Large Spherical Metric Flags and the Moussong Girth Theorem

The function $d((x,y),(x^{\mathrm{\prime }},y^{\mathrm{\prime }})):=(d_X(x,x^{\mathrm{\prime }}{)}^2+d_Y(y,y^{\mathrm{\prime }}{)}^2{)}^{1\mathrm{/}2}$ is a metric on $X\times Y$: symmetry and vanishing exactly on the diagonal are immediate from [F9], and the triangle inequality is the triangle inequality of the Euclidean norm on ${\mathbb{R}}^2$ applied to the vectors $(d_X(x,x^{\mathrm{\prime }}),d_Y(y,y^{\mathrm{\prime }}))$ and $(d_X(x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }}),d_Y(y^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}))$ [F9, F10].

Let $\gamma =({\gamma }_X,{\gamma }_Y)$ be a rectifiable path and put $A:=L({\gamma }_X)$, $B:=L({\gamma }_Y)$. For any $\epsilon >0$, choose partitions whose coordinate polygonal sums exceed $A-\epsilon$ and $B-\epsilon$, and take a common refinement. If $a_i,b_i$ are the two coordinate distances on its successive subintervals, then the product polygonal sum is ${\sum }_i\sqrt{a_i^2+b_i^2}\ge \sqrt{({\sum }_i{a}_i{)}^2+({\sum }_i{b}_i{)}^2}$ by the Euclidean triangle inequality [F9, F10]. Letting $\epsilon \downarrow 0$ gives $L(\gamma )\ge \sqrt{A^2+B^2}$. For endpoints with factor distances $a,b$, this is at least $D:=\sqrt{a^2+b^2}$. If both factors are geodesic, pair constant-speed factor segments, parametrized on $[0,D]$ at speeds $a\mathrm{/}D,b\mathrm{/}D$ when $D>0$, give a product geodesic because their product distance between parameters $s,t$ is $\mathrm{\mid }s-t\mathrm{\mid }$; if $D=0$ use the constant path. Conversely, if $\gamma :[0,D]\to X\times Y$ is a product geodesic, then $L(\gamma )=D$ and this bound forces $A=a$ and $B=b$. For any $t\in [0,D]$, apply the same bound to the two restrictions $[0,t]$ and $[t,D]$. Their product distances are $t$ and $D-t$, so equality must hold in the coordinate-length bounds and in the Euclidean triangle inequality for the two vectors of coordinate lengths. Thus those lengths are $(ta\mathrm{/}D,tb\mathrm{/}D)$ and $((D-t)a\mathrm{/}D,(D-t)b\mathrm{/}D)$; each coordinate distance equals its path length, so both coordinate paths are geodesics with constant proportional speeds. This proves the stated characterization when $D>0$; when $D=0$ both factors are constant.

For a metric space $L$ of diameter at most $\pi$, the formula of [F7] defines a metric on $C(L)$: the case analysis of [F8] uses only that $d_{\pi }$ is a metric of diameter at most $\pi$ (the zero-radius case reduces to the triangle inequality, the case $\alpha +\beta \le \pi$ places three points in the plane and uses monotonicity of $\cos$ on $[0,\pi ]$, and the case $\alpha +\beta >\pi$ uses the projection estimates $r-s\cos \alpha$, $t-s\cos \beta$ and $\cos \alpha +\cos \beta \le 0$), so it applies verbatim to the spaces $L_1$ and $L_2$.

The product is CAT(0) if $X$ and $Y$ are: by 1.2 its geodesic triangles have componentwise geodesic sides, and for a triangle with vertices $z,x,y$ and the point $p_t$ of $[x,y]$ at fraction $t$ the hinged criterion [F4] applied in each factor and added gives $d(z,p_t{)}^2\le (1-t)d(z,x{)}^2+td(z,y{)}^2-t(1-t)d(x,y{)}^2$, because squared product distances add; by [F4] the product is CAT(0).

