Schur orthogonality relation on fusion categories

Question

Let $\mathcal{F}$ be the Grothendieck ring of an abelian fusion category. Let $(M_i)$ be its fusion matrices and $(\mathrm{diag}(\lambda_{i,j}))$ their simultaneous diagonalization. Take $M_1=id$, so that $\lambda_{1,j}=1$. The numbers $$c_j:=\sum_i \vert \lambda_{i,j} \vert^2$$ are usually called the formal codegrees. For the fusion category $Rep(G)$ with $G$ finite group, by the Schur orthogonality relations, $(|G|/c_j)$ are the class sizes and $$ \sum_j \frac{1}{c_j} \lambda_{i,j} \overline{\lambda_{i',j}} = \delta_{i,i'}. $$

Question: Is above equality true for every abelian complex fusion category? If so, is it true for every abelian fusion ring?

Answer 1

By Lemma 2.3 in this paper by V. Ostrik (which uses Proposition 19.2(b) in this paper by G. Lusztig): $$ \sum_i \lambda_{i,j} \overline{\lambda_{i,j'}} = \delta_{j,j'} c_{j} $$ Let $U$ be the matrix $(\frac{1}{\sqrt{c_j}}\overline{\lambda_{i,j}})$. The above equality means that $U^*U = id$, i.e. $U$ is an isometry. But in the finite dimensional case, an isometry is unitary, so $UU^* = id$ also, which means that: $$ \sum_j \frac{\overline{\lambda_{i,j}}}{\sqrt{c_j}} \frac{\lambda_{i',j}}{\sqrt{c_j}} = \delta_{i,i'}. $$ The result follows (for every abelian fusion ring).