Given a ring $R$ with finite additive basis $\{e\_i\}\_{i=1}^{n}$, such that $e\_i e\_j=\sum c\_{ijk}e\_k$ with $c\_{ijk}\in \mathbb{N}$, we define the Perron-Frobenius dimension $FPDim(e\_i)$ of a basis element $e\_i$ to be the maximal positive real eigenvalue of matrix $M\_{e\_i}$, multiplication by $e\_i$. This exists by the Perron-Frobenius theorem, and we extend by linearity to all of $R$.


Now take $B(G)$ to be the burnside ring of a finite group $G$, with basis given by isomorphism classes of transitive actions of $G$. One can directly check for $G\cong C\_p$ that $FPDim(X)=|X|$. Does this hold in general for finite $G$-sets?


Note that our ring $B(G)$ is not necessarily transitive in the sense of Etingof's Tensor Categories (Definition 3.3.1), so this doesn't seem to follow immediately from the results in there.



Is the Perron-Frobenius dimension of a G-Set given by its cardinality?
Lets fix an orbit $X = G/H$. It suffices to determine the asymptotic growth of the trace of multiplication by $X^n$, since the maximal positive eigenvalue(s) dominates the sum $\sum\_i \lambda\_i^n$. 


This trace is the sum $$\sum\_{K \subset G} \langle G/K, X^n \times G/K \rangle,$$ where the sum is over conjugacy classes of subgroups, and the bracket counts the number of disjoint orbits of type $G/K$. 


Notice that the term $\langle G, X^n \times G \rangle$ equals $|X|^n$ (one free orbit for each $x^n \times e$). Conversely, $\langle G/K, X^n \times G/K \rangle$ must have size $\leq |X|^n |G/K|/|G/K| = |X|^n$. 


Thus the exponent of the order growth is $|X|$, and so $|X|$ is the maximal eigenvalue. The class $[G]$ is an eigenvector corresponding to this eigenvalue.


**Edit:** As Darij points out, in this case Perron--Frobenius only guarantees that $$ max\_{\lambda \in {\rm eig}(A)} |\lambda|$$ can be achieved by some positive real $\lambda$. There may be other complex eigenvalues of the same absolute value. In this case more argument is required to ensure that there is not cancellation.


**Edit2:** Here is a strategy to resolve the issue. Let $z\_1, \dots, z\_k$ be the eigenvalues achieving the maximum. Assume we can show that there exists an eplison such that $$S = \{n \in \mathbb N ~|~ Re(z\_i^n) \geq -\lambda + \epsilon ~ \forall i \}$$ is infinite. Then there would be an infinite subset $S \subset \mathbb N$ and a $C \in \mathbb R\_{> 1}$ such that $C \lambda^s \geq Tr(A^s) \geq 1/C \lambda^s$ for all $s \in S$. 


In our case, we have $$|X|^n \leq {\rm Tr}(A^n) \leq \#\{\text{ subgroups }\}|X|^n$$ for all $n$. Restricting to the infinite subset $S$ we would see that $\lambda = |X|$.


