Fdtxjr

I've got $bf v = x^2partial_x$, and I'm trying to find $exp(varepsilonbf v)$, but I'm having some trouble.

If I define $bf v^n+1 = bf vbf v^n$ then I get a different outcome to $bf v^n+1 = bf v^nbf v$.

For example:

$$bf v^2 = bf vv = (x^2partial_x)(x^2partial_x) = x^2(partial_xx^2)partial_x =
x^2(2x)partial_x = 2x^3partial_x$$

$$bf v^3 = bf vbf v^2 =(x^2partial_x)(2x^3partial_x)=x^2(partial_x2x^3)partial_x = x^2(6x^2)partial_x = 6x^4partial_x$$
$$bf v^3 = bf v^2bf v = (2x^3partial_x)(x^2partial_x) = 2x^3(partial_xx^2)partial_x = 2x^3(2x)partial_x = 4x^4partial_x$$

Using $bf v^n+1 = bf vbf v^n$ gives
$$exp(varepsilon bf v)x = fracx1-varepsilon x$$

While using $bf v^n+1 = bf v^nbf v$ gives
$$exp(varepsilon bf v)x = fracx2(1+mathrm e^2varepsilon x)$$

In both cases, when $varepsilon =0$, we get just $x$, i.e. the identity element. Also, in both cases, we get
$$lim_varepsilon to 0 fracmathrm dmathrm dvarepsilon exp(varepsilon bf v)x = bf v$$

The same $bf v in mathfrak g$ can't possible generate two different flows, can it?

If you apply the product rule carefully, you don't get two different answers. In your example, beginalign*
mathbfv(f(x)) &= (x^2 partial_x)(f(x)) \
&= x^2 f'(x) text, and \
mathbfv^2(f(x)) &= (x^2 partial_x)left( (x^2 partial_x)(f(x)) right) \
&= (x^2 partial_x)left( x^2 f'(x) right) \
&= x^2(2x f'(x) + x^2 f''(x) ) text, or \
mathbfv^2(f(x)) &= left((x^2 partial_x) (x^2 partial_x) right) (f(x)) \
&= (x^2(2x partial_x + x^2 partial_x^2))(f(x)) \
&= x^2(2x f'(x) + x^2 f''(x) )
endalign*
so we see $mathbfv^2 = 2x^3 partial_x + x^4 partial_x^2$. Similarly, $mathbfv^3 = 6 x^4 partial_x + 6 x^5 partial_x^2 + x^6 partial_x^3$.

(Of course, if we're linearizing, we project onto the first term in both of those.)

You are looking at the simplest Lie advective flow in perturbation theory (as applied in physics: QFT) and the workhorse example in the 19th century book of Georg Sheffer cited.

v generates a shift operator, and it pays to define suitable canonical coordinates,
$$
y=-1/x, qquad Longrightarrow qquad x^2 partial_x=partial_y ,
$$
so that you are shifting y by $epsilon$,
$$
e^epsilon partial_y ~~f(y)= f(y+epsilon),
$$
which reads
$$
e^epsilon x^2partial_x ~~g(x)= gleft(frac-1y+epsilonright )=gleft (fracx1-epsilon xright ),
$$
a standard formula in the RG advection of QFT.

Your difficulties are traceable to your refusal to perform your Heaviside calculus manipulations of functions of differential operators with a "test-function" f(x) on the right, which keeps track of the proper chain rule action of noncommuting derivative operators.

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 $bf v = x^2partial_x$ 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 $|varepsilon x|<1$
begineqnarray*
sum_n ge 0 fracvarepsilon^nn!bf v^n(x) &=& sum_n ge 0 fracvarepsilon^nn!n!x^n+1 \ \
&=& xsum_n ge 0 (varepsilon x)^n \ \
&=& fracx1-varepsilon x
endeqnarray*