EDIT (3/16/11): When I first read this question, I thought "hmmm, Weyl algebra? Really? I feel like I never hear people say they're going to categorify the Weyl algebra, but it looks like that's what the question is about..." Now I understand what's going on. Not to knock the OP, but there's a much bigger structure here he left out. If you have any $S_n$ representation $M$, you get a functor $$\operatorname{Ind}_{S_m\times S_n}^{S_{m+n}}-\otimes M: S_m-\operatorname{rep}\to S_{m+n}-\operatorname{rep}$$ and these functors all have adjoints which I won't bother writing down. All of these together categorify a Heisenberg algebra, which is what Khovanov proves in the paper linked below (though cruder versions of these results on the level the OP was talking about are much older, at least as far back as Leclerc and Thibon).
There is a much more general story here, though one my brain is not very up to explaining it this afternoon, and unfortunately, I don't know of anywhere it's summarized well for beginners.
So, how you you prove the restriction rule you mentioned above? You note that the restriction of a $S_n$ rep to an $S_{n-1}$ rep has an action of the Jucys-Murphy element $X_n$ which commutes with $S_{n-1}$. The different $S_{n-1}$ representations are the different eigenspaces of the J-M element.
So, one can think of "restrict and take the m-eigenspace" as a functor $E_m$; this defines a direct sum decomposition of the functor of restriction.
Of course, this functor has an adjoint: I think the best way to think about this is as $F\_m=(k[S\_n]/(X\_n-m))
\otimes\_{k[S\_{n-1}]} V$.
These functors E_m,F_m satsify the relations of the Serre relations for $\mathfrak{sl}(\infty)$. Over characteristic 0, these are all different, and you can think of this as an $\mathfrak{sl}(\infty)$. If instead, you take representations over characteristic p, then E_m=E_{m+p} so you can think of them as being in a circle, an affine Dynkin diagram, so one gets an action of $\widehat{\mathfrak{sl}}(p)$.
Similar categorifications of other representations can deconstructed in general by looking at representations of complex reflection groups given by the wreath product of the symmetric group with a cyclic group. So, Sammy, you shouldn't rescale, you should celebrate that you found a representation with a different highest weight (also, if you really care, you should go talk to Jon Brundan or Sasha Kleshchev; they are some of the world's experts on this stuff).
EDIT: Khovanov has actually just posted a paper which I think might be relevant to your question.
This is a nice question. I have never seen this before.
Let us write
$$\mathcal B_u = \{ V_0 \subset V_1 \subset \cdots \subset V_{n-1} \subset V_n = \mathbb C^n : u V_i \subset V_i \}. $$
Let $ \lambda $ be the Jordan type of $ u $. Then we can partition $ B_u $ into locally closed subsets depending on the Jordan type of $ u \rvert_{V_{n-1}} $; such a Jordan type $ \mu$ must necessarily be obtained by removing one box from $ \lambda$. So we have
$$ \mathcal B_u = \bigsqcup_\mu B_u^\mu \quad B_u^\mu = \{ V_\bullet : u\rvert_{V_{n-1}} \text{ has Jordan type $ \mu$ } \}.$$
Let $ G(n-1, n)_u $ denote the set of all $n-1$-dimensional $u$-invariant subspaces of $ \mathbb C^n $ and partition $ G(n-1,n)_u $ into locally closed subsets $ G(n-1,n)_u^\mu $ according to the Jordan type of the restriction of $ u $ to the subspace.
We have $ \mathcal B_u \rightarrow G(n-1,n)_u $ and $ \mathcal B^\mu_u $ is the preimage of $ G(n-1,n)^\mu_u$. Note that the fibre of $ \mathcal B_u \rightarrow G(n-1,n)_u $ is $ \mathcal B_{u\rvert_{ V_{n-1}}}$.
Now, here are two facts that I don't see right away, but which must be
true:
- Each piece $ \mathcal B_u^\mu $ has the same dimension.
- $ G(n-1,n)^\mu_u $ is irreducible and simply-connected (or at least that the bundle $\mathcal B^\mu_u \rightarrow G(n-1,n)^\mu_u$ is topologically trivial on components).
Edit: I think these facts should be contained in the classic paper by Spaltenstein https://www.sciencedirect.com/science/article/pii/S138572587680008X
Assuming these facts, we get
$$H(\mathcal B_u) = \bigoplus_\mu H(\mathcal B_u^\mu) = \bigoplus_\mu H(\mathcal B_{u(\mu)})$$
where again the direct sum ranges over all partitions made by deleting one box from $\lambda$ and where $ u(\mu)$ denotes a unipotent element of Jordan type $ \mu$, and $ H(X) $ denotes top (co)homology. This gives the desired decomposition.
Best Answer
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.