For integer partitions $\mu\subset\lambda$ we can define the skew character $\chi^{\lambda/\mu}$ (for example?) via the Littlewood-Richardson rule.

Many combinatorial gadgets and algorithms extend in a very natural way from the case of partitions to the skew case.

I think it is fair to expect that characters corresponding to partitions are special, because they are irreducible.

I would like to know whether there are (representation theoretic) properties of skew characters that make them special.

  • $\begingroup$ I don’t think that they are really special, beyond the fact that their irreducible decompositions tells you LR coefficients... $\endgroup$ – Sam Hopkins Nov 29 '19 at 14:13
  • $\begingroup$ Maybe this rewording is useful: can we easily recognise a skew character, for example given its decomposition into irreducibles? $\endgroup$ – Martin Rubey Nov 29 '19 at 14:17
  • $\begingroup$ Here’s one idea for that: the decomposition into irreducibles probably satisfies certain convexity conditions that a “random” representation would not. $\endgroup$ – Sam Hopkins Nov 29 '19 at 14:22
  • $\begingroup$ Hi, Martin! I just sent you a private e-mail about the question. Please, look at it and let me know if it makes sense :) $\endgroup$ – fedja Mar 26 at 22:43

Here are a few references to expand on my comment about "convexity" properties of skew Schur functions $s_{\lambda/\mu}$.

As the OP points out, it's maybe most natural for the question to assume that someone has handed you the decomposition of $s_{\lambda/\mu}$ into irreducibles, i.e., they've given you a list of LR coefficients. For a "convexity" conjecture on LR coefficients (which could be translated to say something about the list you've been given) see Conjecture 1 (due to Lam-Postnikov-Pylyavskyy) of https://arxiv.org/abs/math/0608134. Many special cases of that conjecture are known- see the references given in that paper.

But another thing you can do is think of $s_{\lambda/\mu}$ as a polynomial (i.e., do the monomial expansion), and ask questions about its convexity properties. For instance, it is known that the nonzero coefficients of this polynomial are exactly the lattice points of some convex polytope (this is called the "saturated Newton polytope" (SNP) condition); see https://arxiv.org/abs/1703.02583. It is also conjectured that the "normalized" version of $s_{\lambda/\mu}$ is "Lorentzian" (another kind of convexity property) in https://arxiv.org/abs/1906.09633.

These kind of convexity properties are likely enough to tell the character $s_{\lambda/\mu}$ apart from a "random" character of a symmetric group; but it is doubtful they would lead to a procedure for exactly recognizing $s_{\lambda/\mu}$. For instance, the Stanley symmetric functions (mentioned in a previous answer), also satisfy the SNP property (and indeed this is how the SNP property is deduced for skew Schur polynomials).


One possible answer is that skew characters are not really the largest extension, but a special-case of certain characters associated with any subset of boxes in the plane. For example, if this subset is the Rothe diagram of a permutation, the character is a Stanley symmetric function.

It is an open problem, see [Ricky Ini Liu, Specht modules and Schubert varieties for general diagrams. Massachusetts Institute of Technology. 2010] , to describe the Schur expansion of these (generalized Specht modules) combinatorially.

A quick (quick to implement) way to compute the Schur expansion of Stanley symmetric functions, is to sum over all reduced words of the given permutation, and then take the descent set of the reverse word. The Schur expansion is then $$ F_w(x) = \sum_{a \in RW(w)} s_{comp(DES(rev(a))}(x) $$ where $rev(a)$ is the reverse of the reduced word, $DES$ is the descent set, $comp$ is the classical map from subsets of $[n-1]$ to compositions, and a Schur polynomial indexed by a composition is computed via the first Jacob-Trudi identity.

The following code also works in Sage:

G = SymmetricGroup(6)
w = G.from_reduced_word(Permutation([2,4,6,1,5,3]).reduced_word())
f = w.stanley_symmetric_function()
s = SymmetricFunctions(QQ).schur()

This example outputs:

s[3, 2, 1, 1] + s[3, 2, 2]
  • $\begingroup$ Can you produce a Stanley symmetric function which is not the image of a skew character for me? $\endgroup$ – Martin Rubey Nov 29 '19 at 14:32
  • 2
    $\begingroup$ @MartinRubey: if a permutation of rows and columns can transform the Rothe diagram into a skew shape, then the Stanley symmetric function is the corresponding skew symmetric function (see Proposition 2.4 of link.springer.com/article/10.1023/A:1022419800503). It's reasonable to conjecture that this is an iff (although Billey-Jokusch-Stanley don't go that far). Anyways, I know $w = 246153$ fails this combinatorial condition so it very likely does not have a skew Schur function as its Stanley symmetric function. $\endgroup$ – Sam Hopkins Nov 29 '19 at 14:39
  • 1
    $\begingroup$ @MartinRubey I just computed $F_{246153} = s_{(3,2,1,1)} + s_{(3,2,2)}$, if that's helpful. $\endgroup$ – Sam Hopkins Nov 29 '19 at 14:57
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    $\begingroup$ @MartinRubey: By way of comparison, by messing around a little I found that $s_{(4,2,2,1)/(1,1)} = s_{(3,2,1,1)} + s_{(3,2,2)} + s_{(4,1,1,1)} + s_{(4,2,1)}$ is kind of close to $F_{246153}$, but I don't know how you would verify $F_{246153}$ is not a skew Schur function beyond a big exhaustive search. $\endgroup$ – Sam Hopkins Nov 29 '19 at 15:03
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    $\begingroup$ Suppose $s_{(3,2,1,1)} + s_{(3,2,2)} = s_{\lambda/\mu}$. By the lattice condition in the Littlewood–Richardson rule, $\lambda/\mu$ has exactly four non-empty rows. By putting $1$s at the far left of rows, one sees $\lambda_1 \le 3$. So $\lambda/\mu$ fits in a $3 \times 4$ box; by rotating, we can assume the partition missing from the top has size $1$ or $2$. If $1$ then we have $(3,2,2,1)/(1)$ containing $(2,2,2,1)$; if $2$ we have $(3,2,2,2)/(2)$ not having $(3,2,1,1)$ or $(3,3,2,1)/(2)$ having $(3,3,1)$ or $(3,3,2,1)/(1,1)$ having $(3,3,1)$. $\endgroup$ – Mark Wildon Nov 30 '19 at 18:11

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