For a nonlinear, equation \(F(u) = 0\) where \(u \in E = C^{\infty}(\mathbb{R}^n \to \mathbb{R})\) and \(F : \mathbb{R} \to \mathbb{R}\), we think of \(u \in E \mapsto F(u) \in E\) as map \(E \to E\). The linearisation of \(F\) around \(u_0\) is simply \(dF_{u_0}\). It is a linear map acting on tangent vectors to \(E\) based at \(u_0\). Since \(E\) is a (infinite dimensional) vector space, the tangent space \(T_{u_0} E\) may be identified with \(E\) itself via the linear isomorphism \[ v \in E \mapsto \partial_s|_{s=0} (u_0 + sv). \] More generally, we replace the curve \(s \mapsto u_0 + sv\) with any curve \(r(s) \in E\) representing \(v\) (so that for each \(s\), \(r(s) \in E\) with \(r(0) = u_0\) and \(\partial_s|_{s=0} r(s) = v\)) and the result is independent of the chosen curve. Being able to use arbitrary curves rather than linear ones will be important when working with geodesics since the space of curves is not a linear space.

Then we may compute \[ dF_{u_0} (v) = \partial_s|_{s=0} F(u_0 + s v) = \partial_s|_{s=0} F(r(s)). \]

For example, if \(F(u) = u^2\), then for \(u_0(x) = \sin(x) \in E = C^{\infty}(\mathbb{R} \to \mathbb{R})\) and \(v(x) = e^x \in E = T_{u_0} E\) we have \[ dF_{u_0} (v) = \partial_s|_{s=0} (u_0 + s v)^2 = 2u_0 v. \] That is, \(dF_{u_0} (v)\) is the function \[ x \mapsto 2\sin(x) e^x. \]

Notice here that we really interpret \(F : E \to E\) so that \(dF_{\sin(x)} : v \in T_{\sin(x)} E = E \mapsto 2\sin(x) v \in T_{\sin^2(x)} E = E\).

An example closer to the geodesic equation is a non-linear differential operator. Let us define the Newtonian momentum, \[ P(u) = \frac{1}{2} (u')^2. \] for \(u \in C^{\infty}(\mathbb{R} \to \mathbb{R})\). Then \[ dP_{u_0} (v) = u_0' v'. \] So with \(u_0(x) = \sin(x)\) and \(v(x) = e^x\) as in the previous example, \[ dP_{\sin(x)} (e^x) = \cos(x) e^x. \]

Notice that the function \(x \mapsto dP_{u_0(x)} (v(x))\) need not be a linear function of \(x\). The point is that \(dP_{u_0} (v)\) is linear in the tangent vector \(v\).

Except in special cases (e.g. Euclidean space, the sphere), the non-linearity of the geodesic equation precludes us from solving it. Nevertheless, as is typical with non-linear ODE's, in order to obtain some understanding of the solutions, we linearise the equation.

In the context of geodesics, regarding the linearisation above, there are two points to make here:

  1. The space of curves \(C^{\infty}([0, 1] \to M)\) is not a linear space so expressions like \(\gamma_s = \gamma_0 + s \mu\) do note make sense,
  2. The tangent space \(T_{u_0} E\) is infinite dimensional and that can cause some complications. We will discover below, for variations of geodesics, there is a very important finite dimensional subspace, the space of Jacobi fields.

To overcome 1., we note that as remarked above, any curve \(r(s)\) can take the place of \(u_0 + s v\) so that our variations may be taken just as in deriving the geodesic equation, namely we consider maps \(\varphi(t, s)\) with \(\gamma_0(t) := \varphi(t, 0)\) satisfying \(\gamma_0 = \gamma\).

Regarding point 2., instead of considering all possible variations of \(\gamma\), we consider only geodesic variations:

Let \(\gamma: [0, 1] \to M\) be a geodesic. A geodesic variation of \(\gamma\) is a map \[ \varphi : [0, 1] \times (-\epsilon, \epsilon) \to M \] such that for each \(s \in (-\epsilon, \epsilon)\), the curve \[ t \mapsto \gamma_s(t) := \varphi(t, s) \] is a geodesic.

