Rigidity in rauch comparison

First form: the equality hypothesis meets the Rauch comparison. [A1, F1, F5, F8, F9, F4, given] By [F5] the radial sectional hypothesis is $R_{\gamma }(t)\ge k\mathrm{id}$ on $N_t$ for $0\le t\le T$. The model field $J_k(t)={\mathrm{sn}}_k(t){\mathrm{\Phi }}_t{w}_k$ is a normal Jacobi field along ${\gamma }_k$: it is normal because ${\mathrm{\Phi }}_t$ is isometric and $w_k=\mathcal{I}{w}_0\perp {\dot{\gamma }}_k(0)$ and $\mathrm{\mid }{w}_k\mathrm{\mid }=\mathrm{\mid }{w}_0\mathrm{\mid }=a$ [F4], and by [F9] and [F8] $D_t^2{J}_k+R(J_k,{\dot{\gamma }}_k){\dot{\gamma }}_k={\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\Phi }}_t{w}_k+k(g({\dot{\gamma }}_k,{\dot{\gamma }}_k)J_k-g(J_k,{\dot{\gamma }}_k){\dot{\gamma }}_k)=({\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}+k{\mathrm{sn}}_k){\mathrm{\Phi }}_t{w}_k=0,$ since $g(J_k,{\dot{\gamma }}_k)=0$. Its initial data are $J_k(0)=0$ and $D_t{J}_k(0)=w_k=\mathcal{I}(D_{t}J(0))$, so the initial derivatives correspond under the chosen isometry and both have norm $a>0$. All hypotheses of [F1] are met, and [F1] gives $\mathrm{\mid }J(t)\mathrm{\mid }\le \mathrm{\mid }{J}_k(t)\mathrm{\mid }$ on $[0,T]$ together with $J_k(t)\mathrm{\ne }0$ for $0<t\le T$. The equality hypothesis says $\mathrm{\mid }J(t)\mathrm{\mid }=\mathrm{\mid }{J}_k(t)\mathrm{\mid }=a{\mathrm{sn}}_k(t)\mathrm{\ne }0(0<t\le T),$ so ${\mathrm{sn}}_k(b)>0$ for every $b\in (0,T]$. For $k>0$ this forces $T<\pi \mathrm{/}\sqrt{k}$: at $t_0=\pi \mathrm{/}\sqrt{k}$ the model sine vanishes, so were $t_0\le T$ the equality $\mathrm{\mid }J(t_0)\mathrm{\mid }=a{\mathrm{sn}}_k(t_0)=0$ would make $t_0$ a conjugate instant of $\gamma (0)$ along $\gamma$, contrary to the hypothesis that there is none in $(0,T]$; hence $T<\pi \mathrm{/}\sqrt{k}$ and in particular ${\mathrm{sn}}_k>0$ on all of $(0,T]$, which makes every division below legitimate.

First form: index forms of the normalized Jacobi fields. Fix $b\in (0,T]$. By the equality hypothesis and [F1], $\mathrm{\mid }J(b)\mathrm{\mid }=\mathrm{\mid }{J}_k(b)\mathrm{\mid }=a{\mathrm{sn}}_k(b)>0$; normalize $J^b:=\frac{J}{\mathrm{\mid }J(b)\mathrm{\mid }},J_k^b:=\frac{J_k}{\mathrm{\mid }{J}_k(b)\mathrm{\mid }},$ normal Jacobi fields along $\gamma$ and ${\gamma }_k$ with $J^b(0)=J_k^b(0)=0$ and $\mathrm{\mid }{J}^b(b)\mathrm{\mid }=\mathrm{\mid }{J}_k^b(b)\mathrm{\mid }=1$; a scalar multiple of a Jacobi field is a Jacobi field by [F2] and [F11], and $\mathrm{\mid }{J}^b\mathrm{\mid }>0$ on $(0,b]$. By [F2], applied on the segment $[0,b]$ to the Jacobi fields $J^b$ and $J_k^b$, $I_{\gamma }(J^b,J^b)=[g(D_t{J}^b,J^b){]}_0^b=\frac{g(D_{t}J(b),J(b))}{\mathrm{\mid }J(b){\mathrm{\mid }}^2}=\frac{1}{2}(\log \mathrm{\mid }J{\mathrm{\mid }}^2{)}^{\mathrm{\prime }}(b),$ and likewise $I_{{\gamma }_k}(J_k^b,J_k^b)=\frac{1}{2}(\log \mathrm{\mid }{J}_k{\mathrm{\mid }}^2{)}^{\mathrm{\prime }}(b)$ on $[0,b]$ along ${\gamma }_k$; the lower boundary terms vanish because $J^b(0)=J_k^b(0)=0$. The equality hypothesis $\mathrm{\mid }J{\mathrm{\mid }}^2=\mathrm{\mid }{J}_k{\mathrm{\mid }}^2$ on $(0,T]$ makes the two logarithmic derivatives equal at every $b\in (0,T]$, so

