This post continues my series on representation theory in complex rank, begun here with a discussion of Deligne’s interpolation of the representation categories of the symmetric group, introduced in his 2004 paper.

Semisimplicity is the basic structure theorem for Deligne’s categories, and I would be extremely remiss in my discussion of representation theory in complex rank if I did not say something about it.

So, let’s review. In the first post, I explained and motivated the definition of Deligne’s categories ${\mathrm{Rep}(S_t)}$. Incidentally, Deligne did the same for the other classical groups, i.e. ${GL_n, O_n, Sp_{2n}}$, but I shall not discuss them. The categories ${\mathrm{Rep}(S_t)}$, are defined as the pseudo-abelian envelope of the ${\mathbb{C}}$-linear category generated by objects ${\mathfrak{h}^{\otimes p}}$, where the hom-spaces ${\hom( \mathfrak{h}^{\otimes p}, \mathfrak{h}^{\otimes r})}$ are free on the equivalence relations on ${\mathbf{ p+r}}$, and composition is given by a combinatorial expression which is polynomial in the rank ${t}$ (hence interpolable).

Now, we just have an abstract category with formal objects and morphisms corresponding in no obvious way to anything concrete. To prove it is semisimple, we cannot use therefore techniques such as those in the proof of Maschke’s theorem of Weyl’s complete reducibility theorem.

But we can do it by appealing to what I discussed in the second post of this series: by proving that the endomorphism rings are semisimple and the category is nonnilpotent. In fact, since direct products and factor rings of semisimple rings are semisimple, we only need to prove that the algebras ${\hom_{\mathrm{Rep}(S_t)}(\mathfrak{h}^{\otimes p}, \mathfrak{h}^{\otimes p})}$ are semisimple (in addition to nonnilpotence). This endomorphism ring (depending on the size ${p}$ and the rank ${t}$) is an important object, called the partition algebra, and you can look it up e.g. here. But I don’t know how to prove directly that the partition algebra is semisimple. So I will follow Deligne (and Knop) in the (inductive) proof (which will also imply semisimplicity of the partition algebra).

I will do this in two steps. First, I will use a little bit of combinatorics to show that when ${t \notin \mathbb{Z}_{\geq 0}}$, the category ${\mathrm{Rep}(S_t)}$ is nonnilpotent. Next, I will use this to prove semisimplicity.

We will now check that Deligne’s categories are nonnilpotent in the sense of this post. This will be a crucial ingredient in proving semisimplicity.

We will show that

$\displaystyle B: \hom(X,Y) \times \hom(Y,X) \rightarrow \hom(X,X) \rightarrow \mathbb{C}$

for the last map the trace, is nondegenerate. This clearly implies that the category is nonnilpotent, for then composition ${\hom(X,Y) \times \hom(Y,X) \rightarrow \mathbb{C}}$ is nondegenerate (because the trace of zero is zero), and so if ${f: X \rightarrow Y}$ is nonzero, there exists ${g: Y \rightarrow X}$ with ${gf \neq 0}$.

To do this, it reduces to the case of ${X,Y = \mathfrak{h}^{\otimes p}, \mathfrak{h}^{\otimes r}}$ because everything is a direct factor of a direct sum of these. Now we will pull out a bit of general magic for symmetric tensor categories. As is well-known, we have a functorial isomorphism ${\hom(X,Y) \simeq \hom(\mathbf{1}, Y \otimes X^{\vee})}$ and ${\hom(Y,X) \simeq \hom(Y \otimes X^{\vee}, \mathbf{1})}$. The bilinear form ${B}$ corresponds to composition

$\displaystyle \hom(\mathbf{1}, Y \otimes X^{\vee}) \times \hom(Y \otimes X^{\vee}, \mathbf{1}) \rightarrow \hom(\mathbf{1}, \mathbf{1}).$

As a result, we just have to check that

$\displaystyle B_Z: \hom(\mathbf{1}, Z) \times \hom(Z, \mathbf{1}) \rightarrow \mathbb{C}$

is nondegenerate.

