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I am working on a physical problem, where I need to compute the "reduced plethysm" that is all the irreducibles characterised by the Young tableaux of 2 columns or less. The plethysm problem I want to solve is the following,

$$\mathbb s_{(3)}[\mathbb s_{1^p}] \text{ which can be written as } s_{(3)}[e_p],$$ where $e_p$ is the elementary symmetric polynomial of degree $p$, and we work in the permutation group $S_n$, so everything is in $n$ variables.

Edit: Since your partitions are special cases of "hooks", you can use this reference, which gives an expansion in terms of Schur functions. The coefficients are given as certain character values, and are, according to the paper, difficult to compute in general. However, in your special case, it might be easier.

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  • $\begingroup$ Does $\mathbb{S}_{3^1}$ denote the Schur function $s_{(3)}$, the partition with only one part of size 3, (which is how Stanley denote Schur functions in Enumerative combinatorics II)? And in that case, then $\mathbb{S}_{1^p}$ is perhaps the elementary symmetric polynomial, $e_p(x_1,\dots,x_n)$? $\endgroup$ – Per Alexandersson Aug 5 '14 at 14:21
  • $\begingroup$ I am a physicist sorry for perhaps the abuse of notation. $\mathbb S_{3^1}$ corresponds to a row with size 3. $\endgroup$ – vishmay Aug 5 '14 at 14:24
  • $\begingroup$ I guess what you are saying is what I mean. In my naive language $\mathbb S_{1^p}$ corresponds to column with p -boxes. $\endgroup$ – vishmay Aug 5 '14 at 14:26
  • $\begingroup$ Yes, then we agree. The notation you use is different from "our" standard references, (Macdonald, Stanley). I edit it accordingly. $\endgroup$ – Per Alexandersson Aug 5 '14 at 14:28
  • $\begingroup$ Thanks a lot. Do you know if the decomposition is known? $\endgroup$ – vishmay Aug 5 '14 at 14:28
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This follows on from Dan Peterson's answer. For $r,s,a \in \mathbb{N}_0$, let $N_{r,s}(a)$ be the number of partitions of $a$ whose Young diagram fits into a box with $r$ rows and $s$ columns. By the Cayley–Sylvester formula, the multiplicity of $s_{(rs-a,a)}$ in $s_{(r)} \circ s_{(s)}$ is $N_{r,s}(a) - N_{r,s}(a-1)$ for $a$ such that $1 \le a \le rs/2$. See Corollary 2.12 in this paper of Giannelli for a short proof using the symmetric group, or this paper of Manivel for a generalization.

So the multiplicity of $s_{(3p-a,a)}$ in $s_{(3)} \circ s_{(p)}$ is $N_{(3,p)}(a) - N_{(3,p)}(a-1)$ for $a \le 3p/2$.

When $p$ is odd we need instead the multiplicity of $s_{(3p-a,a)}$ in $s_{(1^3)} \circ s_{(p)}$. By the Wronskian isomorphism (see here or Manivel's paper), we have

$$ \langle s_{(1^3)} \circ s_{(p)}, s_{(3p-a,a)} \rangle = \langle s_{(3)} \circ s_{(p-2)}, s_{(3p-a-3,a-3)} \rangle = N_{(3,p-2)}(a-3) - N_{(3,p-2)}(a-4) $$

for $p \ge 2$ and $4 \le a \le 3p/2$. If $a \le 2$ then the multiplicity is zero and if $a=3$ the multiplicity is $1$.

