Spin of a transparent anyon (${\theta }_a\in \{1,-1\}$)

Apply the anyon balancing identity to the vacuum anyon channel in $a\times \bar{a}$. Transparency gives $1=({\theta }_a{\theta }_{\bar{a}}{)}^{-1}={\theta }_a^{-2}$. Thus a transparent anyon has bosonic or fermionic topological spin. This does not require its quantum dimension to equal one.