For the results I’ve dismissed as “well-known”, readers may look at Deligne’s review of what he calls “multilinear algebra” in his paper. I don’t really understand all this abstract nonsense as well as I should, but at any rate it’s not suited for a blog post, even on a blog with Bourbaki in the title. So I will just assume this. If you don’t feel like doing that, you could also use the fact that these equalities become polynomial identities(!) in ${t}$ in the family of categories ${\mathrm{Rep}(S_t)}$, and their holding at the large nonnegative integers implies them everywhere, a theme that will become recurrent…

Anyway. Now, the unital object ${\mathbf{1}}$ is ${\mathfrak{h}^{\otimes 0}}$. A member of ${\hom(\mathbf{1}, \mathfrak{h}^{\otimes p})}$ (or ${\hom(\mathfrak{h}^{\otimes p}, \mathbf{1})}$ can be represented as a formal sum ${\sum (R_i)}$ of requivalence relations ${R_i}$ on ${\mathbf{ p}}$.

So, to prove ${B}$ nondegenerate, it suffices to show it when ${X = \mathbf{1}}$. Now in this case, ${B_Z}$ nondegenerate for ${Z = \mathfrak{h}^{\otimes p}}$, and we will compute ${B_Z((R), (S))}$ for ${R,S}$ equivalence relations on ${\mathbf{ p}}$.

The problem is, we never really worked out the right way to compose these formal symbols ${(R), (S)}$. Or rather, to be honest, I thought I did, and posted it on my blog earlier. But I was wrong. I’ve revised it since to say that the law is “polynomial in the rank,” but the polynomial I initially said it was was wrong. It’s not a huge deal—I’m just going to end up being slightly less complete with respect to some of the more combinatorial aspects of Deligne’s categories. (If you’re mad at me, you can take a look at some of the references I’ve given there. It won’t interfere with the story.)

But now we need to know what ${B_Z( (R), (S))}$ is, and thus the lack of a computation becomes an obstacle! Fortunately, the computation is easier when one of the objects is ${\mathbf{1}}$.

Lemma 1 Let ${R,S}$ be equivalence relations on ${\mathbf{ p}}$ and let ${(R): \mathbf{1} \rightarrow \mathfrak{h}^{\otimes p}, (S): \mathfrak{h}^{\otimes p} \rightarrow \mathbf{1}}$ the associated morphisms in ${\mathrm{Rep}(S_t)}$. Then ${(S)(R) = t(t-1)\dots (t-s+1) \delta_{RS} \in \mathbb{C} = End(\mathbf{1})}$, where ${s \in \mathbb{Z}_{ \geq 0}}$ depends only on ${R}$.

Here ${\delta_{RS}}$ is the Kronecker delta, and this means that ${B_Z}$ is given by a diagonal matrix whose entries on the diagonal are given by this polynomial in ${t}$. And thus ${B_Z}$ will be nondegenerate for all ${p}$ iff ${t \notin \mathbb{Z}_{\geq 0}}$. So the nonnilpotence of Deligne’s categories will follow once we’ve handled this lemma.

The proof of this lemma will consist of a useful trick, namely the reduction to the integral case. Namely, if this is true for ${t \in \mathbb{Z}_{\geq 0}}$, it must be true for all ${t}$ (because it is a polynomial identity). So suppose ${t = n \in \mathbb{Z}_{\geq 0}}$. In fact, we can suppose that ${n \gg p}$, and we can pretend that we’re working on the ordinary category of representations of ${S_n}$, where ${\mathfrak{h}}$ is the regular representation.

The morphism ${(R): \mathbf{1} \rightarrow \mathfrak{h}^{\otimes p}}$ corresponds to the ${S_n}$-invariant element ${\sum_f e_f \in \mathfrak{h}^{\otimes p}}$, where ${f}$ ranges over functions ${\mathbf{ p} \rightarrow \mathbf{ n}}$ inducing the relation ${R}$. The morphism ${(S): \mathfrak{h}^{\otimes p} \rightarrow \mathbf{1}}$ sends each ${e_g}$ for ${g: \mathbf{ p} \rightarrow \mathbf{ n}}$ inducing ${S}$ to 1. It is thus clear that ${(S)(R) =0}$ if ${S \neq R}$.

