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.

  • $\begingroup$ 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 ${\rm SL_2}(\mathbb{C})$. $\endgroup$ Feb 14 '13 at 4:45
  • 1
    $\begingroup$ @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. $\endgroup$ Feb 14 '13 at 14:17
  • $\begingroup$ ...the page 43 is the first of the article. The one where this isomorphism is mentioned is page 46. $\endgroup$ Feb 14 '13 at 14:32
  • $\begingroup$ Thank you for the reference. As you say in your paper, it's a remarkable isomorphism. $\endgroup$ Feb 15 '13 at 10:28

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.

  • $\begingroup$ 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. $\endgroup$ Feb 14 '13 at 14:21
  • $\begingroup$ 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. $\endgroup$ Feb 14 '13 at 14:37
  • 1
    $\begingroup$ 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". $\endgroup$ Feb 14 '13 at 14:55
  • $\begingroup$ (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 $\mathrm{Sym}^k\mathrm{Sym}^n V$. $\endgroup$ Jun 6 '14 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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