Rigidity in bishop gromov on an interval

The extended relative density and the weighted means. [F2, F3, F4, F6, given] Define $W_k(t):={\mathrm{sn}}_k(t{)}^{n-1}$ on the positive domain of ${\mathrm{sn}}_k$ and $W_k(t):=0$ for $t\ge \pi \mathrm{/}\sqrt{k}$ when $k>0$, and extend the radial volume Jacobian by zero past the cut time, ${\tilde{J}}_v(t):=J_p(t,v)(0<t<c_p(v)),{\tilde{J}}_v(t):=0(t\ge c_p(v)),$ and then $Q_v(t):={\tilde{J}}_v(t)\mathrm{/}{W}_k(t)$ where $W_k(t)>0$, $Q_v(t):=0$ where $W_k(t)=0$. By [F4] and [F6] the identity $Q_v{W}_k={\tilde{J}}_v$ holds on all of $(0,\mathrm{\infty })$: where $W_k>0$ it is the definition, and for $k>0$ and $t\ge \pi \mathrm{/}\sqrt{k}\ge c_p(v)$ both sides vanish. On $(0,\min (c_p(v),\pi \mathrm{/}\sqrt{k}))$ or $(0,c_p(v))$ according as $k>0$ or $k\le 0$, the function $Q_v$ equals $q_v$, so by [F2] it is nonincreasing, satisfies $0\le Q_v\le 1$ and $Q_v(t)\to 1$ as $t\downarrow 0$. For $0<s\le R$ define $A_v(s):=\frac{{\int }_0^s{Q}_v{W}_k}{{\int }_0^s{W}_k},\text{so }0\le A_v(s)\le 1,$ a measurable function of $v$ because $Q_v$ is [F3, F2]. The inner integral is positive: ${\int }_0^s{W}_k>0$ for every $s>0$ by [F4]. For $0<s_1<s_2$ the two-interval estimate for a nonincreasing profile gives $A_v(s_2)\le A_v(s_1)$: with $N={\int }_0^{s_1}{Q}_v{W}_k$, $D={\int }_0^{s_1}{W}_k$, $M={\int }_{s_1}^{s_2}{Q}_v{W}_k$, $E={\int }_{s_1}^{s_2}{W}_k$ one has $Q_v(t)W_k(t)D\le W_k(t)N$ for $t\in (s_1,s_2)$ after integrating $Q_v(t)\le Q_v(s)$ against $W_k(s)\ge 0$ over $s\in (0,s_1)$, and integration over $t\in (s_1,s_2)$ gives $DM\le NE$, that is $A_v(s_2)\le A_v(s_1)$. Moreover $A_v(s)\to 1$ as $s\downarrow 0$: given $\epsilon >0$ choose $\delta >0$ with $\mathrm{\mid }{Q}_v-1\mathrm{\mid }\le \epsilon$ on $(0,\delta )$ by [F2], then for $0<s<\delta$, $\mathrm{\mid }{A}_v(s)-1\mathrm{\mid }\le {\sup }_{0<t<s}\mathrm{\mid }{Q}_v(t)-1\mathrm{\mid }\le \epsilon$.

The three model spaces. [F13, F14, given] For the model of part 4: (i) for $k>0$, $M_k=S_{1\mathrm{/}\sqrt{k}}^n$ has constant sectional curvature $k$ and cut time $c_o(v)=\pi \mathrm{/}\sqrt{k}>R$ at every unit $v$, so $B_0(R)\setminus \{0\}\subseteq D_o$; [F7] applies off zero. At zero its differential is the identity, and no other vector of $B_0(R)$ maps to $o$, so it extends to a diffeomorphism onto $B_{M_k}(o,R)$; also $S_{1\mathrm{/}\sqrt{k}}^n$ is simply connected and geodesically complete, hence metrically complete by Hopf–Rinow [F13, F14]; (ii) for $k=0$, $M_k={\mathbb{R}}^n$ is complete and contractible, hence simply connected, with zero curvature [F13]; (iii) for $k<0$, the half-space $(U^n,g_k)$ has constant sectional curvature $k$ [F13], is convex and therefore contractible and simply connected, and is complete by the half-space model proposition [F13]; it has $K=k<0$ and is simply connected, so Cartan–Hadamard [F14] makes ${\exp }_o:T_o{M}_k\to M_k$ a global diffeomorphism. In the spherical case, [F7] and the explicit round-sphere distance formula give $d(o,{\exp }_{o}y)=\mathrm{\mid }y\mathrm{\mid }$ for $\mathrm{\mid }y\mathrm{\mid }<R$. In the Euclidean case this is the Euclidean distance formula; in the nonpositive-curvature case, the cited unique-minimizing-geodesic result gives the same radial distance identity. Thus in all cases ${\exp }_o$ maps $B_0(R)\subseteq T_o{M}_k$ diffeomorphically onto $B_{M_k}(o,R)$, with the pole included.

