# Is the Perron-Frobenius dimension of a G-Set given by its cardinality?

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.

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 $$\lambda = |X|$$.

• Is it guaranteed that the maximal positive eigenvalue dominates the sum even if the matrix is not irreducible? – darij grinberg Apr 1 at 20:09
• 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. – Phil Tosteson Apr 1 at 21:42
• 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. – darij grinberg Apr 1 at 21:53
• 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. – Phil Tosteson Apr 1 at 22:27
• 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! – alpoge Apr 3 at 14:10

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 $$|G/H|$$.

• 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). – Frieder Ladisch Apr 3 at 13:53
• I agree this is a much more satisfying answer. I did identify/use that eigenvector though. – Phil Tosteson Apr 3 at 15:17
• @PhilTosteson: fair enough! Sorry I overlooked this. – Frieder Ladisch Apr 3 at 20:15