Alexandrov comparison: straightening a hinge, gluing comparison triangles, and patchwork

Spherical size and orientation. The model comparison angles agree with geometric angles: on the sphere, the unit tangent towards $V$ at $U$ is $(V-(U\cdot V)U)\mathrm{/}\sin d(U,V)$, whose dot product with the analogous tangent towards $W$ is exactly the cosine formula in [F2]; the Euclidean assertion is the corresponding dot-product formula. Write $u=d(C,B)$, $y=d(C,B^{\mathrm{\prime }})$, $p=d(A,B)$, $q=d(A,B^{\mathrm{\prime }})$, $x=d(A,C)$ and $L=u+y+p+q$. For $\kappa =1$, $L<2\pi$ gives $x<\pi$, since $2x\le (u+p)+(y+q)=L$; every boundary edge is also $<\pi$, since its complementary path has length at least that edge. The polygonal loop $A,B,C,B^{\mathrm{\prime }},A$ lies in an open hemisphere: take points $P,Q$ dividing its arclength into halves of length $L\mathrm{/}2$; every point $Z$ on either half satisfies $d(P,Z)+d(Q,Z)\le L\mathrm{/}2<\pi$, hence $(P+Q)\cdot Z=\cos d(P,Z)+\cos d(Q,Z)>0$. In that hemisphere the projection $Z\mapsto Z\mathrm{/}(h\cdot Z)$, with $h=(P+Q)\mathrm{/}\mathrm{\parallel }P+Q\mathrm{\parallel }$, maps minor great-circle arcs to straight segments and preserves the side of each line and whether an interior angle is reflex. Because $B,B^{\mathrm{\prime }}$ are on opposite sides of $AC$, the projected quadrilateral is simple. Its angles at $B,B^{\mathrm{\prime }}$ are convex, and its angle at $C$ is reflex or straight because $\gamma +{\gamma }^{\mathrm{\prime }}\ge \pi$. Triangulating along $AC$ shows that its angle at $A$ is convex and that $C$ lies in the closed triangle $AB{B}^{\mathrm{\prime }}$; the same conclusions hold before projection. Thus $\alpha +{\alpha }^{\mathrm{\prime }}\le \pi$. Moreover $u+y<\pi$: otherwise $p+q<\pi$, so $(B+B^{\mathrm{\prime }})\cdot A=\cos p+\cos q>0$, and $(B+B^{\mathrm{\prime }})\cdot B=(B+B^{\mathrm{\prime }})\cdot B^{\mathrm{\prime }}=1+B\cdot B^{\mathrm{\prime }}>0$ (the original open hemisphere excludes antipodal $B,B^{\mathrm{\prime }}$). A hemisphere contains the minor geodesic segments joining its points, since their unit vectors are normalized positive linear combinations; hence the hemisphere with pole $B+B^{\mathrm{\prime }}$ contains the spherical triangle $AB{B}^{\mathrm{\prime }}$ and $C$, giving $\cos u+\cos y>0$, equivalently $u+y<\pi$, a contradiction. In the Euclidean case the same planar quadrilateral argument gives $\alpha +{\alpha }^{\mathrm{\prime }}\le \pi$, and there is no size restriction.

A monotonicity lemma for a marked point. Let $V,W,U$ be a triangle in ${\mathbb{M}}_{\kappa }^2$ with $\mathrm{\mid }VW\mathrm{\mid }=s>0$, $\mathrm{\mid }VU\mathrm{\mid }=p>0$ (with $s,p<\pi$ when $\kappa =1$), $\mathrm{\mid }WU\mathrm{\mid }=q$, and angle $\theta$ at $V$; let $C_0\in [V,W]$ satisfy $\mathrm{\mid }V{C}_0\mathrm{\mid }=u$ with $0<u<s$. Then $\mathrm{\mid }U{C}_0\mathrm{\mid }$ is a strictly increasing function of $\theta \in [0,\pi ]$: it is the third side of the model triangle with sides $p$ and $u$ meeting at the angle $\theta$, so the claim is the strict monotonicity in the included angle from [F2], applied with the fixed positive sides $p,u$. Consequently, if $V^{\mathrm{\prime }}{W}^{\mathrm{\prime }}{U}^{\mathrm{\prime }}$ is a second such triangle with $\mathrm{\mid }{V}^{\mathrm{\prime }}{W}^{\mathrm{\prime }}\mathrm{\mid }=s$, $\mathrm{\mid }{V}^{\mathrm{\prime }}{U}^{\mathrm{\prime }}\mathrm{\mid }=p$, $\mathrm{\mid }{W}^{\mathrm{\prime }}{U}^{\mathrm{\prime }}\mathrm{\mid }=q^{\mathrm{\prime }}\le q$, and marked point $C_0^{\mathrm{\prime }}\in [V^{\mathrm{\prime }}{W}^{\mathrm{\prime }}]$ with $\mathrm{\mid }{V}^{\mathrm{\prime }}{C}_0^{\mathrm{\prime }}\mathrm{\mid }=u$, having angle ${\theta }^{\mathrm{\prime }}$ at $V^{\mathrm{\prime }}$, then ${\theta }^{\mathrm{\prime }}\le \theta$ by the strict monotonicity of the angle in the opposite side from [F2] (fixed sides $p,s$), hence $\mathrm{\mid }{U}^{\mathrm{\prime }}{C}_0^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }U{C}_0\mathrm{\mid }$.

