Lie algebra $\implies$ Lie group?

Solution 1:

Just saying $\partial_y$ and $y\partial_y$ are infinitesimal generators is implicitly assuming a smooth structure on your would-be Lie group $G,$ so I don't think this is the right approach. Here's another one.

First we need to notice $y\mapsto \epsilon y$ for $\epsilon=0$ gives a non-invertible transformation, so we want $G$ to have underlying smooth manifold $\mathbb{R}^*\times \mathbb{R},$ not $\mathbb{R}^2$. If we represent $y\mapsto \epsilon y+t$ as $(\epsilon,t)$ then its inverse $y\mapsto \epsilon^{-1}(y-t)=\epsilon^{-1}y-\epsilon^{-1}t$ is given by $(\epsilon^{-1},-\epsilon^{-1}t)$ and the multiplication law realizing $y\mapsto \epsilon y+t\mapsto \epsilon'\epsilon y+\epsilon' t+t'$ is $(\epsilon,t)(\epsilon',t')=(\epsilon\epsilon',\epsilon't+t')$. But these are smooth maps on $\mathbb{R}^*\times\mathbb{R}$ with the usual smooth structure: they're just polynomials and a rational map with its poles outside the domain.