# Symmetric Powers, Tableau and Wreath Products

Let V and W be irreducible representations of $S_n$ and $S_m$ over a field of characteristic 0. Then the Littlewood-Richardson coefficients allow us to compute the isomorphism type of the induced $S_{n+m}$-module V⊗W↑. This induction comes from the inclusion

$S_n\times S_m \rightarrow S_{n+m}$.

Now suppose V=W. Then V⊗V↑ is a $S_{2n}$-module. But actually there's a symmetry coming from the symmetric monoidal category structure, so there is another induction up to an $S_{2n}$-module structure:

Extend the action of $S_n\times S_n$ on V⊗V by including the symmetry $c_V$, this naturally extends the group to the wreath product $S_n\sim S_2$. Induction along

$S_n\sim S_2 \rightarrow S_{2n}$

gives the representation that I want:

$(V\otimes V)_{S_n\sim S_2}\uparrow^{S_{2n}} \hookrightarrow V\otimes V\uparrow^{S_{2n}}$.

Using the Littlewood-Richardson rules we know the structure of the last term in terms of semi-standard skew tableau. My question is, how do we characterise the inclusion?

• David's provided exactly the answer I wanted (a win for MathOverflow!). I've edited the question (removed a confusing wrong bit) so it should be clearer for anyone else interested. – James Griffin Nov 18 '09 at 10:15

## 3 Answers

You want to read Splitting the square of a Schur function into its symmetric and antisymmetric parts, where this question is answered in terms of domino tableaux. I am also a big fan of Domino tableaux, Schützenberger involution, and the symmetric group action which, to my mind, gives the "right" formulation of the domino tableaux rule.

• Thank you David, that's just the thing. The way the first paper presents its results really does make the whole process rather mysterious, I'll try to find time to look at the second paper, especially as I might need S^nV at some point. From the first paper on page 202 I particularly liked: "But Littlewood could not find a general simple method of discriminating the tableaux pertaining to S^2(S_I)." That makes me feel much better :). – James Griffin Nov 18 '09 at 9:29

My understanding was that induction from wreath products was supposed to correspond to plethystic substitution of symmetric functions. But I don't really understand that, so I can't really say.

• Plethysistic substitution corresponds to composition of Schur functors. I have been told that Frobenius reciprocity turns composition of Schur functors into induction from wreath products, but I have never dotted that particular i myself. – David E Speyer Nov 18 '09 at 13:04

Is there a connection to zonal polynomials, or is that what David just said?

• If they are related, I don't know it, and would be interested to learn. – David E Speyer Nov 17 '09 at 22:49