# Parallel vector field along a geodesic is a Jacobi field?

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?

• Hint: What is the simplest manifold of nonzero curvature that you know? Do you know what geodesics look like on this manifold? Can you construct a parallel field along a geodesic $\gamma$ which is not a multiple of $\gamma'$? Commented Jun 29 at 13:31
• @Moishe Kohan : I am considering sphere wtih great circle. Perhaps, Jacobi field along great circle is tangential? Commented Jun 30 at 1:07
• That's the one you do not want to use. Try to find one which is orthogonal to the geodesic. Commented Jun 30 at 1:51
• O.K. Let $\gamma : I=[a,b] \to \mathbb{S}^{2}$ be a great circle. Choose a vector $v \in T_{\gamma(a)}\mathbb{S}^{2}$ which is orthogonal to $\dot{\gamma}(a)$. Then there exists a unique parallel vector field $V$ along $\gamma$ such that $V(a) = v$. Since every parallel transport map along $\gamma$ is a linear isometry, $V$ is orthogonal to $\gamma$. (C.f. John Lee's Introduction to Riemannian manifold, Proposition 5.5. -(f), (g) . ) If $V$ is a Jacobi field and Jacobi field along great circle is tangential as I commented, then it contradics to that $V$ is orthogonal to the $\gamma$. Commented Jun 30 at 4:29
• I guess, you just do not know how to construct Jacobi fields. I cannot help you then. Commented Jun 30 at 4:54

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 $$D_t(D_tV) + R(V,\gamma')\gamma' = V \neq 0.$$
• Elementary question : Why $R(V, \gamma')\gamma' = V$? We have $1 = \operatorname{Rm}(V , \gamma' , \gamma' , V ) := \left\langle R ( V,\gamma')\gamma' , V \right\rangle$ by John Lee's book Proposition 8.29 . Commented Jul 3 at 8:03
• O.K. I think that at least $R(V, \gamma')\gamma'$ is a scalar multiple of $V$ so that it is not zero ( C.f. Proof of the Cauchy-Schwarz Inequality ; Linear Algebra done right, p.172 ) . And this fact is sufficient to show the last nonequality in your answer. Commented Jul 3 at 8:28
• Uhm. O.K. we can furthermore show that the scalar multiplied to $V$ is $1$. So $R( V, \gamma')\gamma' = V$ :) Commented Jul 3 at 8:34