5

Is it true that angles in Riemannian triangles with bounded sides are uniformly bounded?

More precisely, let $M$ be a Riemannian manifold.

Given $0<r <s$, is there a number $\delta$ (depending on $M,r,s$), such that for every triangle in $M$ (composed of geodesic edges), with edge lengths in $[r,s]$, its angles are greater than $\delta$?

Do we need compactness of $M$ for that to hold? Or are curvature bounds suffice?

Edit: The statement is false if we allow arbitrary pairs of $(r,s),0<r<s$. In fact, if we allow $2r \le s$, then one can always take a degenerate triangle whose edge lengths are $r,r,2r$ and then two angles will be zero!

Thus, there are to ways to proceed towards a correct version:

Proposition (1):

Let $M$ be compact, $0<r<s, 2r > s$. Then, there exist a number $\delta=\delta(M,r,s)$, such that for every triangle in $M$ with edge lengths in $[r,s]$, all its angles are greater than $\delta$.

Proposition (2):

Let $M$ be compact, $0<r<s$. Then, there exist a number $\delta=\delta(M,r,s)$, such that for every triangle in $M$ with edge lengths in $[r,s]$, at least one angle is greater than $\delta$.

Below I give proofs for the two propositions stated above.

Open questions:

  1. Do the propositions hold if we only assume lower bound the curvature, without compactnes?

  2. Is there a way to get an effective bound $\delta$ as a function of $r,s$ (depending also in $M$ somehow)? What if we assume some bounds on the curvature? (The above discussion on degenerate triangles implies that in Proposition (1), we should expect $\delta(r,s)$ to tend to zero, when $s \to 2r$).

Proof of Proposition (1):

Suppose by contradiction there is no bound $\delta$.

Then, there exist triangles $\Delta_n=\{[a_n,b_n],[b_n,c_n],[c_n,a_n]\}$ (where $[a_n,b_n]$ is a minimizing geodsic of unit speed connecting $a_n,b_n$ etc) with edge lengths in $[r,s]$, and each $\Delta_n$ has an angle $\theta_n$ which tends to zero when $n \to \infty$. By passing to a subsequence (or relabeling the vertices) we can assume $\theta_n=\angle b_n a_n c_n$.

Using the Arzela-Ascoli theorem (and perhaps again passing to a subsequence) we can assume that

$[a_n,b_n] \to [a,b], [b_n,c_n] \to [b,c],[c_n,a_n] \to [c,a]$, when the convergence is uniform ($[a,b],[b,c],[c,a]$ are paths connecting the corresponding limit points).

By our assumption,

$$d(a_n,b_n)=L([a_n,b_n]),$$ hence by the continuity of the distance function, and lower semi-continuity of length (w.r.t uniform convergence), we get that

$$ L([a,b]) \le \lim_{n \to \infty} L([a_n,b_n])=\lim_{n \to \infty} d(a_n,b_n)=d(a,b),$$

hence $[a,b]$ is a shortest path connecting $a,b$ and similar statements hold for $[b,c],[c,a]$.

Observation (1): Continuity of the distance imlies $d(a,b),d(b,c),d(c,a)$ all lie in $[r,s]$.

Now, since angles in Riemannian manifolds are lower semi-continuous* ,

$$ \angle b a c \le \liminf \angle b_n a_n c_n =0,$$ so $\angle b a c=0$

Hence, $b,c$ lie on the same geodesic emanating from $a$. W.L.O.G we assume $b$ lies on that geodesic between $a,c$. But then

$$ s \ge d(a,c)=d(a,b)+d(b,c)\ge 2r,$$ which contradicts the assumption.

(we have used observation (1) above).

Proof of Proposition (2):

The proof is very similar to the proof of proposition (1), so we only give a sketch:

Assume the claim is false. Then there exist suitable triangles $\Delta_n$, whose all angles are not greater than $\frac{1}{n}$. Thus, following the proof above, we conclude there is a "limit triangle" with all angles zero, which is a contradiction.

*Remrak on semi-continuity of angles:

By theorem 4.3.11 in "A course in metric geometry" (by Burago,Burago,Ivanov), if the curvature of $M$ is nonnegative then angles in $M$ are lower semi-continuous.

The assumption of nonnegative curvature can probably be omitted. The lower semi-continuity of angles which seems to hold in any intrinsic space of bounded curvature (see details in HK Lee's answer);

Locally, every Riemannian manifold has bounded curvature, and since angles are "local" objects (we can restrict our attention to the a neighbourhood of the corresponding vertex) the conclusion follows.

