The one-parameter curves generated by Lie dragging a geodesic by a Killing field are geodesics

Question

Let $(M,g)$ be a manifold with Riemannian or Lorentzian metric $g$ (the signature doesn't matter for our purposes).

Consider $\gamma : (a,b)\to M$ a geodesic and $\phi^\xi_s : M\to M$ the flow of a Killing field $\xi$ with $s\in (-\epsilon,\epsilon)$. Build the family of curves $\gamma_\xi : (a,b)\times (-\epsilon,\epsilon)\to M$ as

$$\gamma_\xi (u,v)=\phi^\xi_v\circ\gamma(u).$$

I want to show that $\gamma_\xi(\cdot, v)$ is a geodesic for each $v\in (-\epsilon,\epsilon)$.

My idea is:

1. Assume $N=\gamma_\xi((a,b)\times (-\epsilon,\epsilon))$ is a two-dimensional embeded submanifold. Define a vector field on $N$ by $X_{\gamma_\xi(u,v)}=(\phi^\xi_v\circ \gamma)'(u)$, this is the field of tangent vectors.

2. Show with the definition of the Lie derivative that $L_\xi X = 0$. Furthermore, if we denote $X\lrcorner g=g(X,\cdot)$ it is easy to show that $L_\xi (X\lrcorner g)=0$ as well.

3. From (2) we find $L_\xi g(X,X)=0$ hence $g(X,X)$ is constant along the flow. This means that the function $g(X,X)$ is constant on each integral line of $\xi$. Since it is constant in $\gamma$, it follows $g(X,X)$ is constant on $N$.

4. From (3) we derive that in local coordinates $\nabla_\mu( X^\nu X_\nu) = 0$ implies $X^\nu \nabla_\mu X_\nu = 0$. Hence $X^\mu \nabla_\mu X_\nu=2X^\mu\nabla_{[\mu}X_{\nu]}$.

5. If $\omega = X\lrcorner g$, then since $d\omega =\operatorname{alt}(\nabla \omega)$ and since by definition of the alternation, we have $\operatorname{alt}(\nabla \omega)_{\mu\nu}=\nabla_{[\mu}\omega_{\nu]}$. On the other hand we can easily show by metric compatibility that $(\nabla_X Y)\lrcorner g=\nabla_X(Y\lrcorner g)$. Thus we have $\operatorname{alt}(\nabla \omega)_{\mu\nu}=\nabla_{[\mu}X_{\nu]}$.

6. In that case we have $X^\mu \nabla_\mu X_\nu = 2X^\mu (d\omega)_{\mu\nu}$. In the end we have that $\nabla_X (X\lrcorner g)=X\lrcorner d\omega$. Finaly taking the Lie derivative on both sides and using $L_\xi X = L_\xi \omega = 0$ we find $L_\xi (\nabla_X (X\lrcorner g))=0$ and this yields $L_\xi (\nabla_X X)=0$. Since $\nabla_X X = 0$ on $\gamma$, it follows $\nabla_X X = 0$ on $N$ and each curve is a geodesic.

There are two issues: (i) this is a rather complicated proof, for such a natural result I expected something simpler. (ii) From (4) onwards I'm unsure of the rigor. I mean, $X$ is a vector field on $N$ which is an embeded submanifold, but not an open set of $M$. The connection $\nabla$ allows differentiation on $M$, not $N$. Furthermore, it seems I'm differentiating with respect to the coordinates of $M$ which is ilegal in $N$. Also, even if all holds in the manifold $N$, it is not clear what $\nabla$ is there, and furthermore this would only show the curves are geodesics on this connection of $N$.

Is my proof really wrong? If not, how to address the points above? If it is wrong, how to correctly and easily prove this?

Answer 1

(i) The idiomatic proof is to verify the following two facts:

• the flow of a Killing field is a family of isometries, and
• isometries preserve geodesics.

The first will involve a similar calculation to the one you describe. The second is essentially tautological, since geodesics are defined purely in terms of the metric, and preserving the metric is the definition of an isometry.

(ii) In the case where $N$ is a submanifold these kind of calculations are all legitimate once you have developed the theory of the induced connection. In general, the rigorous way to do this is working with pullback bundles/connections over $(a,b)\times(-\epsilon,\epsilon)$. You should be able to find elementary treatments of this kind of thing in an intro Riemannian geometry textbook - e.g. look for two-parameter maps in O'Neill or parametrized surfaces in do Carmo. For the version in the modern pullback language see Jost.