# Schur orthogonality relation on fusion categories

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?

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).