Reduction to an interior splitting point in (ii). If $r=q_0$ or $r=q_1$ then one of ${\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2$ has a repeated vertex, its comparison triangle degenerates to a segment, the hypothesis for that index is vacuous, and the claim reduces to the hypothesis for the other index; so assume $r\in (q_0,q_1)$, which makes all six side lengths of ${\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2$ positive, since $r\mathrm{\ne }p$ and the vertices of $\mathrm{\Delta }$ are distinct.

Local CAT angle comparison. For two unit-speed spherical rays meeting at angle $\theta$, let their points at distances $s,t>0$ with $s+t<\pi$ have distance $D$, and put $h=\mathrm{\mid }s-t\mathrm{\mid }$. The cosine rule and addition formulas give $4{\sin }^2(D\mathrm{/}2)-4{\sin }^2(h\mathrm{/}2)=2\sin s\sin t(1-\cos \theta )$. Factoring the left side yields $4\sin ((D+h)\mathrm{/}2)\sin ((D-h)\mathrm{/}2)$. With $\mathrm{sinc}(u)=\sin u\mathrm{/}u$ for $u\mathrm{\ne }0$ and $\mathrm{sinc}(0)=1$, it follows that $(D^2-h^2)\mathrm{/}(2st)=(1-\cos \theta )\mathrm{sinc}(s)\mathrm{sinc}(t)\mathrm{/}(\mathrm{sinc}((D+h)\mathrm{/}2)\mathrm{sinc}((D-h)\mathrm{/}2))$; when $D=h$, both sides are zero. Since $h\le D\le s+t$, The limit of sin x divided by x at zero is one makes the ratio tend uniformly to $1$ as $\max (s,t)\to 0$, with no restriction on $s\mathrm{/}t$. The Euclidean comparison cosine $1-(D^2-h^2)\mathrm{/}(2st)$ thus tends to $\cos \theta$. In a CAT(1) triangle, CAT comparison bounds distances of arbitrarily short initial side points by these spherical model distances, hence bounds their Euclidean comparison angles by the model Euclidean angles. Taking the defining upper limit proves that each actual upper angle is at most its spherical model angle. The Euclidean case is the same cosine argument with the model angle constant.

Local triangles for (iii). Apply a finite cover and the Lebesgue-number property of the compact square to the continuous map $c(s,t)=c_s(t)$. Choose partitions $s_i,t_j$ so that the image of each closed small rectangle is contained in a CAT($\kappa$) ball $U_{ij}$ of radius $<D_{\kappa }\mathrm{/}4$ when $\kappa =1$, with its closure inside the corresponding local CAT chart. Such balls are geodesically convex: their intrinsic short geodesics are ambient minimizing segments, and distance from the center is convex in the Euclidean case and balls of radius $<\pi \mathrm{/}2$ are convex in the spherical case by comparison with the convex model ball [F3]. Thus all restrictions of the radial geodesics and the last-row base segments remain in $U_{ij}$, and the joining geodesics chosen there are ambient geodesics. Put $L_j=c_{s_{i-1}}(t_j)$ and $R_j=c_{s_i}(t_j)$. For each strip take the triangle $(p,L_1,R_1)$ and, for $1\le j<n$, the two triangles $(L_j,R_j,R_{j+1})$ and $(L_j,L_{j+1},R_{j+1})$, choosing the horizontal and diagonal edges within the corresponding ball. Each of these $2n-1$ triangles lies in its local CAT ball and has comparison angles dominating its angles: apply the CAT inequality to arbitrarily short initial side points and take the upper-angle definition; the required uniform two-variable model-angle limit is established in step 1.4. In the Euclidean case the comparison angle is constant on short initial subsegments; in the spherical case that limit transfers the same bound to the Euclidean upper-angle convention. Coincident vertices contribute no angle obligation; collinear comparison triangles have angles $0,0,\pi$ and the same limiting inequalities apply.

