Local CAT(1) of the $l^2$ product from a model $S^2\times S^2$ sine-comparison calculation

Product geodesics. Let $\gamma =({\gamma }_1,{\gamma }_2)$ be a constant-speed geodesic segment in $X_1\times X_2$ with endpoints $P,Q$, so that $d(\gamma (s),\gamma (t))=(t-s)L$ for $0\le s\le t\le 1$ with $L=d(P,Q)$; let ${\mathrm{\ell }}_i$ be the length of ${\gamma }_i$ and put $L_i:=d_i(P_i,Q_i)$. For every partition $0=t_0<\cdots <t_m=1$, the Minkowski inequality of [F7] gives ${\sum }_{j}d(\gamma (t_{j-1}),\gamma (t_j))\ge \sqrt{({\sum }_j{d}_1({\gamma }_1(t_{j-1}),{\gamma }_1(t_j)){)}^2+({\sum }_j{d}_2({\gamma }_2(t_{j-1}),{\gamma }_2(t_j)){)}^2}$, taking common refinements of partitions approximating both component lengths gives $L\ge \sqrt{{\mathrm{\ell }}_1^2+{\mathrm{\ell }}_2^2}$; since ${\mathrm{\ell }}_i\ge L_i$ and $L=\sqrt{L_1^2+L_2^2}$, it follows that ${\mathrm{\ell }}_i=L_i$ for both $i$. Hence each ${\gamma }_i$ has length equal to the distance between its endpoints, so ${\gamma }_i$ minimizes between its endpoints. For the partition $0,s,t,1$, equality in the Euclidean triangle inequality for the nonnegative component-distance vectors is forced by the product-geodesic equality. Its equality case, from [F7], makes each vector proportional to $(L_1,L_2)$; the middle vector has norm $(t-s)L$, hence $d_i({\gamma }_i(s),{\gamma }_i(t))=(t-s)L_i$: the projections are geodesics with constant speeds $L_i$ and $L^2=L_1^2+L_2^2$. Conversely, if the ${\gamma }_i$ are constant-speed geodesics with proportional parametrizations and speeds $L_i$, then $d(\gamma (s),\gamma (t){)}^2={\sum }_i(t-s{)}^2{L}_i^2=(t-s{)}^2{L}^2$, so $\gamma$ is a constant-speed geodesic.

Sine comparison on a short interval. Let $0<V<\pi$ and let $F\in C^2([0,1])$ satisfy $F^{\mathrm{\prime }\mathrm{\prime }}+V^{2}F\le 0$. Put $H(t)=(\sin ((1-t)V)F(0)+\sin (tV)F(1))\mathrm{/}\sin V$ and $G=F-H$; then $G(0)=G(1)=0$ and $G^{\mathrm{\prime }\mathrm{\prime }}+V^{2}G\le 0$. Choose $K=(V+\pi )\mathrm{/}2$ and the positive function $w(t)=\cos (K(t-1\mathrm{/}2))$. If $G$ were negative somewhere, $G\mathrm{/}w$ would attain a negative minimum at an interior $t_0$; there $(G\mathrm{/}w{)}^{\mathrm{\prime }}=0$ and $(G\mathrm{/}w{)}^{\mathrm{\prime }\mathrm{\prime }}\ge 0$ by [F11]. Writing $G=w(G\mathrm{/}w)$ gives $G^{\mathrm{\prime }\mathrm{\prime }}+V^{2}G=w(G\mathrm{/}w{)}^{\mathrm{\prime }\mathrm{\prime }}+2w^{\mathrm{\prime }}(G\mathrm{/}w{)}^{\mathrm{\prime }}+(V^2-K^2)w(G\mathrm{/}w)>0$ at $t_0$, because $K>V$, $w>0$, and $G\mathrm{/}w<0$, a contradiction. Hence $F\ge H$. The required $V<\pi$ in the model follows from the triangle inequality $V\le a+b$ and $a+b+V<2\pi$.

