# Math of anyons: Quantum dimension of 1 implies abelian charge

This question originates from the following statement in Bonderson's thesis: Link to Thesis page 16 or pdf-page 23:

The quantum dimension $$d_a$$ of an anyon of charge $$a$$ satisfies $$d_a \geq 1$$ with equality iff a is Abelian.

Here, the quantum dimension is defined as $$d_a = d_{\bar{a}} = |[F^{a\bar{a}a}_a]_{1,1}|^{-1}$$ and a charge $$a$$ is abelian if $$\sum_c N^c_{ab}=1$$ for all $$b$$.

Of course, if $$a$$ is abelian, this $$F$$-matrix is one-dimensional and by unitarity then $$[F^{a\bar{a}a}_a]_{1,1} \in U(1)$$, so that $$d_a=1$$. I have trouble proving the converse: Given that $$d_a=1$$, how to prove that $$a$$ has to be abelian?

What I know so far: $$d_a$$ is the Frobenius-Perron eigenvalue to $$N_a$$ with components $$(N_a)_{bc}=N^c_{ab}$$. Thus, if for $$d_a=1$$ all eigenvalues of $$N_a$$ are of absolute value smaller or equal 1.

• maybe you could provide a link or a full reference to the thesis as right now there is minimal context to your question. – ZeroTheHero Apr 30 at 11:23
• Of course, I forgot to put it, have done so now. – Marsl Apr 30 at 11:26