MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question has an easy answer, if one replaces free by free abelian: Then the resulting group is always solvable and a solvable subgroup of a CAT(0) group is virtually abelian. If the resulting was CAT(0), then the chosen automorphism $\varphi$ in $\mathbb{Z}^n\rtimes_\varphi \mathbb{Z}$ would have finite order - otherwise the group would not be virtually abelian.

Now one can ask the same question for the free group instead or the free abelian group. I would like to know for which automorphisms $\varphi$ of the free group $F_n$ the group $F_n\rtimes_\varphi \mathbb{Z}$ is CAT(0).

I only know, that $F_n \times \mathbb{Z}$ is CAT(0). I think that if the chosen automorphism has finite order, then the result should be CAT(0) (although I don't have a proof). And I do not know automorphism, that gives a non-CAT(0) group.

share|cite|improve this question
If the automorphism is finite order then it can be realised by a combinatorial automorphism of a graph. (This basically follows from Stallings' Ends Theorem.) The resulting mapping torus is CAT(0). – HJRW Feb 1 '11 at 15:49

An example of a free-by-cyclic group that is not CAT(0) was given by Gersten. It is constructed from the automorphism of $F_3\cong\langle a,b,c\rangle$ that sends

$a\mapsto a,~b\mapsto ba,~ c\mapsto ca^2~.$

The idea of the proof is to think about translation lengths and flats in any CAT(0) space on which it acts. As $\langle a,t\rangle\cong\mathbb{Z}^2$, it stabilises some flat. But $t$, $at$ and $a^2t$ are all conjugate, so have the same translation lengths. A little thought shows that this is impossible in a flat.

Note that, in many respects, (fg free)-by-cyclic groups are difficult to distinguish from CAT(0) groups. For instance, they have quadratic isoperimetric inequality.

As pointed out in this blog post, the question

Which free-by-cyclic groups are CAT(0)?

is Question 7.9 of Bridson's AIM article about `Problems concerning hyperbolic and CAT(0) groups'.

share|cite|improve this answer
Here, $t$ is the stable letter, of course. – HJRW Feb 1 '11 at 15:57

You'll find some examples of CAT(0) free-by-cyclics in

Samuelson, "On CAT(0) structures for free-by-cyclic groups"


Barnard and Brady, "Distorsion of surface groups in CAT(0) free-by-cyclic groups"

share|cite|improve this answer
For future reference, there is a free version of the first paper mentioned <a href="">; here</a> It is essentially identical to the published version. – Peter Samuelson Oct 19 '11 at 14:21
Sorry, I messed up the link... here's one that works: – Peter Samuelson Oct 19 '11 at 14:21
Thanks, Peter, I'll edit to reflect this. – Richard Kent Oct 19 '11 at 14:34

If you take the mapping torus of an automorphism of a surface with boundary, then it has a non-positively curved metric by a result of Leeb. Such automorphisms though will be sparse in the set of all automorphisms of a free group (except in rank 2, as noted in a paper of Brady).

share|cite|improve this answer

In this paper Mark F. Hagen and Daniel T. Wise show that a hyperbolic, free-by-cyclic group whose monodromy is irreducible acts geometrically on a CAT(0) cube complex. Hyperbolicity means for such groups that the automorphism does not fix the conjugacy class of a nontrivial word by Brinkmanns Theorem.

share|cite|improve this answer
It is nice to hear a general statement like this! – Peter Samuelson Dec 5 '13 at 17:52
Not quite - they also assume that the monodromy is irreducible, although they say they are working on removing that hypothesis. – HJRW Dec 5 '13 at 21:46
sorry, Thank you! – HenrikRüping Dec 6 '13 at 0:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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