Straightening construction. Extend the segment $[B,C]$ beyond $C$ by length $y$ to $B_1^{\mathrm{\prime }}$. Step 1.1 gives $u+y<\pi$ in the spherical case, so this extension remains a minimal arc; in both models $C\in [B,B_1^{\mathrm{\prime }}]$, $d(B,B_1^{\mathrm{\prime }})=u+y$, and $\mathrm{\angle }AC{B}_1^{\mathrm{\prime }}=\pi -\gamma$.

Perimeter bounds. The perimeter of each of ${\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2$ is at most the perimeter of $\mathrm{\Delta }$, which is $<2D_{\kappa }$, so the comparison triangles ${\bar{\mathrm{\Delta }}}_1,{\bar{\mathrm{\Delta }}}_2$ exist by [F3]; moreover $2d(q_0,q_1)\le d(q_0,q_1)+d(q_1,p)+d(p,q_0)$ and $2d(p,r)\le d(p,q_0)+d(p,q_1)+d(q_0,q_1)$ by the reverse triangle inequality applied to the two ends of $[q_0,q_1]$, so $d(q_0,q_1)<D_{\kappa }$ and $d(p,r)<D_{\kappa }$; in particular $d(\bar{r},{\bar{q}}_0)+d(\bar{r},{\bar{q}}_1)=d(q_0,q_1)<D_{\kappa }$ and $d(\bar{p},\bar{r})=d(p,r)<D_{\kappa }$, and the four boundary distances in (i) sum to the perimeter of $\mathrm{\Delta }$, $<2D_{\kappa }$, as required.

Angle triangle inequality and the splitting point. For three geodesics issuing from $O$, write $a,b$ for the upper angles between the first and middle, and middle and last. If $a+b\ge \pi$ their angle triangle inequality is automatic. Otherwise take $a_0>a$, $b_0>b$ with $a_0+b_0<\pi$. For sufficiently short endpoints at distances $s,t$ on the first and last geodesics, use the point at distance $v=st\sin (a_0+b_0)\mathrm{/}(s\sin a_0+t\sin b_0)$ on the middle geodesic. This tends uniformly to zero with $\max (s,t)$, and the definition of upper angle bounds the two distances to that point by the Euclidean sides for included angles $a_0,b_0$. In the Euclidean triangle formed by rays at angles $a_0,b_0$ to their middle ray, that value of $v$ is the intersection of the middle ray with the side joining the endpoints; hence the sum of those two bounds is $\sqrt{s^2+t^2-2st\cos (a_0+b_0)}$. The metric triangle inequality and the law of cosines show that the comparison angle between the first and last endpoints is at most $a_0+b_0$. Taking the defining upper limit and then $a_0\downarrow a$, $b_0\downarrow b$ proves the angle triangle inequality. At $r$, the first and last segments are the opposite halves of $[q_0,q_1]$, of angle $\pi$, so the angles of ${\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2$ there sum to at least $\pi$; their comparison angles therefore also sum to at least $\pi$. At $p$ the same triangle inequality bounds the angle of $\mathrm{\Delta }$ by the sum of the two subtriangle angles.

The comparison step. In both models $d(A,B_1^{\mathrm{\prime }})$ is the distance determined by the two sides $x=d(A,C)$ and $y=d(C,B_1^{\mathrm{\prime }})$ and the included angle $\mathrm{\angle }AC{B}_1^{\mathrm{\prime }}=\pi -\gamma$ at $C$, while $d(A,B^{\mathrm{\prime }})$ is determined by the same two sides and the included angle ${\gamma }^{\mathrm{\prime }}=\mathrm{\angle }AC{B}^{\mathrm{\prime }}$; since $\gamma +{\gamma }^{\mathrm{\prime }}\ge \pi$ gives $\pi -\gamma \le {\gamma }^{\mathrm{\prime }}$, the strict monotonicity of the third side in the included angle from [F2] yields $d(A,B_1^{\mathrm{\prime }})\le d(A,B^{\mathrm{\prime }})=q$, with equality if and only if $\gamma +{\gamma }^{\mathrm{\prime }}=\pi$.

