TITLE: When is a finite dimensional real or complex Lie Group not a matrix group
QUESTION [27 upvotes]: I have a smattering of knowledge and disconnected facts about this question, so I would like to clarify the following discussion, and I also seek references and citations supporting this knowledge.  Please see my specific questions at the end, after "discussion".

Discussion and Background
In practice, groups that do not have any faithful linear representation seem to be seldom (in the sense that I believe it was not till the late 1930s that anyone found any).
By Ado's theorem, every abstract finite dimensional Lie algebra over $\mathbb{R}, \mathbb{C}$ is the Lie algebra of some matrix Lie group. All Lie groups with a given Lie algebra are covers of one another, so even groups that are not subsets of $GL\left(V\right)$ ($V = \mathbb{R}, \mathbb{C}$) are covers of matrix groups.
I know that the metaplectic groups (double covers of the symplectic groups $Sp_{2 n}$) are not matrix groups. And I daresay it is known (although I don't know) exactly which covers of semisimple groups have faithful linear representations, thanks to the Cartan classification of all semisimple groups. But is there a know general reason (i.e. theorem showing) why particular groups lack linear representations? I believe a group must be noncompact to lack linear representations, because the connected components of all compact ones are the exponentials of the Lie algebra (actually if someone could point me to a reference to a proof of this fact, if indeed I have gotten my facts straight, I would appreciate that too). But conversely, do noncompact groups always have covers which lack faithful linear representations? Therefore, here are my specific questions:

Specific Questions
Firm answers with citations to any of the following would be highly helpful:
1) Is there a general theorem telling one exactly when a finite dimensional Lie group lacks a faithful linear representation;
2) Alternatively, which of the (Cartan-calssified) semisimple Lie groups have covers lacking faithful linear representations;
3) Who first exhibited a Lie group without a faithful representation and when;
4) Is compactness a key factor here? Am I correct that a complex group is always the exponetial of its Lie algebra (please give a citation for this). Does a noncompact group always have a cover lacking a faithful linear representation?
Many thanks in advance.

REPLY [3 votes]: My knowledge is piecemeal, and this is really a comment rather than an answer. I believe that the following is a good way of organizing the problem.
The problem is which connected Lie groups have faithful representations. 
Let $g$ be a Lie algebra and $G$ its simply connected group. When can a given discrete normal (necessarily central) subgroup of $G$ occur as the kernel of a continuous homomorphism $G\to GL_n(\mathbb R)$ for some $n$, or equivalently as the kernel of a continuous homomorphism $G\to GL_n(\mathbb C)$ for some $n$. 
Such homomorphisms correspond to Lie algebra homomorphisms $g\to gl_n(\mathbb C)$, thus to complex-linear Lie algebra homomorphisms $g_{\mathbb C}\to gl_n(\mathbb C)$ where $g_{\mathbb C}=g\otimes \mathbb C$, thus to complex-analytic homomorphisms $G_{\mathbb C}\to gl_n(\mathbb C)$ where $G_{\mathbb C}$ is the relevant simply connected complex group. 
The canonical map $G\to G_{\mathbb C}$ has a discrete kernel. It can be nontrivial, for example in the case $g=sl_2(\mathbb R)$. So in order for a given discrete $D$ in the center of $G$ to be the kernel of a representation it is necessary for $D$ to contain this.
If that test is passed, then the question arises, for a simply connected complex group such as $G_{\mathbb C}$, whether a given discrete subgroup can be the kernel of a complex-analytic representation. In the semisimple case maybe the answer is always yes? I don't know. In the nilpotent case the answer is always yes for the trivial subgroup. But it's no for some nontrivial discrete central subgroups of the Heisenberg group, as we recently learned at another MO question.