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Definition : A group is CAT(0) if it acts properly, cocompactly and isometrically on a CAT(0) space.

Examples : see this blog.

Remark : A CAT(0) space is uniquely geodesic, but the converse is false (see here).

Remark : Every finitely-presented group is geodesic, i.e. acts properly, cocompactly and isometrically on a geodesic space (for example its Cayley graph related to a finite generating set).

Definition : A group is uniquely geodesic if it acts properly, cocompactly and isometrically on a uniquely geodesic space.

Question : Are torsion-free finitely presented groups, uniquely geodesic ? (see the next remark)

Remark : we add the assumption "with a proper action on a contractible CW complex so that the action is cocompact on each skeleton" (see Misha's comment). In addition, if groups with $\infty$-dimensional classifying space (as Thompson groups), give obvious counter-examples, we add the assumption "with a finite-dimensional classifying space". Perhaps the "cocompact" assumption is not relevant in the $\infty$-dimensional case, and should be replaced by something else.

Remark : The Baumslag-Solitar groups $BS(m,n)$ are torsion-free, finitely generated (with $2$-dim. classifying space) but not CAT(0) for $m \ne n$. Perhaps they are counter-examples, I don't know...

This is an optional part about continuously uniquely geodesic groups :

Definition : A continuously uniquely geodesic space is a uniquely geodesic space whose geodesics vary continuously with endpoints.

Remark : A continuously uniquely geodesic space is contractible.

Remark : A proper uniquely geodesic space is continuously uniquely geodesic, but a complete uniquely geodesic space is not necessarily continuously uniquely geodesic: see here page 38 (3.13 and 3.14)

Definition : A group is continuously uniquely geodesic if it acts properly, cocompactly and isometrically on a continuously uniquely geodesic space.

Question : Is a uniquely geodesic (finitely presented) group also continuously uniquely geodesic ?

Remark: In my opinion, the continuously uniquely geodesic (finitely presented) groups are very convenient with operator algebras conjectures as the Baum-Connes conjecture.

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Sebastien Palcoux
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    You have to assume much more than finite generation: For instance, such groups are automatically finitely presented. Moreover, you should assume that the group admits a a proper action on a contractible CW complex so that the action is cocompact on each skeleton. On the other hand, finite-dimensionality of such groups is unclear to me, since a priori, you can have an action on a locally compact space which is not finite dimensional. – Misha Aug 23 '13 at 14:41
  • @Misha : thank you for your comment. So you say that "finitely presented" and "with a proper action on a contractible CW complex so that the action is cocompact on each skeleton" are necessary conditions for being "uniquely geodesic", and you suggest me to edit it, right ? About $\infty$-dimension case, is the Thompson group obviously "uniquely geodesic" or not ? – Sebastien Palcoux Aug 23 '13 at 16:50
  • @Misha : more, is there "finitely presented groups" without "a proper action on a contractible CW complex so that the action is cocompact on each skeleton" ? Of finite-dimension ? – Sebastien Palcoux Aug 23 '13 at 16:56
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    Yes, direct product of 3 copies of $F_2$ contains a finite-dimensional subgroup which is finitely presented but does not admit a cocompact proper action on a contractible space. – Misha Aug 23 '13 at 17:06
  • I am not sure about Thompson group. – Misha Aug 23 '13 at 17:07
  • @Misha : Your example is very interesting, do you have a reference ? About Thompson group, you're not sure but what do you suspect ? – Sebastien Palcoux Aug 23 '13 at 17:15
  • @Misha: If it's well-view by the users, I will improve this post for having the same structure of the post "Uniquely geodesic spaces", so that you would put your example as an answer. – Sebastien Palcoux Aug 24 '13 at 11:24
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    The original example is due to Stallings. For this and further results see http://people.maths.ox.ac.uk/bridson/papers/BMiller07/BMiller08.pdf and references therein (especially the paper by Baumslag and Roseblade). – Misha Aug 24 '13 at 13:03
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    About Thompson's group: If I were a betting man, I would bet that it does not have such a cocompact action. For the, currently, best results on fundamental groups of Riemannian manifolds without conjugate points, see http://arxiv.org/pdf/1205.4455.pdf Note however that they do need infinite differentiability of the metric. – Misha Aug 24 '13 at 13:07
  • @Misha : thank you so much for your comment. In which category "without conjugate points" is equivalent to "uniquely geodesic" ? Contractible Riemannian manifolds ? Is it true for contractible geodesic space in general ? (these questions seems more adapted for my new post, unfortunately it has many close-votes, I don't understand why...). – Sebastien Palcoux Aug 24 '13 at 13:21
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    These are not quite the same but closely related concepts. No conjugate points plus simple connectivity and completeness, imply uniqueness of geodesics. – Misha Aug 24 '13 at 13:38
  • @Misha : Wonderful !!! do you have a reference ? Perhaps it depends on the definition of "conjugate points", I read here there is no an official definition. – Sebastien Palcoux Aug 24 '13 at 13:45
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    Hmm, as far as I know, there is only one definition, at least in the Riemannian setting (there might be ambiguity in the Finsler case or low-regularity setting). As for the proof, just follow the proof of Cartan-Hadamard theorem for manifolds of nonpositive curvature (see e.g. do Carmo's "Riemannian Geometry"). – Misha Aug 24 '13 at 13:58
  • @Misha : so the result "No conjugate points plus simple connectivity and completeness, imply uniqueness of geodesics", is known only for Riemannian manifolds (already very well...), not for metric spaces in general, right ? Perhaps the class of "uniquely geodesic groups" is strictly larger than the class of "fundamental groups of Riemannian manifolds without conjugate points" as in the paper of Ivanov and Kapovitch you cite. In fact, I have no idea how estimate the difference between this two classes, are they very close ? or, is the first very bigger than the second ? What do you suspect ? – Sebastien Palcoux Aug 24 '13 at 17:15
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    The concept of conjugate points is meaningless once you leave the setting of (smooth) Finsler manifolds. – Misha Aug 24 '13 at 17:17
  • @Misha : Thank you for this information about Finsler manifolds. So perhaps the concept "uniquely geodesic" is a kind of generalization of the concept of "without conjugate points", for simply connected, complete metric (or geodesic ?) spaces in general. The problem is : is there new groups (i.e. is the first class of groups equal or close or bigger or very bigger, than the second class ? What do you suspect ? – Sebastien Palcoux Aug 24 '13 at 17:26
  • @Misha Can you please explain why a uniquely geodesic has to be finitely presented? – Chris Apostol May 16 '17 at 23:26
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    @ChrisApostol: Under the cocompactness assumptions in this question the space will have to be locally compact, hence, proper. Therefore, it is "continuously uniquely geodesic", hence, contractible. Hence, the group $G$ is finitely presented as it acts cocompactly. – Misha May 17 '17 at 13:47

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