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This is probably an easy question, but I don't know any Riemannian geometry and a literature search hasn't helped. Any help (e.g. providing a reference) would be greatly appreciated.

For a triangle $ABC$ in euclidean space, denote by $\alpha$ the angle at $A$ and by $a$ its opposite side. Given $\epsilon >0$, consider all triangles $ABC$ with $\alpha>\epsilon$. Then it is clear (e.g. from the sine law) that the perimeter $p(ABC)$ of any such triangle satisfies $p(ABC)< C_\epsilon \cdot a$, with $C_\epsilon$ a positive constant only depending of $\epsilon$.

The same type of bound holds true, for the same reasons, for triangles in hyperbolic space of constant negative curvature $-\rho$, with a constant $C_\epsilon$ depending on $\rho$. (However, note that it doesn't hold for spaces of positive curvature, as an example with two geodesics on the sphere shows).

My question: Let $X$ be a complete Riemannian manifold with sectional curvatures in $[-\rho,0]$ for some $\rho>0$. Is the analogous statement true? That is, is it true that for every $\epsilon>0$, there exists $C_\epsilon>0$ such that $p(ABC)< C_\epsilon \cdot a$ holds for every geodesic triangle $ABC$ with $\alpha>\epsilon$?

In fact, I am only interested in the case where $X$ is a symmetric space of non-compact type, but it seemed more reasonable to ask the question in a more general setting.

Thank you!

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1 Answer 1

up vote 2 down vote accepted

If the manifold is complete simply connected and curvature $\le 0$ then it is CAT(0) space. In particular, all the triangles are thin; i.e., the flat triangle with the same sides has bigger angles.

So your question can be reduced to the question in plane geometry and the answer is yes.

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This is the OP. Thank you so much, $\epsilon-\delta$! I accepted your answer, but could not upvote it since I don't have enough reputation. But it is very clear and helpful. –  user44031 Dec 11 '13 at 22:03

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