The spherical radius estimate, the quadrilateral separation constant, and the finite midpoint-operation comparison disk

Radius estimate (i). Let $x$ be a point of $\gamma$; it lies on one of the two subarcs between $y$ and $z$, so writing $b:=d(x,y)$, $d:=d(x,z)$, $c:=d(y,z)$ we have $b+d\le r$ (the two pieces of that subarc dominate the two distances), $\mathrm{\mid }b-d\mathrm{\mid }\le c$ (triangle inequality) and $c\le r<\pi$. The triangle $(x,y,z)$ of $Z$ has perimeter $b+d+c\le 2r<2\pi$, so a comparison triangle exists in the round sphere $S^2$ of [F7] and the CAT(1) inequality of [F1] applied to the pair $(a,x)$ gives $d(a,x)\le d_S(\bar{a},\bar{x})$, where $\bar{a}$ is the comparison point of the midpoint $a$ of $[y,z]$. The midpoint identity of [F2] gives $\cos d_S(\bar{a},\bar{x})=(\cos b+\cos d)\mathrm{/}(2\cos (c\mathrm{/}2))$, and the addition formula ([F8]) turns this into $\cos ((b+d)\mathrm{/}2)\cos ((b-d)\mathrm{/}2)\mathrm{/}\cos (c\mathrm{/}2)$; since $\mathrm{\mid }b-d\mathrm{\mid }\le c$ and $\cos$ decreases on $[0,\pi ]$ with $\cos (c\mathrm{/}2)>0$, this is at least $\cos ((b+d)\mathrm{/}2)$, which is at least $\cos (r\mathrm{/}2)$ because $b+d\le r$ and $r\mathrm{/}2<\pi$. Hence $\cos d(a,x)\ge \cos (r\mathrm{/}2)$ with both arguments in $[0,\pi ]$, so $d(a,x)\le r\mathrm{/}2$: the image of $\gamma$ is contained in the closed ball of radius $r\mathrm{/}2$ about $a$, and $r\mathrm{/}2=\pi \mathrm{/}2-\eta$ for $\eta =(\pi -r)\mathrm{/}2$.

