Dear all, I am teaching a course in Riemannian geometry, and I would like to prove some comparison theorems in the next lessons, building on the well-known theory of Jacobi fields, and of Rauch comparison Theorem for Jacobi fields. I would like to stress the fact that several arguments can work in the case of geodesic metric spaces, and I'd like to prove as soon as possible that the metric and differential notion of angle coincide on Riemannian manifolds.

Let me briefly recall the well-known notion of Alexandrov angle. If $X$ is a geodesic metric space and $\gamma_1,\gamma_2$ are geodesics (parameterized by arc-length) exiting from a point $p$, then the Alexandrov angle between $\gamma_1$ and $\gamma_2$ is the quantity $$\angle_p (\gamma_1,\gamma_2)=\limsup_{t,t'\to 0} \overline{\angle}_p (\gamma_1(t),\gamma_2(t'))\ ,$$ where $\overline{\angle}_p (\gamma_1(t),\gamma_2(t))$ is the Euclidean angle (in the point corresponding to $p$) of the Euclidean comparison triangle for the triple $(p,\gamma_1(t),\gamma_2(t'))$.

It is well-known that, in the case when $X$ is a Riemannian manifold, then the above $\limsup$ is in fact a genuine limit, and the Alexandrov angle between two geodesics exiting from the same point coincides with the Riemannian angle between them.

Here is my question: Is there a direct proof of this fact? Here, by direct proof I mean a proof that does not use too much of the theory of CAT(k)-spaces.

In the book by Bridson and Haefliger, the above statement is proved in Corollary II.1A.7, which in turn relies on Proposition II.1.7, where several tools form the preceding sections are used. On the contrary, if we want to prove the easier fact that, in a Riemannian manifold, the limit $$\lim_{t\to 0} \overline{\angle}_p (\gamma_1(t),\gamma_2(t))$$ exists and is equal to the Riemannian angle, then an easy computation using normal coordinates seems to suffice. I was wondering if some local estimates (for example, the expansion of the metric in normal coordinates with the subsequent estimates on the deviation of the exponential from being an isometry) could work for computing the $\limsup$ in the definition of Alexandrov angle (the difficult case being when $t,t'$ convergence to $0$ at a very different speed).


1 Answer 1


Your equality is two inequalities.

To show the upper bound you can use the triangle inequality --- come closer to $p$ along the geodesic and apply the local estimates. (This is the "first variation inequality" it holds in any metric space where angles defined.)

The lower bound follows since the injectivity radius at $p$ is positive. Indeed, if the angle is smaller, the geodesic $[\gamma_1(t)\gamma_2(\tau)]$ converges as $\tau\to0$ to an other geodesic distinct from $\gamma_1$.

This question is the baby case of so called "lemma about strong angle" in Alexandrov geometry it also can be called "first variation formula".

(Hope it helps, sorry if I misinterpret your question.)

  • $\begingroup$ Dear Anton, thank you very much for your answer. A lemma about the strong upper angle appears in the book by Bridson and Haefliger (Proposition I.1.16). Using it I easily proved that the Alexandrov angle is not smaller than the Riemannian one: more precisely, for every fixed t the comparison angle for the triangle with vertices p, γ1(t) and γ2(s) tends to the Riemannian one when s tends to 0 (this obviously provides a lower bound on the Alexandrov angle). However, I still have problems in proving that the Riemannian angle is not smaller than the Alexandrov one... $\endgroup$ Apr 28, 2013 at 21:05
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    $\begingroup$ Roberto, I can not add much to my answer. I did not read Bridson and Haefliger, so I can not comment on this part. If you are not happy with the proof there you may look at 3.3.2, 8.10.2-3 and maybe also 7.5.6-7 in our book math.psu.edu/petrunin/papers/alexandrov-geometry Formally everything is done for Alexandrov spaces, but the same proof (even simpler) works in Riemannian case... $\endgroup$ Apr 29, 2013 at 2:02

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