CAT Comparison, Link Criteria, and Local Globalization

The series of Sine and cosine defined by their real power series contain only even powers of $t$ in $\cos t$ and only odd powers in $\sin t$, so $\cos (-t)=\cos t$, $\sin (-t)=-\sin t$, $\cos 0=1$ and $\sin 0=0$; replacing $y$ by $-y$ in [F4] therefore gives $\cos (x-y)=\cos x\cos y+\sin x\sin y$ and $\sin (x-y)=\sin x\cos y-\cos x\sin y$, and the case $x=y=t$ of the cosine subtraction formula gives the Pythagorean identity ${\cos }^{2}t+{\sin }^{2}t=\cos 0=1$, whence $\cos 2t=2{\cos }^{2}t-1$. Since $\sin \pi =0$ by [F5], ${\cos }^2\pi =1$; $\cos$ is strictly decreasing on $[0,\pi ]$ by [F3] with $\cos 0=1$, so $\cos \pi <1$ and hence $\cos \pi =-1$; then $0=\cos \pi +1=2{\cos }^2(\pi \mathrm{/}2)$ gives $\cos (\pi \mathrm{/}2)=0$, and strict decrease gives $\cos t>0$ for $0\le t<\pi \mathrm{/}2$.

Let $a,b,c>0$ satisfy the three triangle inequalities and put $t:=(b^2+c^2-a^2)\mathrm{/}(2b)$ and $s:=\sqrt{c^2-t^2}$, which is real because $c^2-t^2=\frac{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}{4b^2}\ge 0$; set $\bar{p}:=(0,0)$, $\bar{r}:=(b,0)$ and $\bar{q}:=(t,s)$ in ${\mathbb{R}}^2$. Then $d_2(\bar{p},\bar{r})=b$, $d_2(\bar{p},\bar{q}{)}^2=t^2+s^2=c^2$ and $d_2(\bar{r},\bar{q}{)}^2=(t-b{)}^2+s^2=t^2-2bt+b^2+s^2=c^2+b^2-2bt=a^2$, so the three prescribed distances are realised. This construction includes every positive degenerate triple because then $s=0$. If $b=0$, the inequalities force $a=c$, and $\bar{p}=\bar{r}=(0,0)$, $\bar{q}=(c,0)$ realise the triple; the other zero-side cases follow by relabeling.

For $x,y\in S^{n-1}$ the Cauchy–Schwarz inequality [F6] gives $\mathrm{\mid }x\cdot y\mathrm{\mid }\le 1$, so $d_S(x,y)=\arccos (x\cdot y)$ is defined by [F2]; symmetry is $x\cdot y=y\cdot x$, separation at $x=y$ is $\arccos 1=0$, and if $d_S(x,y)=0$ then $x\cdot y=1$, so $x,y$ are linearly dependent by the equality case of [F6]; as unit vectors they satisfy $y=\pm x$, and $x\cdot y=1$ excludes $y=-x$, so $x=y$.

Let $A,B,C\in S^{n-1}$ with $a:=d_S(B,C)$ and $b:=d_S(C,A)$ both in $(0,\pi )$, and let $c:=d_S(A,B)$, $u:=(B-(B\cdot C)C)\mathrm{/}\mathrm{\mid }B-(B\cdot C)C\mathrm{\mid }$ and $v:=(A-(A\cdot C)C)\mathrm{/}\mathrm{\mid }A-(A\cdot C)C\mathrm{\mid }$. Here $B\cdot C=\cos a$ and $A\cdot C=\cos b$ by [F2], so $\mathrm{\mid }B-(B\cdot C)C{\mathrm{\mid }}^2=1-{\cos }^{2}a={\sin }^{2}a>0$ and likewise $\mathrm{\mid }A-(A\cdot C)C{\mathrm{\mid }}^2={\sin }^{2}b>0$, using $\sin a>0$ and $\sin b>0$ from [F5]; hence $B=(\cos a)C+(\sin a)u$ and $A=(\cos b)C+(\sin b)v$ with $u,v$ unit vectors orthogonal to $C$. Expanding gives the spherical cosine rule $\cos c=A\cdot B=\cos a\cos b+\sin a\sin b(u\cdot v)$, and $\cos \gamma :=u\cdot v\in [-1,1]$ by [F6] defines the vertex angle $\gamma \in [0,\pi ]$ at $C$.