Quadrilateral separation (ii). Let $a,x,y,z$ satisfy the hypotheses, and consider the two comparison triangles ${\bar{T}}_y=(\bar{a},\bar{x},\bar{y})$ and ${\bar{T}}_z=(\bar{a},\bar{x},\bar{z})$ of the triangles $(a,x,y)$ and $(a,x,z)$ in $S^2$; both are admissible because their perimeters are at most $2d(a,x)+\mu \le 2(\pi \mathrm{/}2-\eta )+\mu <2\pi$. Glue them along the common side $[\bar{a},\bar{x}]$ on opposite sides, obtaining an abstract spherical quadrilateral $Q$ (two triangles joined along the isometric side $[\bar{a},\bar{x}]$) with intrinsic distance $d_Q$; the corresponding boundary maps agree on the common side because the geodesic $[a,x]$ is unique ($d(a,x)<\pi$, [F3]); comparison on each triangle boundary, followed by the triangle inequality at the crossings of the common side, gives $d(y,z)\le d_Q(\bar{y},\bar{z})$. Write $\vartheta$ for the angle at $\bar{x}$ between $[\bar{x},\bar{a}]$ and $[\bar{x},\bar{y}]$ inside ${\bar{T}}_y$ and ${\vartheta }^{\mathrm{\prime }}$ analogously in ${\bar{T}}_z$; by the cosine rule and $d(a,y)\le d(a,x)$, $\cos \vartheta =(\cos d(a,y)-\cos d(a,x)\cos \mu )\mathrm{/}(\sin d(a,x)\sin \mu )\ge (\cos d(a,x)-\cos d(a,x)\cos \mu )\mathrm{/}(\sin d(a,x)\sin \mu )=\cot d(a,x)\tan (\mu \mathrm{/}2)$, and since $d(a,x)\le \pi \mathrm{/}2-\eta$ gives $\cot d(a,x)\ge \tan \eta$, we get $\cos \vartheta \ge \tan \eta \tan (\mu \mathrm{/}2)$ and likewise $\cos {\vartheta }^{\mathrm{\prime }}\ge \tan \eta \tan (\mu \mathrm{/}2)$; as the right-hand side is positive and $\mu >0$, both $\vartheta ,{\vartheta }^{\mathrm{\prime }}$ are strictly less than $\pi \mathrm{/}2$, so with $\psi :=\arcsin (\tan \eta \tan (\mu \mathrm{/}2))>0$ the angle $\mathrm{\Phi }:=\vartheta +{\vartheta }^{\mathrm{\prime }}$ satisfies $\mathrm{\Phi }\le \pi -2\psi <\pi$. Put $q:=\min \{1\mathrm{/}2,\tan \eta \tan (\mu \mathrm{/}2)\}$ and $\psi :=\arcsin q$, so the preceding angle bounds still give $\mathrm{\Phi }\le \pi -2\psi$. The triangle inequality gives $\mu =d(x,y)\le d(a,x)+d(a,y)\le 2d(a,x)$, hence the common side has length at least $\mu \mathrm{/}2$. Cut both outgoing edges at distance $e:=\mu \mathrm{/}4$ from $\bar{x}$. The short spherical chord joining the cutpoints lies in the wedge of angle $\mathrm{\Phi }<\pi$ and in the convex closed ball $\bar{B}(\bar{x},e)$, since $e<\pi \mathrm{/}4$. The other two edges are outside this small ball: for $p\in [\bar{a},\bar{y}]$, writing $t=d(a,x)$, the triangle inequality gives $d_S(\bar{x},p)\ge \max \{\mu -d_S(p,\bar{y}),t-d_S(p,\bar{a})\}\ge (\mu +t-d(a,y))\mathrm{/}2\ge \mu \mathrm{/}2$, and likewise for $[\bar{a},\bar{z}]$. It meets the common radial side before its far endpoint (at distance at most $e<\mu \mathrm{/}2$); each half lies in the corresponding convex spherical triangle. Thus this chord really is a path in $Q$, irrespective of whether the chord joining $\bar{y},\bar{z}$ lies in $Q$. Its length is at most ${\mathrm{\ell }}_e:=\arccos ({\cos }^{2}e+{\sin }^{2}e\cos (\pi -2\psi ))<2e$. Adding the two remaining edge pieces gives $d_Q(\bar{y},\bar{z})\le 2(\mu -e)+{\mathrm{\ell }}_e=2\mu -\delta (\eta ,\mu )$, with $\delta (\eta ,\mu ):=2e-{\mathrm{\ell }}_e>0$. This formula is continuous on the whole stated parameter domain, including parameter values for which no configuration exists; the clamp in $q$ keeps the inverse sine defined there. Degenerate comparison triangles, with a zero angle at $\bar{x}$, give the same chord shortcut directly; equality of radial distances was already included in the cosine bound; $d(a,x)=0$ would give $\mu =d(x,y)\le d(x,a)+d(a,y)\le 0$, contradicting $\mu >0$.

The finite disk and its face bounds (iii)(a). In the Euclidean plane take a regular $n$-gon $P_0$, and let $P_{i+1}$ be its midpoint polygon; it is a regular $n$-gon, rotated through $\pi \mathrm{/}n$ and scaled by $\cos (\pi \mathrm{/}n)>0$. The difference between $P_i$ and $P_{i+1}$ consists of the $n$ ears $T(i,j)$ with principal vertices $v(i,j),v(i+1,j-1),v(i+1,j)$; their radial sides meet at the row-$i$ vertices, while each row-$i$ side is subdivided at $v(i+1,j)$. The ears for $0\le i<m$ and the $n-2$ triangles $T(m,j)=(v(m,1),v(m,j-1),v(m,j))$, $3\le j\le n$, of a diagonal fan triangulate the closed disk $P_0$. Thus the prescribed edge identifications yield a topological disk, with boundary vertices $V(0)\cup V(1)$ and interior vertices $V(2)\cup \cdots \cup V(m)$; the row-$i$ polygon is the boundary of the remaining disk after the first $i$ ear layers are removed. Assigning spherical metrics leaves this topology unchanged, since the strict triangle inequalities make each face a nondegenerate closed triangle. The ear sides are ${\xi }_{j-1}^{(i)}\mathrm{/}2$, ${\xi }_j^{(i)}\mathrm{/}2$ and $d(v(i+1,j-1),v(i+1,j))$ in $X$, so by [F4] their perimeter is at most ${\xi }_{j-1}^{(i)}+{\xi }_j^{(i)}\le 2h<2l<\pi$. For a cap triangle, the three consecutive subarcs of the last-row loop between its three assigned vertices have total length $L(f^{m}x)<2l$ and dominate its three distances; hence its perimeter is also $<2l<\pi$. Every face is therefore contained in an open spherical hemisphere and has angles $<\pi$.

