A proof of Toponogov’s theorem

in Alexandrov geometry111Supported by NSFC 11971057 and BNSF Z190003.

Shengqi Hu, Xiaole Su, Yusheng Wang222The corresponding author (E-mail:
wyusheng@bnu.edu.cn).

Abstract. This paper aims to give an elementary proof for Toponogov’s theorem
in Alexandrov geometry with lower curvature bound. The idea of the proof comes
from the fact that, in Riemannian geometry, sectional curvature can be
embodied in the second variation formula.

Key words. Toponogov’s theorem, Alexandrov geometry

Mathematics Subject Classification (2000): 53-C20.

##  0 Introduction

An Alexandrov space $X$ with curvature $\geq k$ is roughly defined to be a
locally complete intrinsic metric space on which Toponogov’s type theorem
holds locally ([BGP]). In fact, if $X$ is complete, then such a type theorem
holds globally on $X$, which is called Toponogov’s theorem in Alexandrov
geometry. In its proofs ([BGP], [AKP]), to simplify the arguments, it is
always assumed that there exists a minimal geodesic (shortest path) between
any two points in $X$ 333If $X$ is both complete and locally compact, there is
a shortest path between any two points in it.. For any two minimal geodesics
$[pq]$ and $[qr]$ in $X$, we can define the angle $\angle pqr$ naturally; and
then, Toponogov’s theorem in Alexandrov geometry can be formulated as follows
([BGP]).

Theorem A Let $X$ be a complete Alexandrov space with curvature $\geq k$. Then
for any triangle $\triangle pqr\subset X$ (a union of three minimal geodesics
$[pq],[pr],[qr]$), we have that

$\angle pqr\geq\tilde{\angle}_{k}pqr.$ $None$

In Theorem A, $\tilde{\angle}_{k}pqr$ denotes the angle
$\angle\tilde{p}\tilde{q}\tilde{r}$ in the comparison triangle
$\triangle\tilde{p}\tilde{q}\tilde{r}\subset\mathbb{S}^{2}_{k}$ of $\triangle
pqr$ (i.e. $|\tilde{p}\tilde{q}|=|pq|,|\tilde{p}\tilde{r}|=|pr|$,
$|\tilde{q}\tilde{r}|=|qr|$), where $\mathbb{S}^{2}_{k}$ denotes the complete
and simply connected $2$-dimensional space form with constant curvature $k$.
And in a proof of Theorem A, one need only to consider the case where
$|pq|+|pr|+|qr|<\frac{2\pi}{\sqrt{k}}$ (and thus each of $|pq|,|pr|,|qr|$ is
less than $\frac{\pi}{\sqrt{k}}$) if $k>0$ ([BGP]) 444Once Theorem A has been
proven, it can be shown that $|pq|\leq\frac{\pi}{\sqrt{k}}$ and
$|pq|+|pr|+|qr|\leq\frac{2\pi}{\sqrt{k}}$ for all $p,q,r\in X$ ([BGP]).
Moreover, if $|pq|=\frac{\pi}{\sqrt{k}}$, it is a convention that
$\tilde{\angle}_{k}pqr=0$..

So far, there are several proofs for Theorem A ([BGP], [Pl], [Sh], [Wa]), each
of which is skillful and the one in [Pl] is quite beautiful. In Riemannian
geometry, sectional curvature can be seen just from geodesic variations (the
second variation formula). However, all known proofs of Theorem A have no
direct relation with such an idea. The main goal of this paper is to provide a
proof of such an idea for Theorem A.

Our strategy is: If $\angle pqr<\tilde{\angle}_{k}pqr$ for some $\triangle
pqr\subset X$, we can split $\triangle pqr$ into two triangles along some
$[ps]$ with $s\in[qr]^{\circ}$ (the interior part of $[qr]$) so that $\angle
psq<\tilde{\angle}_{k}psq$ or $\angle psr<\tilde{\angle}_{k}psr$. By repeating
this process, we can locate an $o\in[qr]$ such that there is information of
“curvature $\not\geq k$” around some $[po]$, i.e. there is $\triangle
pr_{1}r_{2}$ with $r_{i}$ sufficiently close to $o$ such that $\angle
pr_{1}r_{2}<\tilde{\angle}_{k}pr_{1}r_{2}$. Note that Theorem A holds on a
small ball $B(o,\delta)$ (i.e. (0.1) is true for triangles in $B(o,\delta)$)
because $X$ is of curvature $\geq k$. Then via Alexandrov’s Lemma (Lemma 1.3
below) on $\triangle pr_{1}r_{2}$, we can locate another $o^{\prime}\in
B(o,\delta)$ with $|po^{\prime}|\leq|po|-\frac{\delta}{3}$ such that there is
still information of “curvature $\not\geq k$” around some $[po^{\prime}]$.
Step by step, such information can be transmitted to a small neighbourhood of
$p$, a contradiction (because Theorem A holds around $p$).

