Cheng maximal diameter rigidity

Endpoints and the diameter pair. By [F1], $M$ is compact and $\mathrm{diam}(M,g)\le R$; the hypothesis gives $\mathrm{diam}(M,g)=R$. Define $\mathrm{\Phi }(x):=\sup \{d_g(x,y):y\in M\}$ for $x\in M$. By [F10] applied to the continuous function $y\mapsto d_g(x,y)$ on the nonempty compact space $M$ the supremum is a maximum, so $\mathrm{\Phi }(x)={\max }_y{d}_g(x,y)$; the triangle inequality 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 ${\sup }_x\mathrm{\Phi }(x)=\mathrm{diam}(M,g)=R$. By [F10] again, $\mathrm{\Phi }$ attains its maximum at some $p\in M$, and then choosing $q\in M$ with $d_g(p,q)=\mathrm{\Phi }(p)$ gives $d_g(p,q)={\sup }_x\mathrm{\Phi }(x)=R\mathrm{.}$

The model sphere exponential is an isometry onto the punctured sphere. Keep $R=\pi \mathrm{/}\sqrt{k}$ and put $\rho =1\mathrm{/}\sqrt{k}$, so $R=\pi \rho$. For the pole $N\in S_{\rho }^n$ and $v=tu\in T_N{S}_{\rho }^n$, $\mathrm{\mid }u\mathrm{\mid }=1$, the scaled sphere formula of [F11] is ${\stackrel{}{\exp }}_N(tu)=\cos (t\mathrm{/}\rho )N+\rho \sin (t\mathrm{/}\rho )u\mathrm{.}$ For $a=a^{\perp }+⟨a,u⟩u$ its differential at $tu$, $0<t<R$, is $d({\stackrel{}{\exp }}_N{)}_{tu}(a)=⟨a,u⟩(-\sin (t\mathrm{/}\rho )N\mathrm{/}\rho +\cos (t\mathrm{/}\rho )u)+\frac{\rho \sin (t\mathrm{/}\rho )}{t}{a}^{\perp }\mathrm{.}$ The radial vector in parentheses has unit length and is perpendicular to $a^{\perp }$. Since ${\mathrm{sn}}_k(t)=\rho \sin (t\mathrm{/}\rho )$, polarization gives ${\stackrel{}{\exp }}_N^{\ast }{g}_{S_{\rho }^n}=h_k$; at zero the same identity holds by the identity differential. By the sphere model [F11], this exponential is a diffeomorphism from $B_R(0_N)$ onto $S_{\rho }^n\setminus \{S\}$, where $S=-N$, and ${\stackrel{}{\exp }}_N(Ru)=S$ for every unit $u$. Thus it is a Riemannian isometry from $(B_R(0_N),h_k)$ onto the punctured sphere.

Agreement lemma for local isometries. Agreement lemma. 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 $F=G$. Proof. Let $A=\{x\in U:F(x)=G(x)\text{ and }d{F}_x=d{G}_x\}$. It is nonempty and closed, because smooth maps and their differentials are continuous. It is open: if $x\in A$, put $y=F(x)=G(x)$, use Existence of normal neighborhoods to choose a normal neighbourhood $U_0$ of $x$ in the common source metric, small enough for both maps. Write its points as ${\exp }_x(w)$ for $w$ near zero. For $z={\exp }_x(w)\in U_0$ with $w$ small, [F9] and the uniqueness in [F8] give $F(z)={\exp }_y(d{F}_{x}w)$ and $G(z)={\exp }_y(d{G}_{x}w)$, which are equal because $d{F}_x=d{G}_x$; and then $d{F}_z=d{G}_z$ by the chain rule applied to these expressions. Hence $U_0\subseteq A$, and $A$ is open. In the connected space $U$, the only nonempty open and closed subset is $U$, so $F=G$. This proves the lemma.

Cut times are at most $R$ and the radius-$R$ balls have full volume. Let $v\in S_{p}M$ and $0<t<c_p(v)$. By [F6], $d_g(p,{\gamma }_v(t))=t$; since $t\le \mathrm{diam}(M,g)=R$ we get $c_p(v)\le R$ for every $v\in S_{p}M$. Now for a Borel-measurable $f$ and the centre $p$, the polar formula of [F3] reads ${\int }_{M}fd{\mathrm{vol}}_g={\int }_{S_{p}M}{\int }_0^{c_p(v)}f({\gamma }_v(t))\det a_v(t)dtd{\sigma }_p(v)$. Taking $f={\mathbf{1}}_{B(p,R)}$: for $0<t<c_p(v)$ the identity $d_g(p,{\gamma }_v(t))=t$ shows $f({\gamma }_v(t))=1$ exactly when $t<R$, so the inner integral is over $(0,\min \{c_p(v),R\})$, and $c_p(v)\le R$ makes this $(0,c_p(v))$. Hence ${\mathrm{vol}}_g(B(p,R))={\mathrm{vol}}_g(M),$ and the same argument with $q$ in place of $p$ gives ${\mathrm{vol}}_g(B(q,R))={\mathrm{vol}}_g(M)>0$.

