Diameter rigidity from toponogov under a sectional lower bound

Diametral pair. By [F1], $M$ is compact: it is closed in itself and bounded, because $\mathrm{diam}(M,g)=D<\mathrm{\infty }$. For $x\in M$ the function $y\mapsto d_g(x,y)$ is continuous on the nonempty compact space $M$, so [F2] makes $\mathrm{\Phi }(x):={\max }_{y\in M}{d}_g(x,y)$ finite and attained; the Lipschitz bound of [F1] gives $\mathrm{\mid }\mathrm{\Phi }(x)-\mathrm{\Phi }(x^{\mathrm{\prime }})\mathrm{\mid }\le d_g(x,x^{\mathrm{\prime }})$, so $\mathrm{\Phi }$ is continuous, and ${\sup }_x\mathrm{\Phi }(x)=\mathrm{diam}(M,g)$ because the diameter is the supremum over all pairs. By [F2] there is $p\in M$ with $\mathrm{\Phi }(p)=D$, and applying [F2] once more gives $q\in M$ with $d_g(p,q)=\mathrm{\Phi }(p)=D$. Thus $(p,q)$ is a diametral pair and $p\mathrm{\ne }q$.

Model chord limit at the degenerate perimeter. Let $k_{\ast }>0$ and let $a,b,c>0$ satisfy the strict triangle inequalities with $a+b+c=2\pi \mathrm{/}\sqrt{k_{\ast }}$; put $s:=(a-b+c)\mathrm{/}2\in (0,c)$, which lies in $(0,c)$ exactly by $\mathrm{\mid }a-b\mathrm{\mid }<c<a+b$. For $0<k^{\mathrm{\prime }}<k_{\ast }$ each side is smaller than $\pi \mathrm{/}\sqrt{k^{\mathrm{\prime }}}$ and the perimeter is smaller than $2\pi \mathrm{/}\sqrt{k^{\mathrm{\prime }}}$: indeed $2a<a+b+c=2\pi \mathrm{/}\sqrt{k_{\ast }}<2\pi \mathrm{/}\sqrt{k^{\mathrm{\prime }}}$ and likewise for $b,c$, so the ordered side lengths $(a,b,c)$ admit a comparison triangle $(\bar{x},\bar{y},\bar{z})$ in $M_{k^{\mathrm{\prime }}}^2$ by [F3]. Let $\bar{u}\in \bar{x}\bar{y}$ have $d_{k^{\mathrm{\prime }}}(\bar{x},\bar{u})=s$ and put $\beta (k^{\mathrm{\prime }}):=d_{k^{\mathrm{\prime }}}(\bar{u},\bar{z})$. Write $R^{\mathrm{\prime }}:=1\mathrm{/}\sqrt{k^{\mathrm{\prime }}}$. (i) The broken geodesic from $\bar{u}$ to $\bar{x}$ to $\bar{z}$ has length $s+b=(a+b+c)\mathrm{/}2=:\mathrm{\Pi }\mathrm{/}2$, so $\beta (k^{\mathrm{\prime }})\le \mathrm{\Pi }\mathrm{/}2<\pi R^{\mathrm{\prime }}$; also $\bar{u}\mathrm{\ne }\bar{z}$, because otherwise $\bar{z}$ would lie on the side $\bar{x}\bar{y}$ and the comparison angle of [F3] at $\bar{x}$, which lies in $(0,\pi )$, would be $0$ or $\pi$. Hence $0<\beta (k^{\mathrm{\prime }})\mathrm{/}{R}^{\mathrm{\prime }}<\pi$. (ii) By [F3] the data $(s,b,\beta (k^{\mathrm{\prime }}))$ also satisfy the strict triangle inequalities and the required bounds, the perimeter being at most $2(s+b)$, and the angle at $\bar{x}$ between the geodesic directions toward $\bar{u}$ and $\bar{z}$ is the comparison angle $\bar{\alpha }(k^{\mathrm{\prime }})$ of the original triangle, because $\bar{u}$ lies