Berestovskii's cone criterion and the polyhedral link criterion

Cone geodesics and angular projections. The cone metric and the sector geodesics of [F1] apply to any truncated metric: their triangle-inequality proof uses only the triangle inequality and diameter bound of $d_{\pi }$. A minimizing segment from the apex is radial: for a point $(t,z)$ on it, equality $t+d_C((t,z),(r,x))=r$ and the cone formula force $z=x$ when $t>0$. On a minimizing cone segment from $(r,x)$ to $(s,y)$, $r,s>0$, a point $(t,z)$ satisfies equality in the triangle inequality. If $\alpha =d_{\pi }(x,z)$ and $\beta =d_{\pi }(z,y)$ have sum greater than $\pi$ and $t>0$, the strict cone triangle inequality proved in [F1] excludes equality. Otherwise develop the radii in the plane at angles $0,\alpha ,\alpha +\beta$: the Euclidean triangle inequality and the decreasing cosine give the cone triangle inequality, and equality forces the developed middle point onto the straight endpoint segment. If $\gamma =d_{\pi }(x,y)<\pi$, equality also forces $\alpha +\beta =\gamma$, and the segment avoids the apex; its polar angle varies continuously and monotonically from $0$ to $\gamma$ (unless $\gamma =0$, when it is radial). The angular projection, parametrised by that angle, is a minimizing segment in $(L,d_{\pi })$: apply the same equality argument to every subsegment. If $\gamma =\pi$, the developed segment is a diameter, so every nonapex point has direction $x$ or $y$ and the segment is the through-apex path. Assume now $C(L)$ is CAT(0). Its geodesics exist and are unique by [F5], so this argument gives existence of short angular segments; their uniqueness follows since two such segments would, by sector development, give two cone geodesics between the same positive-radius endpoints.

Forward comparison with the actual chord radii. Let a triangle of short angular segments in $L$ have perimeter $<2\pi$, and choose its spherical comparison triangle $({\bar{y}}_1,{\bar{y}}_2,{\bar{y}}_3)$. Take cone vertices $P_i=s{y}_i$, $s>0$, and Euclidean vertices ${\bar{P}}_i=s{\bar{y}}_i\in {\mathbb{R}}^3$. Their pairwise distances agree by the cosine formula, so their affine plane is a Euclidean comparison triangle. If $y$ lies on an angular side, let $P=ty$ be the point on the corresponding cone geodesic whose angular projection is $y$, as supplied by step 1.1. Its radius $t$ is the radius of the point $\bar{P}=t\bar{y}$ on the corresponding straight chord: both sectors have the same opening angle, endpoint radii and polar position. In particular $P$ and $\bar{P}$ have the same side parameter; generally $t\mathrm{\ne }s$. For any two points $y,y^{\mathrm{\prime }}$ on angular sides, use the corresponding chord points $P=ty$, $P^{\mathrm{\prime }}=t^{\mathrm{\prime }}{y}^{\mathrm{\prime }}$ and their comparison points. CAT(0) gives $t^2+t^{\mathrm{\prime }2}-2t{t}^{\mathrm{\prime }}\cos d_{\pi }(y,y^{\mathrm{\prime }})\le \mathrm{\mid }t\bar{y}-t^{\mathrm{\prime }}{\bar{y}}^{\mathrm{\prime }}{\mathrm{\mid }}^2=t^2+t^{\mathrm{\prime }2}-2t{t}^{\mathrm{\prime }}\cos d_S(\bar{y},{\bar{y}}^{\mathrm{\prime }})$. Since $t,t^{\mathrm{\prime }}>0$, decreasing cosine on $[0,\pi ]$ yields $d_{\pi }(y,y^{\mathrm{\prime }})\le d_S(\bar{y},{\bar{y}}^{\mathrm{\prime }})$. With step 1.1 this proves CAT(1).

Reverse comparison, short angular perimeter. Assume $(L,d_{\pi })$ is CAT(1). Short angular segments exist and are unique (the CAT(1) inequality on the degenerate triangle made from two competing short segments forces their equal-parameter points to coincide). Sector development and through-apex paths give cone geodesics, and step 1.1 classifies all of them. To prove CAT(0), use the squared vertex-to-side criterion of Comparison triangles in the Euclidean plane and the round sphere, model spaces, and CAT(0) and CAT(1) consequences (iv)(c). Fix a vertex $P_1=a{y}_1$, and a point $Q$ at fraction $v\in [0,1]$ of the side from $P_2=r{y}_2$ to $P_3=s{y}_3$. If $a=0$, the criterion is equality by the developed chord norm. If $r=0$ or $s=0$, the side is radial and the same equality follows by expanding the cone formula. Otherwise put $\alpha =d_{\pi }(y_1,y_2)$, $\beta =d_{\pi }(y_1,y_3)$ and $\gamma =d_{\pi }(y_2,y_3)$. For $\gamma =0$ the side is radial and again equality holds. Suppose $0<\gamma <\pi$ and write $Q=ty$ with angular position $\tau =d_{\pi }(y_2,y)$. In its developed sector, $A:=\frac{(1-v)r}{t}=\frac{\sin (\gamma -\tau )}{\sin \gamma },B:=\frac{vs}{t}=\frac{\sin \tau }{\sin \gamma }\mathrm{.}$ If $\alpha +\beta +\gamma <2\pi$, the spherical comparison inequality gives $\cos d_{\pi }(y_1,y)\ge A\cos \alpha +B\cos \beta$, since the comparison direction at position $\tau$ is $A{\bar{y}}_2+B{\bar{y}}_3$. Expanding the cone formula therefore gives $d_C(P_1,Q{)}^2\le (1-v)d_C(P_1,P_2{)}^2+v{d}_C(P_1,P_3{)}^2-v(1-v)d_C(P_2,P_3{)}^2\mathrm{.}$