  • Note that we make no restriction that our variations fix the end points.

Now, rather than dealing with arbitrary variations of geodesics, we will work only with geodesic variations. Here's an unmotivated definition:

Let \(\gamma\) be a geodesic. A Jacobi field along \(\gamma\) is a vector field \(J\) along \(\gamma\) such that \[ \nabla^2 J (\gamma', \gamma') + \operatorname{Rm}(J, \gamma') \gamma' = 0. \]

And here's the motivation.

Geodesic variations correspond to Jacobi fields.

More precisely, if \(\varphi\) is a geodesic variation of a geodesic \(\gamma\), then the variation vector field \[ J(t) = \varphi_{\ast} \partial_s (t, 0) \] is a Jacobi field along \(\gamma\).

Conversely, if \(J\) is a Jacobi field along \(\gamma\), then there exists a geodesic variation, \(\varphi\) of \(\gamma\) with variation vector field equal to \(J\).

The proof is given in Comparison Geometry by Eschenburg, Section 2, Remark 2.2.

A few remarks on the proof are in order.

Remarks on the implication variation fields of geodesic variations are Jacobi fields.

Following Eschenburg, let us write \(\gamma(s, t)\) for the geodesic variation instead of \(\varphi\) as in the statement of the lemma above. Let \(J(t, s) = \gamma_{\ast} \partial_s\), be the variation field and \(J_s(t) = J(t, s)\). Let \(V = \gamma_{\ast} \partial_t\) and \(V_s(t) = V(t, s)\). Notice that each \(\gamma_s = \gamma(\cdot, s)\) is a geodesic with velocity field \(V_s\), and \(J_s\) is a vector field along the geodesic \(\gamma_s\). In fact, for each fixed \(s\), \(J_s\) is a Jacobi field along \(\gamma_s\) and not just for \(s = 0\). That is, the proof works for arbitrary \(s\) and not just \(s = 0\).

Then \[ \nabla^2 J(V, V) = \nabla_{V} (\nabla_{V} J) - \nabla_{\nabla_{V} V} J = \nabla_{V} (\nabla_{V} J) \] since each \(\gamma_s\) is a geodesic. Relating our notation to Eschenburg's, we have \[ J'' = \nabla_{V} (\nabla_{V} J) = \nabla^2 J(V, V). \] Notice in particular, that the last equality is only true geodesics.

In (2.1) there are two derivatives commuted. The first commutator of derivatives comes from: \[ \nabla_{V} J = \nabla_J V + [V, J]. \] But \[ [V, J] = [\gamma_{\ast} \partial_t, \gamma_{\ast} \partial_s] = \gamma_{\ast} [\partial_t, \partial_s] = 0 \] so that \[ \nabla_{V} J = \nabla_J V. \] Compared with Eschenburg's notation then, \[ \frac{D}{dt} J = \nabla_{V} J = \nabla_J V = \frac{D}{ds} \frac{\partial \gamma}{\partial t}. \]

Therefore \[ \nabla_V (\nabla_V J) = \nabla_V (\nabla_J V). \]

Using \(\nabla_V V = 0\) and \([V, J] = 0\), the second commutator is \[ \nabla_V (\nabla_J V) = \nabla_V (\nabla_J V) - \nabla_J (\nabla_V V) - \nabla_{[V, J]} V = \operatorname{Rm} (V, J) V = - \operatorname{Rm} (J, V) V. \] Therefore, \[ \nabla^2 J(V, V) = \nabla_V (\nabla_V J) = \nabla_V (\nabla_J V) = - \operatorname{Rm} (J, V) V. \] That is, we obtain equation (2.1): \[ J'' + R(J, V) V = \nabla^2 J(V, V) + \operatorname{Rm}(J, V) V = 0. \]

Remarks on the construction of a geodesic variation from a Jacobi field.

