# Known decomposition of $\bigwedge^k Sym^d \mathbb C^n$ in special cases?

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.

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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
@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

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.

Edit (June 2014). The constituents of $\bigwedge^3 \mathrm{Sym}^{d}(V)$ are determined on page 141 of Macdonald's book, Symmetric functions and Hall polynomials. Remark 3.6(b) in Howe, $(GL_n,GL_m)$-duality and symmetric plethysm, Proc. Indian Acad. Sci. 97 (1987) 85–109, gives a method for computing the plethysm $\bigwedge^4 \mathrm{Sym}^d(V)$.

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

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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
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
(In response to the previous comment.) I did not mention Thrall's paper because, as far as I can see, it only deals with symmetrized plethysm of the form $\mathrm{Sym}^k\mathrm{Sym}^n V$. –  Mark Wildon Jun 6 '14 at 12:48