## Freedman on distinguishing manifolds with quantum topology				October 16, 2007

Posted by Noah Snyder in Category Theory, low-dimensional topology, quantum algebra, talks, tqft, Uncategorized.
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This week Mike Freedman was in Berkeley for the annual series of three Bowen lectures. The first two were about topological quantum computation and the fractional quantum Hall effect. Since I missed one of those because of closing arguments in the case I was on the jury for, I’ll instead only discuss his third talk which dove-tails nicely with my last post.

His motivating question was why do we look for topological quantum computation in 3 space-time dimensions (that is, by restricting to a system stuck in a plane) rather than the full 4 space-time dimensions? His explanation was the following: In 3-dimensions it seems that unitary TQFT can distinguish all manifolds, however in 4-dimensions they can not.

Let’s begin by recalling the definition of a unitary TQFT. This is a unitary monoidal functor from the category of formal d+1-dimensional bordisms to the category of finite dimensional Hilbert spaces. Let’s unpack this definition a bit.

What is the category of formal d+1-dimensional bordisms? Its objects are closed d-manifolds. Given two d-manifolds A and B, the morphisms $\mathrm{Hom}(A,B)$ are linear combinations of d+1-manifolds $X_i$ where the boundary of $X_i$ is $A \cup \bar{B}$ (where $\bar{B}$ denotes the same manifold with the opposite orientation). Extending bar by complex conjugation on the scalars gives a bar structure on the category. The tensor product is just disjoint union extended linearly. Saying that the functor is unitary says that it sends bar to bar. Saying that it is monoidal means that it sends disjoint union to tensor product.

We want to know if there are any formal bordisms which are killed by all unitary TQFT. How might we find such a formal cobordism? Well, the bar structure on formal bordisms gives the space $\mathrm{Hom}(\emptyset,B)$ a Hermitian form. Under any unitary TQFT this Hermition form is mapped to a positive definite inner product. Hence, if $\mathrm{Hom}(\emptyset,B)$ has any isotropic (aka light-like) vectors, they must automatically be killed by any unitary TQFT.

(The converse is not automatically true, that is just because the Hermitian form is positive definite, that doesn’t necessarily imply that its whole structure can be captured by unitary TQFTs.)

What is published so far is that this Hermitian form on formal bordisms is positive definite for d+1 = 0, 1, 2, and has isotropic vectors for d+1 = 4, 5, etc. The d+1 = 3 case is going to be dealt with by Freedman and some of his coauthors in a work in progress that makes heavy use of large amounts of 3-manifold theory including details of Perelman’s work.

In 4-dimensions the construction of isotropic vectors is originally due to Freedman and his coauthors, but there is a nifty argument due to our own Peter Teichner. I’ll sketch that very rapidly. Using some intersection form theory one notes that the theory of even unimodular lattices can be embedded inside of manifold theory. That is, there are manifolds whose intersection forms exhibit an arbitrary unimodular lattice, and that if they have the same lattice then they’re the same manifold. However, the theory of indefinite lattices is easy: all you care about is parity and rank. So if you take the difference of the bordisms (just cut a ball out of the manifold) corresponding to two different definite unimodular lattices, you can easily check that its inner product with itself is (2-2)=0 times a single indefinite lattice.

In low dimensions the idea of the proof is to show “diagonal dominance.” That is you find a way of ranking how complicated manifolds are, and you show that if you glue a bunch of bordisms to each other pairwise that the most complicated ones must appear on the diagonal. On Scott’s request I’ll make this a bit more precise. Let C be our complexity function (it should take values in some poset), we want that for any pair of bordisms M and N that either $C(M \cup \bar{N}) < C(M \cup \bar{M})$ or $C(M \cup \bar{N}) < C(N \cup \bar{N})$.

Now, if you have $\langle\sum_i a_i M_i, \sum_j b_j N_j \rangle$ you see that the glued manifold $M_i \bar{N_j}$ has its largest complexity somewhere on the diagonal. Since the diagonal terms $a_i \bar{a_i} M_i \bar{N_j}$ all have positive coefficients, this means that the most complex term in the inner product cannot cancel. Hence the product is nonzero.

For example, in d+1=1 dimension the bordisms are just arcs and the complicatedness function is just number of connected components. When you glue two of these diagrams together, the number of components of the glued thing is smaller, and it’s strictly smaller unless you’re gluing to your mirror image. (This is a fun exercise, let me know in the comments if I should give you more details).

Freedman also left us all with a fascinating open question (which I think Vaughan asked about during questions). Is this inner product non-degenerate for any dimension? Or could you have a radical?

## Comments»

1. Scott Morrison - October 16, 2007

Perhaps it’s worth saying that the forthcoming result (Freedman, Walker, D. Calegari) is “positive” in d+1=3, that is, there are no isotropic vectors.

This leaves open the possibility that unitary TQFTs detect all 3-manifolds, although this is far from proven.

Their proof actually makes some use of unitary TQFTs along the way. Part of the complexity function has to “control” the fundamental groups of the manifolds. They take advantadge of the fact that $\pi_1(M^3)$ is residually finite (everything turns up in some finite group quotient), and consider the collection of all finite group TQFTs.

Perhaps someone will want to look at the proof, and see if more of the complexity function can actually be written in terms of other known TQFTs. This might be a first step towards proving the big result: that every (formal linear combination of) 3-manifold with boundary is detected by some unitary TQFT.

2. Greg Kuperberg - October 16, 2007

This leaves open the possibility that unitary TQFTs detect all 3-manifolds, although this is far from proven.

You could call it far from proven, but I would say that it is closer to proven than it is a complete mystery. Geometrization is true, so we know that fundamental groups of 3-manifolds are residually finite. This means that even the old-school invariants from finite groups, which are unitary TQFTs, carry a great deal of information.

3. Danny Calegari - October 27, 2007

The finite group TQFT’s are used at an early (but crucial) stage
to deal with compressing disks, which give rise to essential
spheres in the closed manifolds, and contribute to the sphere
decomposition. This is a very “crude” aspect of 3-manifold
topology, and it was therefore surprising to us that we needed
a (relatively) sophisticated tool to take care of it.

Other terms in the complexity function include -vol for the
hyperbolic pieces (in the geometric decomposition). The proof
that this term is diagonal dominant uses a significant amount of
analysis; I think it is plausible that this term could be recovered
by TQFT information (e.g. after Kashaev’s conjecture) but I would
guess that any proof that it satisfies the desired inequality would
be dramatically different from our proof.

4. TQFTs via Planar Algebras « Secret Blogging Seminar - December 6, 2008

[...] of the 3-manifold. Thus for any 3D TQFT we get 3-manifold invariants. In fact, as we discussed before, it might be that these invariants distinguish all 3-manifolds. Exciting [...]

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