Let $E$ be the cosine expression in the join formula [F7]. Its two coefficients are nonnegative and sum to $\cos (\theta -{\theta }^{\mathrm{\prime }})\le 1$, so $E\in [-1,1]$; the function $d_J=\arccos E$ descends to the endpoint quotient and separates its classes by the equality case $E=1$. The unit-radius map $\mathrm{\Psi }$ into $C(L_1)\times C(L_2)$ satisfies $D(\mathrm{\Psi }x,\mathrm{\Psi }y{)}^2=2-2\cos d_J(x,y)$, where $D$ is the product metric: this is the chord distance, not $d_J$. For arbitrary radii, the same expansion transports $D$ along the radial bijection to the cone function built from $d_J$, so that cone function is a metric. The triangle inequality for $d_J$ follows from the radius-$1,s,1$ argument in [F8], in its proof paragraph "The join is a metric space": if $A=d_J(x,y)$, $B=d_J(y,z)$ and $A+B<\pi$, choose $s=\sin (A+B)\mathrm{/}(\sin A+\sin B)$ when $A+B>0$. The intermediate planar point $s(\cos A,\sin A)$ lies on the chord from $(1,0)$ to $(\cos (A+B),\sin (A+B))$, so the cone triangle inequality gives $2\sin (d_J(x,z)\mathrm{/}2)\le 2\sin ((A+B)\mathrm{/}2)$ and hence $d_J(x,z)\le A+B$. The case $A+B=0$ follows from separation, and $A+B\ge \pi$ follows from $d_J\le \pi$. Thus the join formula defines the required angular metric.

The radial map in (ii) is an isometry $C(L_1)\times C(L_2)\to C(L_1\ast L_2)$: for general radii the expansion of 2.2 gives $d_C(\xi (R),\xi (R^{\mathrm{\prime }}){)}^2=R^2+R^{\mathrm{\prime }2}-2R{R}^{\mathrm{\prime }}\cos d$ with $d$ the join distance, which equals the sum of the two squared cone distances, namely the square-sum product distance; the conventions $L\ast \mathrm{\varnothing }=L$ and $\mathrm{\varnothing }\ast \mathrm{\varnothing }=\mathrm{\varnothing }$ of [F7] cover the empty cases.

A Euclidean space ${\mathbb{E}}^n$ is CAT(0) [F3], so replacing either factor in 2.1 by ${\mathbb{E}}^n$ is the special case in which that factor's hinged inequality is an equality; the componentwise description of geodesics of 1.2 and the CAT(0) conclusion of 2.1 therefore yield the clause as stated.

If $L_1$ and $L_2$ are CAT(1) then $L_1\ast L_2$ is CAT(1): by [F6] the cones $C(L_1),C(L_2)$ are CAT(0), by 2.1 their $l^2$-product is CAT(0), by 3.1 it is isometric to $C(L_1\ast L_2)$, and by [F6] again the join is CAT(1); if one factor is empty the join is the other factor [F7], which is CAT(1), and a point is CAT(1) [F1].

For a one-point space $\{p\}$ the spherical cone $\{p\}\ast L$ is CAT(1) exactly when $L$ is: if $L$ is CAT(1) then $\{p\}\ast L$ is CAT(1) by 4.1; conversely if $\{p\}\ast L$ is CAT(1) then $C(\{p\}\ast L)$ is CAT(0) [F6] and is isometric to $C(\{p\})\times C(L)$ by the radial map of (ii), and $C(L)$ is CAT(0) because a geodesic of this product joining two points of a slice $\{0\}\times C(L)$ has constant first coordinate by 1.2, so triangles in the slice lift to the product with their side lengths unchanged and the product comparison inequality of 2.1 restricts to the CAT(0) inequality of $C(L)$; hence $L$ is CAT(1) [F6].

The round sphere $S^k$ is CAT(1): $S^0$ is a two-point space at distance $\pi$ [F11], in which every triangle with two distinct vertices has a side of length $\pi$ and hence perimeter at least $2\pi$, so all admissible tests are degenerate; $S^1=S^0\ast S^0$ of circumference $2\pi$ is CAT(1) [F5, F8]; and $S^k=S^{k-1}\ast S^0$ for $k\ge 1$ [F8], so induction on $k$ with 4.1 gives the claim. Every closed ball $\bar{B}(x,\rho )\subseteq S^k$ with $0<\rho <\pi \mathrm{/}2$ is convex (by [F3] for $k\ge 1$, and because it is a singleton for $k=0$) and hence CAT(1) for the induced metric: a triangle of perimeter $<2\pi$ in a convex subset has its sides, which are ambient geodesic segments of length $<\pi$, contained in the subset, and its comparisons hold in the ambient CAT(1) space.