Let $X$ be a geodesic space with a geodesic triangle $z,x,y$ and $p_t$ on $[x,y]$ at distance $td(x,y)$ from $x$; in the Euclidean comparison triangle ${\bar{p}}_t=(1-t)\bar{x}+t\bar{y}$, so the identity $\mathrm{\mid }\bar{z}-{\bar{p}}_t{\mathrm{\mid }}^2=(1-t)\mathrm{\mid }\bar{z}-\bar{x}{\mathrm{\mid }}^2+t\mathrm{\mid }\bar{z}-\bar{y}{\mathrm{\mid }}^2-t(1-t)\mathrm{\mid }\bar{x}-\bar{y}{\mathrm{\mid }}^2$ holds by expansion, and substituting the comparison distances shows that $d(z,p_t)\le d_2(\bar{z},{\bar{p}}_t)$ is equivalent to its squared form, both sides being nonnegative.

On $S_{\mathrm{\ell }}^1=\mathbb{R}\mathrm{/}\mathrm{\ell }\mathbb{Z}$ the minimum defining $d_{\mathrm{\ell }}$ exists: for $\delta =x-y$, choose $N>2\mathrm{\mid }\delta \mathrm{\mid }\mathrm{/}\mathrm{\ell }$ by [F13]; when $\mathrm{\mid }k\mathrm{\mid }>N$, $\mathrm{\mid }\delta +k\mathrm{\ell }\mathrm{\mid }\ge \mathrm{\mid }k\mathrm{\mid }\mathrm{\ell }-\mathrm{\mid }\delta \mathrm{\mid }>\mathrm{\mid }\delta \mathrm{\mid }$, so a minimum occurs among the finitely many integers $\mathrm{\mid }k\mathrm{\mid }\le N$. Changing representatives just shifts these integers. Thus $d_{\mathrm{\ell }}$ is well defined and symmetric and vanishes exactly on the diagonal, and $d_{\mathrm{\ell }}(x,z)={\min }_k\mathrm{\mid }x-z+k\mathrm{\ell }\mathrm{\mid }\le {\min }_k\mathrm{\mid }x-y+k\mathrm{\ell }\mathrm{\mid }+{\min }_j\mathrm{\mid }y-z+j\mathrm{\ell }\mathrm{\mid }=d_{\mathrm{\ell }}(x,y)+d_{\mathrm{\ell }}(y,z)$ by choosing minimising integers; hence $d_{\mathrm{\ell }}$ is a metric. The map $t\mapsto t+\mathrm{\ell }\mathbb{Z}$ from $[0,\mathrm{\ell }]$ onto $S_{\mathrm{\ell }}^1$ is surjective: a finite integer bound from [F13] permits subtracting an integer multiple of $\mathrm{\ell }$ to place each real representative in $[0,\mathrm{\ell })$ and satisfies $d_{\mathrm{\ell }}(s,t)=\min (\mathrm{\mid }s-t\mathrm{\mid },\mathrm{\ell }-\mathrm{\mid }s-t\mathrm{\mid })$ for $s,t\in [0,\mathrm{\ell }]$, so it is continuous and sends the compact interval onto $S_{\mathrm{\ell }}^1$, which is therefore compact, complete by [F11] and geodesic (the shorter interval between two parameters is mapped isometrically), and each point has a ball of radius $r<\mathrm{\ell }\mathrm{/}4$ isometric to an interval of $\mathbb{R}$, since differences between lifts in that ball are $<\mathrm{\ell }\mathrm{/}2$. A minimizing segment lifts successively through these interval charts, starting at any chosen lift; each chart restriction of its lift is affine of slope $+1$ or $-1$ in unit-speed parameters. Overlaps fix the same slope, so the lift is one straight interval. Therefore all minimizing segments are the shorter arcs, with two choices only at distance $\mathrm{\ell }\mathrm{/}2$; the identity map exhibits an isometrically embedded circle of length $\mathrm{\ell }$ in the sense of [F9].

