Spin of a transparent anyon
= Spin of a transparent anyon
{title2=$\theta_a\in\{1,-1\}$}
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.