Proof of (1). By the triangle inequality and step 2.2, $d(B,A)+d(A,B_1^{\mathrm{\prime }})\ge d(B,B_1^{\mathrm{\prime }})=u+y$, and step 3.1 (together with $d(B,A)=p$) gives $d(B,A)+d(B^{\mathrm{\prime }},A)\ge d(B,A)+d(A,B_1^{\mathrm{\prime }})\ge d(B,C)+d(C,B^{\mathrm{\prime }})$, which is (1).

Proof of (2): the distance assertion. Let ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}$ be as in the Statement and put $x^{\mathrm{\prime }\mathrm{\prime }}:=d(\bar{A},\bar{C})$. The triangle $(A,B,B_1^{\mathrm{\prime }})$ of steps 2.2 and 1.2 has $\mathrm{\mid }AB\mathrm{\mid }=p$, $\mathrm{\mid }B{B}_1^{\mathrm{\prime }}\mathrm{\mid }=u+y$, $\mathrm{\mid }A{B}_1^{\mathrm{\prime }}\mathrm{\mid }=d(A,B_1^{\mathrm{\prime }})\le q$, and its marked point $C\in [B,B_1^{\mathrm{\prime }}]$ has $\mathrm{\mid }BC\mathrm{\mid }=u$; the triangle $(\bar{B},{\bar{B}}^{\mathrm{\prime }},\bar{A})$ has $\mathrm{\mid }\bar{B}{\bar{B}}^{\mathrm{\prime }}\mathrm{\mid }=u+y$, $\mathrm{\mid }\bar{B}\bar{A}\mathrm{\mid }=p$, $\mathrm{\mid }{\bar{B}}^{\mathrm{\prime }}\bar{A}\mathrm{\mid }=q$, and its marked point $\bar{C}\in [\bar{B},{\bar{B}}^{\mathrm{\prime }}]$ has $\mathrm{\mid }\bar{B}\bar{C}\mathrm{\mid }=u$. Both satisfy the hypotheses of step 1.2 with the same $s=u+y$ and $p$, the second with the larger opposite side, so $d(\bar{A},\bar{C})\ge d(A,C)=x$, with equality if and only if $d(A,B_1^{\mathrm{\prime }})=q$, that is, if and only if $\gamma +{\gamma }^{\mathrm{\prime }}=\pi$ by step 3.1.

Proof of (2): the angle at ${\bar{B}}^{\mathrm{\prime }}$. The triangles $(\bar{A},\bar{C},{\bar{B}}^{\mathrm{\prime }})$ and $(A,C,B^{\mathrm{\prime }})$ have two side lengths in common, $y=d(\bar{C},{\bar{B}}^{\mathrm{\prime }})=d(C,B^{\mathrm{\prime }})$ and $q=d(\bar{A},{\bar{B}}^{\mathrm{\prime }})=d(A,B^{\mathrm{\prime }})$, and their third sides are $x^{\mathrm{\prime }\mathrm{\prime }}$ and $x$; the angles ${\bar{\beta }}^{\mathrm{\prime }}$ and ${\beta }^{\mathrm{\prime }}$ lie opposite those third sides, so the strict monotonicity of the angle in the opposite side from [F2] gives ${\bar{\beta }}^{\mathrm{\prime }}\ge {\beta }^{\mathrm{\prime }}$, with strict inequality exactly when $x^{\mathrm{\prime }\mathrm{\prime }}>x$.

Proof of (2): the angle at $\bar{A}$. By step 1.1 the angle between $AB$ and $A{B}^{\mathrm{\prime }}$ is $\alpha +{\alpha }^{\mathrm{\prime }}\le \pi$ in both models. The triangles $(A,B,B^{\mathrm{\prime }})$ and ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}$ have the same two adjacent sides $p,q$; their opposite sides are $d(B,B^{\mathrm{\prime }})$ and $s=u+y$. Since $d(B,B^{\mathrm{\prime }})\le s$, the law of cosines and its strict monotonicity give $\bar{\alpha }\ge \alpha +{\alpha }^{\mathrm{\prime }}$. Equality holds exactly when $d(B,B^{\mathrm{\prime }})=u+y$, equivalently $C\in [B,B^{\mathrm{\prime }}]$, equivalently $\gamma +{\gamma }^{\mathrm{\prime }}=\pi$. Together with steps 4.2 and 5.1 this proves all three comparisons and simultaneous equality.