Two triangles in ${\mathbb{R}}^2$ with the same three side lengths are related by an isometry: when all lengths are zero all vertices coincide, and a translation suffices; otherwise relabel so that the baseline has positive length. If $d_2(P,R)=b>0$, the map $\varphi (x):=P+Bx$, with $B$ orthogonal sending $(1,0)$ to $(R-P)\mathrm{/}b$, is an isometry of ${\mathbb{R}}^2$, and applying the coordinate computation of step 1.2 to ${\varphi }^{-1}$ of a triangle shows that a point $Q$ at distance $c$ from $P$ and $a$ from $R$ has coordinates $(t,\pm s)$ with $t,s$ as in step 1.2; the two solutions differ by the reflection $(x_1,x_2)\mapsto (x_1,-x_2)$, so any two triangles with the prescribed side lengths are obtained from the normal form of step 1.2 by an isometry.

With the notation of step 1.4 and $a,b\in (0,\pi )$: if $a+b\le \pi$ then $\cos c\ge \cos a\cos b-\sin a\sin b=\cos (a+b)$ by [F4] and $\sin a\sin b\ge 0$ from [F5], and since $c\le \pi$ and $\cos$ strictly decreases on $[0,\pi ]$ by [F3], $c\le a+b$; if $a+b>\pi$ then $c\le \pi <a+b$. If $a=0$ then $C=B$ and $c=b=a+b$; if $b=0$ similarly $c=a$. Hence $d_S$ satisfies the triangle inequality, and by step 1.3 it is a metric on $S^{n-1}$.

Let $x,y\in S^{n-1}$ and $\theta :=d_S(x,y)\in (0,\pi ]$; when $\theta <\pi$ put $\xi :=(y-(x\cdot y)x)\mathrm{/}\mathrm{\mid }y-(x\cdot y)x\mathrm{\mid }$, so that $y=(\cos \theta )x+(\sin \theta )\xi$ with $\xi$ a unit vector orthogonal to $x$ by step 1.3 and [F2], and choose any unit $\xi \perp x$ when $\theta =\pi$. Then $c(t):=(\cos t)x+(\sin t)\xi$ satisfies $c(0)=x$, $c(\theta )=y$, and $c(t)\cdot c(t^{\mathrm{\prime }})=\cos (t-t^{\mathrm{\prime }})$ for $t,t^{\mathrm{\prime }}\in [0,\theta ]$ with $\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }\le \theta \le \pi$ by step 1.1, so $d_S(c(t),c(t^{\mathrm{\prime }}))=\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }$ by [F2]; thus minimal great-circle arcs are geodesic segments in the sense of [F12].