If $L$ is CAT(1) of diameter at most $\pi$ and $D$ is a nonempty closed ball of positive radius $<\pi \mathrm{/}2$ in some $S^k$, then $D\ast L$ is CAT(1): $D$ is nonempty, has diameter $<\pi$ and is CAT(1) with the induced metric by 5.2, so the join step 4.1 applies to the pair $(D,L)$.

If $M$ is positive definite with diagonal $1$ and off-diagonal entries at most $0$, deleting its vertex $0$ gives the Schur complement $m_{ij}^{\mathrm{\prime }}=m_{ij}-m_{i0}{m}_{j0}$. For any nonzero vector $x$ on the remaining indices, the vector $w=(-{\sum }_{i\mathrm{\ne }0}{m}_{0i}{x}_i,x)$ is nonzero and satisfies $w^{\mathsf{T}}Mw=x^{\mathsf{T}}{M}^{\mathrm{\prime }}x>0$; hence $M^{\mathrm{\prime }}$ is positive definite. In particular, testing a coordinate vector gives $1-m_{i0}^2>0$ on each diagonal. The off-diagonal entries are nonpositive because $m_{i0}{m}_{j0}\ge 0$; normalizing by $D=\mathrm{diag}((1-m_{i0}^2{)}^{-1\mathrm{/}2})$ preserves positive definiteness and gives diagonal $1$ with off-diagonal entries $m_{ij}^{\mathrm{\prime }}\mathrm{/}\sqrt{(1-m_{i0}^2)(1-m_{j0}^2)}\le 0$.

For a symmetric block matrix $A=\left(\begin{array}{cc}P& Q\\ Q^{\mathsf{T}}& R\end{array}\right)$ with $P$ positive definite, $P$ is invertible: $Pv=0$ for nonzero $v$ would contradict $v^{\mathsf{T}}Pv>0$. Put $S=R-Q^{\mathsf{T}}{P}^{-1}Q$. For $x=(y,z)$, $x^{\mathsf{T}}Ax=(y+P^{-1}Qz{)}^{\mathsf{T}}P(y+P^{-1}Qz)+z^{\mathsf{T}}Sz$. Thus $A$ is positive definite if $S$ is, and if $A$ is positive definite then testing $y=-P^{-1}Qz$ for each nonzero $z$ shows that $S$ is positive definite.

Fix a spherical simplex $\sigma$ and $p$ in the relative interior of its face $F$. In the barycentric ray coordinates of [F2], the local inequalities for $\sigma$ at $p$ are ${\dot{\lambda }}_j\ge 0$ for vertices $j\mathrm{\notin }F$ and no sign restriction on the face-coordinate variations, subject to ${\sum }_i{\dot{\lambda }}_i=0$. The radial normalization chart and its inverse $y\mapsto y\mathrm{/}\phi (y)$ are smooth, so their differentials identify these coordinate tangent cones with the tangent cone in the sphere; its lineality is exactly the subspace with all ${\dot{\lambda }}_j=0$ outside $F$, namely $U:=T_{p}F$. Since $U$ and $-U$ lie in the convex cone $T_p\sigma$, subtracting the orthogonal projection onto $U$ shows $T_p\sigma =U\oplus (T_p\sigma \cap U^{\perp })$.

Work in the component of $p$, with its componentwise chain metric. Let $\mathrm{St}(p)$ be the finite union of cells containing $p$. For each such spherical simplex $\sigma$, every face $\tau$ not containing $p$ is compact and disjoint from $p$, so the minimum of $d_{\sigma }(p,y)$ over $y\in \tau$ is positive. Choose $\delta >0$ below all these finitely many positive minima and with $\delta <\pi \mathrm{/}2$; if there are no such faces, choose any $0<\delta <\pi \mathrm{/}2$. On $\mathrm{St}(p)$ the cellwise radial function $r(x)=d_{\sigma }(p,x)$ is well defined and is $1$-Lipschitz on every incident cell. A chain from $p$ that leaves $\mathrm{St}(p)$ first reaches a face $\tau$ of an incident cell not containing $p$; at that point its prefix has length at least the cellwise radius, which is at least $\delta$. Thus if $d_X(p,x)<\delta$, every cell containing $x$ contains $p$.

