# Infinitesimal generator

• Mar 21st 2014, 11:33 AM
Zoli
Infinitesimal generator
Consider a one-parameter Lie group of transformations $\displaystyle \mathbf{x}^* = \mathbf{X}(\mathbf{x},\varepsilon)$. The infinitesimals are given by $\displaystyle \vec{\xi} = \left.\frac{\partial \mathbf{X}}{\partial \varepsilon}\right|_{\varepsilon=0}$ and the infinitesimal generator: $\displaystyle X = \vec{\xi}\cdot\nabla$. Let $\displaystyle \mathbf{x}^* = (x^*,y^*)$. Now what if we have $\displaystyle x^*=\frac{1}{\varepsilon}x$ and $\displaystyle y^* = \varepsilon y$? If we want to get the first component of $\displaystyle \vec{\xi}$, we need to differentiate $\displaystyle x^*$ with respect to $\displaystyle \varepsilon$ and evaluate it at $\displaystyle \varepsilon=0$ and we should divide by zero. How can it be solved?

Thanks,
Zoli
• Mar 22nd 2014, 04:12 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 $\varepsilon = 0$.
• Mar 22nd 2014, 08:17 AM
Zoli
Re: Infinitesimal generator
I should have thought of the identity transformation. Thank you.