Let $a,b,c\ge 0$ satisfy the triangle inequalities with $a+b+c<2\pi$. Then $a,b,c<\pi$, since $a\ge \pi$ forces $a+b+c\ge 2a\ge 2\pi$. If $a,b,c>0$, put $\gamma :=\arccos \frac{\cos c-\cos a\cos b}{\sin a\sin b}$; the argument lies in $[-1,1]$ directly from the length hypotheses. Indeed $c\ge \mathrm{\mid }a-b\mathrm{\mid }$ gives $\cos c\le \cos (a-b)=\cos a\cos b+\sin a\sin b$. If $a+b\le \pi$, then $c\le a+b$ gives $\cos c\ge \cos (a+b)$; if $a+b>\pi$, the perimeter bound gives $c<2\pi -a-b<\pi$, hence $\cos c>\cos (2\pi -a-b)=\cos (a+b)$. In both cases $\cos c\ge \cos a\cos b-\sin a\sin b$, proving the required range without presupposing a spherical realization, and the construction $C:=(1,0,0)$, $B:=(\cos a)C+(\sin a)(0,1,0)$, $A:=(\cos b)C+(\sin b)(\cos \gamma (0,1,0)+\sin \gamma (0,0,1))$ uses ${\cos }^{2}t+{\sin }^{2}t=1$ (step 1.1) to give unit vectors with $B\cdot C=\cos a$, $A\cdot C=\cos b$ and $A\cdot B=\cos a\cos b+\sin a\sin b\cos \gamma =\cos c$; if one of $a,b,c$ vanishes, say $a=0$, then $b=c$ and $C:=(1,0,0)$, $B:=C$, $A:=(\cos b)C+(\sin b)(0,1,0)$ realise the triple. Any two triples of unit vectors with equal pairwise inner products have spans related by a well-defined inner-product-preserving linear map, extended to an orthogonal map of ${\mathbb{R}}^3$ along orthonormal bases of the orthocomplements; so the comparison triangle is unique up to an isometry of $S^2$ by [F9]. A triangle in $S^2$ of perimeter $<2\pi$ is a comparison triangle for itself, so the CAT(1) inequality holds with equality for it and $S^2$ is CAT(1).

Let $X$ be CAT(0) and let $\gamma ,\delta$ be geodesic segments from $x$ to $y$ of length $L=d(x,y)$; the geodesic triangle with sides $\gamma ,\delta$ and the degenerate third side has side lengths $L,L,0$, and its comparison triangle in ${\mathbb{E}}^2$ is degenerate with the comparison points of $\gamma (t)$ and $\delta (t)$ coinciding, both being at distance $t$ from the comparison vertex $\bar{x}$ along the same comparison side; hence $d(\gamma (t),\delta (t))\le d_2(\bar{\gamma }(t),\bar{\delta }(t))=0$ by the CAT(0) inequality, so $\gamma =\delta$ and geodesic segments are unique.

A geodesic triangle in a metric space has side lengths obeying the triangle inequalities, because the metric satisfies (M3) of Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric; in ${\mathbb{E}}^2$ every geodesic side is a straight segment: equality $\mathrm{\mid }x-w\mathrm{\mid }+\mathrm{\mid }w-y\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }$ along a minimizing segment forces the two displacement vectors to be nonnegative multiples by the equality case of [F6], and their lengths fix $w$. By steps 1.2 and 2.1 every metric triangle therefore has a comparison triangle in ${\mathbb{E}}^2$, unique up to isometry, and if the triangle itself lies in ${\mathbb{E}}^2$ then it is a comparison triangle for itself by step 2.1, so the CAT(0) inequality is an equality for it and ${\mathbb{E}}^2$ is CAT(0).

Let $x,y\in S^{n-1}$, $\theta :=d_S(x,y)\in (0,\pi )$, and let $z\in S^{n-1}$ satisfy $d_S(x,z)+d_S(z,y)=\theta$; write $\alpha :=d_S(x,z)$ and $\beta :=d_S(z,y)$, so $\theta =\alpha +\beta$. The Gram matrix of $x,y,z$ has entries $x\cdot x=y\cdot y=z\cdot z=1$, $x\cdot y=\cos \theta$, $x\cdot z=\cos \alpha$, $y\cdot z=\cos \beta$, and is positive semidefinite because $v^{T}Gv=\mathrm{\mid }{\sum }_i{v}_i{u}_i{\mathrm{\mid }}^2\ge 0$; by step 1.1, $\cos \theta =\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$, so its determinant is $1+2\cos \alpha \cos \beta \cos \theta -{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2\theta ={\sin }^2\alpha {\sin }^2\beta -(\cos \theta -\cos \alpha \cos \beta {)}^2=0$, the bracket being $\sin \alpha \sin \beta$. Hence $x,y,z$ are linearly dependent; since $\theta \in (0,\pi )$ makes $x,y$ independent, $z$ lies in the plane $P:=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{x,y\}$. Parametrise the unit circle of $P$ by $c(s)=(\cos s)x+(\sin s)\xi$ with $\xi$ as in step 2.3, so that $y=c(\theta )$; a unit vector of $P$ with $x\cdot z=\cos \alpha$ is $c(\alpha )$ or $c(-\alpha )$, and if $z=c(-\alpha )$ with $\alpha >0$ then $z\cdot y=c(-\alpha )\cdot c(\theta )=\cos (\theta +\alpha )\mathrm{\ne }\cos (\theta -\alpha )=\cos \beta$ by steps 1.1 and 1.4, since equality would force $\sin \theta \sin \alpha =0$; so $z=c(\alpha )$, a point of the minimal arc from $x$ to $y$.

