Uniqueness of character for Z_+-rings

Question

I have a question about the proof of proposition $3.3.6(3)$ in "Tensor Categories" by Etingof et al..

This part states that for $A$, transitive unital $\mathbb Z_+$-ring, there is a unique character taking non-negative values on the basis elements.

The proof uses the fact that if $\chi$ is a character, and $f$ is a vector with entries $\chi(Y)$ for basis elements $Y$, then $f$ is an eigenvector of the matrix of left multiplication by $\sum_{X\in I}X$, where $I$ is a set of basis elements.

But I don't see why this holds. As I understand we want an equation: $$ \sum_{X,Y\in I} \chi(Y) XY = \lambda \sum_{Z\in I} \chi(Z) Z , $$ to hold, so $\sum_{X,Y\in I} \chi(Y) c_{XY}^Z = \lambda \chi(Z)$. But I do not understand what to do next.

In bibliographical notes there is a link to the paper https://arxiv.org/pdf/math/0203060.pdf (Lemma 8.3). Which uses more or less the same fact with $X$ instead of $\sum_{X\in I} X$ as a matrix, which is even more mysterious to me.

Can someone please clarify this for me?

Answer 1

Hm, I probably figured it out myself, but I won't delete the question since I think the formulation in the proof is slightly misguiding.

Use $b_i$ to denote basis elements ($i \in I$), a take element $y = \sum y_i b_i$ and denote $\chi_i = \chi(b_i)$. We have $\chi(y) = \sum y_i \chi_i$. But if the multiplication by $x$ -- any element of the ring is given by matrix $A_{ij}$ we have $xy = \sum b_iA_{ij}y_j$. Now take character of both sides to obtain: $$ \chi(x) \sum \chi_j y_j = \sum \chi_i A_{ij} y_j \ , $$ since this holds we have $\chi(x) \chi_i = \sum \chi_i A_{ij}$, so the row vector $\chi_i$ is indeed the eigenvector of $A_{ij}$. Now the rest of the proof works the same since the can use the Frobenius-Perron theorem.

Answer 2

Hrm, I can’t seem to get it to work without assuming that the ring is based...

Let’s just do the version without the sum from the paper. So you want: $$\sum_{Y\in I} \chi(Y^*) c_{Z X^*}^{Y^*} = \sum_{Y\in I} \chi(Y) c_{XY}^Z = \lambda \chi(Z)$$

Now take $\lambda = \chi(X)=\chi(X^*)$.