jump to navigation

Deligne’s “La Categorie des Representations du Groupe Symetrique S_t, lorsque t n’est pas un Entier Naturel.” February 25, 2009

Posted by David Speyer in Uncategorized.
trackback

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:

deligne1

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:

deligne2

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).

About these ads

Comments

1. Noah Snyder - February 25, 2009

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. davidspeyer - February 25, 2009

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

3. Noah Snyder - February 25, 2009

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. davidspeyer - February 25, 2009

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

5. Noah Snyder - February 25, 2009

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. A warmup: GL_t for t not an integer « Secret Blogging Seminar - February 25, 2009

[...] Snyder in Uncategorized. trackback This post is meant as a warm-up to my planned follow-up to David’s post.  You don’t have to have read his post to understand this one, but there are a few technical [...]

7. Scott Carnahan - February 25, 2009

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.

8. Noah Snyder - February 25, 2009

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.

9. Urs Schreiber - February 26, 2009

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:

Action of a category on a set.

10. davidspeyer - February 26, 2009

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”?

11. Allen Knutson - February 26, 2009

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?

12. davidspeyer - February 26, 2009

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.

13. Noah Snyder - February 26, 2009

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.

14. Noah Snyder - February 26, 2009

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).

15. Noah Snyder - February 26, 2009

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.

16. Noah Snyder - February 26, 2009

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

17. davidspeyer - March 1, 2009

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.

18. James - March 1, 2009

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)?

19. Noah Snyder - March 2, 2009

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”.

20. James - March 2, 2009

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.

21. Coincidences of tensor categories « Secret Blogging Seminar - March 22, 2009

[...] Please note that in the slides I’ve completely elided the distinctions between a quantum group, its category of representations, and (when q is a root of unity) its semisimplified category of representations (where you quotient out by the kernel of the inner product as in David’s post). [...]

22. Pavel Etingof - June 6, 2009

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).

23. Pavel Etingof - June 6, 2009
24. Deligne’s category Rep(S_t) for t not necessarily an integer « Climbing Mount Bourbaki - June 8, 2010

[...] also recommend looking at these posts of David Speyer and Noah Snyder, which talk about some of Deligne’s work as well (and which I learned a lot [...]

25. Motive-ating the Weil Conjecture Proof « Secret Blogging Seminar - June 10, 2010

[...] used, then the endomorphism rings of our objects will contain negligible morphisms. (See Noah and my posts on this in a different context.) Cohomological equivalence builds into the theory a [...]


Sorry comments are closed for this entry

Follow

Get every new post delivered to your Inbox.

Join 648 other followers

%d bloggers like this: