Radial riccati operator

Let $(M,g)$ be a Riemannian manifold of dimension $n\ge 2$, let $\gamma :I\to M$ be a unit-speed geodesic on an interval with nonempty interior and $0\in I$, let $A(t):N_0\to N_t$ be the radial Jacobi tensor of Radial Jacobi tensor in the notation fixed there, and let $P_t$ be parallel transport along $\gamma$, so that $\bar{A}(t)=P_t^{-1}\circ A(t):N_0\to N_0$ is the matrix family of $A$. By Radial jacobi tensor is invertible before the first conjugate point the map $\bar{A}(t)$ is invertible for every $t\in I$ with $0<t<\tau$, where $\tau$ is the first conjugate instant of $\gamma (0)$ along $\gamma$.

On that interval the radial Riccati operator is defined by $S(t):={\bar{A}}^{\mathrm{\prime }}(t)\bar{A}(t{)}^{-1}\in \mathrm{End}(N_0),$ and, equivalently, as the endomorphism $D_{t}A(t)\circ A(t{)}^{-1}$ of $N_t$ transported back to $N_0$ by $P_t$. Its domain is precisely the set of $t\in I$ with $0<t<\tau$; the operator is smooth there because $\bar{A}$ is smooth and inversion is smooth on invertible endomorphisms.

The definition is independent of the choices made. The parallel identification uses the unique parallel transport of Existence and uniqueness of parallel sections, and for a change of basis of $N_0$ with matrix $C$ the matrix $\bar{A}$ is replaced by $C^{-1}\bar{A}C$, so ${\bar{A}}^{\mathrm{\prime }}{\bar{A}}^{-1}$ is replaced by $C^{-1}({\bar{A}}^{\mathrm{\prime }}{\bar{A}}^{-1})C$; the operator $S(t)$ is therefore a well-defined endomorphism of the abstract normal space, and in an orthonormal parallel frame it is represented by the matrix $M^{\mathrm{\prime }}(t)M(t{)}^{-1}$ for the matrix $M$ of $A$. No symmetry of $S(t)$, no differential equation and no relation to the curvature is asserted by this definition; those are separate results. For $n=2$ the normal space is one-dimensional and $S(t)$ is multiplication by the scalar ${\bar{A}}^{\mathrm{\prime }}(t)\mathrm{/}\bar{A}(t)$.