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?

  • $\begingroup$ Doesn't it seem odd to you that you are not getting iterates (i.e. nested compositions) of $\partial_x$? For instance, if $\mathbf{v} = \partial_x$, what should $\mathbf{v}^2$ be? $\endgroup$ – Eric Towers Mar 14 at 17:37
  • $\begingroup$ If ${\bf v} = \partial_x$ then I would say ${\bf v}^2 = (1\partial_x)(1\partial_x) = 1(\partial_x 1)\partial_x=0$. Similarly ${\bf v}^n =0$ for all $n \ge 2$, meaning that $\exp(\varepsilon {\bf v})x= (1+\varepsilon \partial_x)x=x+\varepsilon$, which is the correct flow. $\endgroup$ – Fly by Night Mar 14 at 18:16
  • $\begingroup$ And yet, "the derivative with respect to $x$ of the derivative with respect to $x$" is not the zero operator (without projection to first order operators). If $\mathbf{v} = \partial_x$, $\mathbf{v}^2 = \partial_x^2$, the second derivative with respect to $x$. (... which projects onto the zero vector on the subspace of first order differentials.) But in short, in your expression "$1(\partial_x 1) \partial_x$", you have made an error: the first $\partial_x$ acts on the rest of the expression; you should have "$1 \partial_x (1 \partial_x) = 1 (1 \partial_x^2 + 0 \partial_x)$". $\endgroup$ – Eric Towers Mar 15 at 14:03

If you apply the product rule carefully, you don't get two different answers. In your example, \begin{align*} \mathbf{v}(f(x)) &= (x^2 \partial_x)(f(x)) \\ &= x^2 f'(x) \text{, and }\\ \mathbf{v}^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 } \\ \mathbf{v}^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) ) \end{align*} so we see $\mathbf{v}^2 = 2x^3 \partial_x + x^4 \partial_x^2$. Similarly, $\mathbf{v}^3 = 6 x^4 \partial_x + 6 x^5 \partial_x^2 + x^6 \partial_x^3$.

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

  • $\begingroup$ Thanks for the reply. Am I right to think that ${\bf v}$ is best understood by its action on a function, i.e. ${\bf v}(f)$, and so "multiplying on the right" doesn't make sense. ${\bf v}[{\bf v}(f)]$ is fine, while $[{\bf v}(f)]{\bf v}$ doesn't make sense because it's not a function. Hence ${\bf v}^{n+1} = {\bf v}{\bf v}^n$ is the only way to go? $\endgroup$ – Fly by Night Mar 14 at 18:22
  • $\begingroup$ @FlybyNight : $[\mathbf{v}(f)]\mathbf{v}$ is an operator as much as $x^2 \partial_x$ is since $\mathbf{v}(f)$ is some function and the $\mathbf{v}$ on the right still has an open slot to consume a function. Note that I showed both $\mathbf{v}(\mathbf{v}(f))$ and $(\mathbf{v}(\mathbf{v}))(f)$ above, so both "wrap another $\mathbf{v}$ on the left" ($\mathbf{v} (\mathbf{v}^n(f))$) and "let $\mathbf{v}^n$ act on $\mathbf{v}(f)$" ($\mathbf{v}^n (\mathbf{v}(f))$), arriving at the same result both ways. $\endgroup$ – Eric Towers Mar 15 at 13:57
  • $\begingroup$ But ${\bf v}(f){\bf v}$ isn't a function, while ${\bf vv}(f)$ is. That's what I meant. $\endgroup$ – Fly by Night Mar 16 at 18:22

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
$$ e^{\epsilon x^2\partial_x} ~~g(x)= g\left(\frac{-1}{y+\epsilon}\right )=g\left (\frac{x}{1-\epsilon x}\right ), $$ 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 ${\bf v} = x^2\partial_x$ 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*}


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