Gluing and straightening. If one comparison triangle is collinear, use the closed cosine inequalities of (i), obtained from noncollinear model triangles by continuity of their side-length formulas; these inequalities remain valid when an included angle is $0$ or $\pi$, and no continuity of angles in the metric space is being assumed. Glue ${\bar{\mathrm{\Delta }}}_1$ and ${\bar{\mathrm{\Delta }}}_2$ along their common side $[\bar{p},\bar{r}]$ so that ${\bar{q}}_0$ and ${\bar{q}}_1$ lie on opposite sides of the line through $\bar{p}$ and $\bar{r}$, which is possible because comparison triangles are unique up to isometry [F3]. In the glued configuration the four points $\bar{p},\bar{r},{\bar{q}}_0,{\bar{q}}_1$ are distinct, the points ${\bar{q}}_0,{\bar{q}}_1$ lie on opposite sides of the line through $\bar{p}$ and $\bar{r}$, the angles at $\bar{r}$ sum to at least $\pi$ by step 2.4, and the spherical perimeter hypothesis of clause (i) holds by step 2.3. Applying clause (i), whose clause (2) is established in steps 4.2, 5.1 and 6.1, with $(A,C,B,B^{\mathrm{\prime }})=(\bar{p},\bar{r},{\bar{q}}_0,{\bar{q}}_1)$ and then with ${\bar{q}}_0$ and ${\bar{q}}_1$ interchanged produces the comparison triangle ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}=\mathrm{\Delta }(\bar{p},{\bar{q}}_0,{\bar{q}}_1)$ of $\mathrm{\Delta }$ together with the point $\bar{C}\in [{\bar{q}}_0,{\bar{q}}_1]$ with $d({\bar{q}}_0,\bar{C})=d(q_0,r)$, and clause (2) gives: the angles of ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}$ at ${\bar{q}}_0$ and at ${\bar{q}}_1$ are at least the comparison angles of ${\bar{\mathrm{\Delta }}}_1,{\bar{\mathrm{\Delta }}}_2$ at ${\bar{q}}_0,{\bar{q}}_1$; the angle of ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}$ at $\bar{p}$ is at least the sum of those comparison angles at $\bar{p}$; and $d(\bar{p},\bar{C})\ge d(p,r)$, $\bar{C}$ being the point at distance $d(q_0,r)$ from ${\bar{q}}_0$ on the side $[{\bar{q}}_0,{\bar{q}}_1]$.

Conclusion of (ii). By step 7.1 the angles of the comparison triangle ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}$ of $\mathrm{\Delta }$ at ${\bar{q}}_0,{\bar{q}}_1$ dominate the comparison angles of ${\bar{\mathrm{\Delta }}}_1,{\bar{\mathrm{\Delta }}}_2$ at those vertices, which dominate the angles of ${\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2$ there by the hypothesis of (ii), and the angle of ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}$ at $\bar{p}$ dominates the sum of the comparison angles at $\bar{p}$, which dominates the angle of $\mathrm{\Delta }$ at $p$; hence every vertex angle of the comparison triangle ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}$ is at least the corresponding angle of $\mathrm{\Delta }$.

Assembling the strips. Starting with $(p,L_1,R_1)$, glue $(L_j,R_j,R_{j+1})$ along $L_j{R}_j$ and then $(L_j,L_{j+1},R_{j+1})$ along $L_j{R}_{j+1}$. The splitting points are respectively $R_j\in [p,R_{j+1}]$ and $L_j\in [p,L_{j+1}]$, so (ii) successively gives the angle comparisons for $(p,L_j,R_{j+1})$ and $(p,L_{j+1},R_{j+1})$. The perimeter of each such intermediate triangle is at most the full strip perimeter: bound its cross side by the remaining lengths on its two radial sides plus its final base side. The full strip perimeter is at most that of $\mathrm{\Delta }$, by bounding both radial endpoint distances through the appropriate endpoints of the base. Thus every application has perimeter $<2D_{\kappa }$. Repeated or collinear triangles are removed or treated with the degenerate comparisons just described, and the same cosine inequalities extend to degenerate model triangles by continuity of the side-length formulas. Finally join the strips successively along $[p,\gamma (s_i)]$; now the splitting point is on the geodesic base and again each intermediate perimeter is at most the perimeter of $\mathrm{\Delta }$. This proves (iii), and the last sentence follows by the same finite induction whenever its stated intermediate-triangle conditions hold.