Now we just have a vector field \(J(t)\) along a geodesic \(\gamma\). Then somehow, we must construct a geodesic variation from \(J\). A naive attempt to try would be something like \[ \varphi(t, s) = \exp_{\gamma(t)} s J(t). \] This variation satisfies, \[ \varphi(t, 0) = \exp_{\gamma(t)} 0 = \gamma(t) \] and \[ (\partial_s|_{s=0} \varphi) (t) = \gamma_{J(t)}'(s=0) = J(t) \] where recall that for any tangent vector \(X\) at \(x \in M\), \(\gamma_X (s) = \exp_x (s X)\) is the unique geodesic through \(x\) with \(\gamma_X'(0) = X\).

In other words, we have indeed managed to construct a variation of \(\gamma\) with variation vector field \(J\). But how do we know it is a geodesic variation? This is rather tricky and would require that for each fixed \(s\), them map \[ \gamma_s : t \mapsto \varphi(t, s) = \exp_{\gamma(t)} s J(t) \] is a geodesic. You might like to think about how you would prove \(\nabla_{\gamma_s'} \gamma_s' = 0\).

Instead of this direct approach, something else is called for in order to obtain a geodesic variation. Then we will need to prove that the variation field is \(J\) which turns out to be easier than proving \(\gamma_s\) is a geodesic. The interesting thing about the construction is that it is obtained only from \(J(0)\) and \(J'(0)\)! The reason is that \(J\) satisfies the second order linear initial value problem:

\begin{cases} \tilde{J}'' + R(\tilde{J}, V) V &= 0 \\ \tilde{J}(0) &= J(0) \\ \tilde{J}'(0) &= J'(0). \end{cases}

Thus if we can construct a geodesic variation \(\varphi\) with \(\varphi_0 = \gamma\) such that \[ \partial_s|_{s=0}(t=0) \varphi = J(0), \quad \partial_t|_{t=0} \partial_s|_{s=0} \varphi = J'(0), \] then the variation field \(\tilde{J}(t) = (\partial_s|_{s=0} \varphi) (t)\) will satisfy the same initial value problem and hence by uniqueness of solutions, must equal \(J\). In other words, we exploit the fact that Jacobi fields are completely determined by their initial value \(J(0)\) and initial first derivative \(J'(0) = \nabla_{\gamma'(0)} J|_{t=0}\).

Now, in order that \(\tilde{J}(0) = J(0)\), we need to define \(\varphi(0, s)\) in such a way that \(\partial_s|_{s=0} \varphi(0, s) = J(0)\). That is we need to construct a curve \(\alpha\) (with parameter \(s\)) passing through \(\gamma(0)\) and with \(\partial_s \alpha(0) = J(0)\). Eschenburg defines \[ \alpha(s) = \exp_{\gamma(0)} s J(0). \] That is \(\alpha\) is the unique geodesic passing through \(\gamma(0)\) and with velocity vector \(J(0)\). So we have the first condition \(\tilde{J}(0) = J(0)\).

Now we need the \(t\) variable and the second condition that \(\tilde{J}'(0) = J(0)\). Notice that if \(X(s)\) is any vector field along \(\alpha\), then for each \(s\), \[ \gamma_s(t) = \exp_{\alpha(s)} t X(s) \] is a geodesic. In terms of the geodesic flow \(\Phi_t\) on \(TM\), \[ \gamma_s (t) = \pi \circ \Phi_t(X(s))) \] is smooth as a function of \(s\) and \(t\). Thus we have easily obtained a geodesic variation! Notice in particular that \[ \partial_t \gamma_s|_{t=0} = X(s). \]

What we need to do now is define a vector field \(X(s)\) along \(\alpha\) so that

  1. \(\gamma_0 = \gamma\),
  2. \(\tilde{J}'(0) = J'(0)\).

