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An $n$-dimensional TQFT is a representation of the category $n$Cob of $n$-dimensional cobordisms. TQFTs are important sources of invariants of manifolds, and such invariants are highly computable by cutting and pasting.

I am curious if TQFTs give a complete set of invariants of manifolds (either in topological category or smooth category). My impression is that the locality of TQFTs, which enables the computation via cutting and pasting, seems like quite a strong property, so it might be possible that there are two manifolds which cannot be distinguished by any TQFTs.

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    $\begingroup$ You're a bit vague about the exact field theories you're allowing. Can you give more details? But a certain variant does not distinguish smooth 4-manifolds: arxiv.org/abs/math/0503054. $\endgroup$
    – skupers
    Commented Jan 13, 2018 at 23:56
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    $\begingroup$ If you allow TQFTs with arbitrary targets, then there is a universal one, namely the identity functor, which is a complete invariant tautologously. So some comstraints are needed on the target, e.g. that it be linear. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 14, 2018 at 10:01
  • $\begingroup$ @skupers Thanks for the reference. I was intentionally vague, as I wanted to know the limitations of TQFTs in general (with a lot of different variants), not a specific version of it, in capturing some information of manifolds. But if I were to be more specific, I am most interested in the TQFTs as defined in the Atiyah's original paper: math.ru.nl/~mueger/TQFT/At.pdf. $\endgroup$
    – Henry
    Commented Jan 14, 2018 at 19:20
  • $\begingroup$ @QiaochuYuan Yes, of course. I was assuming that the target is the category of vector spaces. Thanks for pointing it out. $\endgroup$
    – Henry
    Commented Jan 14, 2018 at 19:24
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    $\begingroup$ @QiaochuYuan, is that so clear? Is isomorphism in the bordism category the same as in the category of smooth manifolds? $\endgroup$
    – Manuel Bärenz
    Commented Apr 27, 2018 at 10:58

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