For a connected, simply connected Lie group what does the Lie algebra tell me? By Lie's third fundamental theorem we know that for each finite dimensional Lie algebra $\mathfrak{g}$ there is a unique simply connected and connected Lie group $G$ such that $T_eG \simeq \mathfrak{g}$.  

Without knowing anything else about $G$, what can we use the Lie algebra for in order to get as much information about $G$ 

In particular, I want to classify the adjoint representation and consequently the adjoint orbits, of $G$, given a Lie algebra.   
For example take $\mathfrak{g}=\mathfrak{so}(3)$. Now I want to pretend I know nothing about the Lie group and classify the adjoint orbits. How can I proceed? What exponential can I use if I know nothing about $G$?
I am trying to do this in as general a way as possible. 
 A: One thing to realize is that the group 'actually acting' in the adjoint representation is often not the group itself, but one of its, not-simply connected quotients. Your example is excellent: the simply connected group is $SU(2)$ but the group acting on $\mathbb{R}^3$ in the adjoint representation is $SO(3)$: the group elements $I$ and $-I$ in $SU(2)$ act the same.
Related example: the Universal Cover of $SL_{2}(\mathbb{R})$ has no faithful finite-dimensional representation so the group acting on $\mathbb{R}^3$ in the adjoint representation is $PSL(2, \mathbb{R})$.
However this can be a blessing in disguise. The group that acts in the adjoint representation can be found by using the ordinary matrix exponential on the adjoint representation of the Lie algebra.
A: This might not be entirely satisfactory, but at least for a semisimple Lie algebra $\mathfrak{g}$ over $\mathbb R$, take its automorphism group $G_0 := Aut(\mathfrak{g})$. That is the adjoint Lie group sitting over $\mathfrak{g}$. If you want the simply connected one, take its universal cover $\widetilde{G_0}$.
I do not know, though, how efficiently one could actually compute or determine the group structure of the automorphisms of, say, $\mathfrak{so}_3$, when pretending to know nothing about the matrix group $SO_3$; and then the same problem when going to the simply connected cover.