For 1., we simply require that \(X(s=0) = \gamma'(t=0)\) for then \[ \gamma_0(t) = \exp_{\alpha(s=0)} t X(s=0) = \exp_{\gamma(t=0)} t \gamma'(t=0) = \exp_{\gamma(0)} t\gamma'(0) = \gamma(t). \]

For 2., we commute derivatives. Let \(V = \partial_t \gamma_s\) so that \[ V(0, s) = \partial_t|_{t=0} \exp_{\alpha(s)} t X(s) = X(s). \] Then, \[ \tilde{J}'(0) = \nabla_V \tilde{J}|_{t,s=0} = \nabla_{\tilde{J}(s=0)} V(t=0)|_{s=0} = \nabla_{\tilde{J}(s=0)} X|_{s=0} = X'(0). \] That is we require, \[ X'(s=0) = J'(t=0). \]

Finally then, we what need to do is show that there exists a vector field \(X\) along \(\alpha\) such that \[ X(s=0) = \gamma'(t=0), \quad X'(s=0) = J'(t=0). \] One way to do that is to let \(X\) be the unique solution of the second order linear ODE,

\begin{cases} X'' &= 0 \\ X(0) &= \gamma'(t=0) \\ X'(0) &= J'(t=0). \end{cases}

Note here again that since \(\alpha\) is a geodesic, \[ X'' = \nabla^2 X (\alpha', \alpha') \] so this really is a tensorial (and hence invariantly defined) ODE.

In the course of the proof in the previous section, we noted that Jacobi fields along a geodesic \(\gamma\) are completely determined by the following linear system of ODE's:

\begin{cases} J'' + R(J, \gamma') \gamma' &= 0 \\ J(0) &= X \\ J'(0) &= Y. \end{cases}

where \(X, Y \in T_{\gamma(0)} M\) are arbitrary tangent vectors.

Let us denote the set of Jacobi fields along \(\gamma\) by \(\mathcal{J}\).

Show that \(\mathcal{J}\) is a vector space of dimension \(n^2\).

Notice that if \(X = \gamma'(0)\) and \(Y = 0\), then \(J = \gamma'\) is a solution! That is, the velocity vector of a geodesic satisfies the Jacobi equation with tangential initial condition and zero initial acceleration. The Jacobi equation in this case follows from \[ J'' = \nabla_{\gamma'} (\nabla_{\gamma'} \gamma') = 0, \quad R(\gamma', \gamma') \gamma' = 0. \]

Show that \[ J(t) = t \gamma'(t) \] satisfies the Jacobi equation with \(J(0) = 0\) and \(J'(0) = \gamma'(0)\).

Conclude that by linearity that \[ J(t) = (a + b t) \gamma'(t) \] is the unique Jacobi field satisfying \[ J(0) = a \gamma'(0), \quad J'(0) = b \gamma'(0). \]

The space of orthogonal Jacobi fields, \(\mathcal{J}^{\perp}\) is defined to be \[ \mathcal{J}^{\perp} = \{J : g(J(t), \gamma'(t)) \equiv 0\}. \]

The space \(\mathcal{J}^{\perp}\) is an \(n^2 - 2\) dimensional linear subspace of the \(n^2\) dimensional space of Jacobi fields along \(\gamma\). That is, \(\mathcal{J}^{\perp}\) is a co-dimension \(2\) subspace of \(\mathcal{J}\).

Let \(J_1, J_2 \in \mathcal{J}\) be Jacobi fields. Then we claim that \[ g(\nabla_{\gamma'} J_1, J_2) = g(J_1, \nabla_{\gamma'} J_2) + \text{const}. \]

This follows by differentiating:

\begin{split} \partial_{\gamma'} [g(\nabla_{\gamma'} J_1, J_2) - g(J_1, \nabla_{\gamma'} J_2)] &= g(\nabla^2 J_1 (\gamma', \gamma'), J_2) - g(J_1, \nabla^2 J_2 (\gamma', \gamma')) \\ &= \operatorname{Rm(J_1, \gamma', \gamma', J_2)} - \operatorname{Rm(J_2, \gamma', \gamma', J_1)} \\ &= 0. \end{split}

Here in the first line we used metric compatibility and in the second the Jacobi equation. The last line follows from the curvature tensor symmetries.

