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$\begingroup$ I added the plethysm tag since this question is about the composition of two Schur functors. $\endgroup$– John Wiltshire-GordonCommented Apr 28, 2012 at 4:07
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$\begingroup$ This is essentially the Foulkes-Howe conjecture. The smallest conterexample would be for i=m=5. $\endgroup$– Abdelmalek AbdesselamCommented Apr 30, 2012 at 19:14
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1 Answer
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The smallest counterexample is when $i=m=2$. The composition $S^2(S^2(-))$ can be decomposed as a sum of Schur functors:
$$S^2( S^2 ) = S^4 \oplus S^{2,2}.$$
In this case, the first space is five dimensional and the second is one dimensional. The element you give, $v_0^2$, lies completely in the first factor. This may be seen by considering these representations as subrepresentations of $\otimes^4$ and using Schur-Weyl duality to project $v_0^2$ onto the second factor.
Checking my work again this morning, I made a mistake. In fact, $i=m=2$ is not a counterexample. If $V_2$ is spanned by $x$ and $y$, the correct components are $$(xy)^2=\left(\frac{1}{3}(x^2)(y^2) + \frac{2}{3}(xy)^2\right) + \left(\frac{-1}{3}(x^2)(y^2) + \frac{1}{3}(xy)^2\right).$$
answered Apr 28, 2012 at 4:01