on the side $\bar{x}\bar{y}$ and that angle is determined by the two initial directions. Applying the cosine law of [F3], solved for the opposite side, to the comparison triangle and to the configuration $(\bar{x},\bar{u},\bar{z})$ gives $\cos \frac{a}{R^{\mathrm{\prime }}}=\cos \frac{b}{R^{\mathrm{\prime }}}\cos \frac{c}{R^{\mathrm{\prime }}}+\sin \frac{b}{R^{\mathrm{\prime }}}\sin \frac{c}{R^{\mathrm{\prime }}}\cos \bar{\alpha }(k^{\mathrm{\prime }}),$ $\cos \frac{\beta (k^{\mathrm{\prime }})}{R^{\mathrm{\prime }}}=\cos \frac{s}{R^{\mathrm{\prime }}}\cos \frac{b}{R^{\mathrm{\prime }}}+\sin \frac{s}{R^{\mathrm{\prime }}}\sin \frac{b}{R^{\mathrm{\prime }}}\cos \bar{\alpha }(k^{\mathrm{\prime }})\mathrm{.}$ Eliminating $\cos \bar{\alpha }(k^{\mathrm{\prime }})$ yields, for every $k^{\mathrm{\prime }}\in (0,k_{\ast })$, $\cos \frac{\beta (k^{\mathrm{\prime }})}{R^{\mathrm{\prime }}}=\cos \frac{s}{R^{\mathrm{\prime }}}\cos \frac{b}{R^{\mathrm{\prime }}}+\frac{\sin (s\mathrm{/}{R}^{\mathrm{\prime }})}{\sin (c\mathrm{/}{R}^{\mathrm{\prime }})}(\cos \frac{a}{R^{\mathrm{\prime }}}-\cos \frac{b}{R^{\mathrm{\prime }}}\cos \frac{c}{R^{\mathrm{\prime }}})\mathrm{.}$ (iii) Let $k_m^{\mathrm{\prime }}\in (0,k_{\ast })$ increase to $k_{\ast }$. Then $R_m^{\mathrm{\prime }}\to R_{\ast }:=1\mathrm{/}\sqrt{k_{\ast }}$ and $a\mathrm{/}{R}_m^{\mathrm{\prime }}+b\mathrm{/}{R}_m^{\mathrm{\prime }}+c\mathrm{/}{R}_m^{\mathrm{\prime }}\to 2\pi$, while $s\mathrm{/}{R}_m^{\mathrm{\prime }}+b\mathrm{/}{R}_m^{\mathrm{\prime }}\to (s+b)\mathrm{/}{R}_{\ast }=(a+b+c)\mathrm{/}(2R_{\ast })=\pi$. Sine and cosine are continuous and the algebra of limits applies by [F16], with $\sin (c\mathrm{/}{R}_m^{\mathrm{\prime }})\to \sin (c\mathrm{/}{R}_{\ast })>0$, so $s\mathrm{/}{R}_m^{\mathrm{\prime }}\to \pi -b\mathrm{/}{R}_{\ast },a\mathrm{/}{R}_m^{\mathrm{\prime }}\to 2\pi -(b+c)\mathrm{/}{R}_{\ast };$ using $\cos (\pi -x)=-\cos x$, $\sin (\pi -x)=\sin x$ and $\cos (2\pi -x)=\cos x$, the right-hand side of (ii) converges to $-{\cos }^2(b\mathrm{/}{R}_{\ast })-{\sin }^2(b\mathrm{/}{R}_{\ast })=-1$. (iv) By (i) and continuity of the principal arccosine in [F16], $\beta (k_m^{\mathrm{\prime }})\mathrm{/}{R}_m^{\mathrm{\prime }}=\arccos (\cos (\beta (k_m^{\mathrm{\prime }})\mathrm{/}{R}_m^{\mathrm{\prime }}))⟶\arccos (-1)=\pi ,$ so $\beta (k_m^{\mathrm{\prime }})\to \pi R_{\ast }=(a+b+c)\mathrm{/}2=\mathrm{\Pi }\mathrm{/}2$. Since $\beta (k^{\mathrm{\prime }})\le \mathrm{\Pi }\mathrm{/}2$ for every $k^{\mathrm{\prime }}<k_{\ast }$ by (i), the values $\beta (k^{\mathrm{\prime }})$ have supremum $\mathrm{\Pi }\mathrm{/}2$ over $k^{\mathrm{\prime }}<k_{\ast }$.

