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As far as I know, when talking about TQFT, one usually means TQFTs on oriented manifolds with boundary (cobordisms)

It appears to me that the Turaev-Viro-Barrett-Westbury state-sum construction can be put on an arbitrary 3-manifold with boundary (cobordism) which needs not be orientable and still yields a valid (unitary) TQFT on those 3-manifolds (cobordisms). The corresponding TQFTs (whose algebraic data is connected to the Drinfeld double of the input fusion category via the Reshetikin-Turaev construction) are said to be non-chiral.

Is that true and is it also the case that (unitary) TQFTs corresponding to modular fusion categories (via Reshetikin-Turaev) that are not Drinfeld doubles (i.e. the chiral ones) cannot be extended to non-orientable manifolds with boundary?

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    $\begingroup$ I'm not sure, but from the cobordism hypothesis I would have thought that non-oriented TQFTs have to contain some extra data as compared to oriented TQFTs. So one should ask which TQFTs admit this extra data (your question), but the choice of this data should matter as well. $\endgroup$ – Manuel Bärenz Apr 27 '18 at 10:05
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    $\begingroup$ The "extra data" would be the tensors associated to all non-orientable manifolds with boundary in addition to all orientable manifolds with boundary, right? What do you mean by "the choice of data should matter as well"? That there might be multiple ways to extend the TQFT to non-orientable manifolds? $\endgroup$ – Andi Bauer May 4 '18 at 13:01
  • $\begingroup$ I just realize that all TQFTs on orientable manifolds can be trivially extended to non-orientable manifolds by setting the tensors associated to those manifolds to zero. This should work as gluing two boundary parts of a non-orientable manifold again yields a non-orientable manifold and the same is true for disjoint union with a non-orientable manifold. Right? $\endgroup$ – Andi Bauer May 4 '18 at 13:05
  • $\begingroup$ Also I just realized that the way I intended to extend Turaev-Viro state-sums on non-orientable manifolds with boundary yields zero for all non-orientable manifolds with boundary. $\endgroup$ – Andi Bauer May 4 '18 at 13:06
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    $\begingroup$ For example, how about the bordism $([-1, +1] \times S^2, \phi_-\colon S^2 \to \{-1\} \times S^2, \phi_+\colon S^2 \to \{+1\} \times S^2)$, with $\phi_-$ the identity map and $\phi_+$ the orientation reversal. This bordism doesn't appear in the oriented bordism category, but it's an isomorphism (since it squares to $1_{S^2}$), hence can't be 0 (if the original TQFT wasn't 0 on $S^2$). $\endgroup$ – Manuel Bärenz May 8 '18 at 10:18

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