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  Khovanov homology and Crane-Yetter TQFT

+ 7 like - 0 dislike

Crane-Yetter(-Kauffman) have constructed 4-dimensional TQFT in such a way that Reshetikhin-Turaev theory lives on the boundary $\partial M$ of a 4-manifold $M$. Therefore, Crane-Yetter TQFT can be thought of a categorification of Chern-Simons theory. If we add surface defect (or foam) in Crane-Yetter theory that is an extension of a knot in Chern-Simons theory, then does it provide a categorification of quantum invariants of the knot which is Khovanov homology?

On the other hand, Witten has proposed in the paper that TQFT obtained by topologically twisting 4d N=4 SYM provides a framework for a categorification of Chern-Simons theory as well as Khovanov homology, which looks different from Crane-Yetter TQFT. I wonder how Crane-Yetter TQFT is related to Khovanov homology.

This post imported from StackExchange MathOverflow at 2017-07-02 10:59 (UTC), posted by SE-user Satoshi Nawata
asked Jun 28, 2017 in Theoretical Physics by Satoshi Nawata (75 points) [ no revision ]
retagged Jul 2, 2017
Categorification of Chern-Simons would be some 4d TFT $Z_{Kh}$ such that $Z_{CS}(\partial M)$ (which depends on $q$) is the character of $Z_{Kh}(\partial M)$ (the latter theory has no $q$-dependence). In the Crane-Yetter case (which depends on $q$) there is a map $f\colon Z_{CY}(\partial M)\rightarrow \mathbf{C}$ and you can recover the partition function $Z_{CS}(\partial M)$ as the image of $Z_{CY}(M)\in Z_{CY}(\partial M)$ under $f$. This looks different from what $Z_{Kh}$ is supposed to be.

This post imported from StackExchange MathOverflow at 2017-07-02 10:59 (UTC), posted by SE-user Pavel Safronov
@Pavel Thanks for your comment. So do you mean that, roughly speaking, $Z_{CY}(M)=Z_{CS}(\partial M)$? In that case, if we consider a defect supported on knotted surface in the Crane-Yetter theory, is it probable that it gives ``quantum'' surface knot invariant depending on $q$?

This post imported from StackExchange MathOverflow at 2017-07-02 10:59 (UTC), posted by SE-user Satoshi Nawata
Yes, but the Crane-Yetter theory by itself is not very interesting (e.g. it's invertible). For instance, on a closed 4-manifold the invariant is expressible in terms of the Euler characteristic and the signature. I don't know if the CY invariant with a surface defect has been computed in the literature.

This post imported from StackExchange MathOverflow at 2017-07-02 10:59 (UTC), posted by SE-user Pavel Safronov

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