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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 $L_K g$?enter image description here Thank you in advance!

C. Bishop
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1 Answers1

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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?

cmk
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Tsemo Aristide
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  • Thank you so much for your answer, I will accept it as soon as possible. Could you please help me what the notation $L_K g$ means and why $inf\vert g(K, K)\vert >0$. Alternatively, also a reference will be well accepted. – C. Bishop Sep 11 '20 at 19:58
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    $L_Kg$ is the Lie derivative of $g$ by $K$. $f(x)=g(K(x),K(x))$ is a function defined on a compact space, this means there exists $x_0$ such that $f(x_0)=inf_{x\in M}f(x)$, $f(x_0)=g(K(x_0),K(x_0))>0$ since $K$ is a timelike vector. – Tsemo Aristide Sep 11 '20 at 20:08
  • Sorry @Tsemo Aristide, I have another question. In the notations of the paper what is $g(K, K)$? And what is the difference with $g(K, \gamma^{\prime})$? I was thinking that $g(K, \gamma^{\prime})$ be the projection of $\gamma^{\prime}$ on the timelike vector field, but I am not sure. If you prefer, I open a new question. – C. Bishop Sep 12 '20 at 16:12
  • $g(K,K)$ is the set of $g(K(x),K(x)), x\in M$ $g(K,\gamma')$ is the set of $g(K(x),\gamma'(x))$. – Tsemo Aristide Sep 12 '20 at 19:13
  • Still not so clear, but why the fact that $\inf\vert g(K, K)\vert >0$ implies that it is enough to show that $g(K, \gamma^{\prime})$ is bounded? – C. Bishop Sep 12 '20 at 19:27
  • If you want to answer here, I'll accept your answer. https://math.stackexchange.com/questions/3823776/help-with-this-proof-if-k-is-a-timelike-killing-vector-field-what-does-the-not – C. Bishop Sep 12 '20 at 19:32