Modularity of the Drinfeld center of the category of G-graded vector spaces

Question

Background: Let $G$ be a finite group, and $\mathrm{Vect}_G$ be the category of finite dimensional $G$-graded vector spaces over some algebraically closed field $k$ of char 0. It is well-known that $\mathrm{Vect}_G$ is a spherical fusion category with a representative set of simple objects given by $I=\{I_x\}_{x\in G}$, where $(I_x)_g=\delta_{xg}k$. The Drinfeld center $Z(\mathrm{Vect}_G)$ of this category is therefore modular, according to a fundamental theorem of Michael Müger, which states that the Drinfeld center of a spherical fusion category is modular. As a result, the S-matrix of $Z(\mathrm{Vect}_G)$ should be invertible.

Problem: To make things easier, we further assume $G$ to be abelian. Without using the theorem of Müger, how to prove the invertibility of the S-matrix of $Z(\mathrm{Vect}_G)$?

Here is my approach.

Approach: Let $\mathcal{C}=\mathrm{GrRep}(G)$ be the category of finite dimensional $G$-graded representations of $G$. A typical element of $\mathrm{GrRep}(G)$ is a $G$-graded vector space $V\in\mathrm{Vect}_G $ together with a group homomorphism $\rho:G\to \mathrm{Hom}_{\mathrm{Vect}_G}(V,V)$. It is easy to see that $\mathcal{C}$ is fusion with a finite representative set of simple objects $$J=\{(I_x,\rho_\phi):x\in G, \phi\in \mathrm{Hom}_{\mathrm{Grp}}(G,k^\times),\rho_\phi(g)=\phi(g)\mathrm{id}_{I_x} \}. $$ Next, I proved that $Z(\mathrm{Vect}_G)\cong \mathcal{C}$. It remains to show that the S-matrix of $\mathcal{C}$ w.r.t. the basis $J$ is invertible. Note that this is a symmetric matrix of size $|G|^2$. It turns out that even if in the simplest case that $G$ is a cyclic group of order $p$, this matrix is not easy to handle for me.

Did I take a wrong way? Or there are some tricks in linear algebra I missed. Any comments or helps would be very much appreciated.

Answer 1

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{\chi})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde{\chi})} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the conjugation matrix (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.