The skeleton map dominates the ambient distance (iii)(b). Map each edge at constant speed to the corresponding unique short geodesic of $X$. This is consistent with subdivision: the row-$(i+1)$ vertex on a row-$i$ side is its assigned midpoint by [F4]. Each ear has all assigned vertex distances $\le h<l$, and each cap face has all such distances $\le L(f^{m}x)\mathrm{/}2<l$. Thus each assigned triple lies in a radius-$l$ CAT(1) ball; its three geodesic sides lie there by comparison convexity. The CAT(1) inequality gives $d(gu,gv)\le d_T(u,v)$ for any two points of the boundary of its spherical comparison face $T$. The intrinsic path metric of the finite complex is the infimum of lengths of chains of face segments; for skeleton endpoints, each segment of such a chain begins and ends on the boundary of its face (split at successive face crossings). Applying the boundary comparison to these segments and the triangle inequality in $X$ gives $d(gu,gv)\le$ the chain length; the infimum gives $d(gu,gv)\le \rho (u,v)$. No extension of $g$ to face interiors is needed.

Shortest paths through a vertex span at least $\pi$ on each side (cone lemma). Let $W$ be a polyhedral surface with an interior vertex $v$ at which the total cone angle is $\mathrm{\Theta }$, let $u_1,u_2$ be points on two edges issuing from $v$ at positive distance from $v$, and suppose the concatenation of the two segments $[u_1,v]$, $[v,u_2]$ is a shortest path from $u_1$ to $u_2$ in $W$. Then the two sectors of $W$ at $v$ cut out by the segments have angles at least $\pi$, so $\mathrm{\Theta }\ge 2\pi$. Indeed, if a sector had angle $\phi <\pi$, then in that sector (a sufficiently small spherical wedge of angle $<\pi$, hence convex) the two points at equal small distance $\epsilon$ from $v$ on the two edges would be joined by a path of length strictly less than $2\epsilon$: the spherical cosine rule gives $\cos \mathrm{\ell }={\cos }^2\epsilon +{\sin }^2\epsilon \cos \phi >\cos (2\epsilon )$ for $\phi <\pi$, so $\mathrm{\ell }<2\epsilon$, and adding the remaining pieces gives a strictly shorter path, contradicting minimality. At a boundary vertex the same shortcut applies to its one available disk-side sector and forces that sector to have angle at least $\pi$; no second sector or $2\pi$ bound is asserted there.

Ear removal. Let $D_i$ be the closed disk remaining after removal of ear layers $0,\dots ,i-1$, with $D_0=D$. Suppose a simple closed local geodesic $\gamma$ of $D$ lies in $D_i$, $i<m$. It is also locally shortest among paths in $D_i$. At a tip $v(i,j)$ the only incident face of $D_i$ is the ear $T(i,j)$, whose angle is $<\pi$; the shortcut argument of step 1.5 excludes passage through that tip. At an interior point of a boundary side of $D_i$, the local space is a spherical half-disk and a local geodesic meeting that boundary must follow its great-circle side. Continuing toward the adjacent tip would reach $v(i,j)$ before any other vertex, which is impossible. Therefore $\gamma$ avoids all boundary side interiors of $D_i$. A component of its intersection with an ear interior must consequently enter and leave through that ear's base, possibly at its endpoints: the other two sides are boundary sides of $D_i$. Inside the ear the arc is a great-circle arc contained in an open hemisphere; its length is $<\pi$, so it cannot meet the same short great-circle base twice unless it coincides with that base. Nor can a whole closed great circle lie in the ear's hemisphere. Thus $\gamma$ misses the ear interiors and lies in $D_{i+1}$. Induction forces $\gamma$ into the final fan $D_m$.

