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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.

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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 $\lambda = |X|$.

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  • $\begingroup$ Is it guaranteed that the maximal positive eigenvalue dominates the sum even if the matrix is not irreducible? $\endgroup$ – darij grinberg Apr 1 at 20:09
  • $\begingroup$ What does irreducible mean? I think you can write any matrix $A$ in Jordan form and compute traces of powers. If the largest generalized eigenvalue of $a$ (in absolute value) is real, then you get a dominant contribution from it: $c \lambda^n$ where $c$ is the dimension of the generalized eigenspace. $\endgroup$ – Phil Tosteson Apr 1 at 21:42
  • $\begingroup$ Hmm. What if several eigenvalues with equal absolute values cancel each other? I suspect they won't be able to do so consistently, but I don't see a good source for that. $\endgroup$ – darij grinberg Apr 1 at 21:53
  • $\begingroup$ Ah, thanks. I misunderstood what the OP said Perron--Frobenius implied. It does seem reasonable that you can fix this, but I don't see how at the moment. $\endgroup$ – Phil Tosteson Apr 1 at 22:27
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    $\begingroup$ Sorry if I’ve misunderstood, but let z_i be all the eigenvalues, with z_1\in \R the real one of maximal absolute value. Write \zeta_i := z_i/|z_i|, so that \zeta_i\in S^1 for all i and \zeta_1 = 1. Now choose n_k\to \infty for which (\zeta_1^{n_k}, \zeta_2^{n_k}, ...) are all within \eps of (1, ..., 1) [via the pigeonhole principle]. Then tr(A^{n_k}) = z_1^{n_k} (1 + \sum_{i > 1} \zeta_i^{n_k} (|z_i|/z_1)^{n_k}) > z_1^{n_k}/2 once \eps is sufficiently small and n_k is sufficiently large. Now Phil’s argument works to show that z_1 = #|X|. Let me know if I’ve overlooked something! $\endgroup$ – alpoge Apr 3 at 14:10
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For $H \subset G$, write $e_H$ for the basis element of $B(G)$ corresponding to the $G$-set $G/H$. Recall that for $K \subset G$, $$ e_H\cdot e_K = \sum\limits_{H g K \in H \backslash G / K} e_{H \cap g K g^{-1}}. $$

Say that $K$ is subconjugate to $H$ if $K$ is conjugate to a subgroup of $H$. This gives rise to a partial order on the set of conjugacy classes of subgroups of $G$, and hence on the basis elements $e_H$. With respect to this partial ordering, $M_{e_H}$ is lower-triangular for each $H \subset G$. The eigenvalues are then the diagonal elements. The diagonal element corresponding to $e_{\{\textrm{id}_G\}}$ is $|G/H|$. The diagonal element corresponding to $e_K$ for $K \subset G$ is bounded above by the number of $(H, K)$-double cosets in $G$, which is bounded above by $|G/H|$.

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    $\begingroup$ The diagonal element corresponding to $K$ is exactly the number of cosets in $G/H$ that are fixed by $K$. I think this answer is better than the other one since it makes explicit the common Perron-Frobenius eigenvector for all the $e_H$'s, namely $e_{\{1\}}$. For nontrivial subgroups $K$, the corresponding eigenvector is not $e_K$, but can be computed by Möbius inversion, this has been done by David Gluck (1981, Illinois J. Math. 25, no.1, pp.63-67). $\endgroup$ – Frieder Ladisch Apr 3 at 13:53
  • $\begingroup$ I agree this is a much more satisfying answer. I did identify/use that eigenvector though. $\endgroup$ – Phil Tosteson Apr 3 at 15:17
  • $\begingroup$ @PhilTosteson: fair enough! Sorry I overlooked this. $\endgroup$ – Frieder Ladisch Apr 3 at 20:15

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