The exponential is a diffeomorphism onto the ball. [F6, F7, F14, given] Assume $c_p(v)\ge R$ for every $v\in S_{p}M$. Then $B_0(R)\setminus \{0\}\subseteq D_p$, and [F7] makes ${\exp }_p{\mathrm{\mid }}_{D_p}$ a diffeomorphism onto $M\setminus (\{p\}\cup \mathrm{Cut}(p))$; restricted to $B_0(R)\setminus \{0\}$ it is injective with nonsingular differential. At zero, [F7] gives the identity differential and image $p$; all other vectors of $B_0(R)$ map to points at positive distance from $p$. Its image is exactly $B(p,R)$: it is contained in $B(p,R)$ because $d_g(p,{\exp }_p(x))\le \mathrm{\mid }x\mathrm{\mid }<R$, and conversely every $q\in B(p,R)$ with $q\mathrm{\ne }p$ is ${\exp }_p(w)$ for a minimizing geodesic $[0,1]\ni s\mapsto {\exp }_p(sw)$ of length $d_g(p,q)<R$, by Hopf–Rinow ([F14]); while $p={\exp }_p(0)$. Hence ${\exp }_p:B_0(R)\to B(p,R)$ is a bijective local diffeomorphism, i.e. a diffeomorphism.

Ball volume as a spherical integral of weighted means. [F1, F2, F3, F4, F6, step 1.1, given] Fix $0<s\le R$ and apply the polar formula [F3] to $f:={\mathbf{1}}_{B(p,s)}$. For $0<t<c_p(v)$ one has $d_g(p,{\gamma }_v(t))=t$ by [F6], hence ${\mathbf{1}}_{B(p,s)}({\gamma }_v(t))={\mathbf{1}}_{\{t<s\}}$ and ${\mathrm{vol}}_g(B(p,s))={\int }_{S_{p}M}{\int }_0^{\min (c_p(v),s)}{J}_p(t,v)dtd{\sigma }_p(v)={\int }_{S_{p}M}\left({\int }_0^s{\tilde{J}}_v\right)d{\sigma }_p(v),$ the last identity by the extension by zero of ${\tilde{J}}_v$ in step 1.1 (for $k>0$ one has $c_p(v)\le \pi \mathrm{/}\sqrt{k}$ by [F6], so the truncation is consistent with the vanishing of $W_k$ beyond the model pole). Since ${\tilde{J}}_v=Q_v{W}_k$ [step 1.1] and $V_k^{\star }(s)=V_k(s)={\omega }_{n-1}{\int }_0^s{W}_k$ by [F1, F4] for $s\le R<\pi \mathrm{/}\sqrt{k}$ or $k\le 0$, division gives the identity of finite real numbers $R_p(s)=\frac{1}{{\omega }_{n-1}}{\int }_{S_{p}M}{A}_v(s)d{\sigma }_p(v),0<s\le R\mathrm{.}$

The model pullback metric. [F4, F8, F9, F12, step 1.2, given] Fix $y=tv\in B_0(R)$ in the model, $\mathrm{\mid }v\mathrm{\mid }=1$, and $a=a^{\perp }+\lambda v$ in $T_o{M}_k$. The model has constant sectional curvature $k$, so by [F12] the normal part of the Jacobi field $Y_a$ with initial data $(0,a)$ is ${\mathrm{sn}}_k(t)P_t{a}^{\perp }$, while its tangential part is $t\lambda P_{t}v$ by [F9]; hence, applying [F8] to this model Jacobi field, $({\exp }_o^{\ast }{g}_k{)}_y(a,b)=\frac{{\mathrm{sn}}_k(t{)}^2}{t^2}⟨a^{\perp },b^{\perp }⟩+⟨a,v⟩⟨b,v⟩$ for all $a,b\in T_o{M}_k$, where the inner products are those of $g_k$ at $o$, identified with those of $T_{p}M$ through the linear isometry $L$.

