I was initially going to talk about why Deligne’s categories of representations of the symmetric group on a nonintegral number of elements are semisimple generically. This is a rather difficult result, and takes quite a bit of preparation in his paper.  However, I got sidetracked. Instead, I will devote this post to a general discussion of semisimple categories.  According to the material here, it follows that in order to show that Deligne’s categories are semisimple, one has to show that the so-called “partition algebra” is a semisimple ring.

1. Review of semisimple categories

Before we specialize to the case of Deligne’s categories, it may help to go through a little abstract nonsense. Suppose ${\mathcal{C}}$ is a semisimple category. This means that ${\mathcal{C}}$ is abelian, and each object in ${\mathcal{C}}$ is a direct sum of simple objects, where simple means that there is no proper subobject. So for instance, the ${A}$-modules for ${A}$ a semisimple algebra form a semisimple category. The finite-dimensional representations of a semisimple Lie algebra form a semisimple category (though the finite-dimensional condition is necessary; the enveloping algebra is not a semisimple algebra generally).

Now, I want to look at the hom-spaces in a semisimple category. But first, in the next lemma, there is no need to have the semisimplicity asumption, so I drop that.

Remember Schur’s lemma—that lemma in group representation theory, that any morphism between irreducible representations over ${\mathbb{C}}$ is a scalar? The proof of it in different textbooks tends to vary between nonintuitive and clean (depending on the extent of the allegiance of said textbook to category theory). Because when thought of categorically, I claim that it is trivial.

Lemma 1 (Schur, categorical version) Let ${X}$ be a simple object in a ${\mathbb{C}}$-linear abelian category with finite-dimensional hom-spaces. Then ${\hom(X,X) \simeq \mathbb{C}}$. Also, ${\hom(X,Y) =0}$ if ${X,Y}$ are simple and nonisomorphic.

So, let’s prove this. We will first prove that any morphism between simple objects ${X,Y}$ is an isomorphism or zero. If one were not zero, it would have either a nontrivial kernel or cokernel. And this would mean either that ${X}$ had a nontrivial subobject or ${Y}$ a nontrivial subobject—two things that can’t happen for simple ${X,Y}$.

It is now clear that ${\hom(X,Y) = 0}$ when ${X,Y}$ are nonisomorphic, because a nontrivial morphism would be an isomorphism by the above.

Well, then ${\hom(X,X)}$ is a ring where every nonzero element is invertible—that is to say, a division algebra. It is also finite-dimensional over ${\mathbb{C}}$by the assumption on the hom-spaces. But every f.d. division algebra over ${\mathbb{C}}$ is ${\mathbb{C}}$ itself; indeed, if ${\alpha \notin \mathbb{C}}$ belonged to such a division algebra, then ${\mathbb{C}(\alpha)}$ would be a finite extension field (yes, commutative—${\alpha}$ commutes with itself!) and this cannot happen since ${\mathbb{C}}$ is algebraically closed.

In particular, ${\hom(X,X) = \mathbb{C}}$. This proves Schur’s lemma. Not entirely trivial, but at least swift.

So that’s done. I claim then that, in a semisimple category ${\mathcal{C}}$, the hom spaces ${\hom(X,X)}$ is ring-isomorphic to a product of matrix algebras over ${\mathbb{C}}$. This is now straightforward: decompose ${X}$ as a sum of simple objects ${S_1 \oplus S_2 \oplus \dots \oplus S_k}$. Partition ${S_1, \dots, S_k}$ into equivalence classes based on isomorphism and take the sums ${T_j, 1 \leq j \leq l}$ of the ${S_i}$ in each equivalence class. Each ${T_j}$ has hom-spaces isomorphic to a matrix algebra, so the claim is clear.

In particular, the hom-rings of a semisimple category are—surprise, surprise—semisimple algebras!

2. What if the hom-spaces are semisimple?

The 45-million-dollar question now arises whether the opposite might be true. In fact, I think it is, with certain hypotheses: this isn’t really about Deligne’s paper anymore, but it’s something that I learned from Friedrich Knop’s very interesting paper “Tensor envelopes of regular categories.” Knop actually generalizes Deligne’s construction and axiomatizes it to constructing large classes of interesting tensor categories (such as representation categories of wreath products ${S_t \ltimes G^{t}}$ for ${G}$ finite and ${t}$ complex. I may talk more about Knop’s paper later, but right now I am just using it as a source of some fun abstract nonsense.

So suppose we have an additive, ${\mathbb{C}}$-linear category ${\mathcal{C}}$, such that the hom-spaces are finite-dimensional over ${\mathbb{C}}$. Suppose it is pseudo-abelian, i.e. every idempotent has an image. (We don’t want to assume abelianness because a priori we don’t have this for Deligne’s categories.) Finally, suppose the hom-spaces ${\hom(X,X)}$ are semisimple algebras. Does it follow that ${\mathcal{C}}$ is semisimple?

