# Deligne’s “La Categorie des Representations du Groupe Symetrique S_t, lorsque t n’est pas un Entier Naturel.”

In the requests section, Noah asks for help understanding Deligne’s paper “La Categorie des Representations du Groupe Symetrique S_t, lorsque t n’est pas un Entier Naturel.” Peter Arndt refers us to two papers of Knop extending the construction. I’ve been reading them, and I think I understand what’s going on. Moreover, this material is making me think of combinatorial question which seem interesting to me, although others may already have thought about them. I should warn everyone that I had never come across this material until Noah’s question, so there is a risk that I will say something completely ignorant. I also have made no attempt to make my notation match anyone else’s, because I don’t like their notation; if someone wants to compile a dictionary, that would be great.

Nonetheless, here is my attempt at explaining Deligne’s construction in English, with only low level category theory, and with pictures!

Let $t$ be a parameter, which for the moment I want to think of as a very large integer. Let $S_t$ be the symmetric group of permutations of $t$ letters. Let $V$ be the standard $t$-dimensional representation of $S_t$, over some ground field $k$. Let $CRep(S_t)$ be the category whose objects are the vectors spaces $k$, $V$, $V^{\otimes 2}$, $V^{\otimes 2}$, $V^{\otimes 3}$ … and where $Hom(V^{\otimes m}, V^{\otimes n})$ is the $S_t$ equivariant homorphisms from $V^{\otimes m}$ to $V^{\otimes n}$. Tensor products, duals and traces are defined in the obvious way on $CRep(S_t)$. (Every object of $CRep(S_t)$ is self-dual.) By the way, the $C$ stands for combinatorial — these are the representations of $S_t$ that are combinatorially obvious.

Let’s try to give a combinatorial description of the category $CRep(S_t)$. First of all, let’s find a basis for the vector space $Hom(V^{\otimes m}, V^{\otimes n})$. This is the space of $S_t$ invariants in $V^{\otimes (m+n)}$. Now, the dimension of $V^{\otimes (m+n)}$ is $t^{m+n}$. It has an obvious basis which I’ll label $e_{a_1 a_2 \cdots a_m}^{b_1 b_2 \cdots b_n}$ where each of $a_1$, $a_2$, …, $a_m$, $b_1$, $b_2$, …, $b_n$ is an index from $\{ 1,2,\ldots, t \}$.

The group $S_t$ permutes the basis elements $e_{a_1 a_2 \cdots a_m}^{b_1 b_2 \cdots b_n}$, and we get an $S_t$-invariant for each orbit. For example, when $m=2$ and $n=1$, the space of $S_t$-invariants is five dimensional, with basis:

$A(121') := \sum_{i} e_{ii}^i$

$A(12, 1'):= \sum_{i \neq j} e_{ii}^j$

$A(11',2):=\sum_{i \neq j} e_{ij}^i$

$A(21', 1):=\sum_{i \neq j} e_{ji}^i$

$A(1,1',2) := \sum_{i\neq j \neq k \neq i} e_{ij}^k$.

Crucial Difficulty: Well, almost. If $t$ is too small, some of these can be zero. For example, $A(1,1',2)$ vanishes for $t=2$, and all but $A(121')$ vanish when $t=1$. For now, we’ll think of $t$ as very large, and ignore this issue.

In general, we get one basis element of $Hom(V^{\otimes m}, V^{\otimes n})$ for each equivalence relation on $\{ 1,2, \ldots m, 1', 2', \ldots, n' \}$. So the dimension of $Hom(V^{\otimes m}, V^{\otimes n})$ is given by the $(m+n)$-th Bell number. Pictorially, we can represent the basis elements above by figures:

It turns out to be useful to work with a slightly different basis. This basis will also be indexed by equivalence relations on $\{ 1,2, \ldots m, 1', 2', \ldots, n' \}$, but we will only impose equalities in the sum, not inequalities. Continuing with the $m=2$, $n=1$ example, our basis is

$B(121') := \sum_{i} e_{ii}^i$

$B(12, 1'):= \sum_{i, j} e_{ii}^j$

$B(11',2):=\sum_{i, j} e_{ij}^i$

$B(21', 1):=\sum_{i, j} e_{ji}^i$

$B(1,1',2) := \sum_{i,j,k} e_{ij}^k$.

