Comparison triangles in the Euclidean plane and the round sphere, model spaces, and CAT(0) and CAT(1) consequences

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.

The case $t=1\mathrm{/}2$ of the squared form in step 1.5, with $m:=p_{1\mathrm{/}2}$ a midpoint of $[y,y^{\mathrm{\prime }}]:=[x,y]$ and $z$ the opposite vertex, gives $d(z,m{)}^2\le \frac{1}{2}d(z,y{)}^2+\frac{1}{2}d(z,y^{\mathrm{\prime }}{)}^2-\frac{1}{4}d(y,y^{\mathrm{\prime }}{)}^2$, which is (d) since midpoints are unique by step 2.5.

Let $\mathrm{\ell }\ge 2\pi$ and let $x,y,z\in S_{\mathrm{\ell }}^1$ be the vertices of a triangle of perimeter $p<2\pi$, so that all three sides are $<\pi$; if vertices repeat, the two nonzero sides coincide by the circle interval charts: a minimizing circle path of length $<\mathrm{\ell }\mathrm{/}2$ has a lift with one fixed direction in the overlapping interval charts, hence is the shorter arc. Such triangles realize a degenerate comparison. For three distinct vertices write their cyclic gaps as $g_1,g_2,g_3>0$. If all $g_i\le \mathrm{\ell }\mathrm{/}2$ then the three pairwise distances are $g_1,g_2,g_3$ and $p=\mathrm{\ell }\ge 2\pi$, contrary to hypothesis; so some $g_3>\mathrm{\ell }\mathrm{/}2$, the complementary arc of length $s:=\mathrm{\ell }-g_3$ contains all three points, and the two adjacent distances are $g_1,g_2$ while the third is $g_1+g_2=s$, so $p=2s<2\pi$ and $s<\pi$. Thus the triangle is degenerate and isometric, side by side, to a configuration on an arc of length $s$; placing three points of a great arc of $S^2$ of the same length $s<\pi$ realises the same three side lengths, so by uniqueness of comparison triangles in step 2.4 the comparison triangle is isometric to the triangle and the CAT(1) inequality holds with equality; hence $S_{\mathrm{\ell }}^1$ is CAT(1) for $\mathrm{\ell }\ge 2\pi$.

Geodesic segments in a CAT(0) space vary continuously with their endpoints. Let $\gamma$ be the geodesic from $x$ to $y$ and ${\gamma }^{\mathrm{\prime }}$ the geodesic from $x^{\mathrm{\prime }}$ to $y^{\mathrm{\prime }}$, let $t\in [0,1]$, and let $\delta$ be the geodesic from $x$ to $y^{\mathrm{\prime }}$, so that $\gamma ,\delta$ are geodesics from the common initial point $x$ with proportional parametrizations. Step 3.4 gives $d(\gamma (t),\delta (t))\le (1-t)d(\gamma (0),\delta (0))+td(\gamma (1),\delta (1))=td(y,y^{\mathrm{\prime }})$, and step 3.4 applied to the reversed geodesics $\bar{\delta }(s)=\delta (1-s)$ from $y^{\mathrm{\prime }}$ to $x$ and ${\bar{\gamma }}^{\mathrm{\prime }}(s)={\gamma }^{\mathrm{\prime }}(1-s)$ from $y^{\mathrm{\prime }}$ to $x^{\mathrm{\prime }}$, which again have a common initial point and proportional parametrizations, gives $d(\delta (t),{\gamma }^{\mathrm{\prime }}(t))=d(\bar{\delta }(1-t),{\bar{\gamma }}^{\mathrm{\prime }}(1-t))\le (1-t)d(x,x^{\mathrm{\prime }})$. Hence $d(\gamma (t),{\gamma }^{\mathrm{\prime }}(t))\le (1-t)d(x,x^{\mathrm{\prime }})+td(y,y^{\mathrm{\prime }})$ for every $t$, which tends to $0$ uniformly in $t$ as $x^{\mathrm{\prime }}\to x$ and $y^{\mathrm{\prime }}\to y$; since the geodesic segments are unique by step 2.5, they vary continuously with their endpoints as asserted in (iv)(a).

Let $\gamma :[0,\theta ]\to S^{n-1}$ be a geodesic from $x$ to $y$ with $\theta <\pi$. If $\theta =0$ it is constant. Otherwise $d_S(x,\gamma (t))+d_S(\gamma (t),y)=\theta$, so step 3.2 puts $\gamma (t)$ on the unique minimal arc at position $t$, and step 2.3 identifies its parametrization. If $\theta =\pi$, choose $z=\gamma (\pi \mathrm{/}2)$, perpendicular to $x$. Each half has length $\pi \mathrm{/}2$ and is the unique short arc just proved; because $y=-x$, both halves lie in the plane spanned by $x,z$ and form one semicircle. This proves the claimed classification of all spherical geodesics.

Let $X$ be CAT(1), $p\in X$, $0<r<\pi \mathrm{/}2$ and $y,z\in B(p,r)$. Since $d(y,z)<2r<\pi$, segments exist. Two competing segments give a triangle with repeated vertex and perimeter $2d(y,z)<2\pi$; its comparison sides coincide, so CAT(1) forces equal-parameter points to coincide. Thus $[y,z]$ is unique. Choose $r^{\mathrm{\prime }}$ with $\max \{d(p,y),d(p,z)\}<r^{\mathrm{\prime }}<r$. The triangle $p,y,z$ has perimeter $<4r^{\mathrm{\prime }}<2\pi$. Its comparison side lies in the convex spherical ball $\bar{B}(\bar{p},r^{\mathrm{\prime }})$ by step 3.3, so CAT(1) gives $d(p,w)\le r^{\mathrm{\prime }}<r$ for every $w\in [y,z]$. Hence $B(p,r)$ is convex, and the same argument with non-strict endpoint bounds proves convexity of closed balls of radius $<\pi \mathrm{/}2$.