Distinguishable ways of splitting/fusing anyons

Question

I have a difficulty in understanding the possibility that two simple anyons can fuse into one simple anyon in distinguishable ways: \begin{eqnarray} a\times b= 2c. \end{eqnarray} Let us put it in the other way and split $c$ into $a$ and $b$. The above fusion relation tells us that we have two distinguishable ways to do this splitting. However, I have no feelings on this "distinguishable".

To be concrete, we could consider a lattice system constructed on a sphere. First, we split the vacuum $1$ into $c$ and $\bar{c}$ by acting on the ground state with a string operator having end points $c$ and $\bar{c}$. Then, we further apply two string operators: one with $a$ at one end point and $\bar{a}$ at the position of $c$ as the other end, the second with $b$ at one end and $\bar{b}$ at the position of $c$. However, I cannot see other distinguishable ways.

Does it mean that the "distinguishable" is due to the choices of string operators, or some other intrinsic degrees of freedom of anyon $c$? Or is it even wrong that two simple anyons cannot fuse into a simple anyon in two distinguishable ways?

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Supplemented:

I make my question more concrete below. If we have an initial state $|c\rangle$ on a sphere with well-separated $c$ and $\bar{c}$ and $N_{ab}^c=2$. Then we can split the $c$ into $a$ and $b$ in two "distinguishable" ways. It means (?) there exist two local operators $U_1$ and $U_2$ (edited: may not be unitary, but satisfy $U^\dagger_iU_{i}|c\rangle=|c\rangle$) such that $U_1|c\rangle$ and $U_2|c\rangle$ are orthogonal and they have $a$ and $b$ in the finite $\text{supp}(U_{1,2})$ around $c$. Next, we drag $a$ from supp$(U_{1,2})$ so that $a$ and $b$ are well-separated, e.g., by a unitary string operator $L_a$. Then two final states would be \begin{eqnarray} |a,b;c;1\rangle=L_a U_1|c\rangle\\ |a,b;c;2\rangle=L_a U_2|c\rangle. \end{eqnarray} We expect these two states should be topologically degenerate, but \begin{eqnarray} \langle a,b;c;2|L_a U_2 U^\dagger_1L^\dagger_a |a,b;c;1\rangle=1\neq0. \end{eqnarray} It would be OK if $L_a U_2 U^\dagger_1L^\dagger_a$ is not local, but it seems that it is local and supported approximately by supp$(U_{1,2})$ as follows.

(*)For example, $L_a$ is a product of spin operators and most of them are geometrically far away from supp$(U_{1,2})$ so we can move them through $U_2 U^\dagger_1$ to cancel with the corresponding part of $L^\dagger_a$.

Then $|a,b;c;1\rangle$ and $|a,b;c;2\rangle$ are not topologically degenerate since the local perturbation [$L_a U_2 U^\dagger_1L^\dagger_a$+h.c.] can open the gap, which is inconsistent with the expectation. I guess my mistake should be in the cancellation of the remote parts of $L_a$ and $L_a^\dagger$ of the paragraph (*) above since there could be a short-range obstruction in decomposition of $L_a$ into a remote part and a supp$(U_{1,2})$-nearby part which commutes with each other, though I have no lattice model in my mind.

Answer 1

If you consider a configuration on a sphere with four anyons $a, b, c$ and $\bar{c}$, the fact that $N_{ab}^c=2$ says that there are actually two degenerate ground states, both of which have the same configuration of anyons, but are otherwise indistinguishable by any local operator. The choice here is really the operator that splits $c$ into $a$ and $b$. Of course, there should exist some non-local observables which can distinguish the two states.

Examples of fusion multiplicity are not so common in anyon theories discussed most often in condensed matter literature, but there are plenty of them if you look at discrete gauge theories with non-Abelian gauge group, for example $S_3$.

Edit: $S_3$ gauge theory actually has no fusion multiplicity. One example that does work is $A_4$ gauge theory. The 3-dimensional irrep of $A_4$ (denoted by 3) has $N_{\mathbf{33}}^\mathbf{3}=2$. Another example is $SU(3)_3$.