# Metric angles in Riemannian manifolds of low regularity

Given three points $a,b,c$ in a (geodesic) metric space $X$, one defines a comparison angle $\angle(a,b,c)$ by the cosine law: $$\angle(a,b,c) = \arccos \frac{|ab|^2 + |ac|^2 - |bc|^2}{2\cdot|ab|\cdot|ac|}$$ where $|ab|$, $|ac|$ and $|bc|$ are distances in $X$. In other words, $\angle(a,b,c)$ is the angle of a planar triangle with the same side lengths as the triangle $abc$ in $X$.

Given two paths $\xi,\eta:[0,\varepsilon)\to X$ starting at the same point $p=\xi(0)=\eta(0)$, define a metric angle $\angle(\xi,\eta)$ by $$\angle(\xi,\eta) = \lim_{t,t'\to 0} \angle(p,\xi(t),\eta(t'))$$ if the limit exists.

If $X$ is a smooth manifold with a $C^2$ Riemannian metric (which defines a geodesic distance in the usual way) and $\xi,\eta$ are $C^1$ regular curves, then a curvature comparison argument shows that the metric angle is well-defined and equals the Riemannian angle between the velocity vectors $\dot\xi(0)$ and $\dot\eta(0)$. But what if the Riemannian metric is only $C^1$, or even $C^0$? More precisely:

• Let $X$ be a smooth manifold with a $C^1$ Riemannian metric. Does the metric angle exist for every pair of $C^1$ regular curves $\xi,\eta$?

• If not, does every $C^1$ regular curve have a well-defined angle with itself? (Of course, this angle is zero if it exists).

• If the answer is affirmative, what about $C^0$ Riemannian metrics?

Remark. The limit in the definition trivially exists (even in the $C^0$ case) if the ratio $t/t'$ is bounded away from zero and infinity as $t$ and $t'$ go to zero. The trouble is with very "thin" triangles $p,\xi(t),\eta(t')$ where $t\ll t'$ or vise versa.

• Just for my education: If I join two truncated cones along a common circle, is the metric inherited from $\mathbb{R}^3$ on this topological cylinder $C^0$? – Joseph O'Rourke Oct 11 '10 at 18:40
• Yes, but you need to define a $C^1$ atlas in order for this to make sense. – Sergei Ivanov Oct 11 '10 at 19:04
• I probably don't understand this question properly. You appear to be assuming that the two velocity vectors are nonzero at the intersection of the two curves. In that case, can't you just set $t' = t$ and take the limit $t \rightarrow 0$? Equivalently, the arclength parameter for each curve is strictly increasing near the intersection, so you can reparameterize both curves in terms of arclength. So you can use the arclength parameterization in both curves, and let both approach zero at the same rate. And I don't see why you need anything more than a $C^0$ Riemannian metric to do all of this. – Deane Yang Oct 12 '10 at 1:18
• Yes, "regular" means nonzero velocity. The problem is whether you get the same limit for e.g. $t'=t^2$. – Sergei Ivanov Oct 12 '10 at 8:54

To Deane Yang, concerning the smoothness of geodesics. These issues have been studied by Hartman and his coauthors more than 40 years ago. Together with Calabi he claimed to prove that in a $C^{\alpha}$ continuous metric all geodesics are uniformly $C^{1,\alpha}$. (I do not have mathscinet access now, but I think the paper was called "On the smoothness of isometries", or something like that). The paper is not quite correct, the computations prove that geodesics are only $C^{1,\alpha /2}$ (this was observed at some point by Reshetnyak). The optimal regularity appears in my paper with Asli Yaman "On Hoelder continuous Riemannian and Finsler metrics" (I apologize for citing my own work). The starting point of all these computations is a metric characterization of $C^{1,\alpha}$ curves in metric terms (in terms of the distance of the midpoint of a pair of points on the curve and the image of the corresponding midpoint of the interval of the definition). It is contained in the work of Hartman, but is probably much older.

To Sergei Ivanov: The questions of extendibility of geodesics have also been studied by Hartman. I again do not remember the correct reference, but can check it on Monday. I think he constructs example of even $C^{1,\alpha}$ Riemannian metrics, for any $\alpha <1$, that do not have extendible geodesics. (For $\alpha =1$, the metric has curvature bounded from both sides).

Now the main point. I am very sorry and have to apologize for my previous short posting. I slightly misunderstood your question. But I hope that my answer was still correct. Is the following more detailed explanation correct?

