# On triangle comparison in Riemannian manifolds with upper sectional curvature bound

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.

• is it $d(Q_t,R_t)\le c(k,t)d(Q,R)$? Oct 23, 2017 at 13:33
• To valeri: Thanks. You are absolutely correct. Oct 23, 2017 at 17:48
• 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. Oct 23, 2017 at 21:07
• 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. Oct 24, 2017 at 0:31
• 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 $c(k,t)$, Oct 24, 2017 at 7:01

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 $P$).
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)\$.