= Solution
Write $c_{u,v}$ for <braiding> and $M_{u,v}=c_{v,u}c_{u,v}$ for <anyon monodromy>. Transparency says $M_{a,x}=1$ for every $x$. The <hexagon identities for braiding> express <braiding> a composite $a\otimes b$ past $x$ as successive <braidings> of its two factors past $x$. If $a,b$ are both <transparent anyons>, move $x$ around $b$ and then around $a$; both full moves are identities, hence
$$
M_{a\otimes b,x}=1.
$$
This is the <fusion closure of transparent anyons> at the level of the composite object.
For completeness, let $i:c\to a\otimes b$ and $p:a\otimes b\to c$ be an inclusion and projection for an allowed <anyon fusion> summand, with $pi=1_c$. Naturality of <braiding> gives
$$
M_{c,x}=(p\otimes1_x)M_{a\otimes b,x}(i\otimes1_x)=1_{c\otimes x}.
$$
Thus
$$
\boxed{a,b\text{ transparent and }N^c_{ab}>0\quad\Longrightarrow\quad c\text{ transparent}.}
$$
The <vacuum anyon> belongs to this collection, and taking duals preserves transparency, so it is also closed under <antianyons>.
There is a further restriction on the <spin of a transparent anyon>. Apply the <anyon balancing identity> to the <vacuum anyon> channel of $a\times\bar a$:
$$
1=R^{a\bar a}_1R^{\bar a a}_1=\frac{\theta_1}{\theta_a\theta_{\bar a}}=\frac1{\theta_a^2}.
$$
Here ribbon duality gives $\theta_{\bar a}=\theta_a$ and $\theta_1=1$. Consequently
$$
\boxed{\theta_a=+1\ \text{or}\ -1.}
$$
Their <topological spins> are those of <bosons> or <fermions>. This does not imply $d_a=1$: a <transparent anyon> can have several <anyon fusion> channels, with nontrivial exchange action within them, even though every full <anyon monodromy> is trivial. If the theory is additionally a <modular tensor category>, the only <transparent anyon> is the <vacuum anyon>; Question 1 need not impose that extra hypothesis.
Back to article page