1. We assume that $t'>>t$ and want to prove that $d(\xi (t), \eta (t') )= d(\eta (t'), p) - cos (\alpha ) t + o(t)$. Here $o(t) /t \to 0$ and $\alpha$ is the "usual Euclidean " angle between $\xi$ and $\eta$.

1. The statement about the uniform "smoothness" of geodesics, show that it suffices to prove the claim in the case, when $\eta$ and $\xi$ are geodesics (since the "usual" angles between $\eta$ and any geodesic between $\eta (t')$ and $p$ goes to $0$).

2. Going on the geodesic $\eta$ first to the point $\eta (rt)$, with large, but fixed $r$, and then to $\xi (t)$, one obtains the right upper bound for $d (\eta (t'),p)$.

3. In a "general" metric space it would be now difficult to obtain the right lower bound of $d(\eta (t'),p)$. One could do it, if one knew that the geodesic $\eta$ was extendible beyond $p$. Then one could use a traingle inequality and the distance from $\xi (t)$ and $\eta (-rt)$. I hope I understood your question correctly and this was what you had in mind.

4. Here one can use another trick to obtain a lower bound. Observe namely, that from the "smoothness" of geodesics the "usual" angle between $\xi$ and any geodesic from $\eta (t')$ to $\xi (t)$ is very close to $\alpha$ (goes to $\alpha$ , with $t' \to 0$). Now we just look at the triangle $\xi (t),p, \eta (t')$ from $\xi (t)$ and not from $p$. The inequality obtained in point 3. above gives us the right upper bound of $d(\eta (t'),p)$.

Just a side remark. Your question is very closely related to the question about the correctness of the first formula of variation. I have thought about that question and proved that the formula holds true in Riemannian manifolds with Hoelder continuous metrics in "Differentiation in metric spaces". I just hope that the above proof is correct. Other wise the corresponding statement in my paper is wrong as well.

• Thanks for the detailed response! I had studied the work of Hartman and others years ago, but have forgotten it all by now. And you should definitely not apologize for citing your own work, especially if it answers the questions being asked. – Deane Yang Oct 16 '10 at 22:04
• Any chance you know how to answer: mathoverflow.net/questions/1554/… – Deane Yang Oct 16 '10 at 23:08

For $C^1$ metrics the answer is yes. In fact, the answer is yes for any Hoelder-continuous Riemannian metric. The key observation is that geodesics are uniformly $C^{1,\alpha}$ in this case. From this one can draw the conclusions about the angles by using a few traingel inequalities.
For $C^0$ metrics there is a closely related question, to which I do not know the answer: Is the angle between any pair of geodesics starting in a point well defined. I do believe that the answer is nbo as well. But one cannot construct a counterexample just by choosing a wrong parametrization of the manifold.

• I should know this stuff, but I don't. How do I see that a geodesic is uniformly $C^{1,\alpha}$? Is there a nice reference for this kind of thing? – Deane Yang Oct 15 '10 at 20:10
• I can see a triangle inequality argument if the geodesics are extensible (and remain locally minimizing after extension). Do we have this in a $C^1$ metric? Or do you mean a different kind of an argument? – Sergei Ivanov Oct 15 '10 at 22:05

For $C^0$, the answer is "NO".

Start with Euclidean plane and choose a function $\alpha(r)$ such that $\alpha(r)\to\infty$ as $r\to 0$, but it does it really slow (so $r\cdot\alpha'(r)\to 0$ sufficiently fast). Consider the induced metric by the map in polar coordinates $(\theta, r)\mapsto (\theta+\alpha(r),r)$; it is continuous. The geodesics from the origin in the new metric are formed by spirals which go around origin infinitely many times.

In particular, if $\xi$ is a smooth curve starting at the origin then $\angle (\xi,\xi)$ is undefined.

• Is the situation potentially better for geodesic rays? I find $C^0$ too weak of a condition on a metric to do geometry. I'm much more comfortable with $C^1$ plus some condition on curvature. This gives one at least some control over how geodesics behave. Anything less, and you lose the ability to study metric properties of the space using families of geodesics. – Deane Yang Oct 15 '10 at 13:27
• I think it is $\alpha'(r)\cdot r$ (the angle between the spiral and radial directions) that should go to 0, not $\alpha'(r)/r$ (which cannot). – Sergei Ivanov Oct 15 '10 at 21:08
• @Sergei, yes, sure. – Anton Petrunin Oct 16 '10 at 16:17
• I think this example is due to Calabi and Hartman, On the smoothness of isometries. Duke Math. J. 37 1970 741–750, see pp. 748-749. – Piotr Hajlasz Jun 7 '18 at 17:24