# 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 '19 at 16:03
• I had similar thoughts … I forgot to mention: an arbitrary environment with $\epsilon > 1$. Thanks! – John Smith Mar 13 '19 at 16:46