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

Let $V = \mathbb C^n$. Consider the plethysm $\bigwedge^k Sym^d V$ as a representation of $GL(V)$. In what special cases (e.g., for what $k$, $d$, and $n$) is this representation's decomposition into irreps known?

The only known nontrivial special case that I am aware of is when $k = 2$: in this case the decomposition is $S_{2d-1,1} \oplus S_{2d-3,3} \oplus S_{2d-5,5} \oplus \cdots$. When $n = 2$, I also know that it is equivalent to find decompositions of plethysms of the form $Sym^k Sym^i V$.

Using the Macaulay2 package SchurRings, I computed all examples with $d \leq 8$ with no obvious patterns jumping out at me.

I would be interested in any other special cases people know about (including ones which only apply to $n = 2$), conjectures along these lines, tables of computed data, or ideas about references that might be fruitful.

share|improve this question
    
When $n=2$ the plethysm ${\rm Sym^k}{\rm Sym^d}V$ is given by the Cayley-Sylvester formula. I can't see why this is equivalent to knowing your plethysm. But I think it should be possible to work out the constituents of $\bigwedge^k ({\rm Sym}^d V)$ using similar arguments with formal characters of ${\rm SL_2}(\mathbb{C})$. –  Mark Wildon Feb 14 '13 at 4:45
1  
@Mark: that symmetric plethysm is isomorphic to wedge^{k} sym^{d+k-1} via the so-called Wronskian isomorphism. See my paper with Chipalkatti in J. Pure App. Algebra vol. 210 p. 43. This is why the Cayley-Sylvester formula gives the answer for that case too. –  Abdelmalek Abdesselam Feb 14 '13 at 14:17
    
...the page 43 is the first of the article. The one where this isomorphism is mentioned is page 46. –  Abdelmalek Abdesselam Feb 14 '13 at 14:32
    
Thank you for the reference. As you say in your paper, it's a remarkable isomorphism. –  Mark Wildon Feb 15 '13 at 10:28
    
Thank you for the feedback; this is all very helpful! –  stepanp21 Feb 16 '13 at 17:51
add comment

1 Answer

up vote 5 down vote accepted

When $d=2$, the decomposition is known for all $k$ and $n$. Given a partition $\lambda$ of $k$ with distinct parts, let $2[\lambda]$ denote the partition of $2k$ whose main-diagonal hook lengths are $2\lambda_1, \ldots, 2\lambda_k$, and whose $i$th part has length $\lambda_i + 1$. Then

$$ \bigwedge^k {\rm Sym}^2 V = \sum_\lambda S^{2[\lambda]}(V) $$

where the sum is over all partitions $\lambda$ with distinct parts such that $2[\lambda]$ has at most $n$ parts and $S^\mu$ is the Schur functor for the partition $\mu$. For a proof using the symmetric group see Lemma 7 in http://arxiv.org/abs/0903.2864.

Apart from the case $k=2$ mentioned in this question (and the trivial cases $k=1$ or $d=1$), I think this is the only case where the complete decomposition is known.

share|improve this answer
    
there is formula by Thrall for Sym^3 Sym^d but I don't know about wedge^3 Sym^d although I would think it is easier than sym-sym. –  Abdelmalek Abdesselam Feb 14 '13 at 14:21
    
the paper by Thrall is in American J. Math vol. 64 p. 371. but I just now saw that you referred to it in your recent work. –  Abdelmalek Abdesselam Feb 14 '13 at 14:37
1  
I just saw an interesting unpublished paper by Agaoka on these plethysms. It can be found on Google Scholar by searching his name and "decomposition formulas of the plethysm". –  Abdelmalek Abdesselam Feb 14 '13 at 14:55
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.