Second form: the Riccati setup and its comparison. [F5, F6, F7, F8, given] Transport the given field to $N_0$: $y(t):=P_t^{-1}J(t)$ satisfies $y^{\mathrm{\prime }\mathrm{\prime }}+R_{\gamma }y=0$ with $y(0)=u\mathrm{\ne }0$ and $y^{\mathrm{\prime }}(0)=D_{t}J(0)=\lambda u$; the operator family $R_{\gamma }$ is continuous and self-adjoint by Algebraic symmetries of the Riemann tensor under [A1], and by the curvature hypothesis $R_{\gamma }(t)\ge k\mathrm{id}$ in the Loewner order. By definition $Y$ is the endomorphism-valued solution of the same matrix initial value problem, so $y(t)=Y(t)u$ and $J(t)=P_{t}y(t)$, with $\mathrm{\mid }J(t)\mathrm{\mid }=\mathrm{\mid }y(t)\mathrm{\mid }$. The first focal time $t_f$ is the first singular time of $Y$ in $[0,T]$ (or $+\mathrm{\infty }$ if none occurs there). Applying [F7] with $E=N_0$, $R_1=R_{\gamma }$, $R_2=k\mathrm{id}$, $Y_1=Y$, $Y_2=f_k\mathrm{id}$, and using [F8], gives for $0<t\le T$ with $t<t_f$: the operator $S:=Y^{\mathrm{\prime }}{Y}^{-1}$ is self-adjoint and solves $S^{\mathrm{\prime }}+S^2+R_{\gamma }=0,$ and, with $g_k:=f_k^{\mathrm{\prime }}\mathrm{/}{f}_k$, $S(t)\le g_k(t)\mathrm{id},\text{that is}\tilde{U}(t):=g_k(t)\mathrm{id}-S(t)\ge 0$ in the Loewner order. Moreover $Y(t)$ is invertible and $J(t)=P_{t}Y(t)u\mathrm{\ne }0$ for $0<t\le T$ with $t<t_f$, and $f_k(t)>0$ there by [F6], so $g_k$ is defined and smooth on the compared interval with $g_k^{\mathrm{\prime }}=-k-g_k^2,$ which follows by differentiating $g_k=f_k^{\mathrm{\prime }}\mathrm{/}{f}_k$ with [F8].

