# Curvature $\geq-1$ but not $\geq1$

(Edited again)

In the following, for brevity, I will say that $$X\ \ \mathrm{has}\ \ \kappa_{\mathrm{max}}=k$$ if $X$ is a compact ($n$-dimensional with $n\geq2$, with empty boundary) Aleksandrov space of curvature $\geq k$, but not of curvature $\geq k+\varepsilon$ for all $\varepsilon>0$. Note that the curvature bound is from below.

I apologize to the experts if the terminology is not standard.

There are many facts like

Fact: If $X$ has $\kappa_{\mathrm{max}}=1$, then $X$ has the property $\mathcal P$.

E.g. $\mathcal P=$ "diameter $\leq\pi$"

Question 1: Does someone know, instead, some property $\mathcal P$ such that the following holds?

Fact 1: If $X$ has $\kappa_{\mathrm{max}}=-1$ then $X$ has the propery $\mathcal P$. On the other hand, there exists $X$ that has $\kappa_{\mathrm{max}}=1$ and the property $\neg\mathcal P$.

Example of answer: $\mathcal P=$ "is not isometric to a spherical join $X_1*X_2$ of two spaces $X_1$ and $X_2$ with $\kappa_{\mathrm{max}}=1$". (I hope in some "less trivial" answer).

Actually, I am interested in a very special class of Aleksandrov spaces: hyperbolic, Euclidean and spherical cone-manifolds with all cone angles $<2\pi$.

$X$ is such a space iff it admits a piecewise hyperbolic (resp. Euclidean, spherical) triangulation (i.e. simplices = non-degenerate geodesic simplices $\subset\mathbb H^n$, simplicial maps = isometries) such that $X$ is a normal pseudo-manifold (in particular the space of directions at every point is a compact connected spherical cone $(n-1)$-manifold) and the link of every $(n-2)$-simplex has length $\leq2\pi$.

In that case, $X$ has $\kappa_{\mathrm{max}}=-1$ (resp. $0, 1$), and the curvature is constant outside the singular locus. Let us assume always compactness, empty boundary, and $n\geq2$.

Question 2 Let $X$ be a cone-manifold with all (if any) cone-angles $<2\pi$. Does someone know some property $\mathcal P$ such that the following holds?

Fact 2: If $X$ is hyperbolic, then $X$ has the propery $\mathcal P$. On the other hand, there exists a spherical $X$ with property $\neg\mathcal P$.

I do not nothing about Aleksandrov spaces. Since I imagine there is a (big) difference between, let's say, spherical and hyperbolic cone-manifolds (with all cone-angles $<2\pi$), I am curious about that differences that depend mainly on $\kappa_{\mathrm{max}}$, in some sense.

If you think that such a list of $\mathcal P$ would be too much long (or that this is just a basic question for the experts about Aleksandrov spaces with lower curvature bounds), also some references containing facts like Fact 1 would be appreciated (and I will consider it as an answer).

Thanks.

• By analogy with the Riemannian picture that would be natural to use another global result: namely Cartan-Hadamard theorem. So the property would be: the universal cover is contractable. – Dmitry K Aug 8 '17 at 8:04
• Probably what you say is true for curvature bounds from above. For example, by doubling a geodesic hyperbolic triangle you get a simply connected (homeomorphic to $S^2$) Aleksandrov space with curvature bounded below that is not contractible. – stefaNo Aug 8 '17 at 12:18
• The question is overly broad (many results in comparison geometry can serve as an answer). Voted to close. – Igor Belegradek Aug 8 '17 at 13:10
• If you provide some motivation (i.e. why do you care), the question would be easier to answer. – Igor Belegradek Aug 8 '17 at 20:29
• I don't understand what "Let us assume that $\kappa=\kappa_{max}$ is the maximum possible" means. Also, I don't see how Fact 1 could hold for any $\mathcal{P}$, since having curvature bounded below by $1$ is stronger than having it bounded by $-1$: if the latter implies something, then certainly the former does. I must have misunderstood your question. – Benoît Kloeckner Aug 11 '17 at 13:05