Rauch comparison theorem first form

Index comparison at a terminal time. [F1, F2, F4, F6, A1, given] Let $0<b\le T$ and let $J$, $\tilde{J}$ be normal Jacobi fields along ${\gamma }_1$, ${\gamma }_2$ with $J(0)=\tilde{J}(0)=0$ and $\mathrm{\mid }J(b)\mathrm{\mid }=\mathrm{\mid }\tilde{J}(b)\mathrm{\mid }=\beta >0$. We claim $I_{{\gamma }_1}(J,J)\le I_{{\gamma }_2}(\tilde{J},\tilde{J}),$ each index form taken over the respective geodesic segment from $0$ to $b$. Choose by [F4] parallel orthonormal frames $(E_1,\dots ,E_n)$ along ${\gamma }_1$ and $({\tilde{E}}_1,\dots ,{\tilde{E}}_n)$ along ${\gamma }_2$ with $E_1={\dot{\gamma }}_1$, ${\tilde{E}}_1={\dot{\gamma }}_2$ and $E_2(b)=\frac{J(b)}{\beta },{\tilde{E}}_2(b)=\frac{\tilde{J}(b)}{\beta };$ this is possible because $J(b)$ and $\tilde{J}(b)$ are normal, unit after division, and a parallel frame is determined by its value at $b$ and the prescription of any orthonormal basis there [F4]. Writing the normal fields in these frames, $J(t)={\sum }_{i=2}^n{a}_i(t)E_i(t),\tilde{J}(t)={\sum }_{i=2}^n{\tilde{a}}_i(t){\tilde{E}}_i(t),$ the coefficients are smooth, and $a_i(0)={\tilde{a}}_i(0)=0$ by $J(0)=\tilde{J}(0)=0$ [F3]. Define the transferred field $X(t):={\sum }_{i=2}^n{\tilde{a}}_i(t)E_i(t)$ along ${\gamma }_1$. Then $X$ is continuous piecewise $C^1$, $X(0)=J(0)=0$ and $X(b)=J(b)$, because $a_i(b)={\tilde{a}}_i(b)$ for $i\ge 2$: in the chosen frames the normal vector $J(b)\mathrm{/}\beta$ has coordinates $(0,1,0,\dots ,0)$, and so does $\tilde{J}(b)\mathrm{/}\beta$. Hypothesis 2 gives no conjugate point in $(0,b]$, so the index lemma [F1] applies on $[0,b]$ and yields $I_{{\gamma }_1}(J,J)\le I_{{\gamma }_1}(X,X)\mathrm{.}$ Now $X$ is normal to ${\dot{\gamma }}_1$, $\mathrm{\mid }X(t){\mathrm{\mid }}^2={\sum }_{i\ge 2}{\tilde{a}}_i(t{)}^2=\mathrm{\mid }\tilde{J}(t){\mathrm{\mid }}^2$ and $\mathrm{\mid }{D}_{t}X(t){\mathrm{\mid }}^2={\sum }_{i\ge 2}{\tilde{a}}_i^{\mathrm{\prime }}(t{)}^2=\mathrm{\mid }{D}_t\tilde{J}(t){\mathrm{\mid }}^2$ whenever both sides are evaluated, because the frames are orthonormal [F4]. At each $t$ where $X(t)\mathrm{\ne }0$, hypothesis 1 applied to the normal vectors $X(t)$ and $\tilde{J}(t)$ gives ${\sec }_{M_1}(X(t)\wedge {\dot{\gamma }}_1(t))\ge {\sec }_{M_2}(\tilde{J}(t)\wedge {\dot{\gamma }}_2(t)),$ and at points where $X(t)=0$ the curvature term of $I_{{\gamma }_1}(X,X)$ vanishes by [F6]. Multiplying by the common value $\mathrm{\mid }X(t){\mathrm{\mid }}^2=\mathrm{\mid }\tilde{J}(t){\mathrm{\mid }}^2$ and integrating, $I_{{\gamma }_1}(X,X)={\int }_0^b(\mathrm{\mid }{D}_{t}X{\mathrm{\mid }}^2-{\mathrm{Rm}}_{M_1}(X,{\dot{\gamma }}_1,{\dot{\gamma }}_1,X))dt\le {\int }_0^b(\mathrm{\mid }{D}_t\tilde{J}{\mathrm{\mid }}^2-{\mathrm{Rm}}_{M_2}(\tilde{J},{\dot{\gamma }}_2,{\dot{\gamma }}_2,\tilde{J}))dt=I_{{\gamma }_2}(\tilde{J},\tilde{J}),$ the last equality being the definition of the index form for the geodesic ${\gamma }_2$ [F2]. Chaining the two inequalities proves the claim.

