1
$\begingroup$

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 $M$..

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.