Extension of spherical isometries. Extension of spherical isometries. Let $U\subseteq S_{\rho }^n$ be a connected nonempty open subset and let $F:U\to S_{\rho }^n$ be a local isometry. Then there is an orthogonal map $\widehat{U}\in O(n+1)$ of ${\mathbb{R}}^{n+1}$ with $F=\widehat{U}{\mathrm{\mid }}_U$. Indeed, fix $x_0\in U$, let $(e_1,\dots ,e_n)$ be an orthonormal basis of $T_{x_0}{S}_{\rho }^n$ and set $f_i:=d{F}_{x_0}(e_i)$; then $(x_0\mathrm{/}\rho ,e_1,\dots ,e_n)$ and $(F(x_0)\mathrm{/}\rho ,f_1,\dots ,f_n)$ are orthonormal bases of ${\mathbb{R}}^{n+1}$, and the linear map sending the first basis to the second is a linear isometry of ${\mathbb{R}}^{n+1}$ by [F12], hence restricts to an isometry $\widehat{U}$ of $S_{\rho }^n$ with $\widehat{U}(x_0)=F(x_0)$ and $d{\widehat{U}}_{x_0}=d{F}_{x_0}$ (the tangent space of $S_{\rho }^n$ at a point is the orthogonal complement of that point, transported by the orthogonal map). By the agreement lemma of step 1.3, $F=\widehat{U}{\mathrm{\mid }}_U$. This proves the claim.

The half-balls are disjoint and the volume ratio is constant on $[R\mathrm{/}2,R]$. The balls $B(p,R\mathrm{/}2)$ and $B(q,R\mathrm{/}2)$ are disjoint: a common point $z$ would give $d_g(p,q)\le d_g(p,z)+d_g(z,q)<R$, contradicting step 1.1. Write $\alpha :={\mathrm{vol}}_g(M)\mathrm{/}{V}_k(R)$; by [F2], $R_p(s)={\mathrm{vol}}_g(B(p,s))\mathrm{/}{V}_k^{\star }(s)$ is nonincreasing with $R_p(R)={\mathrm{vol}}_g(B(p,R))\mathrm{/}{V}_k(R)=\alpha$ by [F2] and step 2.1, so $R_p(R\mathrm{/}2)\ge \alpha ,\text{equivalently}{\mathrm{vol}}_g(B(p,R\mathrm{/}2))\ge \alpha V_k(R\mathrm{/}2),$ and the same inequality holds with $q$. The sequence of inequalities ${\mathrm{vol}}_g(M)\ge {\mathrm{vol}}_g(B(p,R\mathrm{/}2))+{\mathrm{vol}}_g(B(q,R\mathrm{/}2))\ge 2\alpha V_k(R\mathrm{/}2)=\alpha V_k(R)={\mathrm{vol}}_g(M)$ uses disjointness, then the two displayed bounds, then $V_k(R)=2V_k(R\mathrm{/}2)$ from [F4]. Every inequality is therefore an equality: the two balls have equal volume $\alpha V_k(R\mathrm{/}2)={\mathrm{vol}}_g(M)\mathrm{/}2$ and $R_p(R\mathrm{/}2)=R_q(R\mathrm{/}2)=\alpha \mathrm{.}$ By monotonicity of $R_p$ and $R_q$ and $R_p(R)=R_q(R)=\alpha$ [F2, step 2.1], this forces $R_p(s)=R_q(s)=\alpha \text{for every }s\in [R\mathrm{/}2,R]\mathrm{.}$

