2
$\begingroup$

Applying the modularisation/deequivariantisation procedure to the representation category $\operatorname{Rep}_G$ of a finite group $G$ with trivial braiding gives the fibre functor to vector spaces. What happens if we chose a nontrivial braiding?

My vague understanding from this question is that we need to choose an abelian normal subgroup $A \hookrightarrow G$ together with a bilinear form on the function algebra $\beta: \mathbb{C}(A) \otimes \mathbb{C}(A) \to \mathbb{C}^*$ to get a quasitriangular structure on the group algebra $\mathbb{C}[G]$. Supposedly, a skew-linear $\beta$ leads to a triangular braiding (which I don't find obvious, and I can't find a detailed proof).

Can the modularisation functor somehow be expressed as composition $\iota^*$ with a subgroup inclusion $\iota: A' \hookrightarrow G$, and $A'$ another subgroup to be determined from $A$ and $\beta$? If not, how does the modularisation relate to the given data (on which it must depend completely)?

Edit: From Qiaochu Yuan's excellent answer to my question, I can add a little more to my vague conjecture.

The symmetric centre of $\operatorname{Rep}_G$ is a symmetric fusion category. Let's assume for a moment that its twist is trivial (i.e. no supergroups), then we can restrict the fibre functor of $\operatorname{Rep}_G$ to the symmetric centre,so it's the representations of another group, which we'll call $G'$. Intuitively, it should be the group that we are going to deequivariantise away!

We can restrict the automorphisms of the fibre functor of $\operatorname{Rep}_G$ to the fibre functor of $\operatorname{Rep}_{G'}$, which gives us a surjective group homomorphism $\phi: G \twoheadrightarrow G'$ such that $\phi^*: \operatorname{Rep}_{G'} \hookrightarrow \operatorname{Rep}_G$ is the inclusion of the symmetric centre. My conjecture is now that $A'$ is the kernel of $\phi$!

$\endgroup$
1
$\begingroup$

Braided structures on the category of representation of a finite dimensional Hopf algebra $H$ are in correspondence with R-matrices. If $H=k[G]$, non-symmetric braid structures (without fermions) correspond with pairs $(A,b)$, where $A \lhd G$ is a normal subgroup and $b\in \operatorname{Hom}(A^{\otimes 2},U(1))$ is a non-degenerated bicharacter Ad-$G$-invariant, (see arXiv: q-alg/9706007 for details). Note a that a non-degenerated bicharacter is exactly the same as a modular structure on $\operatorname{Corep}(k[A])=\operatorname{Vec}_A$.

Let $\mathcal{C}=(\operatorname{Rep}(G),(A,b))$ be the braided category. The symmetric fusion subcategory $\mathcal{D}=\operatorname{Rep}(G/N)\subset \operatorname{Rep}(G)$ corresponds to the Müeguer's center (transparent objects) and the modularization is just the modular category $(\operatorname{Vec}_A,b)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.