I have trouble understanding the definition for a geodesic curve:

Let $\gamma:I \rightarrow M$ be a curve, $M$ pseudo.Riemannian manifold.

Then $\gamma$ is geodesic, if $\nabla_{\gamma^{'}} \gamma^{'}=0$

Now, if we look at the Levi-Civita connection for example, then $\nabla: X(M) \times X(M) \rightarrow X(M)$, therefore, for the expression $\nabla_{\gamma^{'}} \gamma^{'}=0$ to make sense, $\gamma^{'}$ has to be interpreted as a vector field.

Now I see you can define a vector field in the following way: $X(p):=\gamma^{'}(t)$ where $\gamma(t)=p$ but this is only a definition on $\gamma(I)$, not on all of $M$..

  • $\begingroup$ en.wikipedia.org/wiki/… has a short explanation. Intuitively, the covariant derivative of some field $X$ at a point $p$ in the direction of a vector $v$ only cares about what happens to $X$ along the direction of $v$ from $p$. I don't have a good enough grip on this to give a formal and rigorous answer, though. $\endgroup$ – Arthur May 30 at 13:34

There are two ways to do this.

(1) If you understand pullback connexions and bundles, you pull the tangent bundle $TM$ and the Levi-Civita connexion $\nabla$ back to $I$ via $\gamma\colon I\to M$, and the geodesic equation is defined as $(\gamma^*\nabla)_{d/dt}\dot\gamma=0$.

(2) If you don't know pullback connexions, you can simply extend $\dot\gamma$ arbitrarily to a smooth vector field $V$ on some neighbourhood of your point $p=\gamma(t)\in M$. Then prove that $(\nabla_VV)(p)$ is independent of the choice of extension $V$, so $\nabla_{\dot\gamma}\dot\gamma=0$ makes sense.

  • $\begingroup$ Ok this makes sense, thank you! $\endgroup$ – User1 May 30 at 20:29

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