Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
2  
Welcome to MO, Shrawan ! –  Chandan Singh Dalawat Apr 27 '12 at 21:33
    
I added the plethysm tag since this question is about the composition of two Schur functors. –  John Wiltshire-Gordon Apr 28 '12 at 4:07
    
This is essentially the Foulkes-Howe conjecture. The smallest conterexample would be for i=m=5. –  Abdelmalek Abdesselam Apr 30 '12 at 19:14
add comment

1 Answer

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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