Radial riccati equation

The Riccati equation. [F1, F2, F3, F5, given] In the fixed parallel frame, [F1] gives ${\bar{A}}^{\mathrm{\prime }\mathrm{\prime }}=-R_{\gamma }\bar{A}$. Differentiate $\bar{A}{\bar{A}}^{-1}=I$ to obtain $({\bar{A}}^{-1}{)}^{\mathrm{\prime }}=-{\bar{A}}^{-1}{\bar{A}}^{\mathrm{\prime }}{\bar{A}}^{-1}$. Thus $S^{\mathrm{\prime }}={\bar{A}}^{\mathrm{\prime }\mathrm{\prime }}{\bar{A}}^{-1}-{\bar{A}}^{\mathrm{\prime }}{\bar{A}}^{-1}{\bar{A}}^{\mathrm{\prime }}{\bar{A}}^{-1}=-R_{\gamma }-S^2\mathrm{.}$ This proves the matrix equation wherever $0<t<\tau$ and $t\in I$. The corresponding curvature endomorphism of $N_t$ is ${\mathcal{R}}_t=P_t{R}_{\gamma }(t)P_t^{-1}$, namely $X\mapsto R(X,T)T$. Since $P_t$ is parallel, transporting the matrix equation gives $D_t{S}_t+S_t^2+{\mathcal{R}}_t=0$ for $S_t=D_{t}A\circ A^{-1}$. In particular the correctly typed inverse is $A^{-1}={\bar{A}}^{-1}{P}_t^{-1}$.

$S$ is self-adjoint. [F4, F5, F6, F7, step 1.1] Let $u,v\in N_0$ and put $X=J_u$, $Y=J_v$, so that $X=Au$, $Y=Av$ by [F1]. The Wronskian $W(t)=g(D_{t}X,Y)-g(X,D_{t}Y)$ is constant on $I$ by [F6], and at $t=0$ it equals $g(u,0)-g(0,v)=0$ because $X(0)=Y(0)=0$ and $D_{t}X(0)=u$, $D_{t}Y(0)=v$ by [F1]. Hence $g(D_{t}Au,Av)=g(Au,D_{t}Av)\text{on }I\mathrm{.}$ Writing $A=P_t\bar{A}$ and using that $P_t$ is an isometry and that $D_t{P}_t=0$ by [F4], this becomes $g({\bar{A}}^{\mathrm{\prime }}u,\bar{A}v)=g(\bar{A}u,{\bar{A}}^{\mathrm{\prime }}v)$, that is ${\bar{A}}^{\ast }{\bar{A}}^{\mathrm{\prime }}=({\bar{A}}^{\mathrm{\prime }}{)}^{\ast }\bar{A}$ for the metric adjoint of [F7]. Multiplying on the right by ${\bar{A}}^{-1}$ (invertible by [F2]) gives ${\bar{A}}^{\ast }S=({\bar{A}}^{\mathrm{\prime }}{)}^{\ast }$; taking adjoints with the rules of [F7] gives $S^{\ast }\bar{A}={\bar{A}}^{\mathrm{\prime }}$, hence $S^{\ast }={\bar{A}}^{\mathrm{\prime }}{\bar{A}}^{-1}=S\mathrm{.}$ Thus $S(t)$ is self-adjoint on $N_0$; since $P_t$ is an isometry, $S_t=P_{t}S(t)P_t^{-1}$ is self-adjoint on $N_t$ as well.

