# On the correspondence between vector fields, derivations and 1 parameter groups of diffeomorphisms

Let X be a smooth n-manifold. Then there is a 1-1 correspondence between derivations $$\delta: C^{\infty}(X) \rightarrow C^{\infty}(X)$$ and vector fields $$v\in C^{\infty}(TX)$$. For example given $$\delta$$ we can define a vector field by setting $$v\in C^{\infty}(X)$$ to be

$$v: x \mapsto v_x := (a \mapsto \delta(a)\vert_x)$$

Now suppose we have a 1-parameter group of diffeomorphisms $$\varphi: \mathbb{R} \times X \rightarrow X$$. Then we can define a derivation by letting

$$\delta: a\mapsto \frac{d}{dt}(a\circ\varphi_t(x))\vert_{t=0}$$

My question is this: what is the derivation that corresponds to $$\varphi$$? I found some notes that claimed that the vector field would be

$$v: x\mapsto v_x:= \frac{d}{dt}(\varphi_t(x))|_{t=0}$$

But I fail to see how this would act on smooth maps $$a\in C^{\infty}(X)$$...

This latter acts precisely how you want it to act as a derivation: it is a matter of notations. I will expand them a little bit, writing $$\gamma_x:\mathbb{R}\to M$$ the curve $$\gamma_x(t)=\varphi_t(x)$$. I get:
$$v_x(a)=\frac{d}{dt}|_{t=0}(\gamma_x)\,a=d\gamma_x(\frac{d}{dt}|_{t=0})\,a=\frac{d}{dt}|_{t=0}(a\circ\gamma_x)=\delta(a)(x).$$
The second equality is the definition of $$\frac{d}{dt}|_{t=0}(\gamma_x)$$, and the third one is the definition of how a vector field $$df(X)$$ acts on $$a$$, namely $$df(X)\,a=X(a\circ f)$$.