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](https://en.m.wikipedia.org/wiki/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?



Schur orthogonality relation on fusion categories
By Lemma 2.3 in [this paper](https://mathscinet.ams.org/mathscinet-getitem?mr=2576705) by V. Ostrik (which uses Proposition 19.2(b) in [this paper](https://mathscinet.ams.org/mathscinet-getitem?mr=1974442) 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).