The volume ratio is identically one. Let $0<s<R\mathrm{/}2$. Then $B(p,s)$ and $B(q,R-s)$ are disjoint, since a common point would give $d_g(p,q)\le d_g(p,z)+d_g(z,q)<s+(R-s)=R$; and $R-s\in (R\mathrm{/}2,R)$, so step 3.1 applies at $R-s$ and gives ${\mathrm{vol}}_g(B(q,R-s))=\alpha V_k(R-s)$. Hence ${\mathrm{vol}}_g(M)\ge {\mathrm{vol}}_g(B(p,s))+\alpha V_k(R-s)={\mathrm{vol}}_g(B(p,s))+\alpha (V_k(R)-V_k(s))={\mathrm{vol}}_g(B(p,s))+{\mathrm{vol}}_g(M)-\alpha V_k(s),$ using $V_k(s)+V_k(R-s)=V_k(R)$ from [F4]. Therefore ${\mathrm{vol}}_g(B(p,s))\le \alpha V_k(s)$, that is, $R_p(s)\le \alpha$; monotonicity [F2] and step 3.1 give $R_p(s)\ge R_p(R\mathrm{/}2)=\alpha$, so $R_p(s)=\alpha \text{for every }s\in (0,R),$ and the same holds for $R_q$. Letting $s\downarrow 0$ and using ${\lim }_{s\downarrow 0}{R}_p(s)=1$ from [F2] gives $\alpha =1$. Consequently $R_p(s)=R_q(s)=1,{\mathrm{vol}}_g(B(p,s))=V_k(s)={\mathrm{vol}}_g(B(q,s))(0<s<R),$ and ${\mathrm{vol}}_g(M)=V_k(R)$.

Complementarity of the ball volumes at every radius. For $0\le s\le R$ the two balls $B(p,s)$ and $B(q,R-s)$ are disjoint (same triangle-inequality argument as in step 4.1, with $s=0$ or $s=R$ trivial), and step 4.1 gives ${\mathrm{vol}}_g(B(p,s))+{\mathrm{vol}}_g(B(q,R-s))=V_k(s)+V_k(R-s)=V_k(R)={\mathrm{vol}}_g(M)\mathrm{.}$

Equal complementary half-balls. We claim that $d_g(p,x)+d_g(q,x)=R\text{for every }x\in M\mathrm{.}$ Suppose not. Then $d_g(p,x)+d_g(q,x)>R$, and since $d_g(p,x)=:t$ is finite we may choose $\epsilon >0$ with $d_g(p,x)+d_g(q,x)>R+2\epsilon$ and $\epsilon <t$. Put $t^{\mathrm{\prime }}:=t-\epsilon \ge 0$. The three balls $B(p,t^{\mathrm{\prime }})$, $B(q,R-t^{\mathrm{\prime }})$ and $B(x,\epsilon )$ are pairwise disjoint: a point $z$ of the first two would give $d_g(p,q)\le d_g(p,z)+d_g(z,q)<t^{\mathrm{\prime }}+(R-t^{\mathrm{\prime }})=R$; while $d_g(p,z)\ge d_g(p,x)-d_g(x,z)>t-\epsilon =t^{\mathrm{\prime }},d_g(q,z)\ge d_g(q,x)-\epsilon >R+2\epsilon -t-\epsilon =R-(t-\epsilon )=R-t^{\mathrm{\prime }}$ for $z\in B(x,\epsilon )$, so such a $z$ lies in neither of the first two balls. Hence, using $R-t^{\mathrm{\prime }}\in [0,R]$ and step 5.1, ${\mathrm{vol}}_g(M)\ge {\mathrm{vol}}_g(B(p,t^{\mathrm{\prime }}))+{\mathrm{vol}}_g(B(q,R-t^{\mathrm{\prime }}))+{\mathrm{vol}}_g(B(x,\epsilon ))={\mathrm{vol}}_g(M)+{\mathrm{vol}}_g(B(x,\epsilon )),$ so ${\mathrm{vol}}_g(B(x,\epsilon ))\le 0$, contradicting [F13]. Therefore the claimed identity holds.

Cut time equals the diameter. We claim that $c_p(v)=R$ for every $v\in S_{p}M$, and likewise with $q$ in place of $p$. Let $v\in S_{p}M$ and suppose $t:=c_p(v)<R$. Put $x:={\gamma }_v(t)={\exp }_p(tv)$. By [F6], $d_g(p,x)=t$; by step 6.1, $d_g(q,x)=R-t>0$. By [F8] choose a unit-speed minimizing geodesic $\tilde{\gamma }:[0,R-t]\to M$ from $x$ to $q$, and define $\sigma :[0,R]\to M$ by $\sigma (s)={\gamma }_v(s)$ for $0\le s\le t$ and $\sigma (s)=\tilde{\gamma }(s-t)$ for $t\le s\le R$. Then $\sigma$ is piecewise smooth from $p$ to $q$ and has length $R=d_g(p,q)$; by [F7] its arclength reparametrization is a smooth unbroken geodesic whose trace contains the common trace of $\sigma$ on $[0,R]$. The two pieces have unit speed, so arclength reparametrization leaves the parameter $s$ unchanged; by geodesic uniqueness, the resulting geodesic agrees with ${\gamma }_v$ on $[0,R]$. Every subsegment of a minimizing curve minimizes, hence $d_g(p,{\gamma }_v(s))=s$ for $0\le s\le R$. Hence $c_p(v)\ge R$, contradicting $t=c_p(v)<R$. Therefore $c_p(v)\ge R$, and $c_p(v)\le R$ by step 2.1, so $c_p(v)=R$. The argument with $p$ and $q$ interchanged gives $c_q(w)=R$ for every $w\in S_{q}M$.

