Usefulness of using TQFTs - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:50:58Z http://mathoverflow.net/feeds/question/60550 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60550/usefulness-of-using-tqfts Usefulness of using TQFTs ISH 2011-04-04T13:30:39Z 2011-04-05T11:58:27Z <p>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?</p> http://mathoverflow.net/questions/60550/usefulness-of-using-tqfts/60559#60559 Answer by Daniel Moskovich for Usefulness of using TQFTs Daniel Moskovich 2011-04-04T14:58:03Z 2011-04-04T23:19:40Z <p>The only topological information in 3-manifolds besides homology and homotopy is Reidemeister torsion (see <a href="http://mathoverflow.net/questions/47380/fundamental-group-and-complete-invariant-of-irreducible-3-manifolds" rel="nofollow">this question</a>). TQFT sees <a href="http://en.wikipedia.org/wiki/Analytic_torsion" rel="nofollow">Ray-Singer torsion</a>, which is the same thing. Indeed, it was this discovery <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=676337&amp;loc=fromrevtext" rel="nofollow">by Schwartz</a> (and independently, unpublished, by Singer) <strike>which sparked the subject and</strike> which motivated Witten's work on a <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=990772&amp;loc=fromrevtext" rel="nofollow">TQFT for the Jones polynomial</a>.</p> http://mathoverflow.net/questions/60550/usefulness-of-using-tqfts/60594#60594 Answer by Agol for Usefulness of using TQFTs Agol 2011-04-04T19:11:44Z 2011-04-04T19:11:44Z <p>The <a href="http://www.ams.org/mathscinet-getitem?mr=1230928" rel="nofollow">Tait flyping conjecture</a> was proven by Menasco and Thistlethwaite using knot polynomial invariants (which are a version of TQFT invariants). </p> <p>There is a lower bound on the <a href="http://www.ams.org/mathscinet-getitem?mr=766964" rel="nofollow">braid index of a knot</a> 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 <a href="http://www.ams.org/mathscinet-getitem?mr=1302019" rel="nofollow">tunnel number</a> and <a href="http://www.ams.org/mathscinet-getitem?mr=1480887" rel="nofollow">Heegaard genus</a> in terms of TQFT invariants, but these <a href="http://arxiv.org/abs/1009.1653" rel="nofollow">are not sharp in many cases</a>. 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 <a href="http://www.ams.org/mathscinet-getitem?mr=746538" rel="nofollow">Boileau and Zieschang</a> using algebraic techniques, and this has been done by <a href="http://arxiv.org/abs/0801.0738" rel="nofollow">Helen Wong</a> using TQFT invariants. </p> http://mathoverflow.net/questions/60550/usefulness-of-using-tqfts/60599#60599 Answer by Bruno Martelli for Usefulness of using TQFTs Bruno Martelli 2011-04-04T19:43:30Z 2011-04-04T20:06:37Z <p>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.</p> <p>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 <a href="http://www.dm.unipi.it/pages/petronio/public_html/files/3D/c9/c9_census.html" rel="nofollow">these lists</a>. </p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/60550/usefulness-of-using-tqfts/60620#60620 Answer by Paul for Usefulness of using TQFTs Paul 2011-04-04T23:58:34Z 2011-04-04T23:58:34Z <p>Rasmussen's $s$ invariant detects non-sliceness of some knots that no other method applies to. </p> http://mathoverflow.net/questions/60550/usefulness-of-using-tqfts/60629#60629 Answer by Dylan Thurston for Usefulness of using TQFTs Dylan Thurston 2011-04-05T01:08:44Z 2011-04-05T11:57:07Z <p>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 <em>only</em> 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.</p> <p><strong>Edit:</strong> 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 <em>Monopoles and 3-manifolds</em>.</p> <p>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.</p> <p>Instead, you use the fact that there is an exact triangle <code>$\cdots \longrightarrow \widehat{\mathit{HM}} \longrightarrow \overline{\mathit{HM}} \longrightarrow \check{\mathit{HM}}\longrightarrow \cdots$</code> (with right mapping to left), and the map $\overline{\mathit{HM}}(W)$ is $0$ for $b_2^+(W) \ge 1$.</p> <p>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 $\check{\mathit{HM}}(Y)$, <em>backwards</em> 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 $W$.</p>