Agreement lemma for local isometries. Let $U$ be a connected smooth manifold and let $F,G:U\to M^{\mathrm{\prime }}$ be local isometries into a Riemannian manifold $M^{\mathrm{\prime }}$ with $F(x_0)=G(x_0)$ and $d{F}_{x_0}=d{G}_{x_0}$ at some $x_0\in U$. Then $F=G$. Indeed, the set $A=\{x\in U:F(x)=G(x)\text{ and }d{F}_x=d{G}_x\}$ is nonempty and closed by continuity of smooth maps and their differentials. It is open: for $x\in A$ put $y:=F(x)=G(x)$, use Existence of normal neighborhoods to choose a sufficiently small normal domain $U_0\ni x$ in the common source metric; for $z={\exp }_x(w)\in U_0$ with $w$ small, [F14] and the uniqueness part of [F8] give $F(z)={\exp }_y(d{F}_{x}w)={\exp }_y(d{G}_{x}w)=G(z)$, and the chain rule then gives $d{F}_z=d{G}_z$. Hence $A$ is nonempty, open and closed in the connected space $U$, so $A=U$.

Round-sphere local isometries extend to orthogonal maps. Let $U\subseteq S_R^n$ be a nonempty connected open subset and let $F:U\to S_R^n$ be a local isometry. Then $F=\widehat{U}{\mathrm{\mid }}_U$ for some orthogonal map $\widehat{U}$ of ${\mathbb{R}}^{n+1}$. Indeed, fix $x_0\in U$ and an orthonormal basis $e_1,\dots ,e_n$ of $T_{x_0}{S}_R^n=x_0^{\perp }$ [F13]; by [F14], $f_i:=d{F}_{x_0}(e_i)$ is an orthonormal basis of $T_{F(x_0)}{S}_R^n=F(x_0{)}^{\perp }$. The tuples $(x_0\mathrm{/}R,e_1,\dots ,e_n)$ and $(F(x_0)\mathrm{/}R,f_1,\dots ,f_n)$ are orthonormal bases of ${\mathbb{R}}^{n+1}$, so the linear map $\widehat{U}$ carrying the first to the second is orthogonal by [F14]; it restricts to an isometry of $S_R^n$ with $\widehat{U}(x_0)=F(x_0)$ and $d{\widehat{U}}_{x_0}=d{F}_{x_0}$. By the agreement lemma 1.3, $F=\widehat{U}{\mathrm{\mid }}_U$.