Interior cone angles (iii)(c). Let $u=v(i,j)$ with $2\le i\le m$ and let $w_1:=v(i-1,j)$, $w_2:=v(i-1,j+1)$ be the endpoints of the row-$(i-1)$ edge whose midpoint is $u$; the two radial edges $[w_1,u]$ and $[u,w_2]$ are sides of the ears $T(i-1,j)$ and $T(i-1,j+1)$ and have $\rho$-length equal to the $X$-distances ${\xi }_j^{(i-1)}\mathrm{/}2=d(g{w}_1,gu)=d(gu,g{w}_2)$. Since $gu$ is the midpoint of the geodesic $[g{w}_1,g{w}_2]$, $d(g{w}_1,g{w}_2)=d(g{w}_1,gu)+d(gu,g{w}_2)=\rho (w_1,u)+\rho (u,w_2)$, while $\rho (w_1,w_2)\ge d(g{w}_1,g{w}_2)$ by (iii)(b) and $\rho (w_1,w_2)\le \rho (w_1,u)+\rho (u,w_2)$ by the triangle inequality; hence equality holds and the concatenation $[w_1,u]\cup [u,w_2]$ is a shortest path. By the cone lemma of step 1.5 the cone angle of $D$ at $u$ is at least $2\pi$. (The cap supplies the inner sector when $i=m$.)

Boundary angles (iii)(d). The boundary of $D$ is the subdivided row-$0$ polygon. At a first-row vertex $v(0,j)$ exactly the ear $T(0,j)$ is incident, so the boundary angle is the angle of its spherical comparison triangle at $v(0,j)$; the comparison triangle is nondegenerate with perimeter $<2h<2\pi$, and all angles of a nondegenerate spherical triangle of perimeter $<2\pi$ are strictly less than $\pi$ (an angle $\ge \pi$ would force the opposite side to be at least the sum of the other two by the cosine rule, contradicting strictness), so $\mathrm{\angle }(\mathrm{\partial }D,v(0,j))<\pi$. At the boundary vertex $v(1,j)$ of the last row of the first strip, the incident faces are $T(0,j)$, $T(0,j+1)$ and the faces on the remaining-disk side; the concatenation $[v(0,j),v(1,j)]\cup [v(1,j),v(0,j+1)]$ is a shortest path by the argument of step 2.2 with $w_1=v(0,j)$, $w_2=v(0,j+1)$, using only the disk-side sector of the shortcut argument, so by step 1.5 the sector of $D$ at $v(1,j)$ that lies on the disk side of the path -- which is exactly the disk's sector at its boundary vertex -- has angle at least $\pi$, i.e. $\mathrm{\angle }(\mathrm{\partial }D,v(1,j))\ge \pi$; equivalently the row-$0$ polygon is locally geodesic at $v(1,j)$.

