Making the branching rule for the symmetric group concrete

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)?

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I am fairly certain that you can automatically convert your proof for restriction into a proof for induction using "categorified" Frobenius reciprocity. To me the most natural approach would be through lifting $S_n$-modules to $\mathcal{H}'_n$-modules (i.e. modules over the degenerate, or graded, affine Hecke algebra). A useful reference is Okounkov and Vershik, "A new approach to representation theory of symmetric groups", although I can't recall whether this specific result is there. –  Victor Protsak Jul 27 '10 at 22:07
Interesting! I'm not familiar with "categorified" Frobenius reciprocity. One nice thing about the above proof is that everything in it is defined over $\mathbb{Z}$. I always thought of Frobenius reciprocity as being tied to character theory and thus unable to say much over $\mathbb{Z}$. Was I mistaken? –  Andy Putman Jul 27 '10 at 22:11
I am not sure whether this works in abstract nonsense generality, but it works here. One word of caution: just because Specht modules can be defined over $\mathbb{Z}$ doesn't mean that they $\textit{exhaust}$ all integral representations of $S_n$ (see recent MO posts about the cases $n=2$ and $n=3$, which is about as far as one can get). –  Victor Protsak Jul 28 '10 at 2:06
@Andy: restriction is an exact functor on representations. For finite groups it has a left and right adjoint, induction and coinduction. These are isomorphic, but noncanonically so, and finite group theorists tend to use one functor, while Lie theorists use the other when they say "induction". Frobenius reciprocity, at the level of representations, is just the statement of the adjunction(s), so pretty much the definition of induction, if you are thinking in terms of adjoints. I'm assuming this is what Victor means by "categorified" reciprocity? It works just find over $\mathbb Z$. –  Kevin McGerty Jul 28 '10 at 2:26

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.

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It's useful to remind people that recent books (for mathematicians) tend to emphasize prime characteristic: the basic theory is in good shape already in characteristic 0. Specht modules give the right starting point for reduction mod $p$, but then it's complicated. Books directed at physicists or applied math people tend to emphasize concrete computations with characters and may be closer to the spirit of the question. In any case words like "induction" do need to be used carefully, as pointed out by Kevin. –  Jim Humphreys Jul 28 '10 at 16:42

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.

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No, the iso between coinduction and induction doesn't depend on the choice. Actually I have promised to write some notes about this a week ago but probably they will have to wait yet a few weeks longer. –  darij grinberg Jul 28 '10 at 19:15
While we're at it, two more remarks: (1) Induction are coinduction are isomorphic (as functors) if $\left|G:H\right|<\infty$, even when $G$ and $H$ may be infinite. (2) Restriction is similarly two-faced: it's both $\mathrm{Hom}_{A}\left(M_A,\ _B A_A\right)$ and $M_A \otimes_A \left(_A A_B\right)$ (this time the isomorphism actually works for any $A$ and $B$, methinks). This shows why Frobenius reciprocity is just the adjunction between Hom and $\otimes$, and also that there is a mirror version of Frobenius reciprocity, which is rarely discussed because everybody just talks about characters. –  darij grinberg Jul 28 '10 at 19:23