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Assume that $(M, \nabla)$ is a smooth manifold equipped with an affine connection. In our geometry lectures, we defined a geodesic as a curve $\gamma: I \rightarrow M$ satisfying the following system of differential equations:

$$\ddot\gamma_k(t)+\sum_{i,j=1}^{n}\Gamma_{ij}^{k}(\gamma_1(t), \dots, \gamma_n(t))\dot\gamma_i(t)\dot\gamma_j(t)=0, \;\; k=1,2, \dots, n,$$

where $\gamma_i$ denotes the $i$-th coordinate of $\gamma$ and $\Gamma_{ij}^k$ are the Christoffel symbols, all in suitable local coordinates.

What I fail to see is how the above system of equation ensures that the curves $\gamma$ are regular (which should hold even if they are defined on maximal interval). More specifically, our teacher claims that either the geodesic is "trivial", i.e. a constant map (mapping the whole interval to one point), or it is a regular curve. Can it be seen somewhat directly from the equations above, or how can it be proven?

Thanks in advance for any help.

PS: This question is very similar to this question. However, I do not know how can the datum of the affine connection $\nabla$ be "translated" into the notion of Riemannian metric (or if the notions are somewhat fundamentally different). Therefore the answer there does not really help me.

Edit: My definition of Chritoffel symbols is that they are the bunch of unique functions such that $$ \nabla_{\frac{\partial}{\partial u_i}}\frac{\partial}{\partial u_i}=\sum_{k=0}^{n}\Gamma_{ij}^k\frac{\partial}{\partial u_k}, \;\; i,j, \in \{1,2, \dots, n\}.$$

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  • $\begingroup$ How do you define the Christoffel symbols without a metric? $\endgroup$ – Mateus Sampaio Nov 25 '13 at 22:16
  • $\begingroup$ @Mateus Sampaio: By the equations $\nabla_{\frac{\partial}{\partial u_i}}\frac{\partial}{\partial u_i}=\sum_{k=0}^{n}\Gamma_{ij}^k\frac{\partial}{\partial u_k}$. $\endgroup$ – Pavel Čoupek Nov 25 '13 at 22:28
  • $\begingroup$ What about the definition of a regular curve? $\endgroup$ – Mateus Sampaio Nov 25 '13 at 22:40
  • $\begingroup$ @Mateus Sampaio: The tangent vector of the curve is nonzero (at any point). I believe one can equivalently say that $\dot\gamma(t)\neq 0$, i.e. one can compute it in the above local coordinates. $\endgroup$ – Pavel Čoupek Nov 25 '13 at 22:50
  • $\begingroup$ The notion of the metric is not needed so, but all you have to do to understand the other solution is to switch $\langle \frac{d \gamma}{dt}, \frac{d \gamma}{dt} \rangle$ to $\sum_k \dot{\gamma}_k (t)^2$, because that's is just the lenght of the vector $\dot{\gamma}$. $\endgroup$ – Mateus Sampaio Nov 25 '13 at 22:57
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So looking at why a geodesic is a regular curve?, it seems that regular simply means $\dot\gamma \ne 0$ on the whole curve. There is a theorem that says that if $F:\mathbb R^n \to \mathbb R^n$ is smooth (actually locally Lipschitz should be enough), then if we try to solve the ode: $$ \dot x = F(x), \quad x(t_0) = x_0 $$ and extend it both forwards and backwards in time to a function $x:(t_0-t_1,t_0+t_2) \to \mathbb R^n$ (where $t_1>0$ and $t_2>0$ might be $\infty$), then that solution, if it exists, is unique. This means that if at any time $F(x(t_0)) = 0$, then $x(t) = x(t_0)$ is a solution, and hence the only solution, along the whole interval.

By using an atlas, it should be possible to extend this to manifolds: $$ F:M\times TM \to T(M\times TM), \quad F(x,y) = \left(y,\sum_{ijk} \Gamma^i_{jk}(x) y_j y_k\frac\partial{\partial x_i}\right) .$$

I know this theorem I cited above is somewhere in "Ordinary Differential Equations With Applications" by Carmen Chicone, but I don't have a copy of the book with me right now.

Also, if the function $F$ is smooth (i.e. if the Christoffel symbols are smooth), then you can show that the solution $x(t)$ is also smooth.

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  • $\begingroup$ Also, the theorem says that the only way that $x(t)$ cannot be extended in the positive direction beyond $t=T$ is because $\|x(t)\| \to \infty$ as $t \to T$. But I don't think this is needed for what I am saying here. $\endgroup$ – Stephen Montgomery-Smith Nov 26 '13 at 3:26
  • $\begingroup$ I think the theorem you mention is Pickard's theorem, right? (more precisely, its higher dimension version.) This seems to be working. I actually think this is the theorem which is used in a proof that one can find precisely one geodesic with prescribed point on the manifold which should lie on the geodesic and the tangent vector in that point. One can then use this fact instead and use the same arguments. Thanks. $\endgroup$ – Pavel Čoupek Nov 26 '13 at 21:45
  • $\begingroup$ Yes, it is Pickard's Theorem. $\endgroup$ – Stephen Montgomery-Smith Nov 26 '13 at 22:20

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