In the rest of the paper, $X$ always denotes the space in Theorem A.

##  1 Tools of the proof

Note that a proof of Theorem A has to depend only on its local version, where
we have the following basic property on angles ([BGP]).

Lemma 1.1 Let $[pq],[rr^{\prime}]\subset X$ with $r\in[pq]^{\circ}$. Then
$\angle prr^{\prime}+\angle qrr^{\prime}=\pi$.

Furthermore, we have the following easy observation (cf. [AKP], [Wa]).

Lemma 1.2 Let $[pq],[qr]\subset X$. Then for $q_{i}\in[qr]$ with $q_{i}\to q$
as $i\to\infty$,

$|pq_{i}|\leq|pq|-\cos\angle pqr\ \cdot|qq_{i}|+o(|qq_{i}|)\ .$ $None$

Proof. Since Theorem A holds around $q$, there is $\bar{p}\in[pq]$ near $q$
such that

$|\bar{p}q_{i}|\leq|\bar{p}q|-\cos\angle pqr\ \cdot|qq_{i}|+o(|qq_{i}|).$

Then (1.1) follows from that $|pq_{i}|\leq|p\bar{p}|+|\bar{p}q_{i}|$. $\Box$

Besides Lemmas 1.1 and 1.2, we will also use Alexandrov’s lemma ([BGP]).

Lemma 1.3 Let $\triangle pqr$, $\triangle pqs$, $\triangle
abc\subset\mathbb{S}^{2}_{k}$ (where $\triangle pqr$ and $\triangle pqs$ are
joined to each other in an exterior way along $[pq]$) such that $|ab|=|pr|$,
$|ac|=|ps|$, $|bc|=|qr|+|qs|$, and $|ab|+|ac|+|bc|<\frac{2\pi}{\sqrt{k}}$ if
$k>0$. Then $\angle pqr+\angle pqs\leq\pi$ (resp. $\geq\pi$) if and only if
$\angle prq\geq\angle abc$ and $\angle psq\geq\angle acb$ (resp. $\angle
prq\leq\angle abc$ and $\angle psq\leq\angle acb$).

##  2 Proof of Theorem A

Due to the similarity of proofs for $k<,=,>0$, we only consider the case where
$k=0$. And for simpleness, we denote by $\tilde{\angle}pqr$ the angle
$\tilde{\angle}_{0}pqr$.

In the proof, we will argue by contradiction, and say that an angle $\angle
pqr$ is bad if $\angle pqr<\tilde{\angle}pqr$. If each angle of a triangle is
not bad, we call the triangle a good one.

First of all, we have the following observation about a ‘bad’ angle just via
Lemmas 1.1 and 1.2 (cf. [Wa], [SSW]).

Lemma 2.1 For a triangle $\triangle pr_{1}r_{2}\subset X$, if $\angle
pr_{1}r_{2}$ is bad, then there is $s_{0}\in[r_{1}r_{2}]^{\circ}$ such that
for any $[ps_{0}]$

$\text{ $\angle ps_{0}r_{1}$ or $\angle ps_{0}r_{2}$ is bad};$ (2.1)

in particular, for each $i$,

$\text{if $|r_{i}s_{0}|\leq|r_{i}p|$, then $\angle ps_{0}r_{i}$ is bad}.$
(2.2)

Moreover, we have that

$|ps_{0}|<\max_{i=1,2}\\{|pr_{i}|\\}.$ (2.3)

