By the cobordism hypothesis, there is an $O(2)$-action on the maximal subgroupoid $\hat{\mathcal{C}}$ of the subcategory of fully dualizable objects in a bicategory $\mathcal{C}$. The $SO(2)$-part of this action can equivalently be described by a natural transformation $id\_{\hat{\mathcal{C}}} \to id\_{\hat{\mathcal{C}}}$ which maps an object $C$ to the Serre automorphism $S\_C$ (see Chris Schommer-Pries' lecture notes "Dualizability in Low-Dimensional Higher Category Theory)". As we have a natural isomorphism between 2-functors, given a 1-morphism $f$, we also expect a 2-morphism $S\_f$ satisfying certain properties. In Section 4.1.1. of Jan Hesse's thesis the existence of $S\_f$ is proven, but no explicit expression is given.


Consider the 2-category $Alg$ in which objects are algebras over $\mathbb{C}$, 1-morphisms are bimodules and 2-morphisms are intertwiners. The subcategory of fully dualizable objects consists of finite-dimensional semisimple algebras, finite-dimensional bimodules and intertwiners (ref: lemma 3.2.1 and 3.2.3 of Orit Davidovich' thesis). In $Alg$, the Serre automorphism is given by the $\mathbb{C}$-linear dual $S\_A = A^\*$ as an $(A,A)$-bimodule (Lemma 4.18 of Jan Hesse's thesis).


Now let $M$ be an invertible $(A,B)$-bimodule, where $A,B$ are finite-dimensional semi-simple. The 2-morphism $S\_M$ expresses a canonical filling of the diagram
$\require{AMScd}$
\begin{CD}
A @>M>> B\\
@V A^\* V V @VV B^\* V\\
A @>>M> B
\end{CD}
In other words, it is an $(A,B)$-bimodule isomorphism $S\_M: A^\* \otimes\_A M \to M \otimes\_B B^\*$.


**Question: Is there an explicit expression for $S\_M$, preferably one that does not depend on a lot of choices, such as bases and direct sums into simples?**



An explicit expression for the naturality of the Serre automorphism in the bicategory of algebras
We will use the fact that $M$ is invertible. Let ${}\_BN\_A$ be an inverse to $M$. Thus we have isomorphisms
$${}\_AM \otimes\_B N\_A \cong {}\_AA\_A$$
and
$${}\_BN \otimes\_A M\_B \cong {}\_BB\_B$$
If we make this data part of an adjoint equivalence (as we should, and as I will assume) then the construction I am about to explain won't depend on these choices.


Rather than construct the map you ask for, I will construct an equivalent map:
$$S\_A: {}\_B N \otimes\_A A^\* \otimes\_A M\_B \to {}\_BB^\*\_B$$
This is easier to express since we are not mapping into a tensor product.


Given an element $b \in B$ we can write it as $\sum\_i n\_i \otimes m\_i$ in $N \otimes\_A M$.


Given $n \otimes f \otimes m$ in $N \otimes\_A A^\* \otimes\_A M$, the map $S\_A$ sends it to the following linear map on $B$:


$$b = \sum\_i n\_i \otimes m\_i \mapsto \sum\_if(mn\_i \cdot m\_in)$$


Here $m n\_i$ and $m\_i n$ are taken as elements in $M \otimes\_B N = A$, which are multiplied together before applying the linear functional $f$. It is not too hard to check that this map is well-defined (doesn't depend on the choice of representation $b =\sum\_i n\_i \otimes m\_i$) and also that it is a $B$-$B$-bimodule map.


It is a little harder to see that this is an isomorphism and I don't have time to write it out just now, but notice that the same construction gives a map the other way:
$$M \otimes\_B B^\* \otimes\_B N \to A^\*$$  

I claim that you can use this to show $S\_A$ is an isomorphism.


