# Frobenius-Perron dimension on a fusion category

Let $C$ be a fusion category with simple objects $V_i\in I$. The fusion rule is $V_i\otimes V_j \cong N_{i,j}^k V_k$.

The Frobenius-Perron dimension of a simple object $V_i$, $\mathrm{FPdim}(i)$, is defined as the largest nonnegative real eigenvalue of a matrix $N_i$, where the $(k,j)$ entry is $(N_i)_{k j}=N_{i,j}^k$.

I would like to know if we have the following equality or not.

$\sum_{j\in I} \mathrm{FPdim}(j) N_{i j}^k=\mathrm{FPdim}(i) \mathrm{FPdim}(k)$

for any fixed $i, k \in I$.

If so, I would like to know how to prove this. Thank you.

• Hmm, are there some more assumptions than written here? Or why is this value well-defined (why are we guaranteed any real eigenvalues at all for example). I would assume this is related to the Perron-Frobenius theorem, but that requires that some power of the matrix is positive, which is not clear to me is the case here. – Tobias Kildetoft Feb 4 '16 at 20:44
• @TobiasKildetoft First, $N_{i,j}^k$ is nonnegative and hence the matrix $N_i$ has nonnegative netry. This ensures there is a nonnegative real eigenvalue. – Snow Feb 4 '16 at 20:54
• Ahh, right. I recalled the much stronger conditions needed for the stronger statements of the usual theorem, sorry. – Tobias Kildetoft Feb 4 '16 at 20:58
• I don't see any reason why you would have the equality as stated. But if you change it to $N_{i,k}^j$ on the left hand side, then it will be true at least when the matrices corresponding to tensoring with $V_i$ and $V_k$ share an eigenvector for their largest eigenvalue. – Tobias Kildetoft Feb 4 '16 at 21:05
• This is really just a fact about finite dimensional rings with a positive multiplicative basis, check out proposition 3.3.6 in this book: www-math.mit.edu/~etingof/egnobookfinal.pdf – Nate Feb 4 '16 at 21:14