The answer is no. Here is a counterexample. We will describe a category ${\mathcal{C}}$. The objects are formal direct sums of two objects ${X,Y}$. We set ${\hom(X,X) = \mathbb{C}, \hom(Y,Y) = \mathbb{C}}$, and ${\hom(X,Y) = \mathbb{C} \epsilon_0, \hom(Y,X) = \mathbb{C} \epsilon_1}$. Composition is defined in the obvious way on ${\hom(X,X), \hom(Y,Y)}$, but we set ${\epsilon_0 \epsilon_1 = \epsilon_1 \epsilon_0 = 0}$.

It is evident that this is a pseudo-abelian category where the hom-spaces ${\hom(Z,Z)}$ for ${Z \in \mathbb{C}}$ are semisimple algebras (because ${Z}$ can be represented as a sum of copies of ${X}$ and ${Y}$). However, it is not semisimple. In fact, ${\epsilon_0 : X\rightarrow Y}$ is not an isomorphism, and it would have to be if ${\mathcal{C}}$ were semisimple!

OK. So now that we’ve seen what can go horribly wrong, let’s try to develop something that will save us. The problem with ${\epsilon_0}$ was that it didn’t have a section: for every ${Y \rightarrow X}$, ${\epsilon_0}$ was annihilated. This suggests to us a radical (sorry, this is awful) idea: say that a ${\mathbb{C}}$-linear category ${\mathcal{C}}$ is nonnilpotent if for every nonzero ${f: X \rightarrow Y}$ there exists some ${g:Y \rightarrow X}$ with ${gf \neq 0}$.

We will show:

Theorem 2 (Knop) Let ${\mathcal{C}}$ be nonnilpotent, ${\mathbb{C}}$-linear, pseudo-abelian and suppose the hom-rings are semisimple. Then ${\mathcal{C}}$ is semisimple abelian.

I’m basically using the ideas in Sec. 4 of Knop’s paper here but in a modified form; this modification is probably developed somewhere else (such as perhaps Nilpotence, radicaux, et structures monoidales; but I’m not ready to read another heavy paper yet).

So, let’s prove this theorem. The first step is that we can split any ${X \in \mathcal{C}}$ into a sum ${X_1 \oplus \dots \oplus X_n}$ where ${\hom(X_i, X_i) = \mathbb{C}}$. This is because there are $m^2$ orthogonal idempotents in the matrix algebra $M_m(\mathbb{C})$. The point is that the ${X_i}$ are legit candidates for being simple objects, though it is not immediately obvious.

We shall (following Knop) call any ${X \in \mathbb{C}}$ with ${\hom(X,X) = \mathbb{C}}$ to be ${\epsilon}$-simple. So we can write any ${X \in \mathbb{C}}$ as a direct sum of ${\epsilon}$-simple objects. It turns out that a version of Schur’s lemma still works for these guys:

Lemma 3 Let ${X, Y}$ be ${\epsilon}$-simple. Then any map ${X \rightarrow Y}$ is either zero or an isomorphism.

For if ${f: X \rightarrow Y}$ is nonzero, there exists ${g: Y \rightarrow X}$ with ${gf \neq 0}$ by nonnilpotence, so ${gf}$ is a nonzero multiple of the identity. Thus ${f}$ is left-invertible, and similarly it is right-invertible. It is thus an isomorphism.

So, consider a set ${R}$ of ${\epsilon}$-simple objects which represent them all, i.e. every ${\epsilon}$-simple object is isomorphic to one and only one in ${R}$. Then any ${X \in \mathcal{C}}$ is a direct sum of objects in ${R}$, in a unique manner.

I now claim that kernels (and by duality, cokernels) exist. But if ${S}$ is ${\epsilon}$-simple, then any morphism

$\displaystyle S^{\oplus p} \rightarrow S^{\oplus r}$

admits a kernel. Namely, we think of the map as a ${r}$-by-${p}$ matrix ${M}$, or equivalently a morphism ${\mathbb{C}^{p} \rightarrow \mathbb{C}^{r}}$, take the kernel as a map ${\mathbb{C}^{s} \rightarrow \mathbb{C}^{p}}$ (or a matrix ${M'}$) and the associated map ${S^{\oplus s} \rightarrow S^{\oplus p}}$. It is then clear that the map

$\displaystyle 0 \rightarrow \hom(S, S^{\oplus s}) \rightarrow \hom(S, S^{\oplus p}) \rightarrow \hom(S, S^{\oplus r})$

because this is just the sequence of vector spaces

$\displaystyle 0 \rightarrow \mathbb{C}^{s} \rightarrow \mathbb{C}^{p} \rightarrow \mathbb{C}^{s}!$

It follows that the sequence

$\displaystyle 0 \rightarrow \hom(X, S^{\oplus s}) \rightarrow \hom(X, S^{\oplus p}) \rightarrow \hom(X, S^{\oplus r})$

is exact when ${X}$ is a sum of copies of ${S}$. This is also true when ${X}$ is a sum of copies of other ${\epsilon}$-simple objects (everything’s zero!), so it is exact for all ${X}$. I.e. we have the kernel of ${S^{\oplus p} \rightarrow S^{\oplus r}}$. By taking direct sums, we get kernels in general. And then the category is abelian, so semisimple (the ${\epsilon}$-simple objects are actually simple).
We have proved the theorem.

So, next time, we’ll see how this applies to Deligne’s categories, and what we can do with them.