Radial identities in the model. Let $M=M_1\times \cdots \times M_m$ where each factor $M_i$ is either the round sphere $S^2$ with metric $d_S$ or a Euclidean space ${\mathbb{R}}^{k_i}$ with its metric ([F5], [F6]), and let $v\text{\hspace{0.17em}⁣}:[0,1]\to M$ be a constant-speed geodesic segment with factor curves $v_i$ of speeds $V_i$ and $V^2={\sum }_i{V}_i^2$, as in step 1.1. Fix $p\in M$, write $r_i(t):=d_{M_i}(p_i,v_i(t))$ and $R(t):=({\sum }_i{r}_i(t{)}^2{)}^{1\mathrm{/}2}=d_M(p,v(t))$, and assume first that $r_i(t)>0$ for all $i$ and $t$. Then $R$ is $V$-Lipschitz, so $R(t)\le \min \{a+tV,b+(1-t)V\}$ where $a=R(0)$ and $b=R(1)$, and under the perimeter hypothesis $a+b+V<2\pi$ one has $R(t)\le \frac{1}{2}(a+b+V)<\pi$; also $\mathrm{\mid }{r}_i^{\mathrm{\prime }}\mathrm{\mid }\le V_i$. If $M_i=S^2$ put $f_i(u):=u\cot u$ and $A_i:=f_i(r_i)$, while for a Euclidean factor put $f_i\equiv 1$ and $A_i:=1$. Differentiating the relation $\cos r_i(t)=p_i\cdot v_i(t)$ twice along the great circle $v_i$ gives $r_i^{\mathrm{\prime }\mathrm{\prime }}=\cot r_i(V_i^2-(r_i^{\mathrm{\prime }}{)}^2)$, that is $r_i{r}_i^{\mathrm{\prime }\mathrm{\prime }}=A_i(V_i^2-(r_i^{\mathrm{\prime }}{)}^2)$, by [F9], [F8] and [F1]; in the Euclidean factor the same identity $r_i{r}_i^{\mathrm{\prime }\mathrm{\prime }}=V_i^2-(r_i^{\mathrm{\prime }}{)}^2$ follows by differentiating $r_i^2=\mathrm{\mid }{p}_i-v_i(t){\mathrm{\mid }}^2$ twice for an affine $v_i$ ([F5], [F9]). Summing the identities over $i$ and using $R^2={\sum }_i{r}_i^2$, so that $R{R}^{\mathrm{\prime }\mathrm{\prime }}={\sum }_i(r_i^{\mathrm{\prime }}{)}^2+{\sum }_i{r}_i{r}_i^{\mathrm{\prime }\mathrm{\prime }}-(R^{\mathrm{\prime }}{)}^2$, one gets $R{R}^{\mathrm{\prime }\mathrm{\prime }}={\sum }_i{A}_i{V}_i^2+{\sum }_i(1-A_i)(r_i^{\mathrm{\prime }}{)}^2-(R^{\mathrm{\prime }}{)}^2$.

A differential inequality for $R$. In the situation of step 2.1 put $f(R):=R\cot R$. Subtracting $f(R)(V^2-(R^{\mathrm{\prime }}{)}^2)$ from the identity of step 2.1 gives $R{R}^{\mathrm{\prime }\mathrm{\prime }}-f(R)(V^2-(R^{\mathrm{\prime }}{)}^2)={\sum }_i(A_i-f(R))(V_i^2-(r_i^{\mathrm{\prime }}{)}^2)+(1-f(R))({\sum }_i(r_i^{\mathrm{\prime }}{)}^2-(R^{\mathrm{\prime }}{)}^2)$, as expanding the right-hand side and using ${\sum }_i{V}_i^2=V^2$ reproduces the left-hand side. Each factor is nonnegative: $V_i^2-(r_i^{\mathrm{\prime }}{)}^2\ge 0$ because $r_i$ is $V_i$-Lipschitz; $A_i\ge f(R)$ because either $A_i=f_i(r_i)$ with $r_i\le R$ and $f_i(u)=u\cot u$ decreasing on $(0,\pi )$ — indeed $f_i^{\mathrm{\prime }}(u)=(\sin u\cos u-u)\mathrm{/}{\sin }^{2}u<0$: the function $u-\sin u\cos u$ vanishes at zero and has derivative $2{\sin }^{2}u>0$ on $(0,\pi )$, so [F11]'s mean-value theorem makes it positive — or $A_i=1\ge f(R)$, valid because $\sin u-u\cos u$ vanishes at zero and has derivative $u\sin u>0$ on $(0,\pi )$, so the same theorem gives $u\cot u<1$; and $1-f(R)\ge 0$ together with ${\sum }_i(r_i^{\mathrm{\prime }}{)}^2\ge (R^{\mathrm{\prime }}{)}^2$ by Cauchy–Schwarz [F7]. Hence $R{R}^{\mathrm{\prime }\mathrm{\prime }}-f(R)(V^2-(R^{\mathrm{\prime }}{)}^2)\ge 0$.

