4
$\begingroup$

I am studying the paper "Laminations, trees, and irreducible automorphisms of free groups" of Bestvina, Feighn and Handel. But I found a note in the paper "Stabilisers of $\mathbb{R}$-trees with isometric $F_n$ actions" of Kapovich and Lustig that the proof of the proposition 2.6 of the first paper has a gap. The statement is:

Let $\phi$ be an irreducible outer automorphism of $F_n$ with irreducible powers and $\Lambda = \Lambda_{\phi}$ be the attracting lamination corresponding to $\phi$. Let $\psi \in Stab(\Lambda)$ and $h: H \rightarrow H$ be a relative train track representative of $\psi$. Then $h$ has finite order in every proper $h$-invariant subgraph (without valence one vertices that is a union of strata).

I would like to know if this proposition is finally true. Actually, I am interested to know if there is in the literature anything like this. For example, if any automorphism which belongs to the stabiliser of the lamination has a representative (not necessarily relative train track map) which has finite order in the complement of the top stratum. Thanks a lot in advance.

$\endgroup$
8
$\begingroup$

The gap is easily fixable in the context of the paper. Let me explain the fix after first explaining the critique of Kapovich and Lustig.

The first paragraph of the proof starts by choosing a leaf $\ell$ of $\Lambda$, and choosing a proper $h$-invariant subgraph $H_0$, and then noting that $\ell$ is a concatenation of nondegenerate segments in $H_0$ and in $H \setminus H_0$. Then comes the following quote from the first paragraph, in which Kapovich and Lustig have identified a gap:

Notice that all $H_0$ segments are Nielsen (periodic) or else $h$-iteration will produce arbitrarily long leaf segments contained in $H_0$ contradicting quasiperiodicity.

Comments: In this sentence, and in what I shall write, the property "periodic" could be more fully expressed as "periodic up to path homotopy". Also, recall here that "quasiperiodicity" means that every leaf segment $\alpha$ of length $L$ that occurs in $\ell$ also occurs as a subsegment of every leaf segment of length $L'$ where $L'$ depends only on $L$ and not on $\alpha$.

The gap is that this contradiction does not hold because there is one other possibility: $H_0$ segments can also be preperiodic [up to path homotopy]. Taking that possibility into account, the contradiction now holds as stated, by applying $h$-iteration to any nonpreperiodic $H_0$ segment. The rest of the proof works fine until one reaches the following quote:

A power of any loop in $H_0$ lifts to $X'$ [[which is a certain finite graph equipped with an immersion to $H_0$]]. It follows that this power is a concatenation of paths in $H_0$ each of which has finite $h$-order.

While the latter sentence of this quote no longer immediately follows, what does immediately follow is that this power is homotopic to a concatenation of paths in $H_0$ each of which is $h$-preperiodic. Thus every loop in $H_0$ is $h$-preperiodic [up to free homotopy]. But an $h$-preperiodic loop is $h$-periodic. So the concluding sentence of the paragraph holds as stated:

Thus every loop in $H_0$ has finite $h$-order and hence $h \mid H_0$ has finite order.


To address your question in the final paragraph, Theorem 2.14 of the paper says that $\text{Stab}(\Lambda^+_\phi)$ is virtually cyclic. It follows that if $\psi \in \text{Stab}(\Lambda^+_\phi)$ then either $\psi$ has finite order and so has a finite order representative, or $\psi,\phi$ have a common nonzero powers. In the latter case, the conclusion you ask for does not quite hold because it is also possible to have "preperiodic edges"; nonetheless Proposition 2.6 (1) can be applied to any relative train track representative of $\phi$ and that might give you a desirable conclusion.

$\endgroup$
  • $\begingroup$ Thanks a lot for your detailed explanations, they are very helpful for me. Really thanks for your time. $\endgroup$ – user75691 Aug 21 '14 at 16:27
  • $\begingroup$ It's my pleasure. $\endgroup$ – Lee Mosher Aug 21 '14 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.