Next, let \(J\) be any Jacobi field and recall that \(\gamma'\) is also a Jacobi field. Then the claim implies that \[ g(\nabla_{\gamma'} J, \gamma') = g(J, \nabla_{\gamma'} \gamma') + \text{const} = \text{const}. \] Therefore, \[ \partial_{\gamma'} g(J, \gamma') = g(\nabla_{\gamma'} J, \gamma') + g(J, \nabla_{\gamma'} \gamma') = \text{const} \] and hence \[ g(J, \gamma') = at + b \] for some constants \(a, b \in \mathbb{R}\).

Finally then, the linear map \[ J \in \mathcal{J} \mapsto (\partial_t g(J, \gamma'), g(J, \gamma')) = (a, b) \in \mathbb{R}^2 \] is surjective (because \((a + b t) \gamma'\) is a Jacobi field) with kernel precisely \(\mathcal{J}^{\perp}\). Therefore by the rank-nullity theorem, \[ \operatorname{dim} \mathcal{J}^{\perp} = \operatorname{dim} \mathcal{J} - 2 = n^2 - 2. \]

The following exercise gives some structure to Jacobi fields. It shows that the radial component is linear and that all the interesting behaviour is contained in \(\mathcal{J}^{\perp}\). This simply follows from \(\mathcal{R}_V (V) = 0\) so that radial Jacobi fields satisfy \(J'' = 0\).

Let \(X, Y \in T_{\gamma(0)} M\) and write uniquely \[ X = a \gamma'(0) + X^{\perp}, \quad Y = b \gamma'(0) + Y^{\perp} \] where \(a = g(X, \gamma'(0))\) and \(X^{\perp}\) is the component of \(X\), \(g\)-orthogonal to \(\gamma'(0)\). Likewise for \(Y\).

Show that the Jacobi field \(J\) determined by \(J(0) = X\), \(J'(0) = Y\) is \[ J(t) = (a + b t) \gamma'(0) + J^{\perp}(t) \] where \(J^{\perp}\) is the Jacobi field determined by \(J^{\perp}(0) = X^{\perp}\), \((J^{\perp})'(0) = X^{\perp}\).

Show moreover that \(J^{\perp} \in \mathcal{J}^{\perp}\). Hint: Evaluate a certain equation from the proof of the previous lemma at \(t = 0\).

Now we split the Jacobi equation into a first order system. The interesting thing about the way we do it is that we do not do the obvious thing, introducing a new variable for the derivative of \(J\), say \(K = J'\) which satisfies \(K' = - \operatorname{R}(J, V) V\). Such a formulation is useful for proving existence and uniqueness of Jacobi fields but does not furnish us with a useful comparison theory.

Rather, suppose for a moment that there is an embedding \(\varphi : [0, 1] \times S \to M\) with \(S\) an \(n-1\) dimensional manifold so that \(U = \varphi([0,1] \times S)\) is an open subset of \(M\) and \(\varphi\) is a diffeomorphism onto this subset.

Suppose further that for each \(s\), \(\gamma_s = \varphi(\cdot, s)\) is a geodesic. In other words, \(\varphi\) is an \(n-1\) parameter geodesic variation. Then for each curve \(\alpha\), in \(S\), \((t, u) \mapsto \varphi(t, \alpha(u))\) is a \(1\) parameter geodesic variation with Jacobi field \(J(t) = \partial_u|_{u=0} \varphi(t, \alpha(u))\). Equivalently, for each \(V\) tangent to \(S\), \(J_s(t) = J(s, t) = \varphi_{\ast} V\) is a Jacobi field along \(\gamma_s\).

Then we have \(\nabla_V V = 0\) where \(V = \varphi_{\ast} \partial_t\) and \([V, J] = 0\) for any \(J\) the Jacobi field arising above as the variation in a direction tangent to \(S\).