Logarithmic derivatives of the squared norms. [F2, F3, F5, F7, step 1.1, given] Put $u:=\mathrm{\mid }{J}_1{\mathrm{\mid }}^2$ and $\tilde{u}:=\mathrm{\mid }{J}_2{\mathrm{\mid }}^2$ on $[0,T]$, and let $b_0:=\sup \{t\in (0,T]:\tilde{u}(s)>0\text{ for all }s\in (0,t)\}\mathrm{.}$ By [F7], $J_2(t)=t{D}_t{J}_2(0)+O(t^2)$ and hence $\tilde{u}(t)=t^2\mathrm{\mid }{D}_t{J}_2(0){\mathrm{\mid }}^2+O(t^3)=a^2{t}^2(1+O(t))$, so $\tilde{u}>0$ on some $(0,\epsilon )$ and $b_0>0$; the same expansion holds for $u$. By hypothesis 2 and [F5], $u(t)>0$ for every $t\in (0,T]$. Let $0<b<b_0$ and consider the normalized fields $J_1^b:=\frac{J_1}{\mathrm{\mid }{J}_1(b)\mathrm{\mid }},J_2^b:=\frac{J_2}{\mathrm{\mid }{J}_2(b)\mathrm{\mid }};$ these are normal Jacobi fields [F3] vanishing at $0$ and of unit norm at $b$, and the index comparison of step 1.1 gives $I_{{\gamma }_1}(J_1^b,J_1^b)\le I_{{\gamma }_2}(J_2^b,J_2^b)$. Since $J_1^b$ and $J_2^b$ are Jacobi fields, [F2] turns their index forms into boundary terms, $I_{{\gamma }_i}(J_i^b,J_i^b)=\left[g\right(D_t{J}_i^b,J_i^b){]}_0^b=\frac{⟨D_t{J}_i(b),J_i(b)⟩}{\mathrm{\mid }{J}_i(b){\mathrm{\mid }}^2}=\frac{u_i^{\mathrm{\prime }}(b)}{2u_i(b)},$ where $u_1=u$, $u_2=\tilde{u}$ and the vanishing at $0$ removed the lower boundary term. Multiplying the index inequality by $2$ gives $(\log \tilde{u}{)}^{\mathrm{\prime }}(b)\ge (\log u{)}^{\mathrm{\prime }}(b)$ for every $b\in (0,b_0)$, so $\tilde{u}\mathrm{/}u$ is nondecreasing on $(0,b_0)$. Finally the Taylor expansions give $\frac{\tilde{u}(t)}{u(t)}⟶\frac{a^2}{a^2}=1(t\downarrow 0),$ and therefore $\tilde{u}(t)\ge u(t)$ for all $t\in (0,b_0)$.

The bootstrap and the conclusion. [step 2.1, F5, F7, given] Suppose $b_0<T$. Both $u$ and $\tilde{u}$ are continuous, so the inequality $\tilde{u}\ge u$ on $(0,b_0)$ passes to the limit $t\uparrow b_0$: $\tilde{u}(b_0)\ge u(b_0)>0$, the strict positivity holding by hypothesis 2 and [F5] because $b_0\le T$. By continuity of $\tilde{u}$ there is $\delta >0$ with $\tilde{u}>0$ on $(0,b_0+\delta )$ and $b_0+\delta <T$, which contradicts the definition of $b_0$ as a supremum. Hence $b_0=T$. The inequality $\tilde{u}\ge u$ holds on $(0,T)$ and extends to $T$ by continuity, where $u(T)>0$; thus $\tilde{u}>0$ on $(0,T]$, and $\mathrm{\mid }{J}_1(t){\mathrm{\mid }}^2=u(t)\le \tilde{u}(t)=\mathrm{\mid }{J}_2(t){\mathrm{\mid }}^2(0\le t\le T)\mathrm{.}$ Taking square roots gives $\mathrm{\mid }{J}_1(t)\mathrm{\mid }\le \mathrm{\mid }{J}_2(t)\mathrm{\mid }$ on $[0,T]$, and $\tilde{u}>0$ on $(0,T]$ says exactly that $J_2$ has no zero there. The excluded endpoint $t=0$ is the common zero of both fields, and the first zero of $J_1$, excluded by hypothesis 2, is the point past which no comparison is asserted.