Note that the $B$-basis can be expressed by an upper-triangular matrix in terms of the $A$ basis. For example,

$B(1,1',2)=$

$A(1,1',2) + A(11',2)+ A(12,1')+A(1'2,1)+A(11'2)$.

The entries of the inverse matrix are called the Mobius function of the partition lattice and are given by a certain product of factorials. Pictorially, we will represent the $B$‘s by the same pictures as the $A$‘s, but we will use black ink instead of azure.

Now, how do we multiply the $B$‘s? That is to say, given $B_1$ in $Hom(V^{\otimes i}, V^{\otimes j})$ and $B_2$ in $Hom(V^{\otimes j}, V^{\otimes k})$, how can we expand $B_1 \circ B_2$ in the $B$-basis for $Hom(V^{\otimes i}, V^{\otimes k})$? I urge you to do a few examples, and then check them against the following description:

We’ll abbreviate $I := \{ 1,2,\ldots, i \}$, $J = \{ 1',2',\ldots, j' \}$ and $K = \{ 1'', 2'', \ldots, k''\}$. Let $\sim_1$ and $\sim_2$ be equivalence relations on $I \sqcup J$ and $J \sqcup K$; let $\sim_1 \sim_2$ be the equivalence relation they generate on $I \sqcup J \sqcup K$. Let $d$ be the number of equivalence classes in $\sim_1 \sim_2$ which have no representatives in $I$ or $K$. Then

Key Equation: $B(\sim_1) \circ B(\sim_2) = t^d B( ( \sim_1 \sim_2)|_{I \sqcup K})$.

This is probably easier to explain with a picture:

The factor of $t$ comes from the blob on the left, which is an equivalence class containing only elements of $J$.

