What does a zero topological S matrix element mean? I realize that for nonabelian anyons, their S matrix elements could be zero (eg. the Ising anyons). I'm confused by the meaning of a zero S matrix element. Does it mean that the corresponding braiding process results in a zero amplitude?
 A: Let us consider the S matrix of Ising anyons as an example: $S_{\sigma\sigma}=0$. To be more precise, suppose we have four $\sigma$'s,  labeled as $1,2,3,4$. We assume $1$ and $2$ are in a definite fusion channel. Then we braid $3$ around $1$ (or $2$, no difference). Then the state of $1$ and $2$ must be orthogonal to the initial one: they must be in the opposite fusion channel! This is the meaning of $S_{\sigma\sigma}=0$. One can do a simple calculation using Majorana modes as a "model" of Ising anyons: say we have $\gamma_1,\gamma_2,\gamma_3,\gamma_4$. Braiding $\gamma_3$ around $\gamma_1$ results in the following transformation: $\gamma_{1,3}\rightarrow -\gamma_{1,3}$. Therefore, the fermion parity $i\gamma_1\gamma_2$ which represents the fusion channel of $1$ and $2$, is flipped.
This observation was first made by Bonderson et al in http://arxiv.org/abs/cond-mat/0508616. They devised an interference experiment to see this zero (vanishing of interference pattern). So the physical meaning of this zero is worth a PRL!