Iterating step 1.1 over the vertices of any nonempty face $F$ shows that every simplex of its face link has a positive-definite Gram matrix with diagonal $1$ and nonpositive off-diagonal entries. The link is finite, so it is a finite large spherical complex.

In a simplex $\sigma \supseteq F$, the unit directions in $U=T_{p}F$ form $S^{k-1}$, while the directions in $T_p\sigma \cap U^{\perp }$ are the normal directions of $F$; their face-link Gram matrix is the Schur complement from [F2]. The orthogonal splitting of step 1.3 expresses each cell of the point link as the join of $S^{k-1}$ with the corresponding face-link cell. On each such cell, the inner-product formula is the spherical-join formula [F7], and the product-cone isometry identifies the induced face metrics with the intrinsic join metric [F7]. The identifications agree on common faces, so ${\mathrm{Lk}}_X(p)=S^{k-1}\ast {\mathrm{Lk}}_X(F)$ with the join metric. For a vertex $F$, the first factor is empty and the point link is the face link.

Choose $0<\epsilon <\min \{\delta \mathrm{/}4,\pi \mathrm{/}4\}$. Map a cone point $(\theta ,u)$ to the point at distance $\theta$ on the cellwise spherical geodesic from $p$ in direction $u$; common-face identifications make this map $\mathrm{\Phi }$ well defined. The radial path has length $\theta$, while every chain staying in $\mathrm{St}(p)$ has length at least the variation of $r$, and every chain leaving it has length at least $\delta$. Hence the radial coordinate equals $d_X(p,x)$ for $\theta <\epsilon$, and $\mathrm{\Phi }$ is a bijection from the cone ball onto $B(p,\epsilon )$, with all directions collapsed at the pole. If $d_{\pi }(u,u^{\mathrm{\prime }})<\pi$ and $\theta ,{\theta }^{\mathrm{\prime }}>0$, then for every $\eta >0$ small enough there is a finite chain of link directions with consecutive directions in one link cell and total angle $\alpha <d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(u,u^{\mathrm{\prime }})+\eta <\pi$. Unfold its spherical sectors about their common radial boundaries into a unit-sphere sector of angle $\alpha$. Each sector map is an isometry by the spherical cosine formula, and the great-circle segment between radii $\theta ,{\theta }^{\mathrm{\prime }}<\epsilon$ stays in the radius-$\epsilon$ cap by [F9]; subdividing at its finitely many sector crossings and mapping the pieces back gives a path in $X$ of length $\arccos (\cos \theta \cos {\theta }^{\mathrm{\prime }}+\sin \theta \sin {\theta }^{\mathrm{\prime }}\cos \alpha )$. Letting $\eta \downarrow 0$ gives the upper distance bound by the cone formula. If either radius is $0$, the radial segment gives equality; if $d_{\pi }(u,u^{\mathrm{\prime }})=\pi$, the path through $p$ has length $\theta +{\theta }^{\mathrm{\prime }}$, the cone distance. Conversely, let a chain from $x$ to $y$ have length $L$. If $L\ge 2\epsilon$, then the cone distance is at most $\theta +{\theta }^{\mathrm{\prime }}\le 2\epsilon \le L$. If $L<2\epsilon$, every point of the chain has distance from $p$ less than $3\epsilon <\delta$, so each common cell contains $p$; in that cell the two directions lie in one link cell. Their global truncated link distance is at most their spherical distance in this cell, since the cell segment is an admissible link chain. The cone cosine formula is nondecreasing in that angle, so ${\mathrm{\Phi }}^{-1}$ does not increase this chain step. No global isometry of a link cell is assumed. The cone distance is therefore at most $L$. Taking infima proves the isometry of the open balls.

If $F=\mathrm{\varnothing }$, the link claim is the hypothesis on $X$; take $F$ nonempty. Let $T$ be pairwise adjacent in ${\mathrm{Lk}}_X(F)$. Then $F\cup T$ is pairwise adjacent in $X$. The diagonal link entries are $1$, and each off-diagonal link entry for $s,t\in T$ is the Schur-complement entry from the simplex $F\cup \{s,t\}$; therefore the full link cosine matrix on $T$ is the diagonally normalized Schur complement of $C_F$ in the principal cosine matrix $C_{F\cup T}$. Positive diagonal normalization preserves positive definiteness. Since $C_F$ is positive definite, step 1.2 says this link matrix is positive definite exactly when $C_{F\cup T}$ is. Metric flagness of $X$ makes that equivalent to $F\cup T$ being a simplex, which is equivalent to $T$ being a simplex in the link. The empty set is covered by the empty-matrix convention in [F1], and simplex link matrices are positive definite by step 2.1; hence the link is metric flag and clause (i) follows.