First form: transferring the model field. [F4, F5, F3, step 1.2, given] Fix $b\in (0,T]$ and choose parallel orthonormal frames $(E_1,\dots ,E_n)$ along $\gamma$ and $({\tilde{E}}_1,\dots ,{\tilde{E}}_n)$ along ${\gamma }_k$ with $E_1=\dot{\gamma }$, ${\tilde{E}}_1={\dot{\gamma }}_k$ and $E_2(b)=J^b(b),{\tilde{E}}_2(b)=J_k^b(b);$ this is possible because the two vectors are normal and of unit length, and a parallel frame is determined by its value at the single point $b$ [F4]. Writing $J_k^b(t)={\sum }_{i=2}^n{\tilde{a}}_i(t){\tilde{E}}_i(t),{\sum }_{i=2}^n{\tilde{a}}_i(t{)}^2=\mathrm{\mid }{J}_k^b(t){\mathrm{\mid }}^2,$ define the transferred field $X(t):={\sum }_{i=2}^n{\tilde{a}}_i(t)E_i(t)$ along $\gamma$. Then $X$ is a continuous field that is $C^{\mathrm{\infty }}$ on $[0,b]$, $X(0)=J^b(0)=0$, and $X(b)=J^b(b)$ because the coefficient vector $({\tilde{a}}_2(b),\dots ,{\tilde{a}}_n(b))=(1,0,\dots ,0)$ of $J_k^b(b)$ is also the coefficient vector of $J^b(b)$ in the frame $(E_i)$; and by [F4] $\mathrm{\mid }X(t){\mathrm{\mid }}^2=\mathrm{\mid }{J}_k^b(t){\mathrm{\mid }}^2,\mathrm{\mid }{D}_{t}X(t){\mathrm{\mid }}^2=\mathrm{\mid }{D}_t{J}_k^b(t){\mathrm{\mid }}^2(0\le t\le b),$ because the frames are parallel orthonormal and the coefficients transfer verbatim. The index lemma [F3] applies on the conjugate-free segment $[0,b]$ from step 1.1 and gives $I_{\gamma }(J^b,J^b)\le I_{\gamma }(X,X)\mathrm{.}$ Where $X(t)\mathrm{\ne }0$, [F5] and the curvature hypothesis give ${\sec }_M(\mathrm{span}\{X(t),\dot{\gamma }(t)\})\ge k={\sec }_{M_k}(\mathrm{span}\{J_k^b(t),{\dot{\gamma }}_k(t)\}),$ and where $X(t)=0$ the curvature term of the integrand vanishes; since $\mathrm{\mid }X\mathrm{\mid }=\mathrm{\mid }{J}_k^b\mathrm{\mid }$ and $\mathrm{\mid }{D}_{t}X\mathrm{\mid }=\mathrm{\mid }{D}_t{J}_k^b\mathrm{\mid }$, multiplying the two curvature terms by the common squared norm and integrating over $[0,b]$ yields $I_{\gamma }(X,X)\le I_{{\gamma }_k}(J_k^b,J_k^b)\mathrm{.}$ Chaining with [F3] and the display of step 1.2, $I_{\gamma }(J^b,J^b)\le I_{\gamma }(X,X)\le I_{{\gamma }_k}(J_k^b,J_k^b)=I_{\gamma }(J^b,J^b),$ so both inequalities are equalities.

Second form: equality annihilates the Riccati difference. [F2, F10, step 1.3, given] For $0<t\le T$ with $t<t_f$, put $y:=Y(t)u$. Then $y^{\mathrm{\prime }}=Sy$ and $\mathrm{\mid }y\mathrm{\mid }=\mathrm{\mid }J\mathrm{\mid }=f_k\mathrm{\mid }u\mathrm{\mid }$, with $f_k>0$. Hence $\frac{d}{dt}\log \mathrm{\mid }y\mathrm{\mid }=\frac{⟨Sy,y⟩}{\mathrm{\mid }y{\mathrm{\mid }}^2}=\frac{f_k^{\mathrm{\prime }}}{f_k}=:g_k,$ so $⟨\tilde{U}y,y⟩=0$ for $\tilde{U}:=g_k\mathrm{id}-S$. Since $\tilde{U}$ is self-adjoint and positive semidefinite by step 1.3, the spectral theorem gives $\tilde{U}y=0$.

