According to Gukov et al. in this 2017 paper Seiberg-Witten theory in 4d categorifies Seiberg-Witten theory in 3d. In what sense is this phrase mentioned? I know what the process of categorification is (e.g. how Khovanov homology categorifies Jones polynomial).

What is the exact relation between the 3d and 4d versions of Seiberg-Witten theory and in what sense is the latter the categorification of the former?

  • $\begingroup$ It's just the "TQFT framework" where in 3 dimensions you have homologies (generated by solutions to 3-dimensional SW equations), and in 4-dimensions you have cobordism maps between homologies (by counting solutions to 4-dimensional SW equations that are asymptotic to the 3-dimensional generators), and when you take the cobordism to have empty boundary (i.e. a closed 4-manifold) you get the numerical SW-invariant (roughly speaking). $\endgroup$ – Chris Gerig Mar 25 '17 at 2:44
  • $\begingroup$ @ChrisGerig Firstly homology should be categorified to a 2-category if I am correct. Secondly physically what is this cobordism in 4d? Any references? $\endgroup$ – Gorbz Mar 25 '17 at 3:11
  • $\begingroup$ Cobordism = 4-dimensional manifold with 3-manifold boundaries. Associated to each 3-manifold is the Seiberg-Witten-Floer homology, and a cobordism induces maps between those homologies. Definitive reference: "Monopoles and 3-manifolds" by Kronheimer-Mrowka. $\endgroup$ – Chris Gerig Mar 25 '17 at 3:26
  • $\begingroup$ Hi, thanks. I know that book. My confusion is in what sense the "cobordism" is equal to the SW invariants in 4d. These are polynomial invariants. On the other hand homologies are vector spaces. How are polynomial invariants categorification of a vector space, I thought that that should be a 2 category! $\endgroup$ – Gorbz Mar 25 '17 at 9:40
  • $\begingroup$ The "SW invariant" I refer to is an integer, and it is obtained from a closed 4-manifold (cobordism with empty ends) by stretching out two balls, so you get a cobordism from 3-sphere to 3-sphere and hence a map on homologies which (roughly speaking) is $\mathbb{Z}\to\mathbb{Z}$, where the image of 1 is the SW invariant. $\endgroup$ – Chris Gerig Mar 25 '17 at 16:38

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