How do you get the explicit lie group from the infinitesimal generators (vector fields)? I am doing my thesis on the application of Lie groups to differential equations and I am curious how do you proceed to find the general group that corresponds to the found infinitesimal generators. That is if you have a differential equation (say heat equation) and you applied all the steps to find its infinitesimal generators (as described in Olver's book) that are given as differential operators of the form :

what is the next step to finding the group (say SO(2))? what I was thinking was to find a representation for these infinitesimals generators and take the exponential, however, I am not sure how to do that - especially finding a representation. Can someone recommend a source where they explain that with examples? Moreover, apparently, the Lie Algebra that you obtain is not unique for different groups, how do you also find the correct group then?
Your help and guidance would be very much appreciated!
 A: A Lie algebra (where your infinitesimal generators live) does not correspond to a unique group. Two Lie groups with the same Lie algebra are locally isomorphic but not necessarily globally isomorphic. As an example, $SO(n)$ has the same Lie algebra as $\mathrm{Spin}(n)$ but the second is the double cover of the former.
For each group $G$ there is an exponential map from the Lie algebra to the Lie group. This map has image in the connected component of the identity $G_0$. In general, it isn't surjective even onto $G_0$ but its image will generate $G_0$.
If there is a Lie group representation, we can differentiate to find a Lie algebra representation but the converse doesn't work for each Lie group with that Lie algebra. I think there is always some Lie group with such a representation but it doesn't have to be the only one (even if we restrict ourselves to connected groups). As an example every Lie group has a representation on its own Lie algebra.
In summary, I think you need to make it more clear what "the correct group" means.