First form: equality in the index lemma identifies the fields. [F3, step 2.1, given] The first equality of step 2.1, $I_{\gamma }(J^b,J^b)=I_{\gamma }(X,X)$, is the equality case of the index lemma [F3] for the Jacobi field $J^b$ and the admissible field $X$ with the same endpoints; hence $X=J^b$ on $[0,b]$. By step 2.1 the transferred model field $X$ is the parallel transfer of $J_k^b$, so for every $t\in [0,b]$ $J^b(t)=X(t)={\sum }_{i=2}^n{\tilde{a}}_i(t)E_i(t)\mathrm{.}$

First form: equality in the curvature comparison. [F5, step 2.1, given] Write the difference of the two index forms of step 2.1 as a single integral. Using $\mathrm{\mid }{D}_{t}X\mathrm{\mid }=\mathrm{\mid }{D}_t{J}_k^b\mathrm{\mid }$, $\mathrm{Rm}(X,\dot{\gamma },\dot{\gamma },X)={\sec }_M(\dots )\mathrm{\mid }X{\mathrm{\mid }}^2$ and $\mathrm{Rm}(J_k^b,{\dot{\gamma }}_k,{\dot{\gamma }}_k,J_k^b)=k\mathrm{\mid }{J}_k^b{\mathrm{\mid }}^2$ from [F5], $0=I_{\gamma }(X,X)-I_{{\gamma }_k}(J_k^b,J_k^b)={\int }_0^b\mathrm{\mid }X(t){\mathrm{\mid }}^2(k-{\sec }_M(\mathrm{span}\{X(t),\dot{\gamma }(t)\}\left)\right)dt,$ the integrand at points $X(t)=0$ being $0$ by the same conventions. The integrand is continuous on $[0,b]$ and nonpositive, and its integral vanishes, so it vanishes identically; hence ${\sec }_M(\mathrm{span}\{X(t),\dot{\gamma }(t)\})=k(0<t\le b),$ at every $t$ where $X(t)\mathrm{\ne }0$; and $X(t)\mathrm{\ne }0$ for all $t\in (0,b]$, as shown in step 3.1 above together with $\mathrm{\mid }{J}^b\mathrm{\mid }>0$ on $(0,b]$.

Second form: differentiating the vanishing. [step 1.3, step 2.2, given] On $0<t\le T$ with $t<t_f$, $g_k^{\mathrm{\prime }}=-k-g_k^2$ and the Riccati equation gives ${\tilde{U}}^{\mathrm{\prime }}=g_k^{\mathrm{\prime }}\mathrm{id}-S^{\mathrm{\prime }}=(S-g_k\mathrm{id})(S+g_k\mathrm{id})+(R_{\gamma }-k\mathrm{id})\mathrm{.}$ The operators $S$ and $\tilde{U}=g_k\mathrm{id}-S$ commute. Since $\tilde{U}y=0$ and $y^{\mathrm{\prime }}=Sy$, differentiating yields $0=(\tilde{U}y{)}^{\mathrm{\prime }}={\tilde{U}}^{\mathrm{\prime }}y+\tilde{U}Sy=(R_{\gamma }-k\mathrm{id})y,$ because $(S-g_k\mathrm{id})(S+g_k\mathrm{id})y=0$ and $\tilde{U}Sy=S\tilde{U}y=0$. Thus $R_{\gamma }y=ky$ on the compared interval.

