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?


1 Answer 1


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.

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

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