What if ${S = R}$? Then we just have to count the number of ${f}$ inducing the relation ${R}$. Now ${R}$ corresponds to a partition of ${\mathbf{ p}}$ into equivalence classes ${C_1 \dots C_s}$, and to say that ${f: \mathbf{ p} \rightarrow \mathbf{ n}}$ induces the relation ${R}$ means that ${f}$ is constant on each ${C_i}$ and sends each ${C_i}$ to a different point in ${\mathbf{ n}}$. There are thus ${n(n-1)(n-s+1)}$ different choices of ${f}$, and the lemma follows.

So, we now know that the Deligne categories are nonnilpotent. In fact, Deligne proves a theorem that whenever the trace form is nondegenerate, the category is necessarily semisimple. I won’t go there, though, because I want to do the inductive proof that actually tells you what the simple objects look like.

2. The inductive argument

And now for the proof of semisimplicity. This is the first big result in Deligne’s paper:

Theorem 2 If ${t \notin \mathbb{Z}_{\geq 0}}$, the category ${\mathrm{Rep}(S_t)}$ is semisimple.

In fact, we will show that the simple objects are parametrized by the partitions (equvilanetly, Young diagrams) of arbitrary size. I will explain why this is only natural at the end.

2.1. The filtration on ${\mathrm{Rep}(S_t)}$

In proving this, we should start by noting that ${\mathrm{Rep}(S_t)}$ is a pretty big category because in ${\mathfrak{h}^{\otimes p}}$, ${p}$ is allowed to get large. So we define a full subcategory ${\mathrm{Rep}(S_t)^{(N)}}$ which is the pseudo-abelian envelope of the ${\mathfrak{h}^{\otimes p}, p \leq N}$. Thus, we get a filtration on Deligne’s categories. Note that the elements of the filtration are not tensor categories, only pseudo-abelian ones.

The idea will be to prove that the subcategories ${\mathrm{Rep}(S_t)^{(N)}}$ are semisimple, and this we do by induction on ${N}$. It is clearly true for ${N=-1}$ (the zero category).

2.2. Splitting off a chunk

Now suppose we know that ${\mathrm{Rep}(S_t)^{(N-1)}}$ is semisimple, with simple objects ${\{A_i\}_{i \in Q_N}}$ for some set ${Q_N}$. The ${A_i}$ are not a priori simple when regarded as objects of ${\mathrm{Rep}(S_t)^{(N)}}$ (which may, for all we know, be a non-abelian category…how evil), but they are still ${\epsilon}$-semisimple: that is to say, their endomorphism rings have dimension 1.

So, I claim now that if ${X \in \mathrm{Rep}(S_t)^{(N)}}$, there is a splitting

$\displaystyle X = X' \oplus \bigoplus_{Q_N} A_i^{\oplus n_i }$

where ${\hom(X', W) = \hom(W, X')}$ when ${W \in \mathrm{Rep}(S_t)^{(N-1)}}$. In other words, we can excise the simple objects in the “smaller” category ${\mathrm{Rep}(S_t)^{(N-1)}}$ away. The idea is simple. If ${\hom(X, W) \neq 0}$ for some ${W \in \mathrm{Rep}(S_t)^{(N-1)}}$, then this is true for some ${W}$ simple (by semisimplicity of ${\mathrm{Rep}(S_t)^{(N-1)}}$) and there is a nonzero ${f: X \rightarrow W}$ with ${W}$ simple. Take ${g: W \rightarrow X}$ such that ${fg \neq 0}$ (OK because composition is a nondegenerate pairing of vector spaces); then by ${\epsilon}$-simplicity ${fg}$ is a multiple of the identity. So wlog ${fg}$ is the identity and ${W}$ occurs as a direct factor in ${X}$ (we use pseudo-abelianness to get a complement).

