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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 $a\in C^{\infty}(X)$...

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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 $df(X)\,a=X(a\circ f)$.

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