Let $G$ be a finite abelian group and $q: G \to k^\times$ be a quadratic form with $k$ a field of characteristic $0$. i.e. $(G,q)$ is a pre-metric group. It is known that such pre-metric groups correspond to braided equivalence classes of pointed braided fusion categories (see e.g. EGNO Tensor Categories, Section 8.4). Let $\mathcal{C}(G,q)$ denote such a category corresponding to $(G,q)$.
$\mathcal{C}(G,q)$ has a twist $\theta\in \operatorname{Aut}(\operatorname{id}_\mathcal{C})$ defined by $\theta_X = q(g)\operatorname{id}_X$, where $X$ is a simple object belonging to the isomorphism class corresponding $g\in G$. It is also known that in this setup, $\theta$ is a ribbon structure, i.e. $$\theta_{X\otimes Y} = (\theta_X\otimes \theta_Y)\circ c_{Y,X}\circ c_{X,Y}.$$
I came across the fact that the $S$-matrix of $\mathcal{C}(G,q)$ is $S_{gh} = (b(g,h))_{g,h\in G}$, where $$b(g,h) = \frac{q(gh)}{q(g)q(h)}$$ is the symmetric bilinear form associated to $q$. For example, in EGNO pg. 224, or On Braided Fusion Categories I Exhibit 2.11.5 by Drinfeld, Gelaki, Nikshych, and Ostrik. Why is this?
The $S$-matrix is defined to be $S = (\operatorname{Tr}(c_{Y,X}c_{X,Y}))_{X,Y\in \mathcal{O}(\mathcal{C})}$ the matrix of traces of doing braidings twice. Here, $\mathcal{O}(\mathcal{C})$ is the collection of simple objects of $\mathcal{C}$. Specifically, why is $\operatorname{Tr}(c_{Y,X}c_{X,Y}) = b(g,h)$?
I made a very silly oversight in the very reference I provided, but I will answer for closure anyway. We know $X \otimes Y \cong Y \otimes X$ is a simple object in this category since they are labelled by the group elements. The fact that the twist satisfies the ribbon condition means that $$ q(gh) = q(g)q(h) c_{Y,X}c_{X,Y}$$ where we abuse notation and write $c_{X,Y}$ for the scalar in $k^\times$ picked up by applying the (simple object!) endomorphism $c_{X,Y}$. This concludes.