If ${\mathrm{Lk}}_X(F)$ is CAT(1), step 2.2 and [F6] show that ${\mathrm{Lk}}_X(p)$ is CAT(1), and the one-point join $\{p\}\ast {\mathrm{Lk}}_X(p)$ is CAT(1) as well. For $0<r<\min \{\epsilon ,\pi \mathrm{/}2\}$, step 2.3 identifies ${\bar{B}}_X(p,r)$ with the closed radius-$r$ ball about the pole in that cone; [F8] makes this ball convex and CAT(1) with its induced metric. This proves the local CAT(1) conclusion at $p$.

Let $v$ be a vertex. For a point $x$ in a simplex containing $v$, its cellwise radial distance $\theta =d_{\sigma }(v,x)$ and direction are defined throughout that simplex. If $y$ lies in the opposite face $\tau$ not containing $v$, its barycentric ray coordinates involve only vertices $t\mathrm{\ne }v$, so $⟨v,y⟩$ has the sign of ${\sum }_{t\mathrm{\ne }v}{\lambda }_t{c}_{vt}\le 0$; hence $d_{\sigma }(v,y)\ge \pi \mathrm{/}2$. The radial function is $1$-Lipschitz on every cell containing $v$, so every chain leaving $\mathrm{St}(v)$ has length at least $\pi \mathrm{/}2$. For $x\in \mathrm{St}(v)$ with cellwise radius $\theta <\pi \mathrm{/}2$, chains staying in the star have length at least $\theta$, chains leaving have length at least $\pi \mathrm{/}2$, and the radial segment has length $\theta$; thus $d_X(v,x)=\theta$. If $x\in \mathrm{St}(v)$ has cellwise radius $\theta \ge \pi \mathrm{/}2$, chains staying in the star have length at least $\theta$ and chains leaving have length at least $\pi \mathrm{/}2$, so $d_X(v,x)\ge \pi \mathrm{/}2$; points outside the star also have distance at least $\pi \mathrm{/}2$. Conversely, every unit link direction has a representation $\eta ={\sum }_{t\mathrm{\ne }v}{b}_t(u_t-c_{vt}v)$ with $b_t\ge 0$ in an incident simplex. Its radial ray is $\cos \theta v+\sin \theta \eta =(\cos \theta -\sin \theta {\sum }_{t\mathrm{\ne }v}{b}_t{c}_{vt})v+\sin \theta {\sum }_{t\mathrm{\ne }v}{b}_t{u}_t$; all coefficients are nonnegative for $0\le \theta \le \pi \mathrm{/}2$. Hence every direction extends through that radius within the star, and $B(v,\pi \mathrm{/}2)$ is exactly the open radial cone on ${\mathrm{Lk}}_X(v)$.

Suppose every finite large metric flag complex of dimension less than $d$ is CAT(1), and let $X$ have dimension $d$. For each nonempty face $F$, clause (i) gives a finite large metric flag link with $\dim {\mathrm{Lk}}_X(F)\le d-\dim F-1<d$; a nonempty link is CAT(1) by the hypothesis, and an empty link satisfies the CAT(1) tests vacuously by [F5]. Clause (ii), proved in step 3.2, then gives a CAT(1) neighborhood at every point of $\mathrm{relint}F$. These relative interiors partition $X$, so $X$ is locally CAT(1), proving clause (iii).

A zero-dimensional finite large metric flag complex is a finite set of vertices with distinct vertices at truncated distance $\pi$. There are no distinct pairs at distance less than $\pi$, and every triangle with at least two distinct vertices has perimeter at least $2\pi$, so the CAT(1) tests hold; the empty link is CAT(1) vacuously. More generally, if a triangle in a finite spherical complex has $d_{\pi }$-perimeter less than $2\pi$, none of its sides can equal $\pi$ because the triangle inequality would force the other two sides to sum to at least $\pi$. Each side is therefore the componentwise intrinsic distance, and all vertices lie in one component, so its comparison is unchanged when using the truncated metric. This proves clause (iv).