Exponentials of Representations of Lie Algebras Assume G is a lie group and g is its lie algebra. Consider a representation of G :
D:G->End(V). Then there is a corresponding representation of g : d:g->End(V). My  question is, when you can express every element of D(G) as an exponential of an element of d(g), i.e. :
When it is true that, for every X in G, there is a x in g, such that D(X) = exp(d(x)). Thanks.
 A: This can be rephrased as a question about Lie groups; taking the image of the representation if necessary, you're asking when the exponential map $\exp : \mathfrak{g} \to G$ from a Lie algebra to its Lie group is surjective. I don't know of a clean necessary and sufficient condition; here are some general facts.
It's clearly necessary that $G$ be connected; from now on we'll assume this. 
The exponential map is always surjective, in fact a diffeomorphism, from some neighborhood of the identity in $\mathfrak{g}$ to some neighborhood of the identity in $G$. It's an important topological fact about connected topological groups that they are always generated by a neighborhood of the identity; in particular, every element of $G$ is a finite product of exponentials of elements of $\mathfrak{g}$.
If the exponential map is surjective, then every element of $G$ has an $n^{th}$ root for all $n$, since if $\exp(x) = g$ then $\exp \left( \frac{x}{n} \right)$ is an $n^{th}$ root for $g$. This isn't true for all connected Lie groups, and shows that the exponential map is not always surjective. For example, when $G = \text{SL}_2(\mathbb{R})$, the element 
$$g = \left[ \begin{array}{cc} -4 & 0 \\ 0 & -\frac{1}{4} \end{array} \right] \in \text{SL}_2(\mathbb{R})$$
has no square root in $G$ since on the one hand such a square root must have at least one of the eigenvalues $\pm 2i$ and at least one of the eigenvalues $\pm \frac{i}{2}$, and on the other hand such a square root must have conjugate eigenvalues if they are complex.
The exponential map is always surjective if $G$ is compact. This isn't completely trivial to prove, but follows from the fact that every element of $G$ is contained in a maximal torus. When $G = \text{SO}(n)$ or $G = \text{U}(n)$ this can be proven more directly using the spectral theorem, and if you allow yourself the Peter-Weyl theorem then the general case reduces to the case that $G = \text{U}(n)$. If you only want to look at representations, then it suffices to average an inner product on $V$ over $G$ to get a $G$-invariant inner product on $V$. And Terence Tao observes in this blog post that you can also use the Hopf-Rinow theorem from differential geometry. 
