Help with this proof: compact lorentzian manifold is complete? Could anyone please help me to understand this proof? It is not clear for me, for example, why $g(\gamma^{\prime}, \gamma^{\prime})$ is constant?
And why is enough to show that the projection on the subbundle maps $\gamma^{\prime}$ into a compact subset of span(K)?
Finally, what is $L_K g$?
Thank you in advance!
 A: If $g$ is a Riemannian or Lorentzian metric on the manifold $M$, it induces a covariant derivative on vector field along curves. Let $c:I=[0,1]\rightarrow M$ be a curve.
For every vector field $V$ defines on $c$ which is equivalent to $V(t)\in T_{c(t)}M$ there exists ${{dV}\over{dt}}$ the covariant derivative and $c$ is a geodesic if the covariant derivative of $c'(t)=0$.
If $V,W$ are vector defined on $c$, ${d\over{dt}}<V,W>={d\over{dt}}V,W>+<V,{d\over{dt}}W>$. We deduce that if $c$ is a geodesic, that is ${D\over{dt}}c'=0$, ${d\over{dt}}<c',c'>=0$.
This is any basic book on Riemannian geometry, see the book of Do Carmo for example.
For the second part, you can write $c'(t)=u(t)+v(t)$ where $u(t)$ is the projection on the timelike vector field and $v(t)$ the projection on its orthoronal. We have $<c'(t),c'(t)>=<u(t),u(t)>-<v(t),v(t)>=D$ where $D$ is a constant and $<u(t),u(t)>>0$, $<v(t),v(t)><0$ thus if $<u(t),u(t)>$ is bounded so is $<v(t),v(t)>$ since the restriction of $<,>$ to the orthogonal of the timelike vector field is negative definite. See the reference.
Vector perpendicular to timelike vector must be spacelike?