The exponential maps are isometries onto the punctured manifolds. Fix $0<R^{\mathrm{\prime }}<R$ and $0<r<R^{\mathrm{\prime }}$. By step 4.1, $R_p(r)=R_p(R^{\mathrm{\prime }})=1$; by step 7.1, $c_p(v)=R\ge R^{\mathrm{\prime }}$ for every $v\in S_{p}M$; and $R^{\mathrm{\prime }}<R=\pi \mathrm{/}\sqrt{k}$. So the rigidity proposition [F5] applies and yields, for every such $R^{\mathrm{\prime }}$: ${\exp }_p:B_0(R^{\mathrm{\prime }})\to B(p,R^{\mathrm{\prime }})\text{ is an isometry onto with }({\exp }_p^{\ast }g)=h_k\mathrm{.}$ Taking the union over all $R^{\mathrm{\prime }}<R$ and using $B(p,R)=\{x:d_g(p,x)<R\}=M\setminus \{q\}$, which follows from step 6.1 ($d_g(p,x)=R-d_g(q,x)<R$ exactly when $x\mathrm{\ne }q$), the metric identity holds on all of $B_0(R)$, the map ${\exp }_p$ is an isometry from $(B_0(R),h_k)$ onto the open ball $B(p,R)$ of $M$, and ${\exp }_p(B_0(R))=M\setminus \{q\}$ because every point of $B(p,R)$ lies in some $B(p,R^{\mathrm{\prime }})$ with $R^{\mathrm{\prime }}<R$ and ${\exp }_p(B_0(R^{\mathrm{\prime }}))=B(p,R^{\mathrm{\prime }})$. In particular ${\mathrm{\Phi }}_0:={\exp }_p:(B_0(R),h_k)⟶(M\setminus \{q\},g)$ is an isometry onto, and by the same argument at the centre $q$ ${\mathrm{\Psi }}_0:={\exp }_q:(B_0(R),h_k)⟶(M\setminus \{p\},g)$ is an isometry onto (the tangent balls are identified with $T_{p}M$ and $T_{q}M$ respectively, each carrying $h_k$ defined by the formula of [F5] with the respective centre).

Two chart isometries and the transition. Choose linear isometries $L_1:T_{p}M\to T_N{S}_{\rho }^n$ and $L_2:T_{q}M\to T_N{S}_{\rho }^n$, which exist by [F12], and define ${\mathrm{\Psi }}_1:={\stackrel{}{\exp }}_N\circ L_1\circ {\exp }_p^{-1}:M\setminus \{q\}⟶S_{\rho }^n\setminus \{S\},{\mathrm{\Psi }}_2:={\stackrel{}{\exp }}_N\circ L_2\circ {\exp }_q^{-1}:M\setminus \{p\}⟶S_{\rho }^n\setminus \{S\}\mathrm{.}$ By step 8.1 and step 1.2 each factor is an isometry onto its target, so ${\mathrm{\Psi }}_1$ and ${\mathrm{\Psi }}_2$ are isometries onto $S_{\rho }^n\setminus \{S\}$. In particular ${\mathrm{\Psi }}_1(p)=N,{\mathrm{\Psi }}_2(q)=N,{\mathrm{\Psi }}_1(M\setminus \{p,q\})={\mathrm{\Psi }}_2(M\setminus \{p,q\})=S_{\rho }^n\setminus \{S,N\}=:P\mathrm{.}$ The transition map $T:={\mathrm{\Psi }}_2\circ {\mathrm{\Psi }}_1^{-1}:P⟶P$ is an isometry of $P$ onto itself. The set $P$ is path-connected, hence connected: via the diffeomorphism ${\exp }_N^{-1}$, it corresponds to $B_R(0_N)\setminus \{0_N\}$. Radial segments connect every point of this punctured ball to a fixed radius $r\in (0,R)$, and the sphere of radius $r$ is path-connected for $n\ge 2$; thus the punctured ball, and hence $P$, is path-connected.