Equality of the ratio forces the extended density to be one. [F1, F11, step 1.1, step 2.1, given] Since $R_p$ is nonincreasing [F1] and $R_p(r)=R_p(R)$, one has $R_p(s)=R_p(R)$ for every $s\in [r,R]$; in particular ${\int }_{S_{p}M}(A_v(r)-A_v(R))d{\sigma }_p(v)={\omega }_{n-1}(R_p(r)-R_p(R))=0$ by step 2.1. The integrand is nonnegative because $A_v$ is nonincreasing in the radius [step 1.1], so $A_v(r)-A_v(R)=0$ for ${\sigma }_p$-almost every $v$ by [F11]. Fix such a $v$ and put $c:=A_v(r)=A_v(R)\in [0,1]$. Set $\varphi :=Q_v-c$, which is nonincreasing [step 1.1]. The two equal averages give ${\int }_0^r\varphi W_k=0$ and ${\int }_r^R\varphi W_k=0$. Put $D_1:={\int }_0^r{W}_k>0$ and $D_2:={\int }_r^R{W}_k>0$. Monotonicity yields $0=D_2{\int }_0^r\varphi (s)W_k(s)ds-D_1{\int }_r^R\varphi (t)W_k(t)dt={\int }_0^r{\int }_r^R(\varphi (s)-\varphi (t))W_k(s)W_k(t)dtds,$ whose integrand is nonnegative. Thus $\varphi (s)=\varphi (t)$ for almost every cross-pair, so Fubini implies that $\varphi$ is constant almost everywhere on $(0,R)$. Its average on $(0,r)$ is zero, so that constant is zero. Thus $Q_v=c$ almost everywhere on $(0,R)$, and therefore $A_v(s)=c$ for every $0<s\le R$, because $Q_v-c$ vanishes almost everywhere on each interval $(0,s)$. Passing to the limit $s\downarrow 0$ and using $A_v(0+)=1$ [step 1.1] gives $c=1$. Hence $Q_v=1$ almost everywhere on $(0,R)$; since $Q_v$ is nonincreasing, $Q_v\equiv 1$ on $(0,R)$: if $Q_v(t_0)<1$ for some $t_0\in (0,R)$, then $Q_v(t)\le Q_v(t_0)<1$ for every $t\in (t_0,R)$, contradicting $Q_v=1$ almost everywhere. On $(0,R)$ one has $Q_v=q_v=J_p\mathrm{/}{\mathrm{sn}}_k^{n-1}$ by step 1.1 and [F2, F4], so $J_p(t,v)={\mathrm{sn}}_k(t{)}^{n-1}(0<t<R)$ for ${\sigma }_p$-almost every $v$; and $c_p(v)\ge R$ for such $v$, since $Q_v$ vanishes on $[c_p(v),\mathrm{\infty })$ by step 1.1 and equals $1$ on $(0,R)$.

The traced Riccati equation forces $S_v={\mathrm{ct}}_k\mathrm{id}$. [F2, F4, F5, F10, step 3.1, given] Fix $v$ with $J_p(t,v)={\mathrm{sn}}_k(t{)}^{n-1}$ on $(0,R)$ [step 3.1]. For $0<t<\min (c_p(v),{\tau }_v)$ the logarithmic-derivative identity (Logarithmic derivative of the radial volume jacobian is the distance laplacian) gives $h(t):=\mathrm{tr}{S}_v(t)=\frac{d}{dt}\log J_p(t,v)=(n-1)\frac{d}{dt}\log {\mathrm{sn}}_k(t)=(n-1){\mathrm{ct}}_k(t)$ on $(0,R)$, since ${\mathrm{ct}}_k={\mathrm{sn}}_k^{\mathrm{\prime }}\mathrm{/}{\mathrm{sn}}_k$ and $J_p$ is positive there [F2, F4]; and $(0,R)$ lies inside the domain of definition of $S_v$ because $R\le c_p(v)\le {\tau }_v$ (Cut time does not exceed first conjugate time, [F5, F6]). Taking traces in the Riccati equation of [F5] on $(0,R)$ gives the exact identity $h^{\mathrm{\prime }}(t)+\mathrm{tr}(S_v(t{)}^2)+{\mathrm{Ric}}_{{\gamma }_v(t)}({\dot{\gamma }}_v(t),{\dot{\gamma }}_v(t))=0,$ where $\mathrm{tr}{R}_{{\gamma }_v}=\mathrm{Ric}({\dot{\gamma }}_v,{\dot{\gamma }}_v)$ by the trace definition of the Ricci curvature (Ricci curvature) in a parallel orthonormal frame, and $S_v(t)$ is self-adjoint [F5]. Substituting $h=(n-1){\mathrm{ct}}_k$, $h^{\mathrm{\prime }}=(n-1){\mathrm{ct}}_k^{\mathrm{\prime }}$ and ${\mathrm{ct}}_k^{\mathrm{\prime }}+{\mathrm{ct}}_k^2=-k$ [F4], and using $\mathrm{Ric}({\dot{\gamma }}_v,{\dot{\gamma }}_v)\ge (n-1)k$, gives $\mathrm{tr}\left(S_v^2\right)=(n-1)(k+{\mathrm{ct}}_k^2)-{\mathrm{Ric}}_{{\gamma }_v}({\dot{\gamma }}_v,{\dot{\gamma }}_v)\le (n-1){\mathrm{ct}}_k(t{)}^2=\frac{h(t{)}^2}{n-1}\mathrm{.}$ By [F10] applied to the self-adjoint $S_v(t)$ on the $(n-1)$-dimensional normal space, $(n-1)\mathrm{tr}(S_v^2)-h^2={\sum }_{i<j}({\lambda }_i-{\lambda }_j{)}^2\ge 0$, so equality holds and all eigenvalues of $S_v(t)$ are equal. Hence $S_v(t)$ is a scalar multiple of the identity with trace $h(t)$, that is $S_v(t)={\mathrm{ct}}_k(t){\mathrm{id}}_{N_0}(0<t<R),$ and equality in the displayed chain also gives ${\mathrm{Ric}}_{{\gamma }_v(t)}({\dot{\gamma }}_v(t),{\dot{\gamma }}_v(t))=(n-1)k$; the endpoint $t=0$ and the conjugate endpoint $t={\tau }_v$ are excluded by [F5].