The comparison function satisfies the differential inequality. With $F:=\cos R$ one computes $F^{\mathrm{\prime }}=-\sin R{R}^{\mathrm{\prime }}$ and $F^{\mathrm{\prime }\mathrm{\prime }}=-\cos R(R^{\mathrm{\prime }}{)}^2-\sin R{R}^{\mathrm{\prime }\mathrm{\prime }}$, so $(\sin R\mathrm{/}R)(R{R}^{\mathrm{\prime }\mathrm{\prime }}-f(R)(V^2-(R^{\mathrm{\prime }}{)}^2))=\sin R{R}^{\mathrm{\prime }\mathrm{\prime }}-\cos R(V^2-(R^{\mathrm{\prime }}{)}^2)=-(F^{\mathrm{\prime }\mathrm{\prime }}+V^{2}F)$, because $Rf(R)=R^2\cot R$ and $\sin R>0$; since $0<R<\pi$, step 3.1 gives $F^{\mathrm{\prime }\mathrm{\prime }}+V^{2}F\le 0$ wherever $R>0$.

The model inequality (ii). In the situation of steps 2.1–3.1 with $V>0$ and $r_i>0$ everywhere, step 1.2 applied to $F=\cos R$ with $F(0)=\cos a$, $F(1)=\cos b$ gives $\cos R(t)\ge (\sin ((1-t)V)\cos a+\sin (tV)\cos b)\mathrm{/}\sin V$, whose right-hand side is $\cos d_{S^2}(\bar{p},\bar{v}(t))$ for the spherical comparison triangle with side lengths $a,b,V$ and the comparison point $\bar{v}(t)$ at parameter $t$ on the side of length $V$, by the cosine rule [F2]; since $\cos$ is strictly decreasing on $[0,\pi ]$ and both arguments lie in $[0,\pi ]$ ([F8]), $d(p,v(t))\le d_{S^2}(\bar{p},\bar{v}(t))$. If some factor distance vanishes at some time, choose approximating data $p^{ϵ}$ with every $r_i^{ϵ}>0$ throughout: for a spherical factor the trace of $v_i$ has empty interior in $S^2$, so $p_i^{ϵ}$ can be chosen arbitrarily close to $p_i$ outside it; for a Euclidean factor, first embed ${\mathbb{R}}^{k_i}$ isometrically in ${\mathbb{R}}^{k_i+1}$ and displace $p_i$ by $ϵ$ in the new orthogonal direction, making its distance to the entire trace positive; the inequality proved for the approximants passes to the limit because $R$, $a$, $b$ and $V$ depend continuously on $(p,v(0),v(1))$ and on $t\mapsto v(t)$ ([F3]). For $V=0$ the geodesic $v$ is constant and $d(p,v(t))=a$ for all $t$, which is the asserted inequality.

Vertex-to-side inequality in a product chart. Let $B(x,r)\subseteq X$, $B(y,r)\subseteq Y$ be CAT(1) and convex, $r<\pi \mathrm{/}2$, let $(P,Q,R)$ be a geodesic triangle in the chart $B(x,r)\times B(y,r)$ with perimeter $<2\pi$, and let $U$ be a point of the side $[Q,R]$, say at parameter $t$ from $Q$. By step 1.1 the factor projections of the sides are geodesic segments in $X$ and $Y$ lying in the convex balls $B(x,r),B(y,r)$ ([F4]), and $U_i$ lies on $[Q_i,R_i]$ at the same parameter $t$. The factor triangles $(P_i,Q_i,R_i)$ have perimeter at most the perimeter of $(P,Q,R)$, hence $<2\pi$, so the CAT(1) inequality of [F1] applies to the pair $(P_i,U_i)$: $d_i(P_i,U_i)\le d_{S^2}(P_i^{\mathrm{\prime }},U_i^{\mathrm{\prime }})$, where $P_i^{\mathrm{\prime }},U_i^{\mathrm{\prime }}$ are the corresponding points of the spherical comparison triangle of $(P_i,Q_i,R_i)$. Therefore $d(P,U{)}^2={\sum }_i{d}_i(P_i,U_i{)}^2\le {\sum }_i{d}_{S^2}(P_i^{\mathrm{\prime }},U_i^{\mathrm{\prime }}{)}^2$. By step 1.1 the pairing $t\mapsto (U_1^{\mathrm{\prime }},U_2^{\mathrm{\prime }})$ is a constant-speed geodesic in $S^2\times S^2$ from $(Q_1^{\mathrm{\prime }},Q_2^{\mathrm{\prime }})$ to $(R_1^{\mathrm{\prime }},R_2^{\mathrm{\prime }})$ of length $\sqrt{{\sum }_i{d}_i(Q_i,R_i{)}^2}=d(Q,R)$, while $\sqrt{{\sum }_i{d}_{S^2}(P_i^{\mathrm{\prime }},Q_i^{\mathrm{\prime }}{)}^2}=d(P,Q)$ and likewise for $R$. Since the perimeter hypothesis is unchanged, the model inequality of step 5.1 applies with $p:=(P_1^{\mathrm{\prime }},P_2^{\mathrm{\prime }})$, $a=d(P,Q)$, $b=d(P,R)$ and $V=d(Q,R)$: $d(P,U)\le d_{S^2\times S^2}((P_1^{\mathrm{\prime }},P_2^{\mathrm{\prime }}),(U_1^{\mathrm{\prime }},U_2^{\mathrm{\prime }}))\le d_{S^2}(\bar{P},\bar{U})$, where $(\bar{P},\bar{Q},\bar{R})$ is the spherical comparison triangle of $(P,Q,R)$ and $\bar{U}$ is the comparison point of $U$.

