Consider a Lorentzian manifold of dimension $1+n$ (with $n\geq1$) and a lightlike geodesic $\gamma(t)$ on it. One can define a Jacobi field $J(t)$ along $\gamma$ in the usual way without issues.

In Riemannian geometry one can decompose the Jacobi field into two components: The one parallel to $\dot\gamma$ and the one normal to it. A Jacobi field that starts (to zeroth and first order) parallel/normal stays parallel/normal. Is there a similar decomposition for a lightlike geodesic on a Lorentzian manifold?

Of course one can say that a vector $v\in T_{\gamma(t)}M$ is parallel to $\gamma$ if it is a scalar multiple of $\dot\gamma(t)$ and normal if $\langle v,\dot\gamma(t)\rangle=0$. When $\dot\gamma$ is lightlike, the problem is that the parallel direction is normal. If $n=1$, "parallel" and "normal" are in fact equivalent. It would be convenient if I could somehow naturally extract the part of $J$ which is not parallel so that this part still satisfies the Jacobi equation or something similar enough.

The best I can think of is to take $n$ lightlike vectors $u_1,\dots,u_n$ so that the set $\{\dot\gamma(0),u_1,\dots,u_n\}$ is linearly independent and parallel transport this frame along $\gamma$. Then I can express a Jacobi field $J$ in this basis. This feels clumsy in comparison to the Riemannian version and is less invariant in nature as I need to fix a frame. Is there something better?

It is still true in Lorentzian geometry that $\partial_t^2\langle J(t),\dot\gamma(t)\rangle=0$. This identity underlies the decoupling of parallel and normal Jacobi fields in Riemannian geometry, but I can't seem to be able to project the parallel component of a Jacobi field because $\langle\dot\gamma,\dot\gamma\rangle=0$.


1 Answer 1


Have a look at how Harris does it in §2 in

Harris, Steven G., A triangle comparison theorem for Lorentz manifolds, Indiana Univ. Math. J. 31, 289-308 (1982). ZBL0496.53042.


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