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!