The final fan excludes a closed local geodesic (iii)(e). Write $c=v(m,1)$. On each cap face define $R(p)$ as its spherical distance to $c$. These functions agree on fan diagonals, hence define a continuous function on $D_m$; each face lies in the convex spherical ball of radius $L(f^{m}x)\mathrm{/}2<l<\pi \mathrm{/}2$ about $c$, so $R<\pi \mathrm{/}2$. Suppose the curve forced into $D_m$ by step 2.1 exists, and let $p$ maximize $R$ on it. Its maximum is positive, so $p\mathrm{\ne }c$. If $p$ is in a face interior or a fan diagonal interior, unfold the adjacent faces into $S^2$: the common radial segment to $c$ agrees in the unfolding, and the local geodesic is a great-circle segment. At its radial maximum its two outgoing directions are perpendicular to the radial direction, so the cosine rule gives $\cos R(\gamma (t))=\cos R(p)\cos t<\cos R(p)$ for small nonzero $t$, contradicting maximality. The same argument applies at a boundary side interior if the curve follows that side. At a remaining boundary vertex $p\mathrm{\ne }c$, the one or two incident cap faces form a sector, split by the radial diagonal to $c$ when there are two. If an outgoing direction has angle $\alpha$ to the radial direction, the cosine rule reads $\cos R(\gamma (t))=\cos R(p)\cos t+\sin R(p)\sin t\cos \alpha$. Maximality implies $\cos \alpha \ge 0$, hence both outgoing directions make angles at most $\pi \mathrm{/}2$ with the radial direction. The sector between them consequently has angle at most $\pi$; local minimality requires angle at least $\pi$ by step 1.5. Equality forces both angles to be $\pi \mathrm{/}2$, and the same displayed cosine rule again increases $R$ for small positive $t$, a contradiction. Thus no simple closed local geodesic exists in $D$.

Local CAT(1) via a spherical cone chart. At a vertex let $J$ be its angular link with metric truncated at $\pi$: an interior link is a circle of circumference $\mathrm{\Theta }\ge 2\pi$, and a boundary link is an interval of its finite boundary angle. Both are CAT(1). For the circle use F2; truncation changes neither short segments nor triangles of perimeter $<2\pi$. For the interval every tested triangle lies in a subinterval of length $<\pi$ and is degenerate. Let $p$ be a one-point space. The Euclidean cone criterion Berestovskii's cone criterion and the polyhedral link criterion makes $C(J)$ CAT(0). The product $[0,\mathrm{\infty })\times C(J)$ is CAT(0), by adding the squared vertex-to-side inequalities of F2. The join–product isometry The cone and join metrics and the local product chart of a polyhedral gluing (3) identifies it with $C(\{p\}\ast J)$; the same cone criterion now makes $\{p\}\ast J$ CAT(1). Its polar metric about $p$ is $\cos d=\cos t\cos t^{\mathrm{\prime }}+\sin t\sin t^{\mathrm{\prime }}\cos d_J(u,u^{\mathrm{\prime }})$, by The angular path metric, the Euclidean cone and spherical joins. On each angular subinterval of length $<\pi$ this is exactly the spherical sector metric, and these sectors glue in the same order as the incident faces of $D$. Thus their polar charts identify a sufficiently small neighborhood of the vertex with a neighborhood of $p$ in the join, preserving lengths. This also identifies the induced distances on smaller balls: choose an outer chart radius $\sigma <\pi \mathrm{/}2$ within every incident face; a path leaving it between points of radii at most $\sigma \mathrm{/}4$ has length at least $3\sigma \mathrm{/}2$, whereas the path through the vertex has length at most $\sigma \mathrm{/}2$. Shorter paths remain in the chart. The join's smaller closed balls are convex and CAT(1), so their corresponding balls in $D$ are CAT(1). Face and edge interiors have spherical disk or half-disk charts, which have convex small CAT(1) balls. Hence $D$ is locally CAT(1), including boundary angles greater than $\pi$.

Global CAT(1) (iii)(f). The intrinsic finite spherical complex is compact, being a quotient of finitely many compact model triangles, and geodesic: a minimizing sequence of constant-speed paths has uniformly bounded speed, and compact Arzelà–Ascoli gives a uniformly convergent subsequence whose limit has length at most the infimum, by the finite-partition definition of length. This is the AC use of [F11]. Step 4.1 supplies local CAT(1), and step 3.2 excludes every simple closed local geodesic, hence every isometrically embedded circle of length $<2\pi$. The compact short-circle criterion [F10] now proves that $D$ is CAT(1).