The transition is a global orthogonal map swapping the poles. By step 2.2 applied to the local isometry $T$ on the connected open set $P$, there is $\widehat{U}\in O(n+1)$ with $T=\widehat{U}{\mathrm{\mid }}_P$. We claim $\widehat{U}(N)=S,\widehat{U}(S)=N\mathrm{.}$ For $x\in P$ with $x\to N$: $({\stackrel{}{\exp }}_N^{-1})(x)\to 0$, so ${\mathrm{\Psi }}_1^{-1}(x)={\exp }_p(L_1^{-1}({\stackrel{}{\exp }}_N^{-1}(x)))\to {\exp }_p(0)=p$; and for $z\in M\setminus \{p\}$ with $z\to p$ one has $d_g(q,z)\to d_g(q,p)=R$, so $L_2({\exp }_q^{-1}(z))$ is a tangent vector of norm $d_g(q,z)\to R$ and ${\stackrel{}{\exp }}_N$ of such vectors tends to $S$ by the formula ${\stackrel{}{\exp }}_N(Ru)=S$ (step 1.2). Hence $T(x)={\mathrm{\Psi }}_2({\mathrm{\Psi }}_1^{-1}(x))\to S$ as $x\to N$, and continuity of $\widehat{U}$ gives $\widehat{U}(N)={\lim }_{x\to N}\widehat{U}(x)={\lim }_{x\to N}T(x)=S$. The identity $\widehat{U}(S)=N$ follows by the same computation with $p$ and $q$ interchanged.

Gluing to the global isometry $\mathrm{\Theta }$. [step 1.2, step 8.1, step 9.1, step 10.1] Define $\mathrm{\Theta }:M\to S_{\rho }^n$ by $\mathrm{\Theta }(x)={\mathrm{\Psi }}_1(x)(x\mathrm{\ne }q),\mathrm{\Theta }(x)={\widehat{U}}^{-1}({\mathrm{\Psi }}_2(x))(x\mathrm{\ne }p)\mathrm{.}$ The two formulas agree on $M\setminus \{p,q\}$: there ${\mathrm{\Psi }}_2=T\circ {\mathrm{\Psi }}_1=\widehat{U}\circ {\mathrm{\Psi }}_1$ by step 10.1, so ${\widehat{U}}^{-1}{\mathrm{\Psi }}_2={\mathrm{\Psi }}_1$. Hence $\mathrm{\Theta }$ is well defined, and it is smooth: near any point different from $p$ and $q$ both expressions agree and are compositions of smooth maps, near $p$ (where $p\mathrm{\ne }q$) the first expression ${\mathrm{\Psi }}_1$ is smooth, and near $q$ the second expression ${\widehat{U}}^{-1}\circ {\mathrm{\Psi }}_2$ is smooth. Being locally one of the two isometries onto an open set, $\mathrm{\Theta }$ is a local isometry. It is bijective: on $M\setminus \{q\}$ it equals ${\mathrm{\Psi }}_1$ and maps onto $S_{\rho }^n\setminus \{S\}$; moreover $\mathrm{\Theta }(q)={\widehat{U}}^{-1}({\mathrm{\Psi }}_2(q))={\widehat{U}}^{-1}(N)={\widehat{U}}^{-1}(\widehat{U}(S))=S$ by step 10.1 and ${\mathrm{\Psi }}_2(q)=N$ from step 9.1, so $\mathrm{\Theta }$ is onto. If $\mathrm{\Theta }(x)=\mathrm{\Theta }(y)=w$ with $w\mathrm{\ne }S$, then $x\mathrm{\ne }q$ and $y\mathrm{\ne }q$, so ${\mathrm{\Psi }}_1(x)=w={\mathrm{\Psi }}_1(y)$ and $x=y$ by injectivity of ${\mathrm{\Psi }}_1$; and if $w=S$ then $x=q=y$ because ${\mathrm{\Psi }}_1$ takes values in $S_{\rho }^n\setminus \{S\}$ on $M\setminus \{q\}$. Hence $\mathrm{\Theta }$ is a bijective local isometry between boundaryless manifolds of the same dimension, so $\mathrm{\Theta }$ is a diffeomorphism with ${\mathrm{\Theta }}^{\ast }{g}_{S_{\rho }^n}=g$: a Riemannian isometry, and $M$ is isometric to the round sphere $S_{\rho }^n$ of curvature $1\mathrm{/}{\rho }^2=k$.