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.

  • 2
    $\begingroup$ 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). $\endgroup$ – 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'.

  • $\begingroup$ Here, $t$ is the stable letter, of course. $\endgroup$ – 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"


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).


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.

  • $\begingroup$ It is nice to hear a general statement like this! $\endgroup$ – Peter Samuelson Dec 5 '13 at 17:52
  • $\begingroup$ Not quite - they also assume that the monodromy is irreducible, although they say they are working on removing that hypothesis. $\endgroup$ – HJRW Dec 5 '13 at 21:46

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