The Reshetikhin-Turaev construction take as input a Modular Tensor Category (MTC) and spits out a 3D TQFT. I've been told that the other main construction of 3D TQFTs, the Turaev-Viro State sum construction, factors through the RT construction in the sense that for each such TQFT Z there exists a MTC M such that the RT construction of applied to M reproduces Z. Is this true for all known 3D TQFTs? Does anyone know any counter examples?

Edit: (1) I want to be flexible with what we call a TQFT, so anomalies are okay.

(2) There have been some good answers to the effect that more or less if I have an extended TQFT then it factors through the RT construction. But this is not really what I'm after. Are there any (non-extended) examples that people know about? Ones which might not come from the RT construction. Are all known 3D TQFTs extended TQFTs?

  • $\begingroup$ In view of the primes-knots analogy, I wonder if nt-versions of the Reshetikhin-Turaev construction exist. $\endgroup$ – Thomas Riepe Jan 3 '10 at 21:13

If you have a 3d TQFT, with no anomaly, and which goes down to points, and where things are sufficiently finite and semisimple, then I think you can show that it comes from a Turaev-Viro type construction on the 2-category Z(pt).

If you have a 3d TQFT, possibly with anomaly, which goes down to circles, and where things are sufficiently finite and semisimple, then I agree with Noah: Z(S^1) is a MTC and the RT construction on the MTC reproduces the TQFT.

Relating these two statements, TV(C) = RT(double(C)), where C is a 2-cat, double(C) is the Drinfeld double (or maybe center), TV is the Turaev-Viro construction, and RT is the Reshetikhin-Turaev construction.

  • $\begingroup$ I think I remember reading that, comparing the TV and RT constructions, one requires modularity and the other doesn't. Can someone remind me what the situation is here? $\endgroup$ – Jamie Vicary Apr 13 '10 at 15:32
  • $\begingroup$ More basically, the RT construction requires a braiding, while the TV construction does not. (The modularity condition cannot be formulated without assuming a braiding.) $\endgroup$ – Kevin Walker Apr 14 '10 at 0:56

I think that the answer is "yes" if by TQFT you mean one that extends all the way down to 1-manifolds. The MTC is the thing assigned to a circle by this extended TQFT.

There's also lurking somewhere in here the issue of whether you mean 3d TQFT "with anomaly." The 3d TQFTs coming from modular tensor categories via RT really have a little bit of 4d structure.

Edit: Removed a false sentence. Also note that precise statements (and eventually a proof, once all parts are out) can be found in work of Bartlett-Douglas-Schommer-Pries-Vicary


Kevin's parenthetical about needing things to be sufficiently finite and semisimple suggests that thought of another way the answer is "no." In particular, there are known non-semisimple TQFTs. I know very little about these, but Alexis Virelizier was very into them. His papers (for example, this one) discuss them and should have references to earlier work.

  • $\begingroup$ I thought Virelizier's stuff was just for closed 3-manifolds. I would be (happily) surprised if someone had worked out a non-semisimple example which gave a functor on 3-dimensional bordisms and was well-behaved axiomatically. $\endgroup$ – Kevin Walker Oct 13 '09 at 7:03
  • $\begingroup$ "there are known non-semisimple TQFTs" -- What's the definition of semisimple here? $\endgroup$ – Kevin H. Lin Oct 13 '09 at 11:51
  • $\begingroup$ You're probably right that that only works for closed things. What about the book "Non-semisimple topological quantum field theories for 3-manifolds with corners"? $\endgroup$ – Noah Snyder Oct 13 '09 at 14:33
  • $\begingroup$ I took a look at this book, and it did seem to me that they form a sort of TQFT (with "exotic anomaly") out of non-semisimple MTCs. The anomaly they use is "exotic" (my word) and corresponds to a central extension of the bordism 2-category, not by a group but by a monoid (I think it was the natural numbers). This way they can get a weird TQFT where the value of, say, S<sup>1</sup> x S<sup>2</sup> is zero. (This can't happen in a usual TQFT without this anomaly since the value of S<sup>1</sup> x S<sup>2</sup> is the trace of the identity. ) $\endgroup$ – Chris Schommer-Pries Oct 13 '09 at 16:31
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    $\begingroup$ Rozansky-Witten model for a compact hyperkahler target space (e.g. K3 surface) is an example of a non-semi-simple 3d TQFT. Z(S^1) is the 2-periodic derived category of coherent sheaves on the target. $\endgroup$ – Anton Kapustin Aug 5 '10 at 3:52

There must be lots of fermionic TQFTs behaving in a very different way compared to RT. One feature of these fermionic (based on Berezin integral) theories is that they are quite straightforwardly generalized to any dimensions, not just 3 - which, by the way, provides you with a powerful and intriguing means of studying 3-manifolds using 4-manifolds, etc. These new theories are related to Lie groups (without any obvious restrictions as for semisimplicity etc.) and their homogeneous spaces, see e.g. http://arxiv.org/abs/0907.3787 .

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    $\begingroup$ I would like to add also that, on a big scale, such great achievements of 20th century as Jones polynomial and related stuff, including RT invariants, fall within the framework of Yang-Baxter equation which, in its turn, goes back to Onsager's solution of 1940's of Ising model in math physics. Now fermionic formulas are my attempt to go outside that framework (think it looks natural in 21st century); see also a 4d paper arxiv.org/abs/0911.1395 (and I am of course its author, perhaps I will once make myself a better sort of login here so as this could be seen automatically). Cheers, I.K. $\endgroup$ – Igor Korepanov Jan 3 '10 at 9:34

If your TQFT does not extend to the circle, (EDIT: and you're willing to generalise to lax TQFTs,) then the answer is no. A reference is: https://arxiv.org/abs/1408.0668

The proof goes by an explicit generators and relations presentation of the 3d cobordism category in terms of surgeries and automorphisms of surfaces.

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    $\begingroup$ If I understand the results of that paper correctly, the author only constructs counter examples as lax TQFTs (in the sense of lax symmetric monoidal functor), and not TQFTs in the usual sense. $\endgroup$ – Chris Schommer-Pries Nov 10 '17 at 17:24
  • $\begingroup$ @ChrisSchommer-Pries, oh, I didn't notice that! Is there any discussion anywhere how his results specialise to strong monoidal functors? $\endgroup$ – Manuel Bärenz Nov 13 '17 at 11:20

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