Perimeter bound. Every triple $x,y,z\in M$ joined by minimizing geodesic segments satisfies $d_g(x,y)+d_g(y,z)+d_g(z,x)\le 2D\mathrm{.}$ Suppose not, and let $P>2D$ be the perimeter. Each side is at most $D$, so the triple is nondegenerate: a degenerate triple (one side the sum of the other two) would have perimeter twice its largest side, at most $2D$. Put $a:=d_g(y,z)$, $b:=d_g(z,x)$, $c:=d_g(x,y)$, so that $P=a+b+c$ and $\mathrm{\mid }a-b\mathrm{\mid }<c<a+b$. Let $k_{\ast }:=(2\pi \mathrm{/}P{)}^2$, so $P=2\pi \mathrm{/}\sqrt{k_{\ast }}$ and $k_{\ast }<k$ because $P>2D=2\pi \mathrm{/}\sqrt{k}$. Choose the point $q_0:=\sigma (s)$ on the minimizing geodesic $\sigma :[0,c]\to M$ from $x$ to $y$, where $s:=(a-b+c)\mathrm{/}2\in (0,c)$, and a minimizing geodesic from $x$ to $z$, which exists by [F1]. For every $k^{\mathrm{\prime }}\in (0,k_{\ast })$ the hypotheses of the chord comparison [F4] are met: $K\ge k>k^{\mathrm{\prime }}$, the side lengths satisfy the strict triangle inequalities, and each side is smaller than $\pi \mathrm{/}\sqrt{k^{\mathrm{\prime }}}$ and the perimeter smaller than $2\pi \mathrm{/}\sqrt{k^{\mathrm{\prime }}}$ as in step 1.2. Applying [F4] with the two legs of lengths $c$ and $b$ from $x$, the points $u=q_0$ at distance $s$ on the first leg and $v=z$ at the endpoint of the second, gives $d_g(q_0,z)\ge \beta (k^{\mathrm{\prime }})$, where $\beta (k^{\mathrm{\prime }})$ is the model chord of step 1.2. Hence $d_g(q_0,z)\ge {\sup }_{k^{\mathrm{\prime }}<k_{\ast }}\beta (k^{\mathrm{\prime }})=\frac{P}{2}>D,$ where the supremum was computed in step 1.2. This contradicts the definition of $D$ as the diameter, since $d_g(q_0,z)\le D$.

Distance-sum identity. For every $x\in M$, $d_g(p,x)+d_g(x,q)=D\mathrm{.}$ The triangle inequality gives $d_g(p,x)+d_g(x,q)\ge d_g(p,q)=D$ by [F1] and step 1.1; applying the perimeter bound of step 2.2 to the triple $(p,x,q)$, whose pairs are joined by minimizing segments by [F1], gives $D+d_g(p,x)+d_g(x,q)\le 2D$.

Radial geodesics minimize up to time $D$. Let $\gamma :\mathbb{R}\to M$ be a unit-speed geodesic with $\gamma (0)=p$. Choose $0<\delta <D$ so that $\gamma {\mathrm{\mid }}_{[0,\delta ]}$ minimizes; such a $\delta$ exists because $c_p(\dot{\gamma }(0))>0$ by [F9]. Put $x=\gamma (\delta )$. Step 3.1 gives $d_g(x,q)=D-\delta$, and [F1] supplies a minimizing segment from $x$ to $q$. Its concatenation with $\gamma {\mathrm{\mid }}_{[0,\delta ]}$ has length $D=d_g(p,q)$, so [F7] makes it a smooth unit-speed geodesic. It agrees with $\gamma$ on $[0,\delta ]$; geodesic uniqueness [F8] therefore identifies it with $\gamma$ on $[0,D]$. Thus $\gamma (D)=q$, and every subsegment of this minimizing path minimizes, giving $d_g(p,\gamma (t))=t$ and $d_g(\gamma (t),q)=D-t$ for $0\le t\le D$. The argument with $p$ and $q$ exchanged proves the symmetric assertion.

Uniqueness of minimizing segments from the two poles. Let $x\in M$ with $0<d_g(p,x)<D$, and let $\sigma :[0,t_0]\to M$ and $\tilde{\sigma }:[0,t_0]\to M$ be minimizing unit-speed geodesics from $p$ to $x$, where $t_0:=d_g(p,x)$. By [F1] there is a minimizing unit-speed geodesic $\tau :[0,D-t_0]\to M$ from $x$ to $q$; its length is $D-t_0=d_g(x,q)$ by step 3.1. Both concatenations $\sigma \ast \tau$ and $\tilde{\sigma }\ast \tau$ are piecewise smooth curves from $p$ to $q$ of length $t_0+(D-t_0)=D=d_g(p,q)$, hence minimizing; by [F7] each is, after arclength reparametrization, a smooth unbroken geodesic, so ${\sigma }^{\mathrm{\prime }}(t_0)={\tau }^{\mathrm{\prime }}(0)={\tilde{\sigma }}^{\mathrm{\prime }}(t_0)$. Now $\sigma$ and $\tilde{\sigma }$ are unit-speed geodesics with the same value and the same derivative at $t_0$, so [F8] forces $\sigma =\tilde{\sigma }$. The same argument with $p$ and $q$ interchanged, using step 3.1, gives uniqueness of the minimizing segments from $q$ to points at distance $<D$ from $q$.

