Non-extendable 2D TQFTs correspond to finite dimensional Frobenius algebras [1].


How about 3D TQFTs? While the answer is clear for the extended ones (e.g. (3,2,1) TQFTs almost correspond to modular tensor categories [2]), I have not seen any discussion for (3,2) TQFTs.


More precisely, can one classify the functors


$$Cob\_{3,2}^{oriented} \to (Vect\_\mathbb{C})?$$


**Reference**


* [1] [Cohomology rings and 2D TQFTs](https://mathoverflow.net/questions/8772/cohomology-rings-and-2d-tqfts)
* [2] K. Walker's answer [here](https://mathoverflow.net/questions/386/do-all-3d-tqfts-come-from-reshetikhin-turaev/)
* [3] related - [Example of a non-extendable TQFT?](https://mathoverflow.net/questions/290880/example-of-a-non-extendable-tqft)



Non-extendable 3D TQFTs
Check out Andras Juhasz' paper: <https://arxiv.org/pdf/1408.0668.pdf>


Specifically, Theorem 1.10:



> 
> There is an equivalence between the symmetric monoidal category
> of (2+1)-dimensional TQFTs and the category of J-Algebras.
> 
> 
> 


J-Algebras are somewhat cumbersome to define.
They are graded vector spaces equipped with a Frobenius-like structure, together with an action of the genus $g$ mapping class group on the $g$-th graded piece.


