11
$\begingroup$

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.

Improve this question
$\endgroup$
18
  • 1
    $\begingroup$ 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
    Aug 23, 2013 at 14:41
  • 2
    $\begingroup$ 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
    Aug 23, 2013 at 17:06
  • 1
    $\begingroup$ 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
    Aug 24, 2013 at 13:03
  • 1
    $\begingroup$ 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
    Aug 24, 2013 at 13:07
  • 1
    $\begingroup$ These are not quite the same but closely related concepts. No conjugate points plus simple connectivity and completeness, imply uniqueness of geodesics. $\endgroup$
    – Misha
    Aug 24, 2013 at 13:38

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.