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Lie's third theorem says that every finite-dimensional Lie algebra g over the real numbers is associated to a Lie group G. So say I have an $r-$ parameter group of symmetries whose tangents at the identity form a Lie algebra, can we conclude that the group is a Lie group?

As an example of why I am asking, consider the ODE $y''=y'$. It has a $2-$parameter group of symmetry transformations given by $y\to \epsilon y$, and $y\to y+t$. It's not clear to me that this forms a Lie group since the transformations don't commute and I can't see a smooth structure. However, its generators, given by $\partial_y,y\partial_y$ form a Lie algebra.

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    $\begingroup$ Who said Lie groups should be commutative? $\endgroup$ Mar 7, 2014 at 9:14
  • $\begingroup$ @KevinCarlson I didn't mean that, I just meant that since they aren't commutative I don't see an obvious atlas...if they did commute then I could just identify it with $\mathbb{R}^2$. $\endgroup$
    – JLA
    Mar 7, 2014 at 18:06

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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.

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