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Let F be a finitely generated free group and let $\gamma : F \rightarrow F$ be an automorphism. Is the semidirect product $F \rtimes \mathbb{Z}$ an hyperbolic group? where $\mathbb{Z}$ acts in F via $\gamma$.

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up vote 10 down vote accepted

The Bestvina-Feighn combination theorem says that this is true if and only if $\gamma$ has no nontrivial periodic conjugacy classes. See

MR1152226 (93d:53053) Bestvina, M.(1-UCLA); Feighn, M.(1-RTG2) A combination theorem for negatively curved groups. J. Differential Geom. 35 (1992), no. 1, 85–101.

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Is this semidirect product a CAT(0) group? – Luis Jorge Nov 16 '11 at 20:27
There are free-by-cyclics which are not CAT(0): S.M. Gersten, The automorphism group of a free group is not a CAT(0) group, Proc. Amer. Math. Soc. 121 (4) (1994) 999–1002 – Richard Kent Nov 16 '11 at 21:02
See also… – Richard Kent Nov 16 '11 at 21:02
I have one more question, If $\gamma$ is induced by a diffeomorphism $f: M\rightarrow M$ where M is a compact 2 manifold with boundary and $\pi_1 (M)$ is isomorphic to F, then my initial demidirect product is hyperbolic? – Luis Jorge Nov 17 '11 at 19:47
The answer to this last question is that it never is (hyperbolic) -- in this case f fixes the boundary components (as a set), which easily gives you a torus, which is $\pi_1$-injective, and so you're not hyperbolic, by the easy half of Andy's answer. – Daniel Groves Dec 7 '11 at 22:48

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