Quantum Topology and Classifying Manifolds October 11, 2007Posted by Noah Snyder in low-dimensional topology, quantum algebra, topology, tqft.
In this post I want to explain an old idea of (our frequent commenter) Greg Kuperberg on classifying low-dimensional manifolds using quantum algebra. In particular I want to discuss a 2-dimensional analogue of it that I’ve thought about a lot, but have recently mostly given up on.
Here’s the essence of Kuperberg’s idea:
- To every Heegaard splitting of a 3-manifold there is a corresponding expression in the structure constants of a Hopf algebra.
- The allowable moves of Heegaard splittings correspond to identities which follow from the axioms of an involutory () Hopf algebra.
- Conversely, if you have an identity that follows from the axioms of an involutory Hopf algebra, then you can do the corresponding move on the Heegaard splitting.
Kupperberg then conjectures that if an identity holds for any finite dimensional involutory Hopf algebra, than it must follow from the axioms. This would mean that his invariants distinguish 3-manifolds!
Now let me turn my attention to the 2-dimensional case, which is conceptually the same, but the details are much clearer. For the rest of the paper I’m going to assume that you understand the basics of lattice 2-d tqfts. The best way to understand these is to read John Baez’s lecture notes from the quantum gravity seminar in Winter 2001, track 1, weeks 16 and 17. For a little more detail and some references see my recent paperlet, and for a completely thorough discussion see Lauda-Pfeiffer.
The basic idea of lattice 2-d tqfts is that to every triangulated oriented closed surface you assign an expression in the structure constants of a symmetric (but not necessarily commutative) suitably normalized Frobenius algebra. The standard moves for relating different triangulations translate directly over to identities in symmetric normalized Frobenius algebras. Just as in Kuperberg’s case, the converse is also true. Namely an algebraic identity follows from the axioms if and only if the corresponding topological move is allowed.
In this case, however, because of the classification of surfaces we know that the analogue of Kuperberg’s conjecture is true! Namely the only invariant of oriented surfaces is Euler characteristic, and there are finite dimensional Frobenius algebras (namely matrix algebras) which distinguish them. However, this proof is unsatsifactory, what we would like to do is turn it on its head and use quantum topology to prove the classification of surfaces! (Yes, I know, this is the sort of thing that only I care about.) But now we’re stuck in the same bind Kuperberg was stuck in, how do we know that any identity that holds for finite dimensional algebras follows from the axioms?
Now it turns out this is difficult, but I think I at least understand why it is difficult now. Suppose we had a similar question about, say, groups. We would just take the free group on countably many generators (where the only relations are those coming from the axioms) and show that this group is residually finite.
Why won’t this idea work for us? Because algebraic structures studied in quantum topology typically don’t have a notion of free algebra. What is a free involutory Hopf algebra? What is a free symmetric normalized Frobenius algebra? They don’t behave well under infinite products. (Also see this comment on our blog from Simon Willerton.)
The way I want to think about Kuperberg’s idea is as follows. There is a universal quantum invariant which distinguishes all manifolds. This universal invariant comes from a free construction. Since the relevant free algebra is residually finite, the actual known invariants distinguish everything. To some extent this is still ok, however the universal invariant doesn’t come from an actual free Hopf algebra, it comes from a Hopf algebra object in some diagram category, and hence is much much more difficult to deal with.
If any of you have any ideas about how to complete this idea in 2-dimensions please contact me, as I think this is a very beautiful idea, but I just have been stuck too long to keep thining about a new way to prove such an old result.