# An explicit expression for the naturality of the Serre automorphism in the bicategory of algebras

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?

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.

• Awesome Chris, thanks a lot! It make sense that you have to pick an inverse of M. Oct 11, 2020 at 16:45