Now, I can explain what Deligne does when $t$ is not an integer. We define $CRep(t)$ to be alatex k$-linear category whose objects are finite sets. $Hom(I, J)$ is the $k$-vector space with basis the set of equivalence relations on $I \sqcup J$, and composition is by the Key Equation. Note that I write $CRep(S_t)$ when I am actually thinking of representations of $S_t$, and $CRep(t)$ for the category which exists for any value of $t$. There is an obvious way to define tensor products and duals, and that gives us, by general nonsense, a way to define a trace. I don’t really understand that general nonsense, so I’ll tell you the result: let $\sim_1$ be an equivalence relation on $I_1 \sqcup I_2$, $\sim_2$ an equivalence relation on $I_2 \sqcup I_3$ and so forth, with $I_r$ an equivalence relation on $I_r \sqcup I_1$. Let $\sim_1 \sim_2 \cdots \sim_r$ be the equivalence relation they generate on $I_1 \sqcup I_2 \cdots \sqcup I_r$, and $D$ the number of equivalence classes of $\sim_1 \sim_2 \cdots \sim_r$. Then the trace of $B(\sim_1) \circ B(\sim_2) \circ \cdots B(\sim_r)$ is $t^D$. Pictorially, take the pictures of black blobs I’ve been drawing to encode $\sim_1$, $\sim_2$, …, $\sim_r$ and wrap them around a cylinder. Let $D$ be the total number of blobs you get. Then the trace is $t^D$. This is our basic construction; we now must address two technical issues. These are largely independent of each other so, if you don’t understand the next section, feel free to skip to the one after it. ## Passing to the Karoubi envelope We now have defined the category $CRep(t)$. But we would rather have an analogue of $Rep(S_t)$ — the category of all representations of $S_t$. We can already feel the influence of $Rep(S_t)$ hiding in $CRep(t)$. For example, consider the ring $Hom(V, V)$. (We will return to this example several times.) It is spanned by two elements, $e:=B(11')$, which is the multiplicative identity, and $v:=B(1, 1')$. We have $v^2 = t v$. Set $\pi = (1/t) v$. (Assuming $t$ is invertible.) Then $\pi$ is idempotent, meaning that $\pi \circ \pi = \pi$. In most nice categories, having an idempotent in $Hom(V,V)$ would allow us to split $V$ as $V_1 \oplus V_2$. Indeed, when $t$ is an integer, $V$ has such a splitting in $Rep(S_t)$. The $S_t$-representation $V$ has a trivial summand, and $\pi$ is precisely the projection onto this summand. Exercise: Check this! In order to build our analogue of $Rep(S_t)$, we will discuss a construction due to Max Karoubi. In any category, let $T$ be an object and $\pi$ an element of $Hom(T, T)$ satisfying $\pi \circ \pi = \pi$. By definition, an image of $\pi$ is an object $S$ with morphisms $p: S \to T$ and $q: T \to S$ such that $S \to T \to S$ is the identity and $T \to S \to T$ is $\pi$. It is a pleasant exercise that, if $(S, p, q)$ is an image of $\pi$ then, for any other object $X$, we have canonical bijections $Hom(X,S) \cong \{ f \in Hom(X,T) : f \circ \pi = f \}$ and $Hom(S,X) \cong \{ f \in Hom(T,X) : \pi \circ f = f \}$. Using these equations, a standard argument shows that images are unique up to unique isomorphism. What is more interesting, though, is that there is a way to formally adjoin an image for any idempotent! Given $(T, \pi)$, one formally adds an object $S$ to the category, and uses the above equations to define the morphisms to and from $S$. There are a lot of details to check, but they all work out. Given any category, the Karoubi envelope of that category is the result of formally adding images for all idempotents. (You don’t need to iterate the procedure; taking the Karoubi envelope might create some new idempotents, but they will already have images.) So, Deligne’s analogue of $Rep(S_t)$, when $t$ is an arbitrary parameter, is the Karoubi envelope of $CRep(t)$. I’ll denote this by $Rep(t)$. ## The Quotient by the Radical So far, we have avoided talking about the Crucial Difficulty above. The time has come to face it — for every positive integer value of $t$, the categories $Rep(t)$ and $Rep(S_t)$ are inequivalent! We can already see the difficulty in the case of $CRep$. When $t=1$, we noticed above that we should have $A(1, 1')=0$. In the $B$-basis, this says $B(11')=B(1,1')$ or, using the notation we introduced above, $e=v$.. Similarly, when$t=2\$, we should have $A(1,1',2)=0$, which gives a certain nontrivial linear relations between the $B$‘s in $Hom(V^{\otimes 1}, V^{\otimes 2})$.

If we fix $N$ and only consider the subcategories of $CRep(t)$ and $CRep(S_t)$ induced by the objects $k$, $V^{\otimes 1}$, $V^{\otimes 2}$, …, $V^{\otimes N}$ then, for $t$ sufficiently large, these subcategories will coincide. But, no matter how large $t$ gets, there will always be some relations between the $B$‘s in $CRep(S_t)$ that we don’t see in $CRep(t)$.

If we think about $Rep(S_t)$ and $Rep(t)$ instead of $CRep$, things are worse; we actaully get objects in one category that aren’t in the other. One can (and should!) check that $B(11')-B(1,1')/t$ is idempotent in $CRep(t)$, so it has an image in $Rep(t)$. In $Rep(S_1)$, this image is isomorphic to the zero object, but in $Rep(1)$ it is not.

All of these issues can be fixed by a general construction. I’m not clear on what generality this works in. I believe the details are in “Nilpotence, radicaux et structures monoïdales“, but I haven’t read that paper.

