# A question on the existence of geodesics on smooth manifolds

I study differential geometry and do not understand the choice of environment of $$(-2,2)$$ for the $$\gamma$$ curve in the following proof. Are there geodesics in arbitrarily small intervals? Why is this interval explicitly desired? I would be very happy if I could get an intuitive explanation, because this proof haunts my head and I can't work it out by my own.

Be $$M$$ a smooth manifold and $$\nabla$$ an affine connection on $$M$$. Then for every point $$p \in M$$ there exists an open environment $$V \subseteq TM$$ of $$\mathbb{0}_p \in T_p M \subseteq TM$$, so that for every tangential vector $$\mathfrak{v} \in V$$ there is exactly one geodetic $$\gamma_{\mathfrak{v}}:(-2,2) \rightarrow M$$ with $$\gamma_{\mathfrak{v}}(0)=\pi(\mathfrak{v})$$ (where $$\pi: TU \rightarrow U$$ ist the natural projection) and $$\gamma'_{\mathfrak{v}}(0) = \mathfrak{v}$$.

Proof. Be $$(U,\varphi)$$ a map around $$p$$ with local coordinates $$x_1, \ldots, x_n$$. Then $$T\varphi:TU \rightarrow \varphi(U) \times \mathbb{R}^n$$ identifies each tangential vector $$\mathfrak{v}\in TU$$ with a point $$(x_1,\ldots,x_n,v_1,\ldots,v_n) \in \varphi(U) \times \mathbb{R}^n$$. A curve $$\gamma: (-\epsilon,\epsilon) \rightarrow U$$ is a geodetic with $$\gamma(0) = \pi(\mathfrak{v})$$ and $$\gamma'(0)=\mathfrak{v}$$, if the coordinates $$\gamma_i = x_i \circ \gamma$$ of $$\gamma$$ solve the following initial value problem ($$1 \leq k \leq n$$):

$$\gamma'_k(t) = \eta_k(t) \quad \gamma_k(0) = x_k\\ \eta'_k(t) = - \sum_{i,j=1}^{n} \Gamma_{ij}^k(\gamma(t))\eta_i(t)\eta_j(t) \quad \eta_k(0) = v_k$$

So there is a $$\eta >0$$ and an open environment $$\Omega_1 \subseteq \varphi(U)$$ of $$\varphi(p)$$ and $$\Omega_2 \subseteq \mathbb{R}^n$$ of $$0$$, so that for all $$(x,v) \in \Omega_1 \times \Omega_2$$ exactly one solution $$(\gamma_1,\ldots,\gamma_n):(-\epsilon,\epsilon) \rightarrow \varphi(U)$$ exists. Then

$$\tilde\gamma_k:(-2,2)\rightarrow \mathbb{R}\quad \tilde\gamma_k(t) = \gamma_k(\frac{\epsilon}{2}t)$$

defines a solution $$(\tilde\gamma_1,\ldots,\tilde\gamma_n)$$ with

$$\tilde\gamma_k(0)=x_k \quad \tilde\gamma'_k(0) = \frac{\epsilon}{2}v_k$$

By replacing $$\Omega_2$$ with $$\frac{\epsilon}{2}\Omega_2$$ we can assume that $$\epsilon = 2$$.

Is it true that it doesn't matter what environment I choose for my curve? Can I always get a geodesic this way?

• I think the point of choosing $(-2,2)$ is to get a geodesic which continues for longer than the length of $\mathfrak{v}$, so it's just supposed to be a bigger interval than $(-1,1)$. – quarague Mar 13 at 16:03
• I had similar thoughts … I forgot to mention: an arbitrary environment with $\epsilon > 1$. Thanks! – John Smith Mar 13 at 16:46

## 1 Answer

The point to these existence results is usually that at the core they boil down to questions about the existence of solutions of certain ODEs. I think a lucid explanation of this can be found in Lang's book: Fundamental of differential geometry. As the proof you cite shows we can prolong the intervall of existence by scaling the initial velocity of the geodesic. So in principle there is nothing special about the value 2 in the construction. The reason many books construct it for the intervall (-2,2) is that this intervall contains the closed unit intervall. This is usually quite convenient as one would like to introduce (usually a few oages down the line) an exponential map associated to the connection and this map takes an initial velocity and sends it to the time 1 evaluation of the geodesic associated to it. Now from the above proof it is immediately clear that the domain of this exponential map contains at least an open neighborhood of the zero section in the tangent bundle.