Let $p\in S^{n-1}$, $0<r<\pi \mathrm{/}2$ and $x,y\in \bar{B}(p,r)$; if $x=y$ there is nothing to prove, so assume $x\mathrm{\ne }y$ and put $\theta :=d_S(x,y)\in (0,\pi ]$. If $\theta =\pi$ then $y=-x$ by [F2], so $p\cdot y=-p\cdot x\le -\cos r$ while $p\cdot y=\cos d_S(p,y)\ge \cos r$, a contradiction since $\cos r>0$; hence $\theta <\pi$. For $0\le t\le \theta$ the point $c(t)=\frac{\sin (\theta -t)}{\sin \theta }x+\frac{\sin t}{\sin \theta }y$ of the arc of step 2.3 has nonnegative coefficients by [F5], whose sum is $\cos (t-\theta \mathrm{/}2)\mathrm{/}\cos (\theta \mathrm{/}2)\ge 1$ by the addition formulas and decreasing cosine, so $p\cdot c(t)\ge \min (p\cdot x,p\cdot y)\ge \cos r>0$ and therefore $d_S(p,c(t))\le r$ by [F2] and [F3]; hence $\bar{B}(p,r)$ is convex.

Let $\gamma ,\delta$ be geodesics with common initial point $x=\gamma (0)=\delta (0)$ and proportional parametrisations on $[0,1]$; if $\gamma (1)=\delta (1)$ the claim follows from step 2.5, so let $y:=\gamma (1)$, $z:=\delta (1)$ and consider the geodesic triangle $x,y,z$. In its Euclidean comparison triangle the points $\bar{\gamma }(t)$ and $\bar{\delta }(t)$ are at distance $td(y,z)$ for every $t\in [0,1]$, by similarity; the CAT(0) inequality applied to the pair $\gamma (t),\delta (t)$ gives $d(\gamma (t),\delta (t))\le td(y,z)=(1-t)d(\gamma (0),\delta (0))+td(\gamma (1),\delta (1))$ since $\gamma (0)=\delta (0)$.