So we split of a factor of ${X}$ isomorphic to a simple object in ${\mathrm{Rep}(S_t)^{(N-1)}}$. We can keep on repeating this; the process will not last forever, since the hom-spaces are finite-dimensional. And this proves the claim.

2.3. Action of the symmetric group ${S_N}$

Now, for the next step. We take ${X = \mathfrak{h}^{\otimes N}}$; this object, loosely speaking, is what introduces the additional simple objects in ${\mathrm{Rep}(S_t)^{(N)}}$ that aren’t in ${\mathrm{Rep}(S_t)^{(N-1)}}$. We have a splitting

$\displaystyle \mathfrak{h}^{\otimes N} = V_N \oplus W_N, \quad W_N \in \mathrm{Rep}(S_t)^{(N-1)}$

where an endomorphism ${f: \mathfrak{h}^{\otimes N} \rightarrow \mathfrak{h}^{\otimes N} }$ factors through ${W_N}$ iff it factors through an element of ${\mathrm{Rep}(S_t)^{(N-1)}}$.

We now invoke a fact familiar from Schur-Weyl duality: the symmetric group(!) ${S_N}$ acts on ${\mathfrak{h}^{\otimes N}}$ by permutation—ok because we are dealing with a symmetric monoidal category ${\mathrm{Rep}(S_t)}$. So we have a map

$\displaystyle \mathbb{C}[S_N] \rightarrow End(V_N).$

I claim that it is an isomorphism. Indeed, it easy to see that the action of ${\sigma \in S_N}$ corresponds to the equivalence relation on ${\mathbf{ N} + \mathbf{ N}}$ obtained by identifying each point of ${\mathbf{ N}}$ (here ${\mathbf{ N}}$ denotes a set with ${N}$ elements, not to be confused with the natural numbers ${\mathbb{N}}$—sorry for the choice of notation) with its image under ${\sigma}$. So the ${\sigma \in S_N}$ act linearly independently on ${\mathfrak{h}^{\otimes N}}$. Moreover, any map ${\mathfrak{h}^{\otimes n} \rightarrow \mathfrak{h}^{\otimes N}}$ that factors through a direct sum of ${\mathfrak{h}^{\otimes m}, m must be a sum of relations ${(C)}$ for ${C}$ an equivalence relation on ${\mathbf{ N} + \mathbf{ N}}$ that identifies a pair of distinct points in ${\mathbf{ N}}$ (this can be checked directly). The ${\sigma \in S_N}$ don’t do that, so any nontrivial linear combination of them cannot factor through anything in ${\mathrm{Rep}(S_t)^{(N-1)}}$. And the map is thus an injection. It is even a surjection, because conversely any relation on ${\mathbf{ N} + \mathbf{ N}}$ which is not of the form ${(n, \sigma(n))}$ for some bijection ${\sigma: \mathbf{ N} \rightarrow \mathbf{ N}}$ factors through a sum of smaller ${\mathfrak{h}^{\otimes m}}$.

In particular, the endomorphism ring of ${V_N}$ is semisimple.

But, every object in ${\mathrm{Rep}(S_t)^{(N)}}$ is a direct summand of copies of ${V_N}$ and objects in ${\mathrm{Rep}(S_t)^{(N-1)}}$. All these objects have semisimple endomorphism rings, and thus so does every object in ${\mathrm{Rep}(S_t)^{(N)}}$. Since the category is nonnilpotent, it must be semisimple by the material in this post. (Note incidentally that if ${e}$ is an idempotent in a semisimple algebra ${A}$, then ${eAe}$ is a semisimple algebra too—a fact easily proved from the structure theorem that ${A}$ is a product of matrix algebras.)

Whew! That was a fairly long proof. But we’re now done. We still don’t know what the simple objects look like. For that, I will resume this series in a later post.