Proof. Let $\triangle\tilde{p}\tilde{r}_{1}\tilde{r}_{2}\subset\mathbb{R}^{2}$
be the comparison triangle of $\triangle pr_{1}r_{2}$. By Lemma 1.2 (and the
first variation formula on $\mathbb{R}^{2}$), the badness of $\angle
pr_{1}r_{2}$ implies that the function $|ps|-|\tilde{p}\tilde{s}|$ with
$s\in[r_{1}r_{2}],\tilde{s}\in[\tilde{r}_{1}\tilde{r}_{2}]$ and
$|r_{i}s|=|\tilde{r}_{i}\tilde{s}|$ attains a negative minimum at some
$s_{0}\in[r_{1}r_{2}]^{\circ}$. By Lemma 1.2 again, for any $[ps_{0}]$, we
have that

$\angle ps_{0}r_{i}\geq\angle\tilde{p}\tilde{s}_{0}\tilde{r}_{i},\ \ i=1,2.$

It then has to hold that

$\angle ps_{0}r_{i}=\angle\tilde{p}\tilde{s}_{0}\tilde{r}_{i},\ \ i=1,2$

because $\angle ps_{0}r_{1}+\angle ps_{0}r_{2}=\pi$ by Lemma 1.1. On the other
hand, since $|ps_{0}|<|\tilde{p}\tilde{s}_{0}|$, there is
$\tilde{p}^{\prime}\in[\tilde{p}\tilde{s}_{0}]^{\circ}$ such that
$|\tilde{p}^{\prime}\tilde{s}_{0}|=|ps_{0}|$. It is clear that
$|\tilde{p}^{\prime}\tilde{r}_{i}|<|\tilde{p}\tilde{r}_{i}|$ for at least one
of $i$, which implies
$\angle\tilde{p}\tilde{s}_{0}\tilde{r}_{i}<\tilde{\angle}ps_{0}r_{i}$, i.e.
(2.1) holds. Especially, it is easy to see that
$|\tilde{p}^{\prime}\tilde{r}_{i}|<|\tilde{p}\tilde{r}_{i}|$ if
$|r_{i}s_{0}|\leq|r_{i}p|$ 666For the case where $k>0$ and
$|r_{i}p|\geq\frac{\pi}{2\sqrt{k}}$, it needs
$\angle\tilde{r}_{i}\tilde{p}\tilde{s}_{0}<\frac{\pi}{2}$ besides
$|r_{i}s_{0}|\leq|r_{i}p|$ (note that
$\triangle\tilde{p}\tilde{r}_{1}\tilde{r}_{2}\subset\mathbb{S}_{k}^{2}$ and
$|r_{i}p|<\frac{\pi}{\sqrt{k}}$, cf. the comments following Theorem A). In
fact, ‘$\angle\tilde{r}_{i}\tilde{p}\tilde{s}_{0}<\frac{\pi}{2}$’ holds
obviously if $|r_{i}s_{0}|\ll|r_{i}p|$ (here, it needs only
‘$|r_{i}s_{0}|\leq|r_{i}p|$’ for other cases including $k\leq 0$).; and thus
(2.2) follows. Moreover, it is clear that
$|\tilde{p}\tilde{s}_{0}|<\max\limits_{i=1,2}\\{|pr_{i}|\\}$ 777For the case
where $k>0$ and $\max\limits_{i=1,2}\\{|pr_{i}|\\}\geq\frac{\pi}{2\sqrt{k}}$,
it needs to be modified to
‘$|\tilde{p}\tilde{s}_{0}|<\max\limits_{i=1,2}\\{|pr_{i}|\\}+\tau(|r_{1}r_{2}|)$’,
where $\tau(|r_{1}r_{2}|)/|r_{1}r_{2}|\to 0$ as $|r_{1}r_{2}|\to 0$; moreover,
the function $\tau(\cdot)$ can be chosen to be the same one (i.e. not
depending on $|pr_{i}|$) if $\max\limits_{i=1,2}\\{|pr_{i}|\\}\leq
c<\frac{\pi}{\sqrt{k}}$ for some constant $c$., which implies (2.3). $\Box$

Corollary 2.2 For $\triangle pr_{1}r_{2}\subset X$, if $\angle pr_{1}r_{2}$ is
bad, then there is $\bar{s}\in[r_{1}r_{2}]$ such that

$|p\bar{s}|\leq\max_{i=1,2}\\{|pr_{i}|\\},$ (2.4)

and for any small $\delta>0$ there exist $s_{1},s_{2}\in[r_{1}r_{2}]\cap
B(\bar{s},\delta)$ such that for any $[ps_{i}]$

