# Decomposition of a Jacobi field along a lightlike geodesic

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 $$\langle\dot\gamma,\dot\gamma\rangle=0$$.