4
$\begingroup$

Does there exist a function $\tau(\varepsilon)=\tau(\varepsilon,n,K,\mu)$ such that $\lim_{\varepsilon\to +0}\tau(\varepsilon)=0$ and for any $n$-dimensional complete Riemannian manifold $M^n$ with sectional curvature $\geq K$ and with the injectivity radius $\geq \mu>0$ the following property is satisfied: if in a geodesic triangle of diameter less than $\mu/100$ one of the angles is at least $\pi-\varepsilon$ then the other two angles are less than $\tau(\varepsilon)$?

$\endgroup$
3
$\begingroup$

Yes it is true and the statement follows from Toponogov's comparison.

Assume that triangle $[xyz]$ is small and $\measuredangle [y^x_z]>\pi-\varepsilon$. Extend the side $[zy]$ behind $y$ and marke the point $v$ on the extension such that $|y-v|=\tfrac\mu2$. Note that $\measuredangle [y^x_v]<\varepsilon$. By Toponogov's comparison we get an upper bound on $|x-v|$.

Further extend $[zy]$ behind $z$ and mark a point $w$ on the extension such that $|y-w|=\tfrac\mu2$. If $\measuredangle [z^x_y]>\alpha$ then $\measuredangle [z^x_w]<\pi-\alpha$. From Toponogov's comparison we get an upper bound on $|x-w|$.

Note that $|v-w|=\mu$ and by triangle inequality $$|x-v|+|x-w|\ge\mu.$$ Assuming that $\alpha$ is big, the later contradicts the estimates above.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Following the nice idea you described, I tried to reconstruct the details. I got that the statement is true for triangles of diameter $\leq \varepsilon\mu/100$ rather than $\leq \mu/100$. Did I miss something or this is what you ment? $\endgroup$ – makt Sep 21 '15 at 8:55
  • 1
    $\begingroup$ @sva $\mu/100$ should work, do it more carefully better assume $K=0$ for beginning. You should get that $\alpha<\mathrm{const}\cdot\varepsilon$. The estimates are very close the proof of 7.16 in" A.D. Alexandrov spaces with curvature bounded below" by Burago, Gromov and Perelman. $\endgroup$ – Anton Petrunin Sep 21 '15 at 11:04
  • $\begingroup$ It seems you are right, although I failed to see the relation to the above mentioned reference. Thank you. $\endgroup$ – makt Sep 23 '15 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.