# given the infinitesimal generator can we deduce the symmetry?

for example given the differential operator $\partial _{x}$ we know that it belong to the translation $y=x+a$ for some 'a'

given the differential operator $x\partial _{x}$ we know that we are dealing with dilations $y=ax$

however for any particular first order differential operator

$a(x) \partial _{x} +b(x)$ for some polynomials a(x) and b(x) can we obtain the symmetry ??

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An infinitesimal generator $X$ is a vector field, which satisfies Leibniz rule

$$X[fg]~=~ fX[g]+g X[f].$$

In the $1$-dimensional case, the generator is of the form $X=p(x)\frac{\partial}{\partial x}$, where $p=p(x)$ is some function. Assume furthermore that there exists a bijective smooth function $h=h(x)$, such that

$$p(x) h'(x)~=~1.$$

In other words, that

$$h(x)~=~ \int^x \frac{dx^{\prime}}{p(x^{\prime})}.$$

Then the symmetry flow becomes

$$e^{a X} f(x) ~=~ e^{a p(x) \frac{\partial}{\partial x}} f(x) ~=~ \left. e^{a \frac{\partial}{\partial y}} f(h^{-1}(y)) \right|_{y=h(x)}$$ $$~=~ \left.f(h^{-1}(y+a)) \right|_{y=h(x)}~=~ f(h^{-1}(h(x)+a)).$$

Example 1. Translation:

$$p(x)~=~1, \qquad h(x)~=~x, \qquad e^{a X} f(x)~=~f(x+a).$$

Example 2. Dilation:

$$p(x)~=~x, \qquad h(x)~=~\ln x, \qquad e^{a X} f(x)~=~f(e^ax).$$

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OK :) all clear thank you very much .. for the 2 or 3 dimensional case how would thas been –  Jose Javier Garcia Nov 11 '11 at 16:25