Consider any category $\mathcal{C}$ with tensor products, duals and (hence) traces. I’m not sure what axioms it should obey, so just think that it should be something like $Rep(t)$. For any objects $X$ and $Y$, define $N(X,Y)$ to be the set of $f \in Hom(X,Y)$ such that $Tr(fg)=0$ for all $g \in Hom(Y,X)$. The collection of $N(X,Y)$, as $(X,Y)$ runs through all pairs of objects in $\mathcal{C}$, is called the radical of $\mathcal{C}$. The radical is a two-sided ideal, meaning that $N(X,Y)$ is closed under addition and, for any $f \in Hom(W,X)$ and any $g \in Hom(Y,Z)$, the product $fN(X,Y)g$ is in $N(W,Z)$.  This means that we can form a quotient category $\overline{\mathcal{C}}$, which has the same objects, but where $\overline{Hom}(X,Y) = Hom(X,Y)/N(X,Y)$. Any computation with traces will give the same result in $\mathcal{C}$ and $\overline{\mathcal{C}}$, but some morphisms will be equal in $\overline{\mathcal{C}}$ that were unequal in $\mathcal{C}$.

Let’s see how $Hom(V, V)$ differes from $\overline{Hom}(V, V)$. (It doesn’t matter whether we do this computation in $Rep(t)$ or $CRep(t)$.) The vector space $Hom(V,V)$ is two dimensional, with basis $e=B(11')$ and $v= B(1, 1')$. Let’s write $\langle f,g \rangle$ for $Tr(fg)$. We have (check!)

$\begin{pmatrix} \langle e,e \rangle & \langle e,v \rangle \\ \langle v,e \rangle & \langle v,v \rangle \end{pmatrix} = \begin{pmatrix} t & t \\ t & t^2 \end{pmatrix}.$

When $t \neq 0$, $1$, which matrix is nonsingular, so $N(V,V) = \{ 0 \}$. But when $t=1$, then $e-v$ is in $N(V,V)$. Thus, in $\overline{Hom}(V,V)$, we have $e=v$, which is what should happen for actual representations of $S_t$.

It turns out that, when $t$ is not a nonnegative integer, $Rep(t) = \overline{Rep}(t)$. Moreover, this is a semi-simple abelian category where the isomorphism classes of simple objects are in bijection with partitions (if I understand correctly).  When $t$ is a nonnegative integer, then $\overline{Rep}(t)$ is a nontrivial quotient of $Rep(t)$, and is isomorphic to $Rep(S_t)$.

## Summary

Deligne builds a family of categories, $CRep(t)$. It is a family in the sense that the objects and morphisms $CRep(t)$ can be viewed as independent of $t$, while the multiplication maps $Hom(X,Y) \times Hom(Y,Z) \to Hom(X,Z)$ vary polynomially with $t$. When $t$ is not a nonnegative integer, the category $CRep(t)$ has no radical. For $t$ a nonnegative integer, there is a functor $CRep(t) \to CRep(S_t)$ whose kernel is precisely the radical of $CRep(t)$. Here $CRep(S_t)$ is the subcategory of the representation category of $S_t$ spanned by the tensor powers of $V$.

By the Karoubi envelope construction, Deligne also builds a category $Rep(t)$ which gives us the ability to split $V^{\otimes n}$ into summands. This can also be thought of as a family of categories in some sense, although I don’t know the details because I am not sure how the Karoubi construction plays with taking limits. (Maybe I’ll post some more about this, if people are interested.) Again, $Rep(t)$ has no radical for $t$ not an nonnegative integer. When $t$ is a nonnegative integer, there is a functor $Rep(t) \to Rep(S_t)$ whose kernel is precisely the radical of $Rep(t)$.

## 25 thoughts on “Deligne’s “La Categorie des Representations du Groupe Symetrique S_t, lorsque t n’est pas un Entier Naturel.””

1. I’m a little confused, all your pictures are *planar*. You should be allowing crossings, right? Did you just choose these pictures for simplicity of drawing?

(Incidentally, if you don’t allow crossings than this is related to the “free quantum group S_n” coming from C*-algebras.)

2. Yes, crossings are permitted. The order on the elements in our finite sets plays no role.

3. Hopefully in the next few days I’ll write a follow-up post, but anyone who wants to play along at home your homework is to figure out why David’s pictures here follow from the representation theory of S_t being the “universal symmetric semisimple tensor category with an action on a t-element set” (where t is not necessarily integral).