Quantitative output (iv). Assume the situation of (iii) and (iv), and use (iii)(f), so that $(D,\rho )$ is CAT(1); put $2r=L(x)$, $\eta =(\pi -r)\mathrm{/}2$, $\mu =\min \{r\mathrm{/}(2n),\eta \}$ and $R=r\mathrm{/}2=\pi \mathrm{/}2-\eta$. Apply the radius estimate (i) to the boundary curve $\mathrm{\partial }D$, which has $\rho$-length $L(x)=2r<2\pi$; this gives a point $a\in D$ with $\mathrm{\partial }D\subseteq \bar{B}(a,\pi \mathrm{/}2-\eta )$. Choose a maximum $b$ of $\rho (a,\cdot )$ on $\mathrm{\partial }D$. The boundary lies in a ball of radius $R<\pi \mathrm{/}2$; on a short boundary segment a maximum cannot occur in its interior. Indeed, if its interior midpoint $b$ maximized the distance, choose equal small offsets $y,z$ on that segment. Its triangle with $a$ is admissible, and the midpoint cosine inequality gives $\cos \rho (a,b)\ge (\cos \rho (a,y)+\cos \rho (a,z))\mathrm{/}(2\cos (\rho (y,z)\mathrm{/}2))>\cos \rho (a,b)$, since all three radial distances are at most $R<\pi \mathrm{/}2$. Thus a maximum occurs at a boundary vertex. It cannot occur at a midpoint vertex of $V(1)$: there the two boundary edges concatenate to a local geodesic (step 3.1), and the same cosine comparison on a short segment through that vertex excludes a maximum there. Thus $b=v(0,k+1)$ for some $k$, and $\rho (a,b)\le \pi \mathrm{/}2-\eta$ and $\rho (a,y)\le \rho (a,b)$ for every point $y$ of $\mathrm{\partial }D$. Let $y_1$ and $y_2$ be the points of the boundary edges $[b,v(1,k)]$ and $[b,v(1,k+1)]$ at distance exactly $\mu$ from $b$; these exist because those edges have $\rho$-length ${\xi }_k\mathrm{/}2$ and ${\xi }_{k+1}\mathrm{/}2$, and $\mu \le r\mathrm{/}(2n)\le \xi \mathrm{/}2$ by hypothesis. Then $\rho (b,y_1)=\rho (b,y_2)=\mu$, $\rho (a,y_i)\le \rho (a,b)$ since $y_i\in \mathrm{\partial }D$, and $\mu <2\eta$; the separation estimate (ii) in the CAT(1) space $(D,\rho )$ gives $\rho (y_1,y_2)\le 2\mu -\delta (\eta ,\mu )$. Also $\rho (v(1,k),y_1)={\xi }_k\mathrm{/}2-\mu$ and $\rho (y_2,v(1,k+1))={\xi }_{k+1}\mathrm{/}2-\mu$, because on the edge $[b,v(1,k)]$ the two points $y_1$ and $v(1,k)$ lie on the same side of $b$ at the stated distances. Since $\rho \ge d\circ (g\times g)$ by (iii)(b) and $g(v(1,k))=(fx{)}_k$, $g(v(1,k+1))=(fx{)}_{k+1}$, the triangle inequality in $D$ gives ${\zeta }_k=d((fx{)}_k,(fx{)}_{k+1})\le \rho (v(1,k),y_1)+\rho (y_1,y_2)+\rho (y_2,v(1,k+1))\le ({\xi }_k+{\xi }_{k+1})\mathrm{/}2-\delta (\eta ,\mu )$, the displayed deficit, with the index $k$ adjacent to the first-row vertex $b$; the triangles to which (ii) is applied have perimeter at most $2(\pi \mathrm{/}2-\eta )+\mu <2\pi$ as required there, and longer or unequal midpoint segments are handled by taking the initial equal $\mu$-pieces and adding the remaining lengths, as stated.

Conclusion. Steps 1.1 and 1.2 give (i) and (ii); steps 1.3, 1.4, 2.2 and 3.1 give (iii)(a)–(d). Ear removal and the final-fan radial maximum argument give (iii)(e), the local link test and compact criterion give (iii)(f), and step 6.1 gives (iv). AC enters through the compactness-of-paths argument and the compact short-circle criterion.