Since Ben slept through the first lecture, I guess it falls to me to describe Turaev’s talk. Fortunately it was on a subject near and dear to my heart: quantum invariants of surfaces. There is a well-known invariant of oriented surfaces attached to a finite dimensional semisimple algebra which I’ll take a couple paragraphs to describe.

Let’s pause for a moment to recall some facts about semisimple algebras. Let Tr: A -> k denote the trace of left multiplication. Consider the symmetric billinear form <x, y> = Tr(xy). Much as the nondegeneracy of the Killing form is equivalent to semisimplicity for Lie algebras, the nondegeneracy of this form is equivalent to the semisimplicity of A (at least over C). Dual to this billinear form is a symmetric map p: k -> A (x) A. Similarly we have the map T_3(x,y,z) = Tr(xyz).

Choose a triangulation of your surface. Following Turaev we call a pair (face, edge of that face) a flag. To every flag we assign the algebra A (thought of as a vector space). To each edge we assign the map where the two A’s represent the two flags containing that edge. To each face we assign the map where the three A’s represent the three flags containing that face ordered using the orientation of the surface. Contracting all these maps gives a map

But this map is just given by a scalar! One can easily prove that this construction doesn’t depend on the choice of triangulation. For more details and history see my recent paperlet. This construction can be extended to a TQFT. I’ve given the most elementary discription, however, there’s a much better motivated way of thinking about this which is very nicely explained in the Quantum Gravity Seminar from winter 2001, track 1 weeks 16 and 17.

What is a good choice for the algebra A? One nice option is the group ring of a finite group. There are two bases for the group ring: the basis of grouplike elements and the basis of matrix elements of representations. One can easily compute this invariant in each of these two representations to get a well-known formula due to Mednykh (chi is the Euler characteristic, the sum is over irreps):

(Please check what this formula says for S the sphere or torus, also check it when G is abelian.) See my paperlet for further details and for the nonorientable version of everything.

Turaev’s starting point in his talk and his recent preprint is to consider another algebra A. Let G be a finite group and c a normalized 2-cocycle. We define the twisted group algebra A to be the formal span of elements of G with the multiplication (Associativity follows from the cocycle condition, and 1 is still a unit by the normalized condition.) Just as before, this algebra has two nice bases: the basis of group elements and the basis of matrix elements in the *projective* representations of G associated to c. Turaev computes the invariant in these two bases. The computation in the group basis is much more delicate than it was for c=1, but in his paper (although not in the talk) Turaev proved that it coincides with the Dijkgraaf-Witten invariant (which has a formula using only classical algebraic topology). The computation in the latter basis is again easy, and so Turaev produces a formula for the Dijkgraaf-Witten invariants. In particular, using some well-known integrality properties of projective representations Turaev produces an integrality result for the Dijkgraaf-Witten invariants which is completely unexpected from the point of view of classical topology.

I’d like to add one point to this whole story (which came up in my discussions with Chris Schommer-Pries). Why is it that computing these invariants (say in the trivial c case as in my paper) is so easy? It is because both the group algebra C[G] and the matrix algebra M_n(C) are both examples of *groupoid algebras*! In the latter case the groupoid is the one with n objects and only one morphism between any two objects (we’ll call such examples of groupoids trivializable).

So we start out with some invariant, which a priori depends on G. But suppose we can prove that it depends only on the group algebra of G. Thus we can see that it can’t distinguish say the two nonabelian groups of order 8. This isn’t terribly useful as many groups don’t have any simpler group with the same group algebra. However, if we can associate that invariant to *groupoids* and it still only depends on the groupoid algebra then we note that any group G has an isomorphic groupoid ring to a very nice groupoid which is the union of trivializable groupoids!

So am I allowed to think of “only depends on the groupoid algebra” as secretly meaning “only depends on the representation theory”?

It’s actually weaker than that. The group algebra

only tells you the dimensions of the irreps!In particular, any two abelian groups have the same group algebra (after all its a commutative multimatrix algebra, there’s only one of those of any dimension).“The same” or “isomorphic”? Is it possible to tack on extra information that makes the resulting structures non-isomorphic?

If you remember the Hopf algebra structure on the group algebra, you can get the group back.

Exactly. Now how do we modify that for groupoids?

There is a notion of a “quantum groupoid” which is like a Hopf algebra but includes groupoids as an example. However, this concept only works for the

functionson a groupoid, not for thegroupoid algebra. Perhaps I’ll write a bigger post on this issue sometime later.Who is “we” here?

The category theorist’s technical term for these beats is “codiscrete categories” (which necessarily are codiscrete groupoids). Sometimes also “pair groupoid” over the objects. Some people even say “chaotic groupoids”, but, personally, I don’t like that at all.

Also personally, I wouldn’t like to call these groupoids “trivial” (weakly or not). They are

trivializable— but not even canonically so.Well, I don’t really care about terminology, as long as I know what is meant. But just thought I’d mention this.

I actually made up that term in about 1 second when writing the post because I didn’t know what they’re called. I wanted a term that suggested “equivalent to the trivial groupoid” but not “isomorphic to the trivial groupoid.” Trivializable is definitely better, I’ll change it to that.

By the way, I haven’t thought of a proof, but I would be willing to bet that the fact that codiscrete groupoids are trivializable is just another incarnation of the fact that their groupoid algebras (which are matrix algebras, as you mentioned) are

Morita equivalentto the trivial algebra (the ground field itself).Yes, that’s more or less correct. But I would say it differently. First of all it’s better to look at the whole 2-category of groupoids. i.e. the objects are groupoinds (viewed as categories) the morphisms are functors and the 2-morphisms are natural transformations.

Btw: (for discrete groupoids) this is equivalent to the 2-category which has groupoids as objects, “left-principle bibundles” as 1-morphism and maps of bibundles as 2-morphisms. For thise who don’t know what this means, I don’t really want to get into it here (sorry). It’s more or less like a principle bundle, but where we use groupoids instead of groups and we have both a left and a right action.

Now there is a functor from this 2-category to the 2-category of algebras, bimodules, and bimodule maps, which associates to each groupoid it’s groupoid algebra. Each bibundle naturally becomes a bimodule, etc.

Thus if a groupoid is “weakly trivial” = equivalent as a category to a trivial groupoid = “trivializable”, then since functors (of 2-categories) map equivalences to equivalences, the groupoid algebras will be morita equivalent. Note that the converse is false. The two nonabelian groups of order 8 provide a counter example.

In general, I’d think, you want GrpdBitor: objects are arbitrary groupoids, morphisms are Hilsum-Skandalis morphisms of these (groupoid bitorsors with the right technical conditions) and homomorphisms of these.

I gather that’s what you have in mind, anyway.

Still another alternative would be, I think, to look at the 2-category of stacks presented by these groupoids.

Once I knew a theorem about which kinds of algebras come from which kind of categories as category algebras. Though it seems I forgot the details. I think finite hereditary algebras all come from categories generated from finite graphs. Something like this.

But I guess there are algebras which are not (Morita equivalent to) category algebras for any category?