So I take it you are clearly aware that the big A Adjoint representation is the homomorphism you're after in this case, so you're seeking a more general method.

Also, I'm assuming you know that the homomorphism of Lie algebras can only lift to a group homomorphism if the homomorphism's domain is simply connected, in which case there is a unique group homomorphism with the given algebra homomorphism as its Lie map. In this case, we're in the clear because $SU(2)$ is simply connected. Page 73 through 76 of:

Anthony Knapp, "Lie Groups Beyond an Introduction"

can then help you. Knapp gives you two methods of systematically constructing the simply connected Lie group: the first leaves you with differential equations for the left / right invariant vector fields, the second I believe is the same as V. Moretti's Answer.

A final "method" is to used Ado's theorem, which assures us that we can always realise a Lie algebra as a matrix Lie algebra; there is even an explicit software algorithm for this:

W. A. De Graaf, "Constructing Faithful Matrix Representations of Lie Algebras"

but if you can understand this algorithm, you are doing better than I (this paper has so far defeated me). Once you have a matrix algebra, you can used the matrix exponential to construct a neighbourhood of the identity, indeed the whole group if the latter is compact; as in V. Moretti's Answer the Lie algebra does not exponentiate to the whole group for noncompact groups (as far as I am aware, the problem of exactly what in a noncompact Lie group can be realised as an exponential of a Lie algebra element is to some extent still an open problem).

So, once you have the Lie group, you can in principle construct the universal cover with homotopy classes and carve out the discrete centre $\mathcal{Z}_d$ of the universal cover. Your original group will have as its fundamental group the quotient group of $\mathcal{Z}_d$ and one of its (normal) subgroups.

This post imported from StackExchange Physics at 2014-04-01 17:33 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance