What's the current state of the classification of not-fully-extended TQFTs? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:11:17Z http://mathoverflow.net/feeds/question/31154 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31154/whats-the-current-state-of-the-classification-of-not-fully-extended-tqfts What's the current state of the classification of not-fully-extended TQFTs? Theo Johnson-Freyd 2010-07-09T05:35:19Z 2010-07-18T16:30:34Z <p>Recall that a <strong>$(k,k+1,\dots,k+n)$-TQFT</strong> is (supposed to be) a functor from the $n$-category whose $j$-morphisms are (isomorphism classes of) compact $(k+j)$-dimensional manifolds with boundary to some target category, usually your favorite version of $n$-Vect. When $k=0$, a full "classification" of TQFTs with a given target category is given in:</p> <ul> <li>Lurie, Jacob. On the classification of topological field theories. <em>Current developments in mathematics, 2008</em>, 129--280, Int. Press, Somerville, MA, 2009. 58Jxx (57Rxx) <a href="http://www.ams.org/mathscinet-getitem?mr=2555928" rel="nofollow">MR2555928</a>. <a href="http://arxiv.org/abs/0905.0465" rel="nofollow">arXiv:0905.0465</a>.</li> </ul> <p>Or, rather, Lurie first provides reasonable definitions for a number of things, end then proves that there is an equivalence of $n$-categories between the $(0,\dots,n)$-TQFTs with target $\mathcal V$ and the $n$-groupoid of ("fully") dualizable objects in $\mathcal V$. (The classification is not particularly effective in two ways: given a dualizable object, which is the value the TQFT assigns to a point, it can be still very hard to understand the functor on complicated manifolds; and given a category, it can be still very hard to classify its dualizable objects.) For a review, see <a href="http://ncatlab.org/nlab/show/cobordism+hypothesis" rel="nofollow">nLab: cobordism hypothesis</a>.</p> <p>But Lurie's result does not describe all gadgets that deserve to be called "TQFT"s. For example, it is a classical folk theorem that $(1,2)$-TQFTs are the same as commutative cocommutative <a href="http://en.wikipedia.org/wiki/Frobenius_algebra" rel="nofollow">Frobenius algebras</a>. I think that there are other similar results of this nature, but I don't know of any theory that puts them all into a single framework. Hence:</p> <blockquote> <p><strong>Question:</strong> Is there a classification, similar to Lurie's, for $(k,\dots,k+n)$-TQFTs with a give target $n$-category?</p> </blockquote> http://mathoverflow.net/questions/31154/whats-the-current-state-of-the-classification-of-not-fully-extended-tqfts/31184#31184 Answer by Kevin Walker for What's the current state of the classification of not-fully-extended TQFTs? Kevin Walker 2010-07-09T13:14:09Z 2010-07-09T13:14:09Z <p>Moore and Seiberg's result (Phys. Lett. <b>212B</b> (1988) p.451) on classifying modular functors can be thought of as classification of (1,2,3) theories. (M&amp;S only do the 1 and 2 of (1,2,3), but it's not hard to extend to 3 as well; see "On Witten's 3-manifold Invariants" <a href="http://canyon23.net/math/" rel="nofollow">here</a>.)</p> <p>My guess is that extending this style of classification to any of the adjacent slots (1,2,3,4), (2,3,4) or (2,3) would be very difficult. For (1,2,3,4) one would need to start by describing a categorified action of mapping class groups of surfaces in terms of local data; the uncategorified version is already long and messy (see refs above). For (2,3,4) one would need to characterize mapping class groups of 3-manifolds in terms of local data (Hatcher-Thurston for 3-manifolds).</p> http://mathoverflow.net/questions/31154/whats-the-current-state-of-the-classification-of-not-fully-extended-tqfts/32376#32376 Answer by Jacob Lurie for What's the current state of the classification of not-fully-extended TQFTs? Jacob Lurie 2010-07-18T16:30:34Z 2010-07-18T16:30:34Z <p>When n > 1 the paper that you cite can give you a little bit of traction: the sketch proof of the main result gives a generators-and-relations presentation of (k,k+1,...,k+n)-Bord relative to (k,k+1)-Bord. There are two caveats:</p> <p>1) (k,k+1)-Bord must be interpreted as an (infty,1)-category (or at least as an (n,1)-category), rather than as an ordinary category. Consequently, this is a very complicated object even when k=1 (to my knowledge, there is no concrete description of its representations along the lines of "commutative Frobenius algebras"). Fortunately it is quite easy to understand when k &lt; 0, which is exploited in the treatment of the case of fully extended field theories.</p> <p>2) The presentation is more complicated than in the fully extended case. When increasing the dimension, you need to add generators and relations corresponding to handles and handle cancellations for all indices (in the fully extended case, there is a cancellation phenomenon which ends up telling you that the only data you need to supply is for a handle of index 0).</p>