In the rep theory of Quantum Double, why does the fusion of 2 "pure fluxes" yield a "pure charge"?

Question

Physicist here, so my notation may be different from standard math notation.

For the quantum double $D(G)$ of a group $G$, we may write representations of $D(G)$ in the following way: Consider a conjugacy class $C \in (G)_{cj}$ and a centraliser group $Z_C := \{g \in G|g r_C = r_C g \}$ for some chosen representative element $r_C \in C$. Then the irreps of $D(G)$ are given by $(C, R)$ where $R \in Irrep(Z_C)$.

We may now consider the tensor product of representations, which then decomposes in terms of a direct sum of Irreps with certain multiplicities: $(C_1, R_1) \otimes (C_2, R_2) = \oplus N^{(C, R)}_{(C_1, R_1), (C_2, R_2)} (C,R)$.

Now lets focus on the case where $R = 1$ (trivial representation of the centralizer), with the group $G = S_3$. $S_3$ has 3 conjugacy classes denoted by $C_1 = \{(1)\}, C_{12} = \{(12), (23), (13)\}, C_{123} = \{(123), (132)\}$. Then we have $(C_{123},1) \otimes (C_{123}, 1) = (C_{123}, 1) \oplus (C_{1}, 1) \oplus (C_{1}, s)$ where $1$ is the trivial representation and $s$ is the sign representation of $S_3$.

I call irreps of the type $(C_{1}, R \neq 1)$ as "pure charges" and irreps of the type $(C \neq C_{1}, 1)$ as "pure fluxes".

I want to understand why a pure charged representation $(C_{1}, s)$ arises in the decomposition of the tensor product. Naively I would hope that a tensor product of two pure fluxes should give pure fluxes. I suspect the reason is that in the decomposition of the tensor product, the rep of $D(G)$ with $C=C_{1}$ is still reducible, and can be decomposed in terms of $R = 1,s$. Is this the correct way to think about it? How could I confirm this?

Answer 1

Think of reps of $D(G)$ as $G$-equivariant vector bundles on $G$, where the group acts by conjugation. In this language, the tensor product is push-forward under multiplication $\mu : G\times G \rightarrow G$ of the exterior tensor product: $$U\otimes V = \mu_\ast (U\boxtimes V).$$ Now let's do your calculation. Let $x=(123)$. Then $U\boxtimes V$ has ${\mathbb C}$ at four points $(x,x),(x^{-1},x^{-1}),(x,x^{-1}),(x^{-1},x) \in G\times G$ and zero elsewhere. Push forward is just summation over the inverse image, hence, it has non-zero fibres at 3 points from 2 orbits: $$U\otimes V_{1}= {\mathbb C}_{(x,x^{-1})} \oplus {\mathbb C}_{(x^{-1},x)}, \ U\otimes V_{x^{-1}}={\mathbb C}_{(x,x)}, \ U\otimes V_{x}= {\mathbb C}_{(x^{-1},x^{-1})}.$$ The last two fibres constitute what you call $(C_{123},1)$. The first two make $(C_{1},W)$, where you have to figure the two dimensional representation $W$ of $S_3$. It has an obvious basis: $1_{(x,x^{-1})}$ and $1_{(x^{-1},x)}$. The key calculation $$x\cdot (x,x^{-1}) = (xxx^{-1},xx^{-1}x^{-1})= (x,x^{-1}), \ (1,2)\cdot (x,x^{-1}) = (x^{-1},x)$$ tells us that $x$ fixes the basis elements and $(1,2)$ swaps them. This gives us the character of $W$: $$ \chi_W (x)=2, \ \chi_W ((1,2)) =0$$ that tells us that $W=1\oplus s$.

To summarize, we see that the calculation yields us the permutation representation of $S_3$ on $S_3/C_3$, which has a sign representation as a direct summand.