The expansion at zero. [F1, F2, F3, F4, F8, step 1.1] By [F1] the matrix family $\bar{A}$ is smooth and satisfies ${\bar{A}}^{\mathrm{\prime }\mathrm{\prime }}=-R_{\gamma }\bar{A}$ with $\bar{A}(0)=0$ and ${\bar{A}}^{\mathrm{\prime }}(0)={\mathrm{id}}_{N_0}$. In particular ${\bar{A}}^{\mathrm{\prime }\mathrm{\prime }}(0)=-R_{\gamma }(0)\bar{A}(0)=0$. If $0$ is an endpoint of $I$, extend each entry of $\bar{A}$ to negative $t$ by its cubic Taylor polynomial at $0$. The one-sided derivatives through order three agree at $0$, so each extension is $C^3$ on an open neighborhood of $0$. Thus the Taylor formula of [F8] with $n=3$ applies to each matrix entry at $t=0$: the constant and quadratic terms vanish and $\bar{A}(t)=t\mathrm{id}+\frac{t^3}{6}{\bar{A}}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(0)+o(t^3)=t(\mathrm{id}+t^{2}B(t))$ for the matrix function $B(t):=(\bar{A}(t)-t\mathrm{id})\mathrm{/}{t}^3$ on $t\mathrm{\ne }0$, which is bounded near $0$ because each entry is $\frac{1}{6}{\bar{A}}_{ij}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(0)+o(1)$. Applying the same Taylor–Peano supplier with $n=2$ to the entries of ${\bar{A}}^{\mathrm{\prime }}$, whose linear term vanishes because ${\bar{A}}^{\mathrm{\prime }\mathrm{\prime }}(0)=0$, gives ${\bar{A}}^{\mathrm{\prime }}(t)=\mathrm{id}+t^{2}C(t)$ with $C(t):=({\bar{A}}^{\mathrm{\prime }}(t)-\mathrm{id})\mathrm{/}{t}^2$ bounded near $0$. For $0<t<\delta$ with $\delta >0$ so small that $\bar{A}(t)$ is invertible and the entries of $E(t):=t^{2}B(t)$ are bounded by $\frac{1}{2}$, the determinant $\det (\mathrm{id}+E(t))$ tends to $1$ as $t\downarrow 0$, so it is nonzero for small $t$, and the adjugate formula of [F8] applies: $(\mathrm{id}+E(t){)}^{-1}=\det (\mathrm{id}+E(t){)}^{-1}\mathrm{adj}(\mathrm{id}+E(t)),$ where each entry of $\mathrm{adj}(\mathrm{id}+E(t))$ is a polynomial in the entries of $E(t)$, hence equals ${\delta }_{ij}+O(t^2)$ because $\mathrm{adj}(I)=I$ by [F8] and $E(t)=O(t^2)$. Therefore $(\mathrm{id}+t^{2}B(t){)}^{-1}=\mathrm{id}+O(t^2),$ and from $\bar{A}(t)=t(\mathrm{id}+t^{2}B(t))$ invertible for $0<t<\min (\delta ,\tau )$ by [F2], $\bar{A}(t{)}^{-1}=t^{-1}(\mathrm{id}+O(t^2))=t^{-1}\mathrm{id}+O(t)\mathrm{.}$ Multiplying by ${\bar{A}}^{\mathrm{\prime }}(t)=\mathrm{id}+O(t^2)$ and using $O(t^2)\cdot O(t^{-1})=O(t)$, $S(t)={\bar{A}}^{\mathrm{\prime }}(t)\bar{A}(t{)}^{-1}=(\mathrm{id}+O(t^2)\left)\right(t^{-1}\mathrm{id}+O(t))=t^{-1}\mathrm{id}+O(t)$ in the entrywise sense, equivalently in the operator norm on the finite-dimensional space $\mathrm{End}(N_0)$.

Conclusion and boundary cases. [F3, F8, step 1.1, step 2.1, step 2.2] Steps 1.1, 2.1 and 2.2 prove, respectively, the Riccati equation on the full interval $0<t<\tau$ where $S$ is defined, the self-adjointness of $S$, and the near-zero expansion $S(t)=t^{-1}\mathrm{id}+O(t)$. The endpoint $t=0$ is excluded because $\bar{A}(0)=0$ is not invertible and $S$ is not defined there; the expansion of step 2.2 is one-sided and uses only $t\downarrow 0$. The interval is nonempty for small $t$: $\tau >0$, since $\bar{A}(t)=t(\mathrm{id}+t^{2}B(t))$ is invertible for small $t>0$ by the expansion, and $S$ is defined on the whole of $(0,\tau )\cap I$. In dimension $n=2$ the normal space is one-dimensional and the Riccati equation is the scalar equation $S^{\mathrm{\prime }}+S^2+K_{\gamma }=0$ with $K_{\gamma }$ the sectional curvature along $\gamma$; in the general case no commutation of the terms is asserted, the equation being an identity in $\mathrm{End}(N_0)$. If $\tau =+\mathrm{\infty }$ the interval is $(0,\mathrm{\infty })\cap I$. No constancy of curvature is used beyond the interpretation of $R_{\gamma }$ recorded in [F9], and no choice beyond the inherited [A1] is used: the parallel frame, the matrix family and the expansion are all explicit.