6
$\begingroup$

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 $S^i(S^m(V_m))$?

$\endgroup$
3
  • 2
    $\begingroup$ Welcome to MO, Shrawan ! $\endgroup$ Apr 27, 2012 at 21:33
  • $\begingroup$ I added the plethysm tag since this question is about the composition of two Schur functors. $\endgroup$ Apr 28, 2012 at 4:07
  • $\begingroup$ This is essentially the Foulkes-Howe conjecture. The smallest conterexample would be for i=m=5. $\endgroup$ Apr 30, 2012 at 19:14

1 Answer 1

1
$\begingroup$

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 $$(xy)^2=\left(\frac{1}{3}(x^2)(y^2) + \frac{2}{3}(xy)^2\right) + \left(\frac{-1}{3}(x^2)(y^2) + \frac{1}{3}(xy)^2\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.