As explained in Jacob Lurie's paper on the cobordism hypothesis, we have an action of O(2) on the $\infty $-groupoid $X$ given by considering fully dualizable objects and invertable morphisms in some symmetric monoidal $(\infty ,2)$ category $\mathcal C$. I am interested in the case when $\mathcal C$ is the category where objects are algebras, 1-cells are bimodules, 2-cells maps of bimodules etc... By an algebra, I want to include algebra objects in some $\infty$-category (e.g. chain complexes), so that $\mathcal C$ really has non-trivial 3-cells, 4-cells and so on.

Now, I know (from remark 4.2.7) that the fixed points for the induced action of SO(2) on this space correspond to Frobenius algebras, i.e. smooth, proper algebras $A$, with a non-degenerate and $SO(2)$-equivariant trace

$A \otimes _{A^e} A \to k$.

Here, non-degenerate means that the pairing $A\otimes A \to A\otimes _{A^e} A \to k$ is nondegenerate. Recall also that the Hochschild homology $A \otimes _{A^e} A$ carries and $SO(2)$ action, so it makes sense to talk about $SO(2)$-equivariant map above.

However, the reason why the fixed points are as above is that both can be identified with 2-d oriented TFTs, after theorem 3.1.8 which describes how to extend from a (n-1)-dimensional TFT to an n-dimensional one.

Question: Can we calculate this action and identify the fixed point space directly?

Part of this is not too hard to see (I think): naively, the SO(2) action on $X$ gives a canonical (Morita) automorphism of each (f.d.) algebra $A$. This automorphism is given by the bimodule dual $A^!$ (or the other one $A^\vee$...), so being a fixed point means to give an isomorphism $A^! \to A$, which is the same as a non-degenerate map $A\otimes _{A^e} A \to k$. If the base category where our algebra lives has no higher structure, then I think this is enough, but this does not explain the $SO(2)$ equivariance.

In particular, where does the SO(2)-equivariance data come from?

To see this, I tried to go a little deeper (with help from Takuo Matsuoka): the action is given by a map $B\mathbb Z =SO(2) \to Aut(X)$, which we transformed into a map $\mathbb Z \to \Omega Aut(X)$. The data described above corresponds to picking an algebra $A$ and composing to get a map $\mathbb Z \to \Omega X$ = Invertable $A-A$-bimodules. However, we haven't used that the original map from SO(2) is a group homomorphism - this should correspond to the map $\mathbb Z \to \Omega Aut(X)$ being an $E_2$-map. So we get braiding data attached to each of the $A^!$, which should be trivialized (because the E_2 -structure on $\mathbb Z$ is trivial (?)). Then to give a fixed point we must take into account all this data...

This has only given me a headache so far, but maybe there is an easier way to think about this? I tried (failed?) to be brief, so let me know if anything here is unclear!