Let $M$ be a smooth manifold, $p \in M$, $I \subset \Bbb R$ and $0 \in I$. Let $v = \sum_{i=1}^n vx^i \left(\frac{\partial}{\partial x^i}\right)_p\in T_p M$. Show that there exists a $C^\infty$-path $\gamma : I \to M$ such that $\dot{\gamma}_0=v$. (Here $\dot{\gamma}_t : C^\infty(p) \to \Bbb R$, $\dot{\gamma}_t(f)=(f\circ \gamma)'(t)$ and $C^\infty(p)$ is the set of smooth functions on an arbitary neighborhood $U$ of $p$).
I think that the idea here is to show that the derivation agrees with the tangent vector $v$ in the tangent space? If our manifold would be $\Bbb R^n$, then $\dot{\gamma}_0(f)=\nabla f \cdot \gamma'(0)$ so I would have the partial derivatives that I'm seeking, but I still don't know how to show this. Also the manifold is not neccessarily any subset of an Euclidean space so this gradient might not make sense. Could I have some hints how to approach this?
Edit: The definition for $v \in T_pM$ is that $v$ must satisfy $v(af+bg)=av(f) + bv(g)$ and $v(fg)=g(p)v(f)+f(p)v(g).$
Best Answer
Let me first observe that $\dot{\gamma}_t(f)=(f\circ \gamma)'(t)$ is not defined for all $f$ unless $t = 0$. In this formula $f$ is a smooth real-valued function defined on some open neighborhood $U$ of $p$. The function $f \circ \gamma$ is defined on $I_U = \gamma^{-1}(U) \subset I$. Certainly $0 \in I_U$, but for no $t \in I \setminus \{0\}$ we can be sure that $t \in I_U$. Anyway, we only need $\dot{\gamma}_0(f)$ which is always defined.
Moreover, the formula $$\dot{\gamma}_0(f)=\nabla f \cdot \gamma'(0) \tag{1} $$ does not make sense for a general manifold. You can use it only if $M$ is an open subset of $\mathbb R^n$.
You say that $T_pM$ is the set of all derivations $v : C^\infty(p) \to \mathbb R$. Here you get the next problem: As you have defined $C^\infty(p)$ it is no real vector space and no $\mathbb R$-algebra. The problem is the addition of $f : U \to \mathbb R$ and $g : V \to \mathbb R$: You can only define $f + g$ on $U \cap V$. This addition has as its neutral element the zero-function $0_M : M \to \mathbb R$. But if $f : U \to \mathbb R$ is defined on $U \subsetneqq M$, then $f$ does not have an inverse: No sum $f + g$ is defined on a bigger set as $U$, and hence no such sum can be $0_M$. Therefore one usually understands $C^\infty(p)$ as the set of germs of smooth functions $f : U \to \mathbb R$, a germ being an equivalence class of such functions, where $f : U \to \mathbb R$ and $g : V \to \mathbb R$ are declared as equivalent if they agree on an open $W \subset U \cap V$ with $p \in W$. With this interpretation $C^\infty(p)$ is an $\mathbb R$-algebra and it is easy to see that $\dot\gamma_0$ is a derivation.
Let us first consider the case that $M$ is an open subset of $\mathbb R^n$. Consider $v \in T_pM$. It has the form $v = \sum_{i=1}^n v_i \frac{\partial }{\partial x_i}\mid_p$. Here the $\frac{\partial }{\partial x_i}$ are the standard partial derivatives of a multivariable function. Now let $\mathbf v = (v_1,\ldots,v_n)$ and define $\gamma : (-r,r) \to M, \gamma(t) = p + t\mathbf v$, which is well-defined for sufficiently small $r > 0$. In the present situation we can use the formula $(1)$ and get $$\dot{\gamma}_0(f)=\nabla f \cdot \gamma'(0) = \sum_{i=1}^n\frac{\partial f}{\partial x_i}\mid_p \cdot v_i = v(f).$$ This proves that $v = \dot{\gamma}_0$ for some smooth path $\gamma$.
Next we consider the general case. Choose a chart $\phi : U \to V \subset \mathbb R^n$ on $M$ with $p \in U$. Let $q = \phi(p)$. We know that $\phi$ is diffeomorphism, thus it induces an isomorphism $$d\phi_p : T_pM = T_pU \to T_qV .$$ Recall that $d\phi_p(v)(g) = v(g \circ \phi)$, where $g : W \to \mathbb R$ is defined on an open neigborhood $W \subset V$ of $q$.
Now let $v \in T_pM$. We find a smooth path $\delta : I \to V$ such that $d\phi_p(v) = \dot\delta_0$. Define $\gamma = \phi^{-1} \circ \delta : I \to U \hookrightarrow M$. Then $$\dot\gamma_0(f) = (f \circ \phi^{-1} \circ \delta)'(0) = \dot\delta_0(f \circ \phi^{-1}) = d\phi_p(v)(f \circ \phi^{-1}) = v(f \circ \phi^{-1} \circ \phi) = v(f) .$$