I have a question on Riemannian geometry or CAT(k) geometry, which might be simple for experts. Suppose $M$ is a complete smooth Riemannian manifold with sectional curvature bounded from above by $k>0$. Fix a ball $B\subset M$ with radius less than $\frac{\pi}{2\sqrt{k}}$. Fix three points $P,Q,R\in B$. Denote by $PQ$ the unique geodesic that connects $P$ and $Q$ in $B$ and $QR$ the unique geodesic that connects $P$ and $R$.

Let $Q_t\in PQ$ be the $t$-fractional point, that is, $d(p,Q_t)=td(p,Q)$ and $d(Q_t,Q)=(1-t)d(P,Q)$, and let $R_t\in PR$ be the $t$-fractional point as well. I wonder whether it is true that there exists a constant $c(k,t)>0$ such that $$d(Q_t,R_t)\le c(k,t)d(R,Q).$$ In case $k=0$, we know that we can choose $c(k,t)=t$.

Any comments or references would be greatly appreciated.

  • $\begingroup$ is it $d(Q_t,R_t)\le c(k,t)d(Q,R)$? $\endgroup$
    – valeri
    Oct 23, 2017 at 13:33
  • $\begingroup$ To valeri: Thanks. You are absolutely correct. $\endgroup$ Oct 23, 2017 at 17:48
  • $\begingroup$ to Changyu Guo. Then you might need also that the closed curve $PQR$ to bound some disk (be null homotopic), otherwise it might be wrong. Take very thin torus where $PQR$ is the meridian, where $Q, R$ are very close and almost opposite to $P$. The quotient $d(Q_t,R_t)/ d(Q,R)$ for a fixed $t=1/2$ may be unbouded when $d(Q,R)$ goes to zero. $\endgroup$
    – valeri
    Oct 23, 2017 at 21:07
  • 1
    $\begingroup$ Since all triangles are thin, the sphere is the worse case. On the sphere (as well as on the plane), your inequality does not hold --- so try to correct your question. $\endgroup$ Oct 24, 2017 at 0:31
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    $\begingroup$ to Changyu Guo. Actually, I can modify previous arguments to get counterexample; simply connected compact surface where inequality does not hold. Take thin cylinder instead of torus and glue to its ends first cone-like surfaces (of nonpositive curvature) smoothly connecting end-circles to much bigger circles on unit spheres. We obtaine surface with $k=1$, but $d(Q_t,R_t)\le c(k,t)d(Q,R)$ does not hold for any bounded $c(k,t)$, $\endgroup$
    – valeri
    Oct 24, 2017 at 7:01

1 Answer 1


EDIT: I misread the sense of your inequality at my first read. It seems to me that the inequality you are asking for cannot be true, not even locally. In the spirit of comparison geometry, upper (resp. lower) bounds on curvature imply lower (resp. upper) bounds on the distance in terms of comparison triangles.

1) A similar result, but with reversed inequality holds and follows immediately from Berger's comparison theorem, and provided that the triangle $PQR$ is sufficiently small (I think that it is enough to be contained in a ball with center $P$ of radius equal to the injectivity radius from $P$).

The result I'm referring to is a basic comparison theorem, number 1.29 in:

Cheeger, Jeff; Ebin, David G., Comparison theorems in Riemannian geometry, North-Holland Mathematical Library. Vol. 9. Amsterdam-Oxford: North-Holland Publishing Company; New York: American Elsevier Publishing Company, Inc. VIII, 174 p. Dfl. 50.00; $ 19.25 (1975). ZBL0309.53035.

More precisely, if $PQR$ is sufficiently small and $\mathrm{Sec} \leq k$, then

$$ d(Q_t,R_t) \geq \tilde{d}_k (\tilde{Q}_t,\tilde{R}_t) $$

On the right hand side of your inequality, $\tilde{d}_k$ is the distance function between $\tilde{Q}_t$ and $\tilde{R}_t$ in the model space with constant curvature equal to $k$, where $\tilde{Q},\tilde{R}, \tilde{P}$ is a comparison triangle on the model space with sides equal to the geodesic triangle with $Q,R,P$ as vertex on the originale space.

In the very special case $k=0$ is as you say, and $\tilde{d}_k(\tilde{Q}_t,\tilde{R}_t) = t \tilde{d}_k(\tilde{Q},\tilde{R}) = t d(Q,R)$.

2) If you assume, instead, a lower curvature bound you have Toponogov's theorem, which is an inequality in the sense you ask. See e.g. Theorem 2.2, statement (B) in the aforementioned reference (in this case, you don't need your triangle PQR to be small).


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