TITLE: Non-trivial factor splitting from vacuum in TQFTs. F-move not unity?
QUESTION [6 upvotes]: Consider a unitary modular TQFT, defined by the F and R moves. More specifically, a braided tensor category relevant for anyon models in 2D topologically ordered phases of matter. I am interested in the value of the F-move, $[F^{abc}_1]_{\bar{a}\bar{c}}$, diagrammatically defined by 
 
where $1$ is the vacuum and $\bar{a}$ is the inverse of $a$, i.e. $1 \in \{a\times \bar{a}\}$
I would naively expect this to be the identity. However, in https://thesis.library.caltech.edu/2447/2/thesis.pdf
the author states that this need not be the case.
I think this is related to bending, and/or the $\mathbb{Z}_2$, or possibly the $\mathbb{Z}_3$, Frobenius Schur indicator. Any clarification (or references) on the value of this F-move would be much appreciated!

REPLY [4 votes]: Consider the case $a=b=c$.  Then the fusion space $V_{aaa}$ affords a representation of $\mathbb Z/3$ via a $2\pi/3$ rotation.  The F-move in your question is essentially this $2\pi/3$ rotation, and the rotation need not act trivially (its eigenvalues might be non-trivial 3rd roots of unity).
See for example eqn 470 of https://arxiv.org/pdf/1709.01941.pdf , which cites https://arxiv.org/pdf/0704.0208.pdf .
[edit:]
In response to the questions in the comments:
(1) Yes, this is sometimes called the $\mathbb Z/3$ Frobenius-Schur indicator.
(2) Away from the case $a=b=c$, it does not (strictly speaking) make sense to ask whether the F-moves in question are the identity, because they are linear maps between distinct vector spaces $V_{abc}$, $V_{bca}$ and $V_{cab}$.  [Another edit: I'm using here the fact that $V_{1a\overline a}$ and $V_{1\overline cc}$ (from the figure in the question) are 1-dimensional and have canonical generators, so that in this particular case the F-move is equivalent to a map from $V_{abc}$ to $V_{cab}$.]  However, if we chose bases of those vector spaces appropriately, we can arrange for the matrix representations of the F-moves to be the "identity" matrix (1s down the diagonal).  Note that in general these matrices are only well-defined up to unitary change of bases in the various $V_{xyz}$, so with a different choice of basis we can arrange for the F-matrix in the question to be any unitary matrix that we like.