The curvature and the radial Jacobi tensor. [F4, F5, F9, step 4.1, given] Fix the same $v$. Substituting $S_v={\mathrm{ct}}_k\mathrm{id}$ into the Riccati equation $S_v^{\mathrm{\prime }}+S_v^2+R_{{\gamma }_v}=0$ [F5] and using ${\mathrm{ct}}_k^{\mathrm{\prime }}+{\mathrm{ct}}_k^2=-k$ [F4] gives $R_{{\gamma }_v}(t)=(-({\mathrm{ct}}_k^{\mathrm{\prime }}+{\mathrm{ct}}_k^2))(t){\mathrm{id}}_{N_0}=k{\mathrm{id}}_{N_0}(0<t<R)$ for the parallel-frame curvature operator; since $P_t$ is isometric (Radial Jacobi tensor), $R(w,{\dot{\gamma }}_v(t)){\dot{\gamma }}_v(t)=kw$ for every $w\perp {\dot{\gamma }}_v(t)$, so every tangent two-plane containing ${\dot{\gamma }}_v(t)$ has sectional curvature $k$ (Sectional curvature, Riemann curvature four-tensor). For the Jacobi tensor: $S_v={\bar{A}}_v^{\mathrm{\prime }}{\bar{A}}_v^{-1}$ means ${\bar{A}}_v^{\mathrm{\prime }}={\mathrm{ct}}_k{\bar{A}}_v$ on $(0,R)$; every column $c$ of ${\bar{A}}_v$ therefore satisfies $c^{\mathrm{\prime }}={\mathrm{ct}}_{k}c$, so $c(t)={\lambda }_c{\mathrm{sn}}_k(t)$ for a constant ${\lambda }_c$ because ${\mathrm{sn}}_k>0$ and $\frac{d}{dt}\left(c\mathrm{/}{\mathrm{sn}}_k\right)=({\mathrm{ct}}_{k}c-{\mathrm{ct}}_{k}c)\mathrm{/}{\mathrm{sn}}_k=0$ [F4]. Evaluating the one-sided derivative at $0$ using ${\bar{A}}_v(0)=0$, ${\bar{A}}_v^{\mathrm{\prime }}(0)=\mathrm{id}$ and ${\mathrm{sn}}_k(t)\mathrm{/}t\to 1$ [F4, F5] gives ${\lambda }_c=c^{\mathrm{\prime }}(0)=e_c$, hence ${\bar{A}}_v(t)={\mathrm{sn}}_k(t){\mathrm{id}}_{N_0},A_v(t)={\mathrm{sn}}_k(t)P_t(0<t<R),$ and therefore $A_v(t)w={\mathrm{sn}}_k(t)P_{t}w$ for every $w\in N_0$.

