Making the branching rule for the symmetric group concrete

Question

This question concerns the characteristic $0$ representation theory of the symmetric group $S_n$. I'm a topologist, not a representation theorist, so I apologize if I state it in an odd way.

First, a bit of background. The finite-dimensional irreducible representations of $S_n$ are given by the Specht modules $S^{\mu}$. Here $\mu$ is a partition of $n$, which is best visualized as a Young diagram. There are classical rules for restricting $S^{\mu}$ to $S_{n-1}$ and inducing $S^{\mu}$ to $S_{n+1}$ (these are known as branching rules). Namely, we have the following.

1. The restriction of $S^{\mu}$ to $S_{n-1}$ is isomorphic to the direct sum of the $S_{n-1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of removing a box from the Young diagram for $\mu$ (while staying in the world of Young diagrams).

2. The induction of $S^{\mu}$ to $S_{n+1}$ is isomorphic to the direct sum of the $S_{n+1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of adding a box to the Young diagram for $\mu$ (again while staying in the world of Young diagrams).

These two rules are equivalent by Frobenius reciprocity.

There is a nice concrete proof of the restriction rule (I believe that it is due to Peel, though I first learned about it from James's book "The representation theory of the symmetric groups"). Assume that the rows of the Young diagram for $\mu$ from which we can remove a box are $r_1 < \ldots < r_k$, and denote by $\mu_i$ the partition of $n-1$ obtained by removing a box from the $r_i^{\text{th}}$ row of $\mu$. There is then a sequence $$0=V_0 \subset V_1 \subset \cdots \subset V_k = S^{\mu}$$ of $S_{n-1}$-modules such that $V_i/V_{i-1} \cong S^{\mu_i}$. In fact, recalling that $S^{\mu}$ is generated by elements corresponding to Young tableaux of shape $\mu$ (known as polytabloids), the module $V_i$ is the subspace spanned by the polytabloids in which $n$ appears in a row between $1$ and $i$.

Question : Is there a similarly concrete proof of the induction rule (in particular, a proof which does not appeal to Frobenius reciprocity and the restriction rule)?

Answer 1

Chapter 17 of James' book, The Representation Theory of the Symmetric Groups, is about modules which have Specht filtrations (where the field is arbitrary) and includes induction of the Specht modules as a special case. (See in particular Example 17.16.) It gives a concrete proof of the branching rule for induction without using restriction.

Answer 2

You are interested in characteristic zero and most books are interested in characteristic $p>0$. The simplest construction of the irreducible representations over $\mathbb{Q}$ is Young's orthogonal form. The reason for using the Specht modules is that they are defined over $\mathbb{Z}$.

For the orthogonal form the matrix representing the generator $s_i$ is block diagonal and all blocks have size 1 or 2. Let $T$ be a Young tableau so the aim is to define $T.s_i$. Then $i$ and $i+1$ are in two boxes of $T$ and we can swap them but this may or may not be a Young tableau. If this does not give a Young tableau then $T.s_i=\pm T$. If it does, then these two tableaux span a two dimensional space fixed by $s_i$. This $2\times 2$ matrix only depends on the difference between the hooklengths of the two boxes. In this form the induction and restriction rules are manifest (in fact that is how you arrive at this form).

If you want to work with the Specht modules (or any other form) then you look at the centraliser algebra of $\mathbb{Q}S(n-1)$ in $\mathbb{Q}S(n)$ (group algebras, not groups). This algebra is commutative and has dimension $n$ and over $\mathbb{Q}$ it has a basis of orthogonal idempotents. These idempotents give the decompositions of the induced representation and the restricted representation corresponding to the branching rules explicitly.

The reference Victor gives sets out to develop the theory from this point of view.

Answer 3

I guess this is an old thread now, but I always thought that James' proof was a little indirect and not very explicit. A much nicer proof was given by Steen Ryom-Hansen in his paper “Grading the translation functors in type A.” Journal of Algebra 274, no. 1 (2004): 138–63; see arXiv:math/0301285.

To blow my own trumpet, a "cellular algebra" proof in the cyclotomic case (which includes the symmetric groups) is given in arXiv:0903.4493. For the corresponding result in the graded setting, following Ryom-Hansen, see arXiv:1008.1462.

All of these arguments work over arbitrary rings.

Answer 4

Here is how I understand induction/restriction as this has come up in comments. This is well-known and fundamental in sheaf theory.

Let $A$ be a finite dimensional algebra over a field $K$ and $B$ a subalgebra. Identify representations with right modules. (This is for simplicity; this framework extends to other contexts).

Restriction is a functor from $A$-modules to $B$-modules and is exact (obviously). Induction is the functor $M_B \mapsto M_B \otimes {}_B A_A$ from $B$-modules to $A$-modules. Coinduction is the functor $M_B \mapsto Hom_B({}_A A_B, M_B)$ from $B$-modules to $A$-modules. Then induction is left adjoint to restriction and coinduction is right adjoint to restriction.

For the case $G$ is a finite group and $H$ a subgroup $A=KG$ and $B=KH$ then $A_B$ is a free $B$-module. A basis is given by a set of coset representatives (there is a choice here). This implies that induction and coinduction are isomorphic (depending on the choice?). In general I think the condition is that $A_B$ is projective (please correct me if this is not correct).

Then let $K_0(A)$ be the Grothendieck group of projective $A$-modules; a covariant functor using induction. Let $K^0(A)$ be the Grothendieck group of $A$-modules; a contravariant functor using restriction. Then we have a non-degenerate pairing given by
$$ \langle [P],[M]\rangle = \dim Hom(P,M) $$ Then induction and restriction become adjoint (as linear maps). I have never known whether it is coincidental that the two uses of adjoint are connected this way.

A straightforward, but not in the text books, observation is that if $F$ is any of induction, restriction, coinduction then $End(F)$ is isomorphic to the centraliser of $B$ in $A$. The extreme cases are $B=K$ when the centraliser is $A$ and $B=A$ when the centraliser is the centre of $A$. Hence idempotents in the centraliser decompose these functors. This is the same algebra in all three cases. Does this avoid reciprocity? I can't make my mind up.