All side-point comparisons. With the notation of step 6.1 let now $U$ lie on $[P,Q]$ and $V$ on $[P,R]$. The triangle $(P,U,R)$ has perimeter at most that of $(P,Q,R)$, so step 6.1 applies to it at the vertex $U$ and the point $V$ of the side $[P,R]$: $d(U,V)\le d_{S^2}(U^{\mathrm{\prime }\mathrm{\prime }},V^{\mathrm{\prime }\mathrm{\prime }})$, where $U^{\mathrm{\prime }\mathrm{\prime }},V^{\mathrm{\prime }\mathrm{\prime }}$ are the corresponding points of the comparison triangle of $(P,U,R)$. That comparison triangle and the configuration $(\bar{P},\bar{U},\bar{V})$ of the big comparison triangle have equal legs $\mathrm{\mid }PU\mathrm{\mid }$ and $\mathrm{\mid }PV\mathrm{\mid }$ from the vertex $P$, while their opposite sides satisfy $\mathrm{\mid }{U}^{\mathrm{\prime }\mathrm{\prime }}{R}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }=d(U,R)\le d_{S^2}(\bar{U},\bar{R})$, the latter by step 6.1 applied to the big triangle at the vertex $R$ with the point $U$ of the side $[P,Q]$. For fixed legs the angle at the vertex of a spherical triangle is an increasing function of the opposite side, because the cosine rule of [F2] gives $\cos \gamma =(\cos c-\cos a\cos b)\mathrm{/}(\sin a\sin b)$ and $\cos$ is decreasing ([F8]); hence the angle at $P^{\mathrm{\prime }\mathrm{\prime }}$ is at most the angle at $\bar{P}$, and applying the cosine rule to both configurations gives $\cos d_{S^2}(U^{\mathrm{\prime }\mathrm{\prime }},V^{\mathrm{\prime }\mathrm{\prime }})\ge \cos d_{S^2}(\bar{U},\bar{V})$, that is $d(U,V)\le d_{S^2}(\bar{U},\bar{V})$. Pairs on the same side satisfy equality because the comparison map is an isometry on each side, and pairs on two sides sharing a vertex reduce to the treated case after relabelling the triangle; thus every pair of points of $(P,Q,R)$ satisfies the CAT(1) inequality.

Local CAT(1) of products. Let $B(x,r)\subseteq X$, $B(y,r)\subseteq Y$ be CAT(1) and convex with $r<\pi \mathrm{/}2$: the chart $B(x,r)\times B(y,r)$ is convex in $X\times Y$, because a product geodesic between two of its points has factor geodesics lying in the convex factor balls by step 1.1, so the induced metric on the chart agrees with the restricted product metric. Every pair of points of the chart at distance $<\pi$ has product distance with factor components $<\pi$ and is joined by the product of the factor geodesics inside the chart, and every geodesic triangle of perimeter $<2\pi$ in the chart satisfies the CAT(1) inequality by steps 6.1 and 7.1; hence the chart is a CAT(1) space. Choose $0<\rho <\min \{r,\pi \mathrm{/}2\}$; the closed product ball of radius $\rho$ about $(x,y)$ lies in this CAT(1) chart and is convex by [F2], hence is itself CAT(1). Therefore $X\times Y$ is locally CAT(1) whenever $X$ and $Y$ are, and if $B(x,r),B(y,r)$ are CAT(1) and convex then every triangle in the chart of perimeter $<2\pi$ satisfies the CAT(1) inequality, the same argument applying verbatim when a factor is Euclidean (steps 2.1 and 5.1 treat the flat case). This proves (i), (ii) and (iii).