# Infinitesimal generator

• March 21st 2014, 12:33 PM
Zoli
Infinitesimal generator
Consider a one-parameter Lie group of transformations $\mathbf{x}^* = \mathbf{X}(\mathbf{x},\varepsilon)$. The infinitesimals are given by $\vec{\xi} = \left.\frac{\partial \mathbf{X}}{\partial \varepsilon}\right|_{\varepsilon=0}$ and the infinitesimal generator: $X = \vec{\xi}\cdot\nabla$. Let $\mathbf{x}^* = (x^*,y^*)$. Now what if we have $x^*=\frac{1}{\varepsilon}x$ and $y^* = \varepsilon y$? If we want to get the first component of $\vec{\xi}$, we need to differentiate $x^*$ with respect to $\varepsilon$ and evaluate it at $\varepsilon=0$ and we should divide by zero. How can it be solved?

Thanks,
Zoli
• March 22nd 2014, 05:12 AM
Jester
Re: Infinitesimal generator
This is not a Lie group. You need the identity meaning when $\varepsilon = 0$ then $x^*=x$. However, $\varepsilon = 1$ does work but you can re-parameterize by let $\varepsilon \rightarrow e^\varepsilon$ so that you can use $\varepsilon = 0$.
• March 22nd 2014, 09:17 AM
Zoli
Re: Infinitesimal generator
I should have thought of the identity transformation. Thank you.