Growth rate of an outer automorphism of a free product

Question

$\DeclareMathOperator\Out{Out}$Let $G=G_1\ast\cdots\ast G_k\ast F_p$ be a Grushko decomposition of a finitely generated group $G$, $\mathcal{O}$ the outer space relative to this decomposition, $[\phi]\in \Out(G)$ an $\mathcal{O}$-irreducible outer automorphism and $f:X\to X,\ X\in\mathcal{O}$ a train track representative of $[\phi]$. Let also $M$ the transition matrix of $f$ with Perron-Frobenius eigenvalue $\mathrm{PF}(f)=\lambda_f$. If $\lambda([\phi],c)$ is the growth rate of the conjugacy class $c$ under $[\phi]$, i.e. $\lambda([\phi],c)=\limsup\limits_{n\to+\infty}\dfrac{\log|\phi^n(c)|}{n}$, then I would like to know whether $\lambda([\phi],c)\not=0$ implies that $$\lambda([\phi],c)=\log\lambda_f.$$

I guess that the proof in the case of free products should imitate the one in the case of automorphisms of free groups. Do I miss anything? Is there any reference with a proof in the case of automorphisms of free products.

Answer 1

I guess it probably depends on how you measure the growth rate? Certainly it should be true that if $\phi$ is irreducible and $c$ is represented by a loop (in a graph of groups) that is not a Nielsen path, then the length of the tightened representative of that loop under iteration will have exponential growth rate $\lambda$. This ought to follow from the work in my paper, CTs for Free Products, to prove Proposition 4.13. Now, for an irreducible $\phi$ things ought to be a lot simpler, so this feels like a big hammer...

Anyway, the problem is that this measure of growth rate is not necessarily the growth rate of the conjugacy class in a finite generating set for your free product. Like, imagine one of your atomic free factors has an automorphism with exponential growth rate larger than $\lambda$?