# Understanding of differential of a curve as a vector field

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 $$M$$..

• 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. – Arthur May 30 at 13:34

(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.