Reverse comparison, large angular perimeter. Retain $0<\gamma <\pi$ but suppose $\alpha +\beta +\gamma \ge 2\pi$. Put $g(\tau )=A\cos \alpha +B\cos \beta$, $l=\pi -\alpha$ and $u=\beta +\gamma -\pi$, so $0\le l\le u\le \gamma$. Since $\beta \ge 2\pi -\alpha -\gamma$ and $\alpha +\gamma \in [\pi ,2\pi ]$, reflection $\cos (\alpha +\gamma )=\cos (2\pi -\alpha -\gamma )$ and decreasing cosine on $[0,\pi ]$ give $\cos \beta \le \cos (\alpha +\gamma )$; the addition formulas consequently give $g(\tau )\le \cos (\alpha +\tau )$ on $[0,l]$. Likewise $g(\tau )\le \cos (\beta +\gamma -\tau )$ on $[u,\gamma ]$. In particular $g(l),g(u)\le -1$. If $l<u$, sine interpolation on this interval (length $u-l<\pi$) gives $g(\tau )=\frac{\sin (u-\tau )}{\sin (u-l)}g(l)+\frac{\sin (\tau -l)}{\sin (u-l)}g(u)\le -1(l\le \tau \le u),$ because the coefficients are nonnegative and their sum is $\cos (\tau -(l+u)\mathrm{/}2)\mathrm{/}\cos ((u-l)\mathrm{/}2)\ge 1$; if $l=u$ use $g(l)\le -1$. The triangle inequality bounds $d_{\pi }(y_1,y)\le \min \{\pi ,\alpha +\tau ,\beta +\gamma -\tau \}$. Thus on the two outer intervals its cosine is at least the corresponding cosine above, and on the middle interval it is at least $-1\ge g(\tau )$. Hence $\cos d_{\pi }(y_1,y)\ge g(\tau )$ everywhere, and the expansion in step 2.2 proves the same squared comparison inequality.

Reverse comparison, antipodal side. Suppose $\gamma =\pi$. Step 1.1 makes the side the through-apex path and the triangle inequality gives $\alpha +\beta \ge \pi$. Thus $\cos \alpha +\cos \beta =2\cos ((\alpha +\beta )\mathrm{/}2)\cos ((\alpha -\beta )\mathrm{/}2)\le 0$. If $Q$ is on the $y_2$ branch, its radius is $t=r-v(r+s)\ge 0$. The difference between the right side of the squared inequality in step 2.2 and $d_C(P_1,Q{)}^2$ is exactly $-2avs(\cos \alpha +\cos \beta )\ge 0$, by expansion. On the $y_3$ branch it is $-2a(1-v)r(\cos \alpha +\cos \beta )\ge 0$. Both include the apex. Together with the preceding two paragraphs, this case covers every side and every vertex; the vertex-to-side criterion [F3] therefore proves CAT(0), completing the reverse implication.

Cone equivalence. Steps 1.1 and 2.1 prove the forward implication and steps 2.2, 3.1 and 4.1 the reverse implication, with the empty cone a single point. This proves (i) in full.

Local charts and cone dilation. Put $Y={\mathbb{R}}^k\times C({\mathrm{Lk}}_X(F))$. By [F1], a small ball at $p$ is isometric to a ball at $(0,o)$ in $Y$. If $Y$ is CAT(0), its convex balls are CAT(0). Conversely if a ball about $(0,o)$ is CAT(0), the maps $(u,(r,x))\mapsto (\lambda u,(\lambda r,x))$, $\lambda >0$, are bijective similarities of $Y$, since both coordinate metrics scale by $\lambda$. Any two points can be scaled into that ball, joined there and scaled back, so $Y$ is geodesic; any chosen geodesic triangle is bounded and can likewise be scaled wholly into the ball, where its CAT(0) inequality holds, and scaled back. Thus local CAT(0) at the origin is equivalent to CAT(0) of $Y$. A square-sum product is CAT(0) when both factors are, by adding their squared vertex-to-side inequalities; conversely the slices of each factor are convex (a minimizing product segment with equal endpoints in one coordinate must keep that coordinate constant), so CAT(0) of the product implies CAT(0) of each factor. Since ${\mathbb{R}}^k$ is CAT(0), (i) now proves that $X$ is locally CAT(0) at $p$ exactly when its normal face link is CAT(1).

Vertex reduction. Necessity follows from step 6.1 with $F=\{v\}$. Conversely suppose every vertex link is CAT(1); then its entire cone is CAT(0) by step 5.1. For any point $p\in \mathrm{relint}F$, choose a vertex $v$ of $F$. In the tangent-cone gluing at $v$, the ray representing the cell segment from $v$ to $p$ has a point $q$ at small positive radius. Its incident cells are exactly those containing $F$: in every incident cell the relative interior of the ray is the relative interior of the tangent face corresponding to $F$. The active facet inequalities at $q$ are therefore precisely those containing $F$, the same as at $p$, and their Euclidean linear parts and gluing maps agree. Thus the tangent-cone gluing at $q$ is isometric to that at $p$. The finite-facet proof of the local chart [F4] applies equally to the finitely many incident polyhedral cones, so it identifies sufficiently small balls at both points with balls in this same tangent gluing (choose radii below the finitely many inactive facet distances, as in [F1]). Since the cone at $v$ is CAT(0), its small convex ball at $q$ is CAT(0), so the corresponding ball at $p$ is CAT(0). Hence $X$ is locally CAT(0) everywhere, proving the vertex form.