4. What is the definition of a category acting on a set?

5. That’s the tricky part…

It’s not that the category is acting on the set, it’s that you figure out what the representation theoretic shadow is of a group having an action on a set.

Exercise: Given a transitive action of a group G on a set X, construct a commutative frobenius algebra object in the category of G-modules. This frobenius algebra object should have the property that module objects over it are H-modules where H is the point-stabilizer.

In general a (not necessarily commutative) frobenius algebra object in a category is a pretty good notion of “quantum group action.” Requiring commutativity here is a bit more subtle and I still don’t fully understand it.

The pictures for commutative frobenius algebra objects in categories are exactly the pictures that you’re drawing. If you relax the conditions for your pictures a little and allow your black colors to reach the bondary, then you end up with a bi-oidal category whose A-A, A-B, B-A, and B-B parts are G-modules, H-modules, H-modules, and H\G/H-modules. In Subfactor land this all goes by the name of “Q-systems.”

Another name for S_t in the literature is the “partition algebra” in the work of Jones, Martin, Ram+Halverson, etc. It’s another interesting exercise to translate your pictures (which are the same as the ones I was drawing) into the pictures in Ram+Halverson.

6. Scott Carnahan says:

Now I’m confused for several reasons. I still don’t know how a category acts on a set, I was unaware that Frobenius algebras could be noncommutative (at least from what I’ve heard about 2D TQFTs), and as far as I can tell, if you have an algebra object in G-modules, then its modules should be automatically G-modules. Next, you’ll probably say that quantum group actions aren’t what I think they are.

7. Frobenius algebras in 2D TQFTs constructed through Morse theory are always commutative. Frobenius algebras giving 2D TQFTs constructed through triangulations and state sums are symmetric (meaning the Frobenius pairing is symmetric) but not necessarily commutative. In general you could have a Frobenius algebra that was neither symmetric nor commutative. In particular, in a general tensor category without a braiding it doesn’t even make sense to ask whether the Frobenius algebra object is commutative or not!

I hope Reid Barton is going to write a guest post at some point explaining what he explained to me about how to construct group/subgroup subfactors without talking about subfactors. But the vaguest version of this idea is that dual Hopf algebras *cancel each other out*. What on earth do I mean by that? Well if you look at H-H* bimodules, that’s the same as a Hopf module. By the celebrated Nichols-Zoeller theorem Hopf modules are always free, hence the category of Hopf modules is equivalent to the category of vector spaces!

I don’t want to steel Reid’s thunder though, so I’ll leave it at that.

8. Urs Schreiber says:

In reply to the question “What is an action of a category on a set?” I have now typed the following explanation into the nLab:

9. Noah — so would it be more accurate to say that we are really categorifying the notion of “having a subgroup of index t” rather than the notion of “acting on a set of size t”?

10. Allen Knutson says:

Moreover, this is a semi-simple abelian category where the isomorphism classes of simple objects are in bijection with partitions (if I understand correctly).

I’m glad they’re not of size t. But when t is a natural number, how do we lose the others? I’m surprised that the radical would contain both the partitions of size > t and size < t. Does it really?

11. Allen — I don’t understand yet how to get the labeling by partitions in an elegant way. However, I can find a labeling by partitions such that the following works:

(1) The summands of $V^{\otimes k}$ are precisely the $V_{\lambda}$ for which $|\lambda| \leq k$.

(2) For any integer $t$ and any partition $\lambda$, define $\lambda^t$ to be the partition $(t-|\lambda|, \lambda_1, \ldots, \lambda_r)$. If $t-|\lambda| < \lambda_1$, we say that $\lambda^t$ is undefined. So the map $\lambda \mapsto \lambda^t$ provides a bijection between {partitions such that $\lambda^t$ is defined} and {partitions of size $t$}, but this map does not simply take a partition to itself.

For every positive integer $t$, the representation $V_{\lambda}$ is zero in $Rep(t)$ if and only if $\lambda^t$ is undefined. If $\lambda^t$ is defined, then $V_{\lambda}$ is the representation of $S_t$ usually indexed by the partition $\lambda^t$.

