I am considering the following set up. Let $\mathcal{A}$ be a fractional Calabi-Yau category and denote by $S$ the Serre functor and $S^m=[n]$. Now I consider a finite group action $G$ on $\mathcal{A}$ and consider the equivariant category $\mathcal{A}_G$. Then by the results in the paper https://link.springer.com/article/10.1007/s10485-016-9432-4 say Proposition 5.7 and Corollary 5.9, if the center $Z(\mathcal{A})=k$, then
the category $\mathcal{A}_G$ is a fractional Calabi-Yau category of dimension $\frac{n|G|}{m|G|}$, where $|G|$ is the order of $G$, and $(S^G)^m\cong(\kappa\otimes-)\circ [n]$, where $\kappa$ is the morphism $G\rightarrow Z(\mathcal{A})$. ($Z(\mathcal{A})$ is the center of $\mathcal{A}$, consisting of natrural transformations $\mathrm{Hom}(\mathrm{Id},\mathrm{Id})$, which can be identified with degree 0 Hochschild cohomology of $\mathcal{A}$). Thus $\kappa$ can be regarded as the character of $G$. My question is how to compute the $\kappa$ explicitly. In this paper, the author considers an example(which is 5.10 in his paper), he consider a smooth elliptic curve $E$ and then consider the involution $\sigma$ on $E$ given by $(x,y,z)\mapsto (x,-y,z)$, then he consider the equivariant derived categroy $D^b_{G}(E)$, where $G=\mathbb{Z}_2$, generated by the involution. It is easy to see the center $Z(D^b(E))\cong\mathrm{HH}^0(E)=k$, then by Corollary 5.9 in his paper, $(S^{G})^2=[2]$ and $S^G=(\kappa\otimes-)\circ [1]$, so we need to know if $\kappa$ is trivial or not. Here, it seems that he already know the category $D^b_{\mathbb{Z}_2}(E)$ is not a CY-1 category, so he knows $\kappa$ must be non-trivial. Thus $S^G\cong [1]\rho_1$, where $\rho_1$ is the sign character of $G=\mathbb{Z}_2$. But how do we compute this explicitly without knowing the knowledge of $D^b_{\mathbb{Z}_2}(E)$ in advance? I think this must be related to the action $\sigma$, but I am not sure how to argue this.
1 Answer
I don't think you can understand $\kappa$ abstractly. In some sense, the computation of $\kappa$ is a more precise version (taking into account the group action) of computation of the Serre functor.
In the particular example that you mention, the Serre functor is given by tensor product with $\omega_E[1]$, where $\omega_E$ is the canonical line bundle with its natural equivariant structure. Since $E$ is an elliptic curve, the underlying line bundle is trivial, so the only question is whether the equivariant structure is trivial or not.
To see that it is not trivial one can, for instance, observe that the action of the involution on the tangent spaces at the fixed points is non-trivial, and since $\omega_E$ is the cotangent bundle, its equivariant structure is not trivial.
