Killing field defines a jacobi field

Question

I have a Killing vector field $X$ on $(M,g)$. Why, if $\gamma$ is a geodesic, is $X_{\gamma(t)}$ a Jacobi field?

Any hints to get started would be welcome

Answer 1

What have you tried? A good place to start is the following: let $\gamma$ be a geodesic, and consider a geodesic variation of $\gamma$; that is a map $\gamma(\cdot,\cdot):[0,T]\times(-\varepsilon,\varepsilon)\to M$ such that $\gamma(t,0)=\gamma(t)$ and $\gamma(\cdot,s)$ is a geodesic for all $s\in(-\varepsilon,\varepsilon).$ Try and show that the vector field along $\gamma(t)$ given by $$Y(t):=\left.\frac{\partial}{\partial s}\gamma(t,s)\right|_{s=0}$$ is a Jacobi field along $\gamma$. Once you have this your result will follow if you show that the Killing field $X$ induces a geodesic variation of $\gamma$.

Answer 2

Hint: Assume that $X$ is a Killing vector field, then $$g\left(D_YX,Z\right)+g\left(D_ZX,Y\right)=0$$ for any vector fields $Y,Z$. Now, let $\gamma$ be a geodesic with tangent vector $T$, then $D_TT=0$. To show that $X$ is a Jacobi vector field, you have to show that $$D_TD_TX+R(X,T)T=0$$