Example: Let’s take $t=4$. The partitions for which $\lambda^4$ is defined are $\emptyset$, $(1)$, $(2)$, $(1,1)$ and $(1,1,1)$. The corresponding partitions of $4$ are $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$ and $(1,1,1,1)$. These representations first appear in $V^{\otimes 0}$, $V^{\otimes 1}$, $V^{\otimes 2}$, $V^{\otimes 2}$ and $V^{\otimes 3}$ respectively.

12. 10. Yes David, saying subgroup is another possibility. The misleading thing about saying subgroup though is that you’d naively guess that every quantum subgroup is actually a quantum group. This turns out not to be true. So, for me it’s a bit more intuitive to think of this as saying that there are quantum actions for which the point stabilizer isn’t a quantum group.

13. Another source for how and when killing the radical works (that has the advantage of being in English) is Barrett and Westbury’s Spherical Categories. This construction I think has its roots in mathematical physics and the fusion product for representations of affine lie algebras (where apparently Ben’s snarky “non-physical” remark was the motivation).

14. 11. Allen, the way I’ve thought about this is the following. Representations of S_t for t generic are given by Young diagrams where the first row is infinite. Induction/restriction (between S_t and S_{t-1}) are given by the usual add/remove a block.

(In particular, note that for generic t, if I’m not mistaken, S_t does not have a sign representation. So there’s no alternating group A_t, for generic t. Similarly, GL_t for t generic won’t have a determinant representation, so you can’t define SL_t for t generic.)

Now when t is an integer, the radical kills off diagrams where the rest of the diagram after the first row is too large. Turns out that the precise condition is that it kills it exactly when you can’t shrink the first row so that the whole diagram has exactly n blocks. I haven’t done this calculation (and am not entirely sure how to do it), but its a lot more plausible than killing off large and small partitions.

15. Oh, and David’s partitions are what you get if you take mine and remove the first (infinite) row entirely.

16. By the way, I just realized that there is an easy proof that the radical vanishes for $t$ not a natural number. In the $A$-basis, the pairing between $Hom(V^{\otimes a}, V^{\otimes b})$ and $Hom(V^{\otimes b}, V^{\otimes a})$ is diagonal. The diagonal entry corresponding to an equivalence relation with $c$ classes is $t(t-1)(t-2)\cdots (t-c+1)$. So, if $t$ is not a natural number, these products are all nonzero.

17. James says:

This is a great post.

I haven’t gone through it on a detailed level yet, but I still have a question: Is it possible to build a natural machine that takes as input some kind of category (say a Tannakian category) and outputs another kind of category such that on the input Rep(S_t), the output is Rep(t)?

What I’m getting at is that Rep(t), if I understood correctly, has nice deformation properties, unlike Rep(S_t). It would be nice to find a general class of categories which have “good” nilpotent thickenings, like Rep(S_t) does. Is there some sense in which Rep(t) is a universal nilpotent thickening of Rep(S_t)?

18. James, I’d be really surprised if this turns out to be a functor rather than an art. Knop’s papers have some interesting generalizations of this process from S_n to groups that look a lot like S_n (wreath products and iterated wreath products), but I see no reason to expect them to work outside of “situations that have the same flavor as S_n”.

19. James says:

That may be true. But in that case, one should find all the categories that are to the given Tannakian category as Rep(t) is to Rep(S_t). If there isn’t a unique one or even a preferred one, then the only natural thing to do is study them all. Perhaps there is a moduli space of them. But the first thing to do is to make sense of this–that is, to figure out what the key properties of Rep(t) relative to Rep(S_t) are.

20. Pavel Etingof says:

Hi everybody,

This is a great discussion! In this connection, I’d like to mention my videotaped talk at the Newton Institute in March 2009, where I discussed these things and suggested possible applications
(development of representation theory of various “non-semisimple” structures in complex rank, such as affine Hecke algebras, superalgebras, real groups, Cherednik algebras, Yangians etc. I feel that there are a lot of opportunities here to define and study new interesting representation categories (some of which are tensor categories).

21. Pavel Etingof says: