Let $X$ be a finite dimensional Alexandrov space with curvature bounded below. Let $p\in X$ be a fixed point.

Is it true that for any $\varepsilon >0$ there exists $\delta>0$ such that for any $x,y\in B(p,\delta)$ one has $$\angle xpy -\tilde \angle xpy<\varepsilon?$$ Here $B(p,\delta)$ denotes the ball of radius $\delta$ centered at $p$, $\angle xpy$ denotes the angle at p in $X$, and $\tilde \angle xpy$ denotes the corresponding angle in the comparison triangle in the fixed model plane.


1 Answer 1


Yes it is true.

Assume contrary, then there are two sequences of points $(x_n)$ and $(y_n)$ such that $$|p-x_n|=|p-y_n|=\tfrac1n$$ and $$\measuredangle [p^{x_n}_{y_n}]-\tilde\measuredangle (p^{x_n}_{y_n})\ge \epsilon$$ for some fixed $\varepsilon>0$. We can assume that the directions of $[px_n]$ and $[py_n]$ converge correspondingly to $\xi$ and $\upsilon$ in $\Sigma_p$. Set $$\alpha=\measuredangle(\xi,\upsilon)=\lim_{n\to\infty}\measuredangle [p^{x_n}_{y_n}]$$

Choose two geodesics $[pa]$ and $[pb]$ in the directions close to $\xi$ and $\upsilon$ correspondingly. Let $a_n\in[pa]$ and $b\in[pb]$ be the points such that $|p-a_n|=|p-b_n|=\tfrac1n$.

Note that for large $n$ we have $$\lim_{n\to\infty}\tilde\measuredangle (p^{a_n}_{b_n})\approx \alpha.$$ By comparison, the values $n\cdot |a_n-x_n|$ and $n\cdot|y_n-b_n|$ are small. It follows that $$\tilde\measuredangle (p^{x_n}_{y_n})\approx\tilde\measuredangle (p^{a_n}_{b_n}).$$ Therefore $$\lim_{n\to\infty}\tilde\measuredangle (p^{x_n}_{y_n})\approx\alpha,$$
a contradiction.

  • 1
    $\begingroup$ I do not understand why one may assume from the vary beginning that $|p-x_n|=|p-y_n|$. Actually the case when say $\lim_n\frac{|p-x_n|}{|p-y_n|}=0$ was my main difficulty. $\endgroup$
    – asv
    May 13, 2016 at 10:38
  • $\begingroup$ Yes, I oversimplified things, but the same proof works --- one only has to chosse $a_n$ and $b_n$ in a smarter way. $\endgroup$ May 13, 2016 at 15:53
  • $\begingroup$ My problem is to prove the claim that $\tilde\measuredangle(p^{x_n}_{y_n})\approx \tilde\measuredangle(p^{a_n}_{b_n})$ in full generality (I can prove it in the special case $|p-x_n|=|p-y_n|$). The same proof I know does not work in general. $\endgroup$
    – asv
    May 13, 2016 at 16:21
  • $\begingroup$ Unfold three hinges $[p^a_{x_n}]$, $[p^{x_n}_{y_n}]$ and $[p^{y_n}b]$ on the plane and choose $a_n$ and $b_n$ so that the broken line $a_nx_ny_nb_n$ is nearly straight. $\endgroup$ May 13, 2016 at 19:17
  • $\begingroup$ I still have a problem: if the broken line $a_nx_ny_nb_n$ is nearly straight then $|p-b_n|$ does not have to tend to 0 provided $|p-x_n|\ll |p-y_n|$. In that case one cannot conclude that $\lim_n\tilde\measuredangle(p^{a_n}_{b_n})\approx \alpha$. It would be very helpful to have a more detailed argument. $\endgroup$
    – asv
    May 14, 2016 at 8:56

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