Let $X$ be a geodesic space in which the CAT(0) inequality holds for every geodesic triangle and every pair consisting of a vertex and a point of the opposite side. Then it holds for all pairs. If $x,y$ lie on one side the comparison preserves distances along that side; if one of $x,y$ is a vertex of the triangle, the claim is either the hypothesis (when the other point lies on the opposite side) or the equality of a side length with its comparison length (when the other point is a vertex or lies on an adjacent side and equals the shared vertex); and degenerate triangles, where some side has length $0$, reduce to these cases. It remains to treat a triangle with sides $[p,q],[q,r],[r,p]$, points $x\in [p,q]$ and $y\in [p,r]$ with $x\mathrm{\ne }p\mathrm{\ne }y$, $x\mathrm{\ne }q$ and $y\mathrm{\ne }r$, comparison triangle $(\bar{p},\bar{q},\bar{r})$, comparison points $\bar{x}\in [\bar{p},\bar{q}]$ and $\bar{y}\in [\bar{p},\bar{r}]$, and the positive numbers $a:=d(p,x)$, $b:=d(p,y)$; write $\widehat{d}:=d(x,y)$ and let $\widehat{\alpha }$ be the vertex angle at the point corresponding to $p$ in the comparison triangle of $(p,x,y)$, so that ${\widehat{d}}^2=a^2+b^2-2ab\cos \widehat{\alpha }$. Apply the hypothesis to the geodesic triangle $(p,x,r)$ with vertex $x$ and the point $y$ of the opposite side $[p,r]$: if $({\widehat{p}}^{\mathrm{\prime }},{\widehat{x}}^{\mathrm{\prime }},{\widehat{r}}^{\mathrm{\prime }})$ is the comparison triangle of $(p,x,r)$ and ${\widehat{y}}^{\mathrm{\prime }}\in [{\widehat{p}}^{\mathrm{\prime }},{\widehat{r}}^{\mathrm{\prime }}]$ is the comparison point of $y$, then $\widehat{d}\le D:=d_2({\widehat{x}}^{\mathrm{\prime }},{\widehat{y}}^{\mathrm{\prime }})$. The triangles $({\widehat{p}}^{\mathrm{\prime }},{\widehat{x}}^{\mathrm{\prime }},{\widehat{y}}^{\mathrm{\prime }})$ and the comparison triangle of $(p,x,y)$ have two sides in common, of lengths $a$ and $b$, and opposite sides $D\ge \widehat{d}$, so the angle at ${\widehat{p}}^{\mathrm{\prime }}$ of the former is at least $\widehat{\alpha }$: for fixed $a,b>0$ the quantity $(a^2+b^2-s^2)\mathrm{/}(2ab)$ is strictly decreasing in $s$ and $\cos$ is strictly decreasing on $[0,\pi ]$ (F3), so the included angle is increasing in the opposite side. As ${\widehat{y}}^{\mathrm{\prime }}$ lies on the side $[{\widehat{p}}^{\mathrm{\prime }},{\widehat{r}}^{\mathrm{\prime }}]$, that angle is exactly the vertex angle ${\widehat{\alpha }}^{\mathrm{\prime }}$ of the comparison triangle of $(p,x,r)$ at ${\widehat{p}}^{\mathrm{\prime }}$, so ${\widehat{\alpha }}^{\mathrm{\prime }}\ge \widehat{\alpha }$. Apply the hypothesis again, to $(p,q,r)$ with vertex $r$ and the point $x$ of the opposite side $[p,q]$: $d(r,x)\le d_2(\bar{r},\bar{x})$, so the triangles $({\widehat{p}}^{\mathrm{\prime }},{\widehat{x}}^{\mathrm{\prime }},{\widehat{r}}^{\mathrm{\prime }})$ and $(\bar{p},\bar{x},\bar{r})$ have two sides in common, of lengths $a$ and $d(p,r)$, and opposite sides $d(x,r)\le d_2(\bar{x},\bar{r})$; the same monotonicity of the included angle gives ${\widehat{\alpha }}^{\mathrm{\prime }}\le \bar{\alpha }$, the vertex angle of $(\bar{p},\bar{q},\bar{r})$ at $\bar{p}$, and $\bar{x}\in [\bar{p},\bar{q}]$ with $\bar{x}\mathrm{\ne }\bar{p}$ makes the ray from $\bar{p}$ to $\bar{x}$ the ray to $\bar{q}$, so this angle is exactly the vertex angle of $(\bar{p},\bar{x},\bar{r})$ at $\bar{p}$. Hence $\widehat{\alpha }\le \bar{\alpha }$, and the two laws of cosines $d(x,y{)}^2=a^2+b^2-2ab\cos \widehat{\alpha }$ and $d_2(\bar{x},\bar{y}{)}^2=a^2+b^2-2ab\cos \bar{\alpha }$ give $d(x,y)\le d_2(\bar{x},\bar{y})$ because $\cos$ is decreasing on $[0,\pi ]$ (F3). Thus a geodesic space is CAT(0) if and only if every vertex-opposite-side comparison holds.