# The Tits alternative for $\operatorname{Out}(F_n)$

Not sure if this is the right place to ask this, but the paper I am reading seems to be too specialised for mathstack (if you do not agree, pleas let me know and I will take down this question)

I am reading the paper 'The Tits alternative for $\operatorname{Out}(F_n)$ I: dynamics of exponentially growing automorphisms' by Bestvina, Feighn and Handel. In it, the following lemma occurs:

There is a homomorphism $PF_{\Lambda^+}: \operatorname{Stab}(\Lambda^+) \to \mathbb{Z}^k$ such that $\psi \in \ker PF_{\Lambda^+}$ if and only if $\Lambda^+ \notin \mathcal{L}(\psi)$ and $\Lambda^+ \notin \mathcal{L}(\psi^{-1})$ ($\psi \in \operatorname{Out}(F_n)$).

This lemma can be found on page 546 of the paper. The proof is the following: and proposition 3.3.3. is question I do not understand why each $\mu(\Psi)$ other than $1$ occurs as the Perron-Frobenius eigenvalue of some irreducible matrix and why the set of Perron-Frobenius eigenvalues is discrete. Any chance someone knows this paper/has read this paper or could point me to some extra lecture which could clarify this proof for me?

My reasoning My confusion came from the following reasoning, which dr. Bestvina (through mail) told me was false (I can not figure out why): I think I could show from the definition of $\mu$ that $\mu(\operatorname{Id}) = 1$. Let $\psi \in \operatorname{Stab}(\Lambda^+)$ be an outer automorphism with $\mu(\psi) > 1$, then $\mu(\psi)$ is indeed a Perron-Frobenius eigenvalue (proposition 3.3.3.(4)). However, this would imply that $\mu(\psi^{-1})$ is not, by proposition 3.3.3.(2): $mu(\psi^{-1}) < 1$ and Perron-Frobenius eigenvalues $\lambda$ of non-negative, integer irreducible matrices satisfy $\lambda \geq 1$. I can not see where my reasoning is flawed, nor can I see why then every $\mu$ has to be a Perron-Frobenius eigenvalue

• Why don't you ask Bestvina, Feighn or Handel. You would get an answer in a few minutes. – Mark Sapir May 6 '18 at 23:32
• @MarkSapir I did ask dr. Bestvina. He replied very quickly, but I couldn't understand his answer, so I thought that their might be someon on here who had the same struggle as I did and found the answer him/herself... – Student May 7 '18 at 6:15
• If you did not understand Bestvina's answer, you could ask followup questions. Posting the same question here is a very strange and indirect way to understand Bestvina's answer. – Mark Sapir May 7 '18 at 10:15
• @MarkSapir: I did, but his answers were very short (which I understand, because I can imagine that he has other stuff to do) and they made me very confused, so I posted this question as a plan B (the correspondence on this question was about $\pm 10$ emails)... Perhaphs I should take down the question, as I can see your point – Student May 7 '18 at 10:20
• Could you rename your question to something more descriptive? Your question is not about the Tits alternative for OutFn, it is about a very particular question about part of a (long) paper on the Tits alternative. – Paul Plummer May 7 '18 at 20:00

I think the first sentence in the proof of 3.3.1, with "$...PF_\Lambda^*(\psi^{-1}) = -PF_\Lambda^*(\psi)...$" was meant to tell you to just think about when that is positive because, as you mention, $PF^*_\Lambda(\psi)$ could not have come from a Perron-Frobenius matrix if it was negative, but up to taking inverse it will, unless $\mu(\psi)=1$. This also makes sense since the theorem is about whether or not $\Lambda^+ \in \mathcal{L}(\phi^\pm)$, and proposition 3.3.3 implies that $\mu(\psi^{-1})<1$ then $\Lambda^+ \in \mathcal{L}(\psi).$
• Bound the eigenvalue $\lambda$, of $M=(m_{ij})$, from below by the minimum row sum of the matrix.
• Show that there is a uniform $k$ so that $M^k$ will have row sums larger than the largest entry of the original matrix $M$. I would use the graph theoretic definition of irreducible(which is in the paper) and there is a uniform bound on the dimensions to get the "mixing of the largest term"
• The above shows $\lambda^k \geq$ min row sum of $M^k$ $\geq \max{m_{ij}}$, which bounds the eigenvalues with the terms of a matrix, so only finitely many eigenvalues below any $K$, so the set of the eigenvalues is discrete.
• I have one more question: I do not understand what is meant with "uniform bound", do they mean that there is a $k$ which works for every possible dimension? – Student May 8 '18 at 17:12
• @Student No, there is a bound on the dimensions of the matrices (at most $m \times m$), and once you have that you can find a $k$ which works for all matrices of dimension less than $m \times m$. The way that this is used is to bound the size of a graph (if using the graph theory definitions) so that you can say something about how long it will take for the largest term to "mix" throughout the matrix(this is length of some oriented path in the graph). – Paul Plummer May 8 '18 at 17:24
• There is a subtlety in bounding the dimension. Although the "$3n-3$" passage in the proof of Theorem 1.7 is sufficient to obtain that bound for train track maps, the "comment about the more general setting on page 37" explains the subtleties that occur for a general exponentially growing stratum. These subtleties occur because of general dangers that arise when one tries to get rid of all valence 2 vertices. It is a fact of life that in a general relative train track map, valence 2 vertices are allowed to exist. – Lee Mosher May 14 '18 at 14:15