Let us set \[ A = \nabla V. \] Then \(A : \Gamma(U, TM) \to \Gamma(U, TM)\) is a \(C^{\infty}\) linear (tensorial) map taking vector fields \(X\) along \(U\) to vector fields along \(U\). Explicitly, \[ A(X) = \nabla V (X) = \nabla_X V. \]

Since \([V, J] = 0\), we get \[ A(J) = \nabla_J V = \nabla_V J = J'. \] This is the first of our equations. It is a first order linear equation in \(J\).

For the second equation, we recall the derivative \(\nabla A\) obtained by requiring \(\nabla\) to commute with traces and satisfy the Leibniz rule with respect to tensor products is \[ (\nabla_X A) (Y) = \nabla_X (A(Y)) - A(\nabla_X Y) = \nabla_X \nabla_Y V - \nabla_{\nabla_X Y} V. \] for vector fields \(X, Y\) along \(\gamma\). Or in other words, the last term on the right is a correction term accounting for the variation in \(Y\).

So we compute the variation of \(A\) along \(\gamma\) (i.e. \(\nabla_V A\)) applied to a Jacobi field \(J\), and commute derivatives in order to obtain an equation for \(A\):

\begin{split} (\nabla_V A) (J) &= \nabla_V \nabla_J V - \nabla_{\nabla_V J} V \\ &= \operatorname{Rm}(V, J) V + \nabla_J \nabla_V V + \nabla_{[V, J]} V - \nabla_{\nabla_J V + [V, J]} V \\ &= -\operatorname{Rm}(J, V) V - \nabla V(\nabla V (J)). \end{split}

In the last line we recall the notation \(\nabla V(J) = \nabla_J V\), and use that each \(\gamma_s = \varphi(\cdot, s)\) is a geodesic so that \(\nabla_V V \equiv 0\).

Notice that if we just took an arbitrary vector field \(X\) (i.e. \(X = \varphi_{\ast} \partial_s\) where \(\varphi\) is not necessarily arising from a geodesic variation), then although \(\nabla_V V = 0\) at \(s = 0\), it need not be true that \(\nabla_V V = 0\) for other \(s\). Then since \(X\) is the variation in the \(s\) direction, \(\nabla_X \nabla_V V\) may not be zero.

Now, since \(A = \nabla V\), rearranging gives, \[ \nabla_V A (J) + A^2 (J) + \operatorname{Rm}(J, V) V = 0. \] Writing \(\operatorname{R}_V (J) = \operatorname{Rm} (J, V) V\) and noting that the equality is true for all \(J\) we obtain our equation for \(A\): \[ \nabla_V A + A^2 + \operatorname{R}_V = 0. \] Such an equation is an example of a non-homogeneous Riccati equation.

Let's rigorously obtain the relationship between Jacobi fields and \(A\) satisfying the Riccati equation. That is, rather than as in the derivation of the Riccati equation where we assumed (without proof) that we had a \(n-1\) parameter geodesic variation. Now we establish the correspondence only assuming \(A\) satisfies the Riccati equation and without assuming \(A = \nabla V\).

Let \(\gamma\) be a geodesic. Then the the Jacobi equation along \(\gamma\) is equivalent to

\begin{cases} \nabla_V J &= A (J) \\ \nabla_V A + A^2 + \operatorname{R}_V &= 0 \\ J(0) &= J_0 \\ A(0) &= A_0. \end{cases}

That is, if \(J\) is a Jacobi field, then there exists an \(A\) satisfying the Riccati equation and such that \(J' = A (J)\). Conversely, if \(A\) satisfies the Riccati equation, then the solution to \(J' = A(J)\) with \(J(0) = X\) is the unique Jacobi field such that \(J(0) = X\) and \(J'(0) = A(0) (X)\).

