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Assume we glue an $n$-dimensional simplicial complex $K$ from copies of an $n$-simplex $\Delta$ with fixed spherical metric. We may think that $\Delta$ has colored vertices and we glue so that the colors match.

If $\Delta$ is right angled then $K$ forms a $\mathrm{CAT}(1)$ space if and only if $K$ is flag; (i.e., $K$ and all its links have no triangles).

Is there a similar condition (if and/or only if) in the case if $\Delta$ is a spherical Coxeter simplex? (say for the quotient $\Delta=\mathbb S^{n}/S_{n+2}$)

Comments.

  • It is easy to make a similar condition if $\Delta$ is 1-dimensional. After that you can get an iff condition for the joints of finite number of 1-simplexes. Note that right-angles simples is joint of finite number of 0-dimensional simplexes, so this is a bit more general and the proof is almost identical.

  • At the moment I do not see a condition for the spherical triangle with angles $\tfrac\pi2$, $\tfrac\pi3$ and $\tfrac\pi3$, which is $\mathbb S^{2}/S_4$. This should be something like this: "no quadrilateral for the vertices with angle $\tfrac\pi3$ and yet something". The "no quadrilateral" condition guarantees* that $K$ is locally $\mathrm{CAT}(1)$ and something has to forbid a finite number of finite subcomplexes. (It should be possible to list such subcomplexes, and likely they are known, but I am interested in $n$-dimensional case.)

(*) In this case the "no triangle" condition is automatic since colors match along the gluing,

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  • $\begingroup$ I guess you know Theorem II.5.4 in Metric spaces of non-positive curvature by Martin Bridson and André Haefliger. So we really just have to get a lower bound on the distance of two points with two different geodesics between them. In the case where the simplices are all right angled, this reduces to the flagness. In general I have no idea. $\endgroup$ Aug 12, 2013 at 22:44
  • $\begingroup$ @HenrikRüping: yes sure. Also you may say that there are no periodic geodesics of length $<2{\cdot}\pi$ in all the links. So from $(n-1)$-dimensional case you get a condition at each vertex and then you need to forbid a finite number of finite subcomplexes. $\endgroup$ Aug 12, 2013 at 23:23

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