Cut time and the exponential map off the poles. For every unit $v\in S_{p}M$ the cut time is $c_p(v)=D$: step 4.1 gives $d_g(p,{\gamma }_v(t))=t$ for all $t\in [0,D]$, so $c_p(v)\ge D$, while $c_p(v)\le D$ because minimizing times are at most the diameter. Hence the cut point in every direction is ${\gamma }_v(D)=q$, the cut locus is $\mathrm{Cut}(p)=\{q\}$, and $D_p=\{tv:0<t<c_p(v)\}=B_0(D)\setminus \{0\}$. By [F9], ${\exp }_p:B_0(D)\setminus \{0\}⟶M\setminus \{p,q\}$ is a diffeomorphism onto, and the same holds with $p$ and $q$ interchanged, by the symmetric statement of step 4.1.

Radial sectional curvature equals $k$. Fix a unit-speed geodesic $\gamma :[0,D]\to M$ from $p$, which minimizes by step 4.1, and a parallel unit normal field $E$ along $\gamma$, which exists by [F11]. Put $V(t):={\mathrm{sn}}_k(t)E(t)$. Then $V(0)=V(D)=0$ by [F5], and $V$ is the variation field of a fixed-endpoint variation of $\gamma$: indeed $F(s,t):={\exp }_{\gamma (t)}(sV(t))$ is defined for small $s$ and is smooth by [F8], its variational field at $s=0$ is $V$ because the differential of ${\exp }_{\gamma (t)}$ at $0$ is the identity [F12], and the endpoints are fixed because $V(0)=V(D)=0$. Write $e(s)$ for the energy of $t\mapsto F(s,t)$; then $e(0)=D\mathrm{/}2$, because $\gamma$ has unit speed and length $D$. Every curve $t\mapsto F(s,t)$ joins $p$ to $q$, so its length is at least $d_g(p,q)=D$, and the length-energy inequality of [F10] gives $e(s)\ge L^2\mathrm{/}(2D)\ge D\mathrm{/}2=e(0)$: the geodesic $\gamma$, of unit speed and length $D$, minimizes energy among these curves. The function $e$ is $C^2$ near $0$, its derivative being computed by differentiating under the integral sign [F16] for the smooth family $F$ of [F8]; and $e^{\mathrm{\prime }}(0)=0$, because the difference quotients of $e(s)-e(0)\ge 0$ are nonnegative for $s>0$ and nonpositive for $s<0$. Were $e^{\mathrm{\prime }\mathrm{\prime }}(0)<0$, the second-derivative test [F16] would make $0$ a strict local maximum, contradicting $e(s)\ge e(0)$; hence $e^{\mathrm{\prime }\mathrm{\prime }}(0)\ge 0$. By the second-variation formula of [F10] applied to the fixed-endpoint two-parameter family $\tilde{F}(s,r,t):=F(s+r,t)$, whose energy is $e(s+r)$, the second derivative $e^{\mathrm{\prime }\mathrm{\prime }}(0)={\mathrm{\partial }}_s{\mathrm{\partial }}_{r}E(\tilde{F}){\mathrm{\mid }}_{(0,0)}$ equals $I_{\gamma }(V,V)$; the endpoint-acceleration term vanishes because the endpoints are fixed. Hence, using $D_{t}V={\mathrm{sn}}_k^{\mathrm{\prime }}E$, ${\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}=-k{\mathrm{sn}}_k$ [F5], the fundamental theorem [F16] and $V(0)=V(D)=0$, $0\le I_{\gamma }(V,V)={\int }_0^D({\mathrm{sn}}_k^{\mathrm{\prime }2}-K(\mathrm{span}(E,\dot{\gamma })){\mathrm{sn}}_k^2)dt\le {\int }_0^D({\mathrm{sn}}_k^{\mathrm{\prime }2}-k{\mathrm{sn}}_k^2)dt=0,$ where the middle inequality uses $K\ge k$ and ${\mathrm{sn}}_k^2\ge 0$. Both endpoint inequalities are equalities, so the continuous nonnegative function $(K(\mathrm{span}(E,\dot{\gamma }))-k){\mathrm{sn}}_k^2$ has zero integral over $[0,D]$ and vanishes identically by [F16]; since ${\mathrm{sn}}_k>0$ on $(0,D)$ by [F5], this gives $K(\mathrm{span}(E(t),\dot{\gamma }(t)))=k$ for every $t\in (0,D)$. Because the initial unit normal $E(0)$ was arbitrary, the same index argument gives this equality for every unit normal $W$ at time $t$; by homogeneity the quadratic form $W\mapsto g(R(W,\dot{\gamma })\dot{\gamma },W)$ equals $k\mathrm{\mid }W{\mathrm{\mid }}^2$ on $\dot{\gamma }(t{)}^{\perp }$; the operator $W\mapsto R(W,\dot{\gamma })\dot{\gamma }$ is self-adjoint there by [F11], and a symmetric bilinear form is determined by its diagonal [F11], so $R(W,\dot{\gamma })\dot{\gamma }=kW\text{for every }W\perp \dot{\gamma }\mathrm{.}$