Community
  • 1
Asaf Shachar
  • 25,967
  • 2
    You need a lower curvature bound, otherwise this is false. I am not sure that a lower curvature bound suffices, but compactness of $M$ is surely enough. – Moishe Kohan Feb 21 '17 at 18:17
  • 1
    @MoisheCohen Thanks. I also thought a curvature bound will be needed here (compactnees ensures this of course). However, even in the compact case, I found I needed to assume in addition that the curvature is nonnegative (please see my added proof in the question). Perhaps there is a more general version of lower semi-continuity of angles, in the presence of any lower curvature bound (not just non-negative?). Also, it would be interesting to know whether curvature bound suffices, even if $M$ is not compact (My proof use compactness for Arzela-Ascoli). – Asaf Shachar Feb 22 '17 at 10:12
  • 1
    Of course, if you have an easier argument than the one I gave, I would be happy to know. – Asaf Shachar Feb 22 '17 at 10:21
  • 1
    BTW, I had a mistake in my proof. It seems it only work for certain pairs $r <s$ which bound the edge lengths (You can see my edited proof in the question). It is interesting to know if the assertion hold for all pairs? – Asaf Shachar Feb 22 '17 at 15:33
  • 1
    The correct claim (in the case of a compact manifold) is that if all three side-lengths are in the interval $[r,s]$ then at least one angle is $\ge \delta>0$, where $\delta$ depends on $M$, $r, s$. Otherwise the claim is false even if $M=S^1$ (two angles are zero and one $=\pi$). The proof of the above claim is based on a compactness argument and the fact that a geodesic triangle cannot have all three angles equal to $0$. – Moishe Kohan Feb 22 '17 at 18:55
  • Thanks! You are right of course. The conclusion is that we can achieve two different statements (on all the angles, or at least one angle, depending if we limit our bounds or not). I have incorporated this in the question, and mentioned some questions left open) . – Asaf Shachar Feb 23 '17 at 08:00

1 Answers1

1

I will enumerate things related with semi-continuity of angles, which can be found in the paper "A.D. Alexandrove spaces with curvature bounded below - Burago, Gromov, and Perelman"

Assume that $X$ is locally compact metric space

Def : $X$ has a curvature $\geq k$ if for any quadruple $(a;b,c,d)$ in $X$ has the property - $(d)$ :

Property - $(d)$ : In $k$-plane, corresponded angles satisfies $\angle BAC +\angle BAD +\angle CAD\leq 2\pi$

Property - $(c)$ : An interior angle in geodesic triangle in $X$ is larger than that of corresponded geodesic triangle in $k$-plane

Property - $(c_1)$ : Sum of adjacent angles in $X$ is equal to $\pi$

EXE : $(d)\Leftrightarrow (c)$ and $(c_1)$

EXE : $(c)$ implies semi-continuity of angles

Open question : $(c)\Rightarrow (d)$ ? If $X=S^2$, then it hold ? (cf. footnote in 108p. in your reference)

HK Lee
  • 20,532
  • Thanks. So, essentially what you are saying is that indeed, semi-continuity of angles holds under any curvature lower bound given, thus my proof can be adapted to the general case of compact manifold (in that case some lower bound on the curvature is always available since the sectional curvature is a continuous function on the Grassmannian of $TM$ which is compact). Did I understand you correctly? BTW, this still leaves open the question of what happens when $M$ is not compact, but a lower bound on the curvature is given. – Asaf Shachar Feb 22 '17 at 14:42
  • (1) Yes. Locally in any Riemannian manifold, semi-continuity of angles holds under curvature bounded below (2) Locally if Riemannian manifold $M$ satisfies property $(c)$, then semi-continuity of angles holds. – HK Lee Feb 22 '17 at 14:44
  • Of course, you are right! Angles are "local" objects, since we can restrict our attention to the a neighbourhood of the corresponding vertex. But doesn't it imply lower semi-continuity of angles in Riemannian manifold hold in general? (I do not understand now why do we need assumption $(c)$ in the Riemannian case). – Asaf Shachar Feb 22 '17 at 14:53
  • Yes. You are right. In any Riemannian manifold, semi-continuity of angles holds. – HK Lee Feb 22 '17 at 14:59
  • 1
    Thanks. BTW, I had a mistake in my proof. It seems it only work for certain pairs $r <s$ which bound the edge lengths (You can see my edited proof in the question). – Asaf Shachar Feb 22 '17 at 15:32