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Recall that a $(k,k+1,\dots,k+n)$-TQFT 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:

  • Lurie, Jacob. On the classification of topological field theories. Current developments in mathematics, 2008, 129--280, Int. Press, Somerville, MA, 2009. 58Jxx (57Rxx) MR2555928. arXiv:0905.0465.

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 nLab: cobordism hypothesis.

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 Frobenius algebras. 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:

Question: Is there a classification, similar to Lurie's, for $(k,\dots,k+n)$-TQFTs with a give target $n$-category?

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I'm really tempted to just write "no," but I'll let a more knowledgeable person do that. The summary I got was that Lurie's technique really depended strongly on going all the way to the point, and one can't hope to get the classification you want with some new developments. –  Ben Webster Jul 9 '10 at 11:41
    
You should expect the non-fully extended cases to be substantially harder. For example, 012 TQFTs are rigid (have no deformations) while 12 TQFTs can come in families. –  Noah Snyder Jul 9 '10 at 16:52
    
@Ben: Well, the answer "no" is more or less what I expected the answer to be :) –  Theo Johnson-Freyd Jul 9 '10 at 17:24
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2 Answers 2

up vote 17 down vote accepted

Moore and Seiberg's result (Phys. Lett. 212B (1988) p.451) on classifying modular functors can be thought of as classification of (1,2,3) theories. (M&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" here.)

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).

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hey Kevin, Can you expand upon what you mean by your last sentence (Hatcher-Thurston for 3-manifolds)? –  Ian Agol Jul 18 '10 at 22:19
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Sure. Hatcher and Thurston constructed a simply connected 2-complex for a surface Y, with vertices the pants decompositions of Y, edges "F" and "S" moves between pants decompositions, and some 2-cells (see math.cornell.edu/~hatcher/Papers/pantsdecomp.pdf). The Moore-Seiberg equations correspond to the Hatcher-Thurston 2-cells, or rather to lifts of those 2-cells to a similar 2-complex for pants decompositions "with seams". By "Hatcher-Thurston for 3-manifolds" I meant doing this when Y is a 3-manifold, maybe with vertices corresponding to compression-body decomps of Y. –  Kevin Walker Jul 18 '10 at 23:43
    
Maybe I shouldn't have said "characterize mapping class groups" in the original answer. The Hatcher-Thurston 2-complex was used to derive a presentation of the MCG, but for the TQFT application you need the 2-complex, not the presentation of the MCG. –  Kevin Walker Jul 18 '10 at 23:45
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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:

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 < 0, which is exploited in the treatment of the case of fully extended field theories.

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).

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Welcome to MathOverflow! I assume the "you" in the first paragraph is in reference to Kevin Walker's answer above? –  Theo Johnson-Freyd Jul 18 '10 at 18:19
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@ Theo: No, Jacob is referring to his own paper, which you mentioned in the question statement. –  Chris Schommer-Pries Jul 18 '10 at 21:47
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