The exponential maps are diffeomorphisms from the model ball. We claim that ${\exp }_p:B_0(D)⟶M\setminus \{q\},{\exp }_q:B_0(D)⟶M\setminus \{p\}$ are diffeomorphisms onto. Surjectivity: for $x\mathrm{\ne }q$ the identity of step 3.1 gives $d_g(p,x)=D-d_g(x,q)<D$, and [F1] supplies a minimizing geodesic from $p$ to $x$ whose initial vector $v$ satisfies ${\exp }_p(v)=x$ with $\mathrm{\mid }v\mathrm{\mid }=d_g(p,x)<D$. Injectivity on $B_0(D)$: if ${\exp }_p(v)={\exp }_p(w)$ with $\mathrm{\mid }v\mathrm{\mid }=:t$, $\mathrm{\mid }w\mathrm{\mid }=:t^{\mathrm{\prime }}<D$, then step 4.1 gives $d_g(p,{\exp }_p(v))=t$ along the geodesic $s\mapsto {\exp }_p(sv\mathrm{/}\mathrm{\mid }v\mathrm{\mid })$, and likewise $d_g(p,{\exp }_p(w))=t^{\mathrm{\prime }}$, so $t=t^{\mathrm{\prime }}$; if $t>0$ the two unit-speed minimizing geodesics from $p$ to the common point agree by step 4.2, hence $v=w$, while $t=0$ gives $v=w=0$. Local triviality: at $0$ the differential of ${\exp }_p$ is invertible by [F12], and on $B_0(D)\setminus \{0\}$ the map is a diffeomorphism by step 5.1; a bijective local diffeomorphism of boundaryless manifolds of the same dimension is a diffeomorphism by [F14]. The statements for ${\exp }_q$ are identical with $p$ and $q$ interchanged, using steps 4.2 and 5.1.

