Let $M$ be a smooth manifold and $X$ a vector fields on $M$. Let $\{x_i\}_{i=1}^n$ be local coordinates, so $X=X^i \partial_i$

The equation for auto parallel transport (geodesic) of $X$ is, in coordinates,

$$(\nabla_X X)^i= X^k(\partial_k X^i + \Gamma^{i}_{\,jk}X^j)=0. $$

So far so good. The problem comes when $X$ is defined to be a vector field along a curve. Let $\dot{\gamma}(t)$ be the velocity vector of the curve $\gamma$, in principle nothing should change, and simply $X^k=(x^k\circ \gamma)'(t)$, the problem is that above $X$ was a vector field on the manifold, so $X^k\in C^\infty(M)$, but now $X^k: \mathbb{R}\rightarrow\mathbb{R}$, $$X^k = \frac{\mathrm{d}}{\mathrm{d}t}\gamma^k$$ where $\gamma^k=x^k\circ \gamma$. Naïvely substituting term by term I get $$\frac{\mathrm{d}\gamma^k}{\mathrm{d}t}\partial_k \frac{\mathrm{d}\gamma^i}{\mathrm{d}t} =-\frac{\mathrm{d}\gamma^k}{\mathrm{d}t} \Gamma^{i}_{\,jk}\frac{\mathrm{d}\gamma^j}{\mathrm{d}t} $$

This doesn't make much sense to me, the problem is I don't know how to interpret $$\partial_k \frac{\mathrm{d}\gamma^i}{\mathrm{d}t}$$

As $\partial_k$ is a map on $C^\infty(M)$ and $\frac{\mathrm{d}\gamma^i}{\mathrm{d}t}:\mathbb{R}\rightarrow\mathbb{R}$. I don't know if I can exchange the two derivatives and interpret it as

$$ \frac{\mathrm{d}}{\mathrm{d}t}(\partial_k x^i)(\gamma(t))=\frac{\mathrm{d}\gamma^k}{\mathrm{d}t}$$ but even if I did I would end up with a derivative squared, not a second derivative. Basically I'm lost and I feel that I seriously misunderstood something: I think the core problem is my definition of a vector field on a manifold and one along a curve don't agree.

How can I solve this problem? I want to manipulate the first equation to get the geodesic equation

$$\frac{\mathrm{d}^2\gamma^i}{\mathrm{d}t^2}+\Gamma^i_{\,jk}\frac{\mathrm{d}\gamma^k}{\mathrm{d}t}\frac{\mathrm{d}\gamma^j}{\mathrm{d}t}=0 $$

  • 1
    $\begingroup$ So... I understand you are having difficulties with vector fields along curves and geodesics. Is there a specific question you would like to ask? $\endgroup$ – Amitai Yuval Nov 2 '18 at 14:02
  • $\begingroup$ @AmitaiYuval I'm sorry if my question wasn't clear enough. I hope that someone could point out the fault in my reasoning and how can I treat the term $\partial_k \frac{\mathrm{d}\gamma^i}{\mathrm{d}k}$, assuming my reasoning is correct until that point $\endgroup$ – user438666 Nov 2 '18 at 14:11
  • $\begingroup$ Maybe it would be helpful to point out that the Riemannian connection satisfies a very strong property: $\nabla{X}{Y}(p)$ depends only on $X_p$ and $Y\circ\gamma$, for $\gamma$ a curve with $\gamma(0)=p$, $\gamma^{'}(0)=X_p$. So you can define a reduced version of derivative, the covariant derivative. In the geodesic equation, the coordinate vector fields and the Christoffel symbols are taken along the curve. This is thoroughly explained in do Carmo's Riemannian Geometry (Chapter 2). $\endgroup$ – Laz Nov 2 '18 at 16:05
  • $\begingroup$ @Laz thank you for the comment, I don't quite get what you mean. My first equation already is, as far as I know, a covariant derivative. And did you mean $\nabla_X Y(p)$ ? $\endgroup$ – user438666 Nov 2 '18 at 16:21
  • $\begingroup$ Oh yes, it was a typo and now I can't edit, but yes. Your first equation is not in terms of covariant derivative. Just read carefully the reference I recommended you, it is very well written. $\endgroup$ – Laz Nov 2 '18 at 16:25

Before finding the geodesic equation, you should be comfortable with the covariant derivative of a vector field along a curve. So let $\gamma=(\gamma^1,\ldots,\gamma^n)$ be a curve, and let $X(t)=X^i(t)\partial_i$ be a vector field along $\gamma$. You are right, to begin with; as $X$ is only defined along $\gamma$, there isn't much sense in expressions like $\partial_kX^i$. The way to solve this is first compute the covariant derivative of $\partial_i$ (which is defined on a coordinate neighborhood) and then use the Leibniz rule. So we start with $$\frac{D}{dt}\partial_i=\nabla_{\dot{\gamma}}\partial_i=\dot{\gamma}^j\Gamma_{ji}^k\partial_k.$$ Then, by forcing the Leibniz rule to hold, $$\begin{align}\frac{D}{dt}X&=\frac{D}{dt}X^i\partial_i\\&=\dot{X}^i\partial_i+X^i\dot{\gamma}^j\Gamma_{ji}^k\partial_k\\&=\left(\dot{X}^k+X^i\dot{\gamma}^j\Gamma_{ji}^k\right)\partial_k.\end{align}$$ Substituting $X=\dot{\gamma}$, $$\frac{D}{dt}\dot{\gamma}=\left(\ddot{\gamma}^k+\dot{\gamma}^i\dot{\gamma}^j\Gamma_{ji}^k\right)\partial_k.$$


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