$\angle ps_{1}s_{2}$ or $\angle ps_{2}s_{1}$ is bad. (2.5)

Proof. By Lemma 2.1, there is $[ps_{0}]$ with $s_{0}\in[r_{1}r_{2}]^{\circ}$
such that $\angle ps_{0}r_{1}$ or $\angle ps_{0}r_{2}$ is bad, say $\angle
ps_{0}r_{1}$. Then we can apply Lemma 2.1 to $\triangle ps_{0}r_{1}$ again. By
repeating this infinite times, we can locate a desired $\bar{s}$ if we take
into account (2.2) and (2.3). $\Box$

In our proof of Theorem A, Lemmas 1.3 and 2.1 shall be the mere keys. For
simpleness, we first consider the case where $X$ is, in addition, locally
compact.

Proof of Theorem A where $X$ is locally compact.

We argue by contradiction. Assume that there is a $\triangle pqr$ such that
$\angle pqr<\tilde{\angle}pqr$. Then by Corollary 2.2 (see (2.5)), we can
consider the nonempty set

$\mathcal{S}\triangleq\\{x\in X|\ \forall\ \delta>0,\ \exists\ \triangle
pr_{1}r_{2}\text{ with }r_{i}\in B(x,\delta)\text{ s.t. }\text{$\angle
pr_{1}r_{2}$ is bad}\\}.$

It is clear that $p\not\in\mathcal{S}$ because Theorem A holds around $p$, and
that $\mathcal{S}$ is closed. Note that a closed and bounded subset in $X$ is
compact because $X$ is complete and locally compact (cf. Chapter 2 in [BBI]).
So, there is $\bar{o}\in\mathcal{S}$ such that
$|p\bar{o}|=\min_{o\in\mathcal{S}}\\{|po|\\}>0$.

On the other hand, Theorem A holds around any $o\in\mathcal{S}$, and thus we
can define a positive function
$\delta(o)\triangleq\min\\{\frac{|po|}{2},\delta_{o}\\}$, where $\delta_{o}$
is the maximal number such that

$\text{any $\triangle xyz\subset B(o,\delta_{o})$ is good}.$ $None$

Claim: In $B(o,\delta(o))$, there is another point $o^{\prime}\in\mathcal{S}$
such that $|po^{\prime}|\leq|po|-\frac{\delta(o)}{3}$.

It is obvious that the claim contradicts the existence of $\bar{o}$. Thereby,
we just need to verify the claim. Due to $o\in\mathcal{S}$, there is a
$\triangle pr_{1}r_{2}$ with $r_{i}\in B(o,\delta(o))$ such that

$|or_{i}|\ll\delta(o),\text{ and $\angle pr_{1}r_{2}$ is bad}.$

Let $\bar{r}_{1}\in[pr_{1}]$ with $|p\bar{r}_{1}|=|po|-\frac{\delta(o)}{3}.$
Note that for any $[\bar{r}_{1}r_{2}]$, $\triangle r_{2}r_{1}\bar{r}_{1}$ is
good because it is contained in $B(o,\delta(o))$; so via Lemmas 1.3 and 1.1 on
$\triangle pr_{1}r_{2}$, the badness of $\angle pr_{1}r_{2}$ implies

$\text{$\angle p\bar{r}_{1}r_{2}$ is bad}.$

Then we can apply (2.2) in Lemma 2.1 (note that
$|\bar{r}_{1}r_{2}|<\delta(o)<|p\bar{r}_{1}|$ because
$\delta(o)\leq\frac{|po|}{2}$) to $\triangle p\bar{r}_{1}r_{2}$ to locate an
$s\in[\bar{r}_{1}r_{2}]^{\circ}$ such that

$\text{$\angle ps\bar{r}_{1}$ is bad for any $[ps]$}.$

Put $\bar{r}_{2}\triangleq s$ if $|ps|\leq|po|-\frac{\delta(o)}{3}$;
otherwise, similar to $\bar{r}_{1}$, we can select $\bar{r}_{2}\in[ps]$ with

$|p\bar{r}_{2}|=|po|-\frac{\delta(o)}{3},\text{ and $\angle
p\bar{r}_{2}\bar{r}_{1}$ is bad for any $[\bar{r}_{1}\bar{r}_{2}]$}.$

