It is known that 3d TQFTs are classified by modular tensor categories, and 2d TQFTs by Frobenius algebras.

A 3d TQFT on a manifold of the form $S^1\times M_2$ induces a 2d TQFT on $M_2$. So there must be a map from the MTC of the former to the FA of the latter. What is this map, explicitly? In particular,

  1. How are the $\{\lambda_i\}$ that appear in the partition function of the 2d TQFT (cf. this PSE post) fixed in terms of the modular data of the 3d TQFT?

  2. Is this map bijective?

These are my thoughts regarding these subquestions:

  1. I can answer this subquestion by reverse-engineering the Verlinde formula, but I am looking for a more conceptual answer rather than the solution itself. So as far as this question goes, I would appreciate if Verlinde is not used. I know what the answer is, I would like to know where it comes from.

  2. I don't expect the map MTC$\to$FA to be injective. My intuition is that higher-dimensional theories contain more information than their dimensionally-reduced/compactified versions. A 3d object, after all, is more complex than a 2d one. From a more formal POV, MTCs carry braiding, and FAs do not, I believe. Conversely, I would expect the map MTC$\to$FA to be surjective (more generally, I feel one should be able to reach any $d$-dimensional QFT by reducing a certain $(d+1)$-QFT). This is probably wrong, for the following reason: any set of real numbers $\{\lambda_i\}$ determines a 2d TQFT, but 3d TQFTs only induce a discrete subset of sets $\{\lambda_i\}$ (in particular, these are always algebraic, I believe). 3d theories are very rigid (cf. Ocneanu), while 2d ones are not.

  • $\begingroup$ I'm not sure your expectation in 2. is entirely "obvious", since the setting of $S^1\times M_2$ is a very specific 3d setting - it would be equally reasonable to me to suspect that the 3d theory only becomes richer - i.e. non-injective - when you allow arbitrary - or at least more - 3d bases. For the converse, surjectivity would mean that the structure of the 3d theory on $S^1\times M_2$ does not impose any additional restrictions on the 2d theory on $M_2$, which I also don't think is obvious. $\endgroup$
    – ACuriousMind
    Aug 28, 2020 at 18:02


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