Do Lie algebras "know" about their Lie groups? An undergraduate in physics asked me this question, and I did not know the answer, so I thought I would ask here. It is well-known that the Lie algebra to a Lie group is the tangent space to the identity of the group. Furthermore, via the exponential map, we know that if we have a covering of Lie groups $p: G_1 \to G_2$, then $G_1$ and $G_2$ have isomorphic Lie algebras. My question is, to what extent can this be reversed? If I have a (finite dimensional, real) Lie algebra (in characteristic $0$, does this matter?) that I know explicitly, say, I know the brackets of all pairs of basis vectors, then I know there is a simply connected Lie group which has this algebra as its Lie algebra, by the correspondence theorem. Can I also find other Lie groups with this same algebra? To what extent does knowing the algebra give me knowledge of the groups?
My guess is the answer is 'not at all,' but maybe there is some hidden information in the algebra that I don't know about, since I am not an expert on Lie theory.
 A: In principle it is possible to determine every (edit: connected) Lie group having Lie algebra some finite-dimensional real Lie algebra $\mathfrak{g}$. As radekzak says in the comments, first we find the unique simply connected Lie group $\widetilde{G}$ with Lie algebra $\mathfrak{g}$, and then every other Lie group $G$ with Lie algebra $\mathfrak{g}$ must admit $\widetilde{G}$ as its universal cover. Standard arguments can be used to show that the covering map $\widetilde{G} \to G$ is given by quotienting by a discrete central subgroup of $\widetilde{G}$, which can be identified with $\pi_1(G)$; conversely any such quotient gives a Lie group with Lie algebra $\mathfrak{g}$.
The conclusion is that (edit: connected) Lie groups with Lie algebra $\mathfrak{g}$ are classified by discrete central subgroups of $\widetilde{G}$ (up to the action of the automorphism group $\text{Aut}(\widetilde{G}) \cong \text{Aut}(\mathfrak{g})$). For classical Lie algebras it's usually not hard to identify $\widetilde{G}$ and its center explicitly; for example if $\mathfrak{g} = \mathfrak{su}(n)$ then $\widetilde{G} = SU(n)$, whose center is the subgroup $\mu_n$ of $n^{th}$ roots of unity. The conclusion is that the Lie groups with Lie algebra $\mathfrak{su}(n)$ are precisely the quotients $SU(n)/\mu_k$ where $k \mid n$. Most of these are not very familiar Lie groups except for $SU(2)/\mu_2 \cong SO(3)$ and maybe $SU(n)/\mu_n \cong PSU(n)$.
You might not consider this completely satisfying since it's not clear a priori how to find $\widetilde{G}$ given $\mathfrak{g}$ in general. But this isn't so bad: using Levi decomposition we can reduce this problem to the case that $\mathfrak{g}$ is semisimple and the case that $\mathfrak{g}$ is solvable. The case that $\mathfrak{g}$ is semisimple can be done explicitly using the classification of semisimple Lie algebras and I am pretty sure we can explicitly compute the center in all of these cases.
The case that $\mathfrak{g}$ is solvable is more annoying. In principle if we write $\mathfrak{g}$ as an iterated extension of abelian Lie algebras then $\widetilde{G}$ is correspondingly determined as an iterated extension of abelian Lie groups but this seems messy. Alternatively, after a single extension $0 \to [\mathfrak{g}, \mathfrak{g}] \to \mathfrak{g} \to \mathfrak{g}/[\mathfrak{g}, \mathfrak{g}] \to 0$ we can pass to the derived subalgebra $[\mathfrak{g}, \mathfrak{g}]$ which is nilpotent. For nilpotent Lie algebras the Baker-Campbell-Hausdorff formula terminates so can be used to explicitly write down a polynomial multiplication on $\mathfrak{g}$ making it isomorphic to $\widetilde{G}$.
Computing the center in general could get tricky. The issue is that we know that $Z(\widetilde{G})$ has Lie algebra $Z(\mathfrak{g})$, which we can compute in principle, but $Z(\widetilde{G})$ may not be connected. I don't know if there's a nice systematic way to figure out the elements of $Z(\widetilde{G})$ not connected to the identity in general.
Re: your comment about integrating the Lie algebra, this can actually be done: there is a uniform definition of $\widetilde{G}$ in terms of a certain space of paths in $\mathfrak{g}$ ("Lie integration"). I don't think it's very easy to use for computations, though, since it constructs $\widetilde{G}$ as a quotient of a messy infinite-dimensional thing; as far as I know it has only theoretical applications.
So, in principle it's possible to do this up to perhaps some difficulty computing the center. In typical cases $\mathfrak{g}$ is probably semisimple or reductive and in this case one can appeal to the classification of such Lie algebras and their corresponding Lie groups to compute the center explicitly.