(Here, $\bar{r}_{1},s,\bar{r}_{2}$ all lie in $B(o,\epsilon)$ with $\epsilon$
very close to $\frac{\delta(o)}{3}$. Note that $|r_{2}x|+|xp|$ with
$x=\bar{r}_{1},s,\bar{r}_{2}$ is very close to $|po|$ because
$|or_{i}|\ll\delta(o)$.) Then by Corollary 2.2, the badness of $\angle
p\bar{r}_{2}\bar{r}_{1}$ enables us to locate a point
$o^{\prime}\in[\bar{r}_{1}\bar{r}_{2}]\cap\mathcal{S}$, which lies in
$B(o,\delta(o))$, such that $|po^{\prime}|\leq|po|-\frac{\delta(o)}{3}$ (see
(2.4) and note that $|p\bar{r}_{i}|\leq|po|-\frac{\delta(o)}{3}$ 888For the
case where $k>0$ and
$\max\limits_{i=1,2}\\{|p\bar{r}_{i}|\\}\geq\frac{\pi}{2\sqrt{k}}$,
$\delta(o)$ should be additionally so small that $\tau(\delta(o))\ll\delta(o)$
(e.g. $\tau(\delta(o))<0.01\delta(o)$), where $\tau(\cdot)$ is just the
function in Footnote 7.). $\Box$

Eventually, we prove Theorem A for general $X$, i.e. $X$ might not be locally
compact.

Proof of Theorem A where $X$ is not locally compact.

When $X$ is not locally compact, there might be no $\bar{o}\in\mathcal{S}$
with $|p\bar{o}|=\min_{o\in\mathcal{S}}\\{|po|\\}$ in the proof right above.
However, starting with an $o_{1}\in\mathcal{S}$ and by the claim below (2.6),
we can step by step obtain $\\{o_{i}\\}_{i=1}^{\infty}\subset\mathcal{S}$ such
that

$|po_{i+1}|\leq|po_{i}|-\frac{\delta(o_{i})}{3}\text{ and
}|o_{i}o_{i+1}|<\delta(o_{i}).$

It follows that

$|po_{i+1}|\leq|po_{1}|-\sum_{j=1}^{i}\frac{\delta(o_{j})}{3},\text{ and thus
}\sum_{j=1}^{i}\frac{\delta(o_{j})}{3}\leq|po_{1}|.$

This implies that $\lim\limits_{i\to\infty}\delta(o_{i})=0$, and that
$\\{o_{i}\\}_{i=1}^{\infty}$ is a Cauchy sequence (note that
$|o_{i}o_{i+1}|<\delta(o_{i})$). Then by the completeness of $X$ and the
closedness of $\mathcal{S}$, $\\{o_{i}\\}_{i=1}^{\infty}$ has a limit point
$\bar{o}\in\mathcal{S}$. However, it follows that
$\lim\limits_{i\to\infty}\delta(o_{i})=\delta(\bar{o})>0$, a contradiction.
$\Box$

## References

  * [AKP] S. Alexander, V. Kapovitch and A. Petrunin, Alexandrov Geometry, 2010.
  * [BBI] D. Burago, Y. Burago and S. Ivanov, A Course in Metric Geometry, Graduate Studies in Mathematics Vol. 33, AMS, 2001.
  * [BGP] Y. Burago, M. Gromov and G. Perel’man, A.D. Aleksandrov spaces with curvature bounded below, Uspeckhi Mat. Nank 47(2): 3-51, 1992.
  * [Pl] C. Plaut, Spaces of Wald-Berestovskii curvature bounded below, J. Geom. Anal. 6: 113-134, 1996.
  * [Sh] K. Shiohama, An introduction to the geometry of Alexandrov spaces, 1992.
  * [Wa] Y. Wang, A Schur-Toponogov theorem in Riemannian geometry & a new proof of Toponogov’s theorem in Alexandrov geometry, arXiv:1809.09818.
  * [SSW] X. Su, H. Sun and Y. Wang, Pythagorean Theorem & curvature with lower or upper bound, to appear in Chinese Ann. of Math. (B), 2022.

School of Mathematical Sciences (and Lab. math. Com. Sys.), Beijing Normal
University, Beijing, 100875 P.R.C.

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