First let \(A\) be any solution the Riccati equation and let \(J\) be any solution of \(J' = A(J)\). Then we need to show \(J\) is a Jacobi field. For this, we compute \[ J'' = \nabla^2 J (\gamma', \gamma') = \nabla_{\gamma'}(A(J)) = (\nabla_{\gamma'} A) (J) + A(\nabla_{\gamma'} J) = -\operatorname{R}_V (J) - A^2(J) + A(\nabla_{\gamma'} J). \] But \[ -A^2(J) + A(\nabla_{\gamma'} J) = A(-A(J) + J') = A(0) = 0 \] by construction hence \[ J'' = - \operatorname{R}_V (J) \] and \(J\) satisfies the Jacobi equation.

Conversely, let \(J\) be any Jacobi field. We need to construct \(A\) satisfying the Riccati equation and such that \(\nabla_{\gamma'} = A(J)\).

Let \(A_0 : T_{\gamma(0)} M \to T_{\gamma(0)} M\) be any linear transformation such that \(J'(0) = A_0 (J(0))\) and let \(A(t)\) be the unique solution of \(A' + A^2 + \operatorname{R}_V = 0\) with \(A(0) = A_0\). Then as in the previous part the solution \(\tilde{J}\) to \(\tilde{J}' = A(\tilde{J})\) with \(\tilde{J}(0) = J(0)\) is a Jacobi field such that \(\tilde{J}(0) = J(0)\) and \[ \tilde{J}'(0) = A(0) (\tilde{J}(0)) = A_0(J(0)) = J'(0). \] That is, \(\tilde{J}\) is a Jacobi field with the same initial conditions as \(J\) hence \(\tilde{J} = J\) and \[ J' = \tilde{J}' = A(\tilde{J}) = A(J). \]

Notice that in each implication of the proof we have \(J' = A(J)\). That is \(A(J) = \nabla_{\gamma'} J\). If we let \(\varphi\) be a geodesic variation corresponding to \(J\), then \([V, J] = 0\) so that \[ A(J) = \nabla_V J = \nabla_J V = \nabla V (J). \] But note here that \(\nabla V\) is only defined here on \(V\) and on \(J\). In other words, we don't need to construct an \(n-1\) dimensional variation of \(\gamma\) and set \(A = \nabla V\) - the construction essentially works for one parameter variations of \(\gamma\).

Note that while the equation for \(J\) is linear, the equation for \(A\) is nonlinear. This may seem like a step backward, but this formulation allows us to develop comparison theorems for \(A\) (and then in turn \(J\)) based on comparisons of the curvature \(\operatorname{R}_V\) along the geodesic \(\gamma\). What's more, the equation for \(A\) is completely decoupled from that of \(J\) so that we may solve for \(A\) independently of any Jacobi field \(J\). There is much geometric information contained in \(A\) and the equation is of the same form for all geodesics and all manifolds with the only difference occurring in the \(\operatorname{R}_V\). In other words, we may think of the geometry along \(\gamma\), expressed through \(\operatorname{R}_V\), as a forcing term in \[ A' + A^2 = - \operatorname{R}_V. \] That is, all the varying behaviour for different manifolds and geodesics is contained only in the forcing term \(\operatorname{R}_V\)!

Compare with the obvious idea mentioned at the start of this section which would give the coupled system \(J' = K\), \(K' = - \operatorname{R}_V (J)\).

Note that \[ \operatorname{R}_V (V) = \operatorname{Rm}(V, V) V = 0. \] Then \(A\) satisfies \(A' + A^2 = 0\) in the \(V\) direction. Notice that this equation is the same for all geodesics in all manifolds! As already noted: there is no geometry in the radial direction.

In the previous section, we observed that the only interesting behaviour occurs along \(\mathcal{J}^{\perp}\). For this reason, sometimes only the restriction of \(A\) to \(\mathcal{J}^{\perp}\) is considered.

Finally, note that \(A\) really encodes for an \(n\)-dimensional family of Jacobi fields. That is if \(J_1, \cdots, J_n\) are linearly independent Jacobi fields, then \(A\) is determined by \[ A(J_i) = J_i' \] or indeed, \(A\) is determined by \[ A_0(J_i(0)) = J_i'(0) \] and the Riccati equation.