The radial pullback metric is the model metric. Define the model metric $h_k$ on the open tangent ball $B_0(D)$ of a Euclidean space by declaring that at the origin $h_k$ is the ambient inner product, and that at $u=te$ with $t\in (0,D)$, $\mathrm{\mid }e\mathrm{\mid }=1$ and vectors $w=ae+w^{\perp }$, $w^{\mathrm{\prime }}=a^{\mathrm{\prime }}e+w^{\mathrm{\prime }\perp }$ decomposed with $w^{\perp },w^{\mathrm{\prime }\perp }\perp e$, $h_k(w,w^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}+\frac{{\mathrm{sn}}_k(t{)}^2}{t^2}⟨w^{\perp },w^{\mathrm{\prime }\perp }⟩\mathrm{.}$ For the geodesic $\gamma (t)={\exp }_p(te)$, along which the curvature operator $W\mapsto R(W,\dot{\gamma })\dot{\gamma }$ equals $k$ on the normal bundle by step 5.2, the Gauss lemma and the Jacobi-field formula of [F12] give $d({\exp }_p{)}_u(w)=a\dot{\gamma }(t)+\frac{{\mathrm{sn}}_k(t)}{t}{P}_1{w}^{\perp },$ where $P_1$ is parallel transport along $s\mapsto {\exp }_p(su)$, $s\in [0,1]$. Indeed the tangential part is the radial derivative $d({\exp }_p{)}_u(ae)=a\dot{\gamma }(t)$ by the Gauss lemma of [F12], while the normal part is $d({\exp }_p{)}_u(w^{\perp })=J(1)$ for the Jacobi field $J(s)=\frac{1}{t}{\mathrm{sn}}_k(ts)P_s{w}^{\perp }$, which satisfies $J(0)=0$, $D_{s}J(0)=w^{\perp }$ and $D_s^{2}J=-t^{2}kJ=-R(J,\dot{\sigma })\dot{\sigma }$ by [F5], with $\sigma (s)={\exp }_p(su)$; uniqueness of Jacobi fields [F12] identifies it. Since parallel transport preserves inner products and $\dot{\gamma }(t)\perp P_1{w}^{\perp }$ by [F11], $g(d({\exp }_p{)}_{u}w,d({\exp }_p{)}_u{w}^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}+\frac{{\mathrm{sn}}_k(t{)}^2}{t^2}⟨w^{\perp },w^{\mathrm{\prime }\perp }⟩=h_k(w,w^{\mathrm{\prime }})\mathrm{.}$ At the origin both sides are the inner product because $d({\exp }_p{)}_0$ is the identity [F12]. Hence ${\exp }_p^{\ast }g=h_k$ on all of $B_0(D)$, and both sides are smooth Riemannian metrics there: $(ex{p}_p^{\ast }g)$ is positive definite at the origin, where the differential is invertible [F12], and at every nonzero point by the displayed formula, in which ${\mathrm{sn}}_k(t)>0$ for $0<t<D$ by [F5]. Repeating step 5.2 for radial geodesics from $q$, which minimize by the symmetric statement of step 4.1, gives the same radial curvature equality there; the preceding Jacobi computation then gives ${\exp }_q^{\ast }g=h_k$ under a linear identification of $T_{q}M$ with the Euclidean model [F14]. For the round sphere, [F13] gives that $S_R^n$ has constant sectional curvature $k$ and that ${\exp }_N$ is injective on $B_0(\pi R)$ and surjective onto $S_R^n\setminus \{S\}$: for $x\mathrm{\ne }S$ the distance formula gives $d_g(N,x)=R\arccos (⟨N,x⟩\mathrm{/}{R}^2)<\pi R$, and the geodesic formula realizes $x$ as ${\exp }_N$ of the initial vector of a minimizing geodesic from $N$. Repeating the displayed computation along the geodesics of $S_R^n$ with curvature operator $k$ gives ${\exp }_N^{\ast }g=h_k$ as well.

The three exponential maps are isometries onto. By step 6.1, ${\exp }_p:B_0(D)\to M\setminus \{q\}$ is a diffeomorphism, and by step 6.2 its pullback metric is $h_k$; therefore ${\exp }_p$ is a Riemannian isometry from $(B_0(D),h_k)$ onto $(M\setminus \{q\},g)$ [F14]. The same two steps give that ${\exp }_q:(B_0(D),h_k)\to (M\setminus \{p\},g)$ is an isometry onto, and, together with the bijectivity in [F13], that ${\exp }_N:(B_0(\pi R),h_k)\to (S_R^n\setminus \{S\},g)$ is an isometry onto; note $D=\pi R$.