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  • $\begingroup$ @DanPetersen: Dear Dan Petersen and Mark Wildon, Thanks a lot for your answers and clarifying the problem. I also found the following paper that gives a direct proof for my case: citeseerx.ist.psu.edu/viewdoc/…. I am trying to figure out the link between the language that uses symmetric polynomials and abstract tensors for the representation of the perm group. Can you pls see mathoverflow.net/questions/178014/… $\endgroup$ – vishmay Aug 7 '14 at 13:41
  • $\begingroup$ Dear Mark Wildon, I am a little confused about the multiplicity given by your formula. Assuming I can use this as multiplicities of the irreps, I took an example say p=3 and a=3, the number of standard young tableaux I can construct for the diagram (6,3) setting a canonical order of the indices is more than one? Does that mean the multiplicity is >1. Where am I going wrong. Help much appreciated and thanks for your previous suggestions. $\endgroup$ – vishmay Aug 18 '14 at 16:28
  • $\begingroup$ Dear Prag1, the multiplicity of $s_{(6,3)}$ in $s_3 \circ s_3$ is $N_{(3,3)}(3) - N_{(3,3)}(2) = 3-2=1$. The relevant partitions (not tableaux) are $(3), (2,1), (1,1,1)$ and $(2), (1,1)$. The multiplicity of $s_{(6,3)}$ in $e_3 \circ s_3$ is also $1$, by the final line in my answer. Please email me if you have further questions. $\endgroup$ – Mark Wildon Aug 18 '14 at 16:58
  • $\begingroup$ Dear Mark Wildon, given the wronskian isomorphism of the plethysm, $S_{1^m}\circ S_p \text{isomorphic to} S_m \circ S_{p-m+1}$. What is the isomorphism between irre components, $S_{mp-a,a}$ in the decomposition of $S_{1^m}\circ S_p$. I want to find the multiplicity $< S_{1^m}\circ S_p,S_{mp-a,a}>$ using Wronskian? Thanks. $\endgroup$ – vishmay Aug 25 '14 at 13:00
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You are asking about $s_3 \circ s_{1^p}$. Recall that there is an involution $\omega$ on the ring of symmetric functions for which $\omega(s_\lambda) = s_{\lambda'}$ (conjugate partition). We use results from Macdonald's Symmetric functions and Hall polynomials: By Chapter 1, §8, Example 1, $ \omega( s_3 \circ s_{1^p})$ equals $s_3 \circ s_p$ if $p$ is even and $s_{1^3} \circ s_p$ if $p$ is odd. Both these plethysms are given completely explicitly in Example 9(b) in the same section, note that $s_p=h_p$ and $s_{1^p} = e_p$.

Your suspicion that there are no summands with only two columns (i.e. only two rows, after applying $\omega$) is incorrect for all $p>1$.

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  • $\begingroup$ Hi Dan, many thanks. I am not at all familiar with plethysms(physicist) can you please explain what you mean by duality operation. Or give some reference. Also I suspect that the expansion of this does not exist if restricted to just two columns. Please see this ref[1] page 12, identity right after theorem 4.5. There the decomposition is restricted to hook + row. My case is hook +column, can we deduce something from this formula by means of conjugation. Thanks. $\endgroup$ – vishmay Aug 5 '14 at 17:09
  • $\begingroup$ @Prag1: Dan is not referencing that paper. You need to see the book "Symmetric Functions and Hall Polynomials" by Macdonald. $\endgroup$ – Per Alexandersson Aug 5 '14 at 18:41
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I am no expert in plethysm calculation, but a quick google on the involved terms leads to this paper, where combinatorial formulas for $s_{\lambda}[s_{\mu}]$ is given. Since your formula is a very special case of this, it is very likely that you can simplify the formula in the paper.

The formula in the paper expresses the plethysm as a sum over certain integer matrices. I think, with some combinatorial work, you can get an explicit formula.

EDIT: This paper give some coefficients in the Schur expansion in your case, since your partitions are hooks. However, it seems that even these coefficients are quite hard to compute in general, as they are certain character values.

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    $\begingroup$ That's quasi-symmetric expansion. I'm pretty sure people don't know Schur positive expansions. $\endgroup$ – DES-SupportsMonicaAndTransfolk Aug 5 '14 at 15:45
  • $\begingroup$ Not even when they are restricted as in my case ? $\endgroup$ – vishmay Aug 5 '14 at 15:56
  • $\begingroup$ @DavidSpeyer: Found another reference, that partially gives the Schur expansion. $\endgroup$ – Per Alexandersson Aug 5 '14 at 16:09
  • $\begingroup$ @PerAlexandersson Thanks that's great reference. Forgive my ignorance, but am I right in saying that eq 1 in page 12, right after theorem 4.5, means that the expansion of my special case is NULL. Help much appreciated. $\endgroup$ – vishmay Aug 5 '14 at 16:20
  • $\begingroup$ @Prag1 No, the paper only consider what hook-shape indices that appear in the expansion. That is, the reference I gave only consider a very small part of the Schur expansion. I suggest to look at Dan's answer. $\endgroup$ – Per Alexandersson Aug 5 '14 at 16:24

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