Usefulness of using TQFTs - MathOverflow

Usefulness of using TQFTs

2011-04-04T13:30:39Z

What is a topological feature, that a (some) tqft (e.g. in 3 or 4 dim) sees but homology/cohomology/homotopy groups dont? Or: what is an example where using classical theories is hard, but using a tqft is comparatively easy?

Answer by Daniel Moskovich for Usefulness of using TQFTs

2011-04-04T14:58:03Z

The only topological information in 3-manifolds besides homology and homotopy is Reidemeister torsion (see this question). TQFT sees Ray-Singer torsion, which is the same thing. Indeed, it was this discovery by Schwartz (and independently, unpublished, by Singer) ~~which sparked the subject and~~ which motivated Witten's work on a TQFT for the Jones polynomial.

Answer by Agol for Usefulness of using TQFTs

2011-04-04T19:11:44Z

The Tait flyping conjecture was proven by Menasco and Thistlethwaite using knot polynomial invariants (which are a version of TQFT invariants).

There is a lower bound on the braid index of a knot in terms of the Jones polynomial. I don't think that there is an efficient algorithm to compute the braid index of knots in general using geometric techniques, so sometimes this works better. There are related estimates of tunnel number and Heegaard genus in terms of TQFT invariants, but these are not sharp in many cases. However, computing TQFT invariants is straightforward, but exponential, so I'm not sure these estimates are necessarily "easier". Estimates of Heegaard genus for Seifert spaces were given Boileau and Zieschang using algebraic techniques, and this has been done by Helen Wong using TQFT invariants.

Answer by Bruno Martelli for Usefulness of using TQFTs

2011-04-04T19:43:30Z

At a very concrete level, Turaev-Viro invariants of a compact 3-manifold (with or without boundary) can be easily computed by a computer from a triangulation and very often (although not always) distinguish non-homeomorphic manifolds.

To calculate a Turaev-Viro invariant you need to fix a level $r=3,4,\ldots$: for $r=5, 7$ you already obtain a quite powerful (and mysterious) invariant, which works on any kind of compact 3-manifold. For instance, it helped to distinguish immediately most of the non-homeomorphic manifolds in these lists.

So, distinguishing many triangulated 3-manifolds is maybe "an example where using classical theories is hard, but using a tqft is comparatively easy". The "classical theory" here would involve recognizing prime summands, decomposing along tori, finding a hyperbolic structure, etc. etc.

Note however that the cost of calculating Turaev-Viro invariants increases exponentially with $r$ and the number of tetrahedra, so I don't know if they can be effectively used to distinguish -- say -- two manifolds having 20 tetrahedra.

Answer by Paul for Usefulness of using TQFTs

2011-04-04T23:58:34Z

Rasmussen's $s$ invariant detects non-sliceness of some knots that no other method applies to.

Answer by Dylan Thurston for Usefulness of using TQFTs

2011-04-05T01:08:44Z

All the answers so far have focused on 3 dimensions, but the answer is much more striking in 4 dimensions. Freedman's theorem tells you that classical homology invariants give you complete information about topological, simply-connected 4-manifolds. These classical invariants cannot, however, distinguish between distinct smooth structures on the same topological 4-manifold, and essentially our only technique for distinguishing smooth 4-manifolds is Donaldson's invariant or the Seiberg-Witten invariant or their relatives. These do not quite form a TQFT, but are related to TQFTs.

Edit: On request, a little about how the 4-manifold invariants are related to a TQFT. This is all nicely explained in the beginning of Kronheimer and Mrowka's book Monopoles and 3-manifolds.

There are actually three different theories, denoted $\widehat{\mathit{HM}}$ ("HM-from"), $\check{\mathit{HM}}$ ("HM-to", unfortunately typeset badly here), and $\overline{\mathit{HM}}$. All are close to satisfying axioms for a TQFT assigning a vector space to a 3-manifold and maps to a 4-manifold, at least for connected manifolds. (The vector spaces are infinite dimensional, but finite in each graded piece.) Unfortunately, however you slice it, in each case the invariant associated to a closed 4-manifold in the usual TQFT way (when defined) is zero.

Instead, you use the fact that there is an exact triangle \[ \cdots \longrightarrow \widehat{\mathit{HM}} \longrightarrow \overline{\mathit{HM}} \longrightarrow \check{\mathit{HM}}\longrightarrow \cdots \] (with right mapping to left), and the map $\overline{\mathit{HM}}(W)$ is $0$ for $b_2^+(W) \ge 1$.

If you have a 4-manifold $W$ with $b_2^+(W) \ge 2$, you factor it as two cobordisms $W = W_1 \cup_Y W_2$ for some 3-manifold $Y$, with $b_2^+(W_i) \ge 1$. Then the properties above let you map from $\check{\mathit{HM}}(S^3)$, to $\check{\mathit{HM}}(Y)$, backwards in the exact triangle to $\widehat{\mathit{HM}}(Y)$, and then forwards to $\widehat{\mathit{HM}}(S^3)$. The resulting map (from $\check{\mathit{HM}}(S^3)$ to $\widehat{\mathit{HM}}(S^3)$) gives the interesting Seiberg-Witten invariants of $W$.