Lemma 2.3. of Riemannian Geometry by do Carmo; Existence/Uniqueness of the Geodesic field $G$ on $TM$

Question

Lemma 2.3 states:

There exists a unique vector field $G$ on $TM$ (+) whose trajectories are of the form $t\mapsto (\gamma(t),\gamma'(t))$, where $\gamma$ is a geodesic on $M$.

Uniqueness: Suppose that there exists a vector field $G$ on $TM$ with (+). Choose $q\in TM$. By assumption, we then know that the trajectory $\alpha:(-\delta,\delta)\to M$ with $\alpha'(t)=G(\alpha(t))$ and $\alpha(0)=q$ is of the form $\alpha(t)=(\gamma(t),\gamma'(t))$, where $\gamma$ is a geodesic of $M$. Now $\gamma$, in local coordinates, has to satisfy a first order system with initial conditions given by $q$, for which there is a unique solution. Since $\alpha'(0)=G(\alpha(0))=G(q)$ and $\alpha$ is completly determined by this unique solution $\gamma$, we conclude that $G(q)$ is also uniquely determined. Since this holds for every $q\in TM$, $G$ has to be unique.

In order to show existence, do Carmo locally defines $G$ by the system mentioned above. It follows that the trajectories of $G$ are of the form $(\gamma,\gamma')$. He finishes the proof by stating:

Using the uniqueness, we conclude that $G$ is well-defined on $TM$.

My idea: In the uniqueness argument, we have shown that $G(q)$ is uniquely determined for every $q\in TM$. It follows that for every open subset $O\subset TM$, there also exists at most one vector field with the property (+). Now let $x:U\subset \mathbb{R}^n \to M$ and $y:V\subset \mathbb{R}^n \to M$ be two parametrizations with $x(U)\cap y(V)\neq \phi$ and denote by $G_x$ and $G_y$ their corresponding vector fields on $x(U)$ and $y(V)$ which satisfy (+) (which we know exist by the construction above). Then $G_x$ and $G_y$ also satisfy (+) on the open set $x(U)\cap y(V)$. It follows that $G_x\bigg|_{x(U)\cap y(V)}=G_y\bigg|_{x(U)\cap y(V)},$ which shows that $G$ is well-defined.

I am not sure if this argument is correct. It also seems a little too complicated to me/not what do Carmo intended. I would be grateful for any other (easier) proof of this fact which uses the uniqueness-part.

Without relying on the uniqueness, the well-definedness is shown in this answer https://math.stackexchange.com/a/1359927/776794e by a direct calculation using the formula for the coordinate change. In Riemannian Manifolds, Lee proves that

$$Gf(p,v)=\frac{d}{dt}\bigg|_{t=0} f(\gamma_v(t),\dot{\gamma}_v(t)),$$
where $\gamma_v$ denotes the unique geodesic with $\gamma(0)=p, \dot{\gamma}(0)=v$, which also implies that $G$ is well-defined.

Answer 1

I don't think do carmo had a better proof then what you suggested. The reason is that he had an identical proof to yours in lemma 5.2 of chapter 0 differential manifold. The last paragraph of his proof reads

To show existence, define $Z_{\alpha}$ in each coordinate neighborhood $\mathbf{x}_{\alpha}\left(U_{\alpha}\right)$ of a differentiable structure $\left\{\left(U_{\alpha}, \mathbf{x}_{\alpha}\right)\right\}$ on $M$ by the previous expression. By uniqueness, $Z_{\alpha}=Z_{\beta}$ on $x_{\alpha}\left(U_{\alpha}\right) \cap x_{\beta}\left(U_{\beta}\right) \neq \phi$, which allows us to define $Z$ over the entire manifold $M .$

In the proof of of lemma 2.3 he suggests using \begin{array}{l} \frac{d x_{k}}{d t}=y_{k} \\ \frac{d y_{k}}{d t}=-\sum_{i, j} \Gamma_{i j}^{k} y_{i} y_{j} \quad k=1, \ldots, n, \end{array} to define the vector fields locally, and my best guess to use his/your strategy to define it globally on $M$.