2d topological field theories $Z : \mathrm{Cob}(2) \to \mathrm{Vect}$ are classified by commutative Frobenius algebras.


What can be said about $(\infty,1)$ 2d TFTs $Z: \mathrm{Cob}(2) \to \mathcal{S}$ with values in a symmetric monoidal $(\infty,1)$-category $\mathcal{S}$? I am interested in different targets $\mathcal{S}$.


I know that $Z(\mathbb{S}^1)$ is naturally an $E\_2$-algebra in $\mathcal{S}$, but certainly it has more structure.


Please note that I do not refer to fully extended (i.e. 2-1-0-dimensional) TFTs.



$(\infty,1)$ 2d TFTs
As you point out, if you just look at the operad of bordisms with exactly one output disc, you get the *framed* $E\_2$-operad (framed here means you can rotate the discs) and so the value of the circle is a framed $E\_2$ algebra. (Aside: if we were working in categories an $E\_2$-algebra is a braided monoidal category, while a framed $E\_2$-algebra would also be *balanced*). Similarly it's a framed $E\_2$-coalgebra. So it's natural to try to defined a framed Frobenius $E\_2$-algebra to be a framed $E\_2$-algebra and an $E\_2$-coalgebra with some kind of "Frobenius" compatibility structure. The problem is that I can't really imagine what description of this Frobenius structure one could possibly hope for that won't make this statement just a tautology... That is, the definition of $E\_2$-algebra is already highly geometric, and so one should expect the definition of Frobenius $E\_2$-algebra to also be highly geometric, and hence the best way to *define* Frobenius $E\_2$-algebra is to say its the value of the circle under a 2d TFT. One way one could hope to make this approach less tautological would be to say what kind of more "combinatorial" description of $E\_2$-algebras you'd like the putative combinatorial description of Frobenius $E\_2$-algebras to extend. For example, you could define an $E\_2$ algebra to be an $A\_\infty$-algebra in $A\_\infty$ algebras to give a Stasheff-like combinatorial description. But you'd still need to also put in the balanced structure somehow.


Perhaps better would be to take an approach that generalizes a different definition of Frobenius algebra. Instead of saying "algebra and coalgebra with Frobenius condition" one can instead say "commutative algebra A with trace such that Tr(xy) is non-degenerate." This has a natural generalization: 



> 
> Def: A framed Frobenius $E\_2$-algebra is, a framed $E\_2$-algebra $A$ together with an $\mathrm{SO}(2)$-invariant map $\mathrm{Tr}: A\rightarrow \mathbf{1}$ such that $X \boxtimes Y \mapsto \mathrm{Tr}(X \cdot Y)$ is the evaluation of a duality between $A$ and itself.
> 
> 
> 


So now we have the following precise question:



> 
> Question: Is the map from the space of 2-dimensional TFTs valued in a symmetric monoidal $(\infty,1)$-category equivalent to the space of framed Frobenius $E\_2$-algebras under the functor which evaluates the circle?
> 
> 
> 


I don't know how to prove this and am not totally sure if it's true, but since duals are defined like adjoints, you can hope that the "uniqueness of coherent adjoints" lets you recover the TFT just from the framed Frobenius $E\_2$-algebra structure.


I'm not even sure if this is true in the case where the target is the $(2,1)$-category of categories... My understand there from Bakalov-Kirillov is that this is asking about the classification of modular functors, which are classified by weakly rigid balanced tensor categories. But this seems to me to be saying that the trace has to be given by $\mathrm{Hom}(-,\mathbf{1})$, and I don't see how that matches up with my definition here of a Frobenius $E\_2$-algebra. Maybe I'm just misunderstanding Bakalov-Kirillov's assumptions though.