The two charts and the transition isometry. Choose linear isometries $L_1:T_{p}M\to T_N{S}_R^n$ and $L_2:T_{q}M\to T_N{S}_R^n$ identifying the copies of $h_k$, which exist by [F14], and define ${\mathrm{\Psi }}_1:={\exp }_N\circ L_1\circ {\exp }_p^{-1},{\mathrm{\Psi }}_2:={\exp }_N\circ L_2\circ {\exp }_q^{-1}\mathrm{.}$ By step 7.1 the maps ${\mathrm{\Psi }}_1:M\setminus \{q\}\to S_R^n\setminus \{S\}$ and ${\mathrm{\Psi }}_2:M\setminus \{p\}\to S_R^n\setminus \{S\}$ are isometries onto. Since ${\mathrm{\Psi }}_1(p)={\exp }_N(0)=N={\mathrm{\Psi }}_2(q)$, both map $M\setminus \{p,q\}$ onto $P:=S_R^n\setminus \{N,S\}$, so the transition map $T:={\mathrm{\Psi }}_2\circ {\mathrm{\Psi }}_1^{-1}:P⟶P$ is a well-defined isometry of $P$ onto itself. The set $P$ is connected by [F15].

The transition extends to an orthogonal map and swaps the poles. By step 2.1 applied to the local isometry $T$ on the connected open set $P$ of [F15], there is an orthogonal map $\widehat{U}$ of ${\mathbb{R}}^{n+1}$ with $T=\widehat{U}{\mathrm{\mid }}_P$. We claim $\widehat{U}(N)=S,\widehat{U}(S)=N\mathrm{.}$ Let $x\in P$ with $x\to N$. Then ${\exp }_N^{-1}(x)\to 0$, because ${\exp }_N$ is a diffeomorphism from $B_0(\pi R)$ onto $S_R^n\setminus \{S\}$ with ${\exp }_N(0)=N$ [F13]; hence ${\mathrm{\Psi }}_1^{-1}(x)={\exp }_p(L_1^{-1}({\exp }_N^{-1}(x)))\to {\exp }_p(0)=p$ by continuity of $L_1^{-1}$ and of ${\exp }_p$, which is smooth on all of $T_{p}M$ [F8]. For $z\in M\setminus \{p\}$ with $z\to p$, the vector $v_z:={\exp }_q^{-1}(z)\in B_0(D)$ satisfies $\mathrm{\mid }{v}_z\mathrm{\mid }=d_g(q,z)$ because ${\exp }_q$ is an isometry by step 7.1, and $d_g(q,z)\to d_g(q,p)=D=\pi R$; so $\mathrm{\mid }{L}_2{v}_z\mathrm{\mid }\to \pi R$, and the explicit formula ${\exp }_N(w)=\cos (\mathrm{\mid }w\mathrm{\mid }\mathrm{/}R)N+(R\mathrm{/}\mathrm{\mid }w\mathrm{\mid })\sin (\mathrm{\mid }w\mathrm{\mid }\mathrm{/}R)w$ of [F13] gives ${\mathrm{\Psi }}_2(z)={\exp }_N(L_2{v}_z)\to -N=S$. Therefore $T(x)={\mathrm{\Psi }}_2({\mathrm{\Psi }}_1^{-1}(x))\to S$ as $x\to N$, and continuity of the linear map $\widehat{U}$ yields $\widehat{U}(N)={\lim }_{x\to N}\widehat{U}(x)={\lim }_{x\to N}T(x)=S\mathrm{.}$ Applying the same computation to $T^{-1}={\mathrm{\Psi }}_1\circ {\mathrm{\Psi }}_2^{-1}={\widehat{U}}^{-1}{\mathrm{\mid }}_P$, whose orthogonal extension is ${\widehat{U}}^{-1}$, gives ${\widehat{U}}^{-1}(N)=S$ and hence $\widehat{U}(S)=N$.