The metric in normal coordinates on the good directions. [F8, F9, step 3.1, step 5.1, given] Let $G\subseteq S_{p}M$ be the set of directions $v$ satisfying the conclusions of steps 3.1 and 5.1; it has full ${\sigma }_p$-measure. Fix $v\in G$, $t\in (0,R)$ and $a\in T_{p}M$, and decompose $a=a^{\perp }+\lambda v$ with $a^{\perp }\in N_0$ and $\lambda =⟨a,v⟩$. Let $Y_a$ be the Jacobi field along the unit-speed geodesic ${\gamma }_v$ with $Y_a(0)=0$, $D_t{Y}_a(0)=a$. The field $t\mapsto {\mathrm{sn}}_k(t)P_t{a}^{\perp }+t\lambda P_{t}v$ is a sum of two Jacobi fields: the first is $A_v(t)a^{\perp }$ by step 5.1 (Radial Jacobi tensor), and the second is the tangential field of [F9] with the same initial data $(0,\lambda v)$; the sum has initial data $(0,a^{\perp }+\lambda v)=(0,a)$, so by uniqueness of Jacobi fields with prescribed initial data the two fields agree: $Y_a(t)={\mathrm{sn}}_k(t)P_t{a}^{\perp }+t\lambda P_{t}v\mathrm{.}$ By [F8], $d({\exp }_p{)}_{tv}(a)=\frac{1}{t}{Y}_a(t)$, and therefore, using that parallel transport is isometric and that normal and tangential fields are orthogonal, $({\exp }_p^{\ast }g{)}_{tv}(a,b)=\frac{1}{t^2}⟨Y_a(t),Y_b(t){⟩}_g=\frac{{\mathrm{sn}}_k(t{)}^2}{t^2}⟨a^{\perp },b^{\perp }⟩+⟨a,v⟩⟨b,v⟩$ for all $a,b\in T_{p}M$ and every $v\in G$. Equivalently, with $⟨a,b⟩=⟨a^{\perp },b^{\perp }⟩+⟨a,v⟩⟨b,v⟩$, $({\exp }_p^{\ast }g{)}_{tv}(a,b)=⟨a,b⟩+(\frac{{\mathrm{sn}}_k(t{)}^2}{t^2}-1)⟨a^{\perp },b^{\perp }⟩\mathrm{.}$

The formula extends to the whole ball. [F4, F7, step 6.1, given] Define on $B_0(R)$ the tensor field $F_x(a,b):=⟨a,b⟩+({\mathrm{sn}}_k(t{)}^2\mathrm{/}{t}^2-1)⟨a^{\perp },b^{\perp }⟩$ for $x=tv$, $\mathrm{\mid }v\mathrm{\mid }=1$, $t>0$, and $F_0:=⟨\cdot ,\cdot ⟩$; the coefficient ${\mathrm{sn}}_k(t{)}^2\mathrm{/}{t}^2$ extends smoothly to $t=0$ with value $1$, as the explicit branch formulas ${\mathrm{sn}}_k(t)=\sin (\sqrt{k}t)\mathrm{/}\sqrt{k}$, $t$, $\sinh (\sqrt{-k}t)\mathrm{/}\sqrt{-k}$ show (Comparison sine, cosine and cotangent functions), so $F$ is a smooth tensor field on $B_0(R)$; and $({\exp }_p^{\ast }g{)}_0=⟨\cdot ,\cdot ⟩=F_0$ because $d({\exp }_p{)}_0=\mathrm{id}$ [F7]. By step 6.1 the identity ${\exp }_p^{\ast }g=F$ holds at every $x=tv$ with $v\in G$, $0<t<R$. The set of such $x$ is dense in $B_0(R)$: every nonempty open subset of $S_{p}M$ has positive surface measure, so the full-measure set $G$ meets each such subset and is dense; a dense set of directions is dense in every sphere of radius $t<R$ under the diffeomorphism $v\mapsto tv$ of the unit sphere onto itself. Both sides of the identity are continuous tensor fields on $B_0(R)$: the left side because ${\exp }_p$ is smooth [F7] and $g$ is smooth, the right side because it is given by smooth formulas, has the normal-coordinate expression defining $F$. Hence ${\exp }_p^{\ast }g=F$ on all of $B_0(R)$, which is the displayed formula of part 3. In polar coordinates, an angular tangent vector $\xi \in T_v{S}_{p}M$ maps under $(t,v)\mapsto tv$ to $t\xi$, so its squared metric length is ${\mathrm{sn}}_k(t{)}^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$; this converts the normal-coordinate factor ${\mathrm{sn}}_k(t{)}^2\mathrm{/}{t}^2$ into the angular factor ${\mathrm{sn}}_k(t{)}^2$.