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 $\varepsilon = 0$.
Consider a one-parameter Lie group of transformations . The infinitesimals are given by and the infinitesimal generator: . Let . Now what if we have and ? If we want to get the first component of , we need to differentiate with respect to and evaluate it at and we should divide by zero. How can it be solved?
Thanks,
Zoli
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 $\varepsilon = 0$.