15
$\begingroup$

Are there simple examples of $n+1$D TQFT that assign 1-dimensional Hilbert spaces to both $n$-torus and $n$-sphere but higher dimensional Hilbert spaces to some other $n$-manifolds? Here I am assuming that the manifolds under consideration are all orientable.

For $n=2$, the requirement of 1-dimensional Hilbert spaces on both 2-torus and 2-sphere implies that the corresponding modular tensor category associated to the TQFT has only 1 simple object. Such TQFT have 1-dimensional Hilbert spaces on all oriented 2-manifold. Therefore, examples I am looking for should not exist with $n=2$. For $n>2$, I cannot think of any reason why such example cannot exist.

Improve this question
$\endgroup$
4
  • 5
    $\begingroup$ You may be interested in the work of Dan Freed and Constantin Teleman on invertible field theories. They have theorems that say that a (fully extended) TQFT is invertible whenever the value assigned to certain spheres, or products of spheres are invertible. See e.g. ma.utexas.edu/users/dafr/Aspects.pdf This does not seem to give the result you want directly, but is in the same spirit. $\endgroup$
    – Sam Gunningham
    Commented Sep 23, 2015 at 3:56
  • $\begingroup$ @SamGunningham Thanks for the comment. As far as I understand, an invertible $n+1$ TQFT has 1-dimension Hilbert spaces on all n-manifolds, which is not quite what I wanted. The criteria for invertibility in ma.utexas.edu/users/dafr/Aspects.pdf also seems much stronger than what I required. I think it is interesting to see to what extend the requirements I posed constrain the corresponding TQFT. $\endgroup$
    – Chao-Ming Jian
    Commented Sep 23, 2015 at 19:20
  • $\begingroup$ Interesting question! Modular tensor categories correspond to partially extended tqfts, assigning data to 1,2, and 3-manifolds. Non-extended 3D tqfts are much more complicated (see arxiv.org/abs/1509.06811 and arxiv.org/abs/1408.0668). Now if we are just considering partially extended TQFTs (assigning data to d, d-1, and d-2 manifolds) then there are no such examples in dimension four, and I think I can show that there are no such examples for all dimensions d. The general case is a little more involved/original than a typical MO answer. Maybe I'll write it up as a note? $\endgroup$
    – Chris Schommer-Pries
    Commented Sep 25, 2015 at 10:52
  • $\begingroup$ @ChrisSchommer-Pries That sounds very exciting. Could you give a little hint how your proof works? In 2+1D, I can consider the "pants" decomposition of the 2-manifold to relate the Hilbert spaces on a 2-sphere and 2-torus to the Hilbert spaces on a general 2-manifolds. Does a similar reasoning exist in a general n+1 dimensions? It seems that if we have a partially extended n+1D TQFTs (assigning data to n+1, n, and n-1 manifolds), some arguments generalizing the 2+1D one can be made. $\endgroup$
    – Chao-Ming Jian
    Commented Sep 27, 2015 at 6:15

1 Answer 1

Reset to default
7
$\begingroup$

If your topological field theory is at least once-extended, by which I mean it assigns values to $(n+1)$-manifolds, $n$-manifolds, and also to $(n-1)$-manifolds, than this cannot happen.

More precisely we have the following result:

Theorem: Suppose that $ Z: Bord_{n+1} \to C$ is an $(n+1)$-dimensional once-extended (oriented) topological field theory valued in the symmetric monoidal 2-category $C$. Then the value $Z(T^n)$ assigned to the $n$-torus is invertible if and only if the whole field theory is invertible (assigns invertible values to all manifolds in all dimnesions).

So in particular suppose $C$ is any 2-category that deloops the category of vector spaces (like linear categories, functors, transformations or algebras, bimodules, maps, etc). Then if the vector space assigned to the n-torus is 1-dimensional (hence invertible) than every $n$-manifold is assigned an invertible (i.e. one-dimensional) vector space. This also holds if you replace vector spaces with super vector spaces.

A version of this theorem also holds for bordism equipped with arbitrary tangential structures. You can read about it here:1511.01772.

The main ingredients in the proof are dimensional reduction and surgery. Dimensional reduction lets you relate theories of different dimensions and so you can attack the problem inductively. This already lets you prove that many bordisms take invertible values. Then next idea is that handle decompositions use handles which have at most codimension two corners. This means that if you are at least once-extended, then you can implement handle decompositions of (n+1)-manifolds, and hence surgery for n-manifolds, entirely in the bordism 2-category in categorical terms.

That is not the complete argument, but those are two of the main ideas.

I don't know what happens if your field theory is not extended. Even in dimension n=2 (n+1 =3) the OP's argument used that we assign categories to 1-manifolds. So if the theory is not extended, perhaps it is possible to assign an a 1-dimensional vector space to the 2-torus, but higher dimensional spaces to other surfaces? András Juhász has a classification of non-extended (2+1)-TQFTs, but it is a bit complicated. Perhaps it is possible to use his work to prove or disprove this in dimension (2+1)? I think that is an interesting question to explore.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you very much for this very interesting and thorough answer!! As you said in the answer and also your paper 1511.01772, the dimension reduction and surgery play key roles in proving this theorem. I am wondering if you have some insights in directly proving this theorem using the handle decomposition of manifolds. In hindsight, it seems that an analysis of the handle decomposition of T^n should provide enough information of the once-extended TQFT to constrain the invertibility on other manifolds. $\endgroup$
    – Chao-Ming Jian
    Commented Nov 9, 2015 at 22:30
  • 1
    $\begingroup$ Small addition: Meanwhile, the published version arXiv:01772 has appeared <cite authors="Schommer-Pries, Christopher J.">_Schommer-Pries, Christopher J._, Tori detect invertibility of topological field theories, Geom. Topol. 22, No. 5, 2713-2756 (2018). ZBL1405.18009.</cite>, and the review includes the discussion here. $\endgroup$
    – Olaf Teschke
    Commented Apr 15, 2019 at 14:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .