Jacobi fields II: connection with curvature

We continue our study of Jacobi fields. We get a closed form expression for Jacobi fields vanishing at a point, and make a connection with curvature.

Jacobi fields vanishing at a point

Let $J: [0,\pi] \to TM$ such that $J(0) = 0$. Then the geodesic variation $\Gamma(s,t) = \exp_{\sigma(s)}(tW(s))$ has a certain form which gives a nice formula for $J$.

Here we shall make use of the identification \(T_v T_p M \simeq T_p M\) for $p\in M$ and $v\in T_pM$.

For the differential of a function $f:M\to N$ between manifolds we shall use the notation at each point $(df)p:T_pM\to T{f(p)}N$.

Lemma 2 (explicit form of Jacobi fields)

Let $\gamma: [0,1] \to M$ be a geodesic, $p = \gamma(0)$, $v = \gamma’(0)$. Let $J: [0,\pi] \to TM$ be a Jacobi field such that $J(0) = 0$, $J’(0) = \frac{D}{dt}J|_{t=0} = w.$

Then $J(t) = (d\exp_p)_{tv}(tw).$

Proof

We have $\gamma(t) = \exp_p(tv)$. Consider the variation $(s,t)\to \exp_p(t(v + sw))$.

Since each $\gamma_s := \exp_p(\cdot(v+sw))$ geodesic, it’s a geodesic variation.

Recall that the exponential map at a point $p$ is a function $\exp_p: T_pM\to M$. Thus by the chain rule, \(J = \partial_s \exp_p(t(v+sw))|_{s=0}\)

\(= (d\exp_p)_{tv}(sw),\) is a Jacobi field.

Gauss Lemma

Let $p \in M$, $v \in T_p M$ such that $\exp_p tv =: \gamma(t)$ defined for $t \in [0,1]$. Then,