First form: conclusion. [step 1.1, step 3.1, step 3.2, F8, F11, given] By step 3.1 and step 3.2, for every $b\in (0,T]$ and every $t\in (0,b]$ $J^b(t)=\frac{{\mathrm{sn}}_k(t)}{{\mathrm{sn}}_k(b)}{P}_t{w}_b,\mathrm{\mid }{w}_b\mathrm{\mid }=1,$ for a unit vector $w_b\in N_0$: indeed $J_k^b$ is the normal Jacobi field along ${\gamma }_k$ with $J_k^b(0)=0$ and ${\mathrm{\Phi }}_b^{-1}{J}_k^b(b)=w_b$, hence equals ${\mathrm{sn}}_k(t){\mathrm{\Phi }}_t{w}_b\mathrm{/}{\mathrm{sn}}_k(b)$ by [F11] and [F8], and the transfer of this field is ${\mathrm{sn}}_k(t)P_t{w}_b\mathrm{/}{\mathrm{sn}}_k(b)$. Multiplying by $\mathrm{\mid }J(b)\mathrm{\mid }=a{\mathrm{sn}}_k(b)$ from step 1.1, $J(t)=a{\mathrm{sn}}_k(t)P_t{w}_b(0<t\le b\le T)\mathrm{.}$ The left side is independent of $b$, so for fixed $t$ and two parameters $b,b^{\mathrm{\prime }}\ge t$ the unit vectors $w_b,w_{b^{\mathrm{\prime }}}$ satisfy $P_t{w}_b=P_t{w}_{b^{\mathrm{\prime }}}$, hence $w_b=w_{b^{\mathrm{\prime }}}$. Writing $w\in N_0$ for this common unit vector, $J(t)=a{\mathrm{sn}}_k(t)P_{t}w$ for every $t\in (0,T]$, and passing to the derivative at $t=0$ (the expansion ${\mathrm{sn}}_k(t)=t+O(t^2)$ and $P_{t}w\to w$ as $t\downarrow 0$) gives $D_{t}J(0)=aw$, so $w=D_{t}J(0)\mathrm{/}a$ and therefore $J(t)={\mathrm{sn}}_k(t)P_t(D_{t}J(0))(0\le t\le T),$ the formula also holding at $t=0$ where both sides vanish. Finally the conclusion of step 3.2, applied to the span of $X(t)=J^b(t)\propto J(t)$, gives ${\sec }_M(\mathrm{span}\{J(t),\dot{\gamma }(t)\})=k(0<t\le T),$ which is the asserted radial curvature equality wherever $J$ is nonzero.

Second form: conclusion. For $0<t\le T$ with $t<t_f$, step 3.3 gives $R_{\gamma }y=ky$, where $y=P_t^{-1}J$. By [F5], this is the actual curvature equation $R(J(t),\dot{\gamma }(t))\dot{\gamma }(t)=kJ(t)\mathrm{.}$ Also $y=Y(t)u\mathrm{\ne }0$, since $Y(t)$ is invertible and $u\mathrm{\ne }0$; hence the radial sectional curvature of $\mathrm{span}\{J(t),\dot{\gamma }(t)\}$ is $k$. In a fixed basis of $N_0$, each coordinate $y_i$ satisfies $y_i^{\mathrm{\prime }\mathrm{\prime }}+k{y}_i=0$, with $y_i(0)=u_i$ and $y_i^{\mathrm{\prime }}(0)=\lambda u_i$. Set $z_i:=y_i-u_i{\mathrm{cs}}_k-\lambda u_i{\mathrm{sn}}_k$. Then $z_i^{\mathrm{\prime }\mathrm{\prime }}+k{z}_i=0$ and $z_i(0)=z_i^{\mathrm{\prime }}(0)=0$. The Wronskian $z_i^{\mathrm{\prime }}{\mathrm{sn}}_k-z_i{\mathrm{sn}}_k^{\mathrm{\prime }}$ has derivative $(z_i^{\mathrm{\prime }\mathrm{\prime }}+k{z}_i){\mathrm{sn}}_k=0$, so it is identically zero. The comparison sine is positive at every $t\in (0,T]$ with $t<t_f$: for $k\le 0$ this follows from its formula; for $k>0$, $f_k>0$ on the compared interval and $f_k(\pi \mathrm{/}\sqrt{k})=-1$, so that interval lies before the first positive zero of ${\mathrm{sn}}_k$. Thus $(z_i\mathrm{/}{\mathrm{sn}}_k{)}^{\mathrm{\prime }}=0$, and its limit at zero is $z_i^{\mathrm{\prime }}(0)=0$, so $z_i=0$. Therefore $y(t)=f_k(t)u,J(t)=f_k(t)P_{t}u(0\le t\le T,t<t_f)\mathrm{.}$ If $t_f\le T$, these equalities and the curvature-vector equation extend to $t_f$ by continuity. Before $t_f$, division by $f_k>0$ gives $R(P_{t}u,T)T=k{P}_{t}u$, so this equation also extends to $t_f$. Thus the plane spanned by the nonzero vector $P_{t_f}u$ and $T(t_f)$ has curvature $k$. The plane spanned by $J(t_f)$ and $T(t_f)$ has the same conclusion only when $J(t_f)\mathrm{\ne }0$.