Parallel vector field along a geodesic is a Jacobi field?

Question

Let $M$ be a manifold with a linear connection $\nabla.$ A vector field $V$ along a curve $\gamma$ is said to be parallel along $\gamma$ with respect to $\nabla$ if $D_tV =0$.

Let $\gamma : I \to M$ be a geodesic and $t_0 \in I$. Fix a vector $V_0 \in T_{\gamma(t_0)}M$ such that $V_0$ is orthogonal to $\dot{\gamma}(t_0)$. Then by the Parallel Translation ( John Lee's Introduction to Riemannian manifolds, Theorem 4.32. ), there exists a unique parallel vector field $V$ along $\gamma$ such that $V(t_0) =V_0$.

Note next theorem ( John Lee's book Theorem 10.1 ) :

Theorem 10.1. ( The Jacobi Equation ). Let $\gamma$ be a geodesic and $V$ a vector field along $\gamma$. If $V$ is the variation field of a variation through geodesics, then $V$ satisfies the following equation, called the Jacobi equation : $$ D_t^2V + R(V, \dot{\gamma})\dot{\gamma} = 0. \tag{10.2}$$

A smooth vector feild along a geodesic that satisfies the Jacobi equation is called a Jacobi field.

Here, 'variation through geodesics' is defined as follows :

Suppose that $I, K \subseteq \mathbb{R}$ are intervals, $\gamma : I \to M$ is a geodesic, and $\Gamma : K \times I \to M$ is a variation of $\gamma$ ( as defined in the John Lee's book Chapter 6 ). We say that $\Gamma$ is a variation through geodesics if each of the main curves $\Gamma_s(t)=\Gamma(s,t)$ is also geodesic.

Q. Then my question is, the above parallel vector field $V$ is a jacobi field? It suffices to show that $V$ is the variation field of a variation through geodesics.

I found next Lemma in the John Lee's book :

Lemma 6.1. If $\gamma$ is an admissible curve and $V$ is a piecewise smooth vector field along $\gamma$, then $V$ is the variation field of some variation of $\gamma$. If $V$ is proper, the variation cna be taken to be proper as well.

This lemma only gaurantees that the above $V$ is the variation field of some variation of $\gamma$, not the variation field of 'a variation through geodesics'. For the case that $V$ is a 'parallel' vector field along $\gamma$, such condition is satisfied?

This question originates from Existence of a parallel orthonormal frame $(E_1, \dots , E_n)$ along $\gamma$ such that $E_n = \dot{\gamma}$. In the question I answered myself and not convince that each $E_1, E_2 ,\dots, E_{n-1}$ are normal along $\gamma$ ( In my argument of the answer in my linked question, for $E_1, E_2, \dots, E_{n-1}$ to become normal, they need to be Jacobi fields. )

EDIT : Uhm.. I think that our above question is not true in general. For the original linked question below, I think that $E_1 , \dots E_{n-1}$ are normal along $\gamma$ since $(E_1, \dots , E_{n-1} , E_n = \dot{\gamma} )$ forms an orthonormal frame. Anyway, can we construct counter-example?

Can anyone help?

Answer 1

This is false, and here is a sketch of a counterexample. Consider the two sphere $S^2 = \{(x,y,z) \in \Bbb R^3 \mid x^2+y^2+z^2= 1\}$ and the parametrized great circle $\gamma(t) = (\cos t, \sin t, 0)$, which is a geodesic. Consider the vector field $V$ along $\gamma$ given by $V(t) = (0,0,1)$. Then one easily checks that:

• $V$ is tangent to $S^2$ along $\gamma$.
• $V$ has unit norm.
• $V(t)$ is orthogonal to $\gamma'(t)$ for all $t$.

Since the sphere has constant sectional curvature $1$, it follows that $R(V,\gamma')\gamma' = V$. To conclude, I let you check as an exercise that $V$ is parallel along $\gamma$, so that $D_t(D_tV) = 0$, and thus $$D_t(D_tV) + R(V,\gamma')\gamma' = V \neq 0.$$