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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.

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Can't you get the statement that compact Lie groups are linear more or less from Peter-Weyl? This is at least writes your compact Lie group as an inverse limit of Lie groups $G_{n}.$ It seems that once $\dim G_{n}=\dim G_{m}$ for $m\geq n,$ the maps $G_{m+1}--->G_{m}--->G_{m-1}--->...--->G_{n},$ will have to be covers and this process can't be nontrivial for very long. –  Benjamin Hayes May 7 '11 at 14:16
    
In 4) I suppose you mean "compact group" and not "complex group"? –  José Figueroa-O'Farrill May 7 '11 at 14:59
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I'm not sure about 3) but the Heisenberg group and also $SL(2,\mathbb{R})$ were known early on not to possess faithful finite-dimensional representations. The answers to 4) are Yes, Yes (if connected) and No, respectively. The exponential map is surjective for a compact connected Lie group and $SL(2,\mathbb{C})$ is noncompact (also complex), simply connected and yet it's a matrix group. –  José Figueroa-O'Farrill May 7 '11 at 15:03
    
In my comment, when I said $SL(2,\mathbb{R})$ I meant, of course, its universal cover! (How I wish I could edit comments!) –  José Figueroa-O'Farrill May 7 '11 at 15:41
    
@José: you can delete a comment, and repost the edited version. Also what Heisenberh group are you talking about? The usual one (see en.wikipedia.org/wiki/Heisenberg_group) is by definition is a group of matrices. –  Igor Belegradek May 7 '11 at 16:33
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3 Answers 3

up vote 13 down vote accepted

Most of the answers can be found in Hochschild's book on the structure of Lie group.

a) Every complex semisimple group has a faithful rep (Thm 3.2 in Chap. XVII)

b) A connected Lie group with Levi decomposition $G=RS$ ($R$ the solvable radical, $S$ a semisimple Levi factor) is linear iff both $R$ and $S$ are linear (Thm 4.2 in Chap. XVIII)

c) A solvable Lie group $G$ is linear iff its commutator subgroup $G'$ is closed, and $G'$ has no non-trivial compact subgroup (Thm 3.2 in Chap. XVIII)

Now, let $G$ be a semi-simple Lie group. Assume that $G$ is simply connected. Then $G$ admits a greatest linear quotient. Indeed, let $G_{\mathbb{C}}$ be the simply connected complex group corresponding to the complexified Lie algebra of $G$. Let $L$ be the kernel of the canonical homomorphism $G\rightarrow G_{\mathbb{C}}$; so $L$ is a finite index subgroup of the center of $G$. Then $G/L$ is the greatest linear quotient of $G$, in the sense that, if $H$ is locally isomorphic to $G$ and $p:G\rightarrow H$ is a universal covering, the group $H$ s linear iff $p$ factors through $G/L$.

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@ Tom: thanks for your editing! –  Alain Valette May 7 '11 at 19:03
    
@ Alain: Thanks heaps for your answer - that's two of my questions I owe you on now! –  WetSavannaAnimal aka Rod Vance May 8 '11 at 23:52
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In fact "most" simple groups have such covers: if and only if there is a long real root. There is a long and rich history of this topic, going back at least to a paper by Jacque Tits from 1967 (as well as some older work which appeared only in Russian). For a uniform (case-free) treatment see Nonlinear covers of real groups, IMRN 2004, Math Reviews 2112326. Also see the references in the paper and in the Mathscinet review. A good basic reference is Onischik and Vinberg, Lie Groups and Algebraic Groups; the answer can be read off from Table 10 in the Reference Chapter.

Suppose $G(\mathbb C)$ is (connected) simple, complex, simply connected, and $G(\mathbb R)$ is a real form. Then $G(\mathbb R)$ has a covering group with no faithful finite dimensional representation if and only if $G(\mathbb R)$ is not simply connected. Let $K(\mathbb R)$ be a maximal compact subgroup, with complexification $K(\mathbb C)$. Then $G(\mathbb R), K(\mathbb R)$ and $K(\mathbb C)$ all have the same fundamental group. So the question becomes: is $\pi_1(K(\mathbb C))=1$?

Examples where $G(\mathbb R)$ is simply connected are rare. Here is the complete list: compact, complex, $SL(n,\mathbb H)$, $Spin(n,1)$ $(n\ge 3)$, $Sp(p,q)$, $E_6(F_4)$ and $F_4(B_4)$. (By complex I mean a complex group, viewed as a real Lie group). If G is simply laced then $\pi_1(G(\mathbb R))=1$ if and only if $G(\mathbb R)$ has only one conjugacy class of Cartan subgroups. [Real forms of exceptional groups may be identified by their maximal compact subgroups; $E_6(F_4)$ is the real form of $E_6$ with maximal compact subgroup of type $F_4$, known in another classification as $E_{6(-20)}$; $F_4(B_4)=F_{4(-20)}$ has maximal compact $B_4$.]

Example: $G(\mathbb C)=SL(n,\mathbb C)$, $G(\mathbb R)=SL(n,\mathbb R)$, $K(\mathbb R)=SO(n,\mathbb R)$, $K(\mathbb C)=SO(n,\mathbb C)$. If $n\ge 3$ then $SO(n,\mathbb C)$ has a two-fold cover $Spin(n,\mathbb C)$, and the universal cover of $SL(n,\mathbb R)$ is two-fold, with maximal compact subgroup $Spin(n)$. If $n=2$, $SO(2,\mathbb C)=\mathbb C^*$, $\pi_1=\mathbb Z$, and the universal cover of $SL(2,\mathbb R)$ is infinite.

Example: $G(\mathbb C)=Spin(n+1,\mathbb C)$, $G(\mathbb R)=Spin(n,1)$ ($n\ge 3$), $K(\mathbb R)=Spin(n)$ is simply connected, so $Spin(n,1)$ is simply connected. For example $Spin(3,1)\simeq SL(2,\mathbb C)$ is simply connected.

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What do $E_6(F_4)$ and $F_4(B_4)$ mean (or is it too long for the comment to contain :-) )? –  L Spice May 8 '11 at 21:13
    
thanks for a very useful and detailed answer! –  SGP May 9 '11 at 1:50
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Thanks for the clarification about the notation. We $p$-adic folks aren't used to such an abundance of compact groups. –  L Spice May 9 '11 at 1:55
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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.

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