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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).1

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.

1M.R. Bridson and A. Haefliger, Metric spaces of non-positive curvature, volume 319 of Grundlehren der mathematischen Wissenschaften. Springer-Verlag, Berlin, 1999; DOI: 10.1007/978-3-662-12494-9, see also here.

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    $\begingroup$ 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. $\endgroup$
    – Misha
    Commented Aug 23, 2013 at 14:41
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    $\begingroup$ 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. $\endgroup$
    – Misha
    Commented Aug 23, 2013 at 17:06
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    $\begingroup$ The original example is due to Stallings. For this and further results see people.maths.ox.ac.uk/bridson/papers/BMiller07/BMiller08.pdf and references therein (especially the paper by Baumslag and Roseblade). $\endgroup$
    – Misha
    Commented Aug 24, 2013 at 13:03
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    $\begingroup$ 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 arxiv.org/pdf/1205.4455.pdf Note however that they do need infinite differentiability of the metric. $\endgroup$
    – Misha
    Commented Aug 24, 2013 at 13:07
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    $\begingroup$ These are not quite the same but closely related concepts. No conjugate points plus simple connectivity and completeness, imply uniqueness of geodesics. $\endgroup$
    – Misha
    Commented Aug 24, 2013 at 13:38

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