Vector Field Exponential Map I've got ${\bf v} = x^2\partial_x$, and I'm trying to find $\exp(\varepsilon{\bf v})$, but I'm having some trouble.
If I define ${\bf v}^{n+1} = {\bf v}{\bf v}^n$ then I get a different outcome to ${\bf v}^{n+1} = {\bf v}^n{\bf v}$.
For example:
$${\bf v}^2 = {\bf vv} = (x^2\partial_x)(x^2\partial_x) = x^2(\partial_xx^2)\partial_x = 
x^2(2x)\partial_x = 2x^3\partial_x$$
$${\bf v}^3 = {\bf v}{\bf v}^2 =(x^2\partial_x)(2x^3\partial_x)=x^2(\partial_x2x^3)\partial_x = x^2(6x^2)\partial_x = 6x^4\partial_x$$
$${\bf v}^3 = {\bf v}^2{\bf v} = (2x^3\partial_x)(x^2\partial_x) = 2x^3(\partial_xx^2)\partial_x = 2x^3(2x)\partial_x = 4x^4\partial_x$$
Using ${\bf v}^{n+1} = {\bf v}{\bf v}^n$ gives 
$$\exp(\varepsilon {\bf v})x = \frac{x}{1-\varepsilon x}$$
While using ${\bf v}^{n+1} = {\bf v}^n{\bf v}$ gives
$$\exp(\varepsilon {\bf v})x = \frac{x}{2}(1+\mathrm e^{2\varepsilon x})$$
In both cases, when $\varepsilon =0$, we get just $x$, i.e. the identity element. Also, in both cases, we get
$$\lim_{\varepsilon \to 0} \frac{\mathrm d}{\mathrm d\varepsilon} \exp(\varepsilon {\bf v})x = {\bf v}$$
The same ${\bf v} \in \mathfrak g$ can't possible generate two different flows, can it?

The action of a vector field as a derivation on the space of functions means that it only makes sense to multiply on the left because ${\bf v}(f)$ is a function and so ${\bf v}[{\bf v}(f)]$ is also a function, while $[{\bf v}(f)]{\bf v}$ is a function times a vector field, i.e. a vector field.
The differential operator ${\bf v} = x^2\partial_x$ is best understood as differentiate with respect to $x$, and then multiply by $x^2$.
Applied to $x$, we get $x \mapsto 1 \mapsto x^2$, meaning ${\bf v}(x) = x^2$.
Applied to $x^2$, we get $x^2 \mapsto 2x \mapsto 2x^3$, meaning ${\bf v^2}(x) = 2x^3$.
Applied to $2x^3$, we get $2x^3 \mapsto 6x^2 \mapsto 6x^4$, meaning ${\bf v}^3(x) = 6x^4$.
In general, ${\bf v}^n(x) = n!x^{n+1}$, and so for all $|\varepsilon x|<1$
\begin{eqnarray*}
\sum_{n \ge 0} \frac{\varepsilon^n}{n!}{\bf v}^n(x) &=& \sum_{n \ge 0} \frac{\varepsilon^n}{n!}n!x^{n+1} \\ \\
&=& x\sum_{n \ge 0} (\varepsilon x)^n \\ \\
&=& \frac{x}{1-\varepsilon x}
\end{eqnarray*}
