6
$\begingroup$

Given a fusion category $\mathcal C$, the Grothendieck Ring $K_0(\mathcal C)$ is the $\mathbb Z$-based ring whose basis elements correspond to isomorphism classes of simple objects and whose multiplication is given by

$$ X\times Y = \sum_Z N_{XY}^Z Z, N_{XY}^Z=\vert \mathcal C(X\otimes Y,Z)\vert $$

Two fusion categories $\mathcal C$ and $\mathcal D$ are said to be Grothendieck equivalent if $K_0(\mathcal C)\cong K_0(\mathcal D)$.

Given $\mathcal C$, the adjoint subcategory $\mathcal C_{ad}$ is the full fusion subcategory of $\mathcal C$ generated by $X\otimes X^*$, where $X$ is simple.

$Rep(D(S_3))$ has eight simple objects $\{1,\epsilon, \phi_{i=1,\ldots,4},\psi_\pm\}$ and $Rep(D(S_3))_{ad}$ is the subcategory generated by $\{1,\epsilon,\phi_{i=1,\ldots,4}\}$. Its Grothendieck ring is commutative and determined by

$$ \begin{align*} \epsilon \otimes \epsilon &\cong 1 \\ \epsilon \otimes \phi_i &\cong \phi_i \\ \phi_i \otimes \phi_i &\cong 1 \oplus \epsilon \oplus \phi_i \\ \phi_i \otimes \phi_j &\cong \phi_k \oplus \phi_l & i\neq j \neq k \neq l \\ \end{align*} $$

$Rep(D(S_3))$ is modular so $Rep(D(S_3))_{ad}$ is braided(properly premodular). $Rep(D(S_3))_{ad}$ also admits a braiding with S-matrix

$$ \left( \begin{array}{cccccc} 1 & 1 & 2 & 2 & 2 & 2 \\ 1 & 1 & 2 & 2 & 2 & 2 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ 2 & 2 & 4 & 4 & 4 & 4 \\ \end{array}\right). $$

This braiding is symmetric $(s_{ab}=d_a d_b)$ and so $Rep(D(S_3))_{ad}$ is equivalent as a fusion category to $Rep(G)$ for some finite group $G$. What is this $G$?

$\endgroup$
4
  • $\begingroup$ Can you tell us the dimensions of the simples? That would narrow the search considerably. I can't extract that information at a glance from what you've written. $\endgroup$ May 15, 2015 at 22:43
  • 1
    $\begingroup$ They are {1,1,2,2,2,2}. $\endgroup$ May 15, 2015 at 22:44
  • $\begingroup$ I know that this is not $D_9$. It has objects of the same dimension, but different tensor structure on the two dimensional objects. There, only one object gives a $Rep(S_3)$ subcategory. $\endgroup$ May 15, 2015 at 22:46
  • 2
    $\begingroup$ In that case it looks like the only possible candidate among the groups of order 18 (groupprops.subwiki.org/wiki/Groups_of_order_18) is $(C_3 \times C_3) \rtimes C_2$, with $C_2$ acting by inverse. $\endgroup$ May 15, 2015 at 23:13

1 Answer 1

6
$\begingroup$

Despite my love of the finite group game, let me give an argument that doesn't use the classification of groups of order 18. The 1-dimensional objects correspond to representations of the abelianization. So your group must have abelianization $C_2$, and so its commutator subgroup must be a group of size 9. Note that this splits as a semidirect product because there's a 2-sylow subgroup.

A full tensor subcategory of $\mathrm{Rep}(G)$ which is closed under summands must be of the form $\mathrm{Rep}(G/N)$. (The proof of this is roughly the same as the proof that faithful representations tensor generate.) Thus your group must have $S_3$ as a quotient in four different ways. Hence the commutator subgroup must be elementary abelian and the $C_2$ must act on each of the factors by inversion (and thus on the whole thing by inversion).

$\endgroup$
2
  • $\begingroup$ Thanks! Is there a good reference on the relationship between the group and its representation category? It has been a while since I read Serre or Fulton and Harris but I don't recall either of these being written from a terribly categorical perspective. $\endgroup$ May 16, 2015 at 22:50
  • $\begingroup$ I don't know of any references that are particularly better than Serre. $\endgroup$ May 17, 2015 at 1:59

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.