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Complex Lie group without faithful real representations?

We know that for a matrix (linear) Lie group $G$, we define it to be a closed subgroup of $GL(n,\mathbb{C})$. But Lie groups are defined as manifolds in $\mathbb{R}^n$ for some $n$, in general. The question is that, do we know any Lie group which is not a matrix Lie group? Thank you very much.

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marked as duplicate by Tom Leinster, Dan Petersen, S. Carnahan Mar 21 '12 at 11:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I googled it, and find at the introduction of the paper (Denis Luminet, Alain Valette, Faithful Uniformly Continuous Representations of Lie Groups,J. London Math. Soc. (1994) 49 (1): 100-108.), said the following: Although any connected real lie group G is locally isomorphic to some linear group, No nontrivial covering group of $SL_2(R)$ is linear. – Xiaolei Wu Mar 21 '12 at 4:20
For reference this was asked (and answered in the same traditional form!) at – Mariano Suárez-Alvarez Mar 21 '12 at 4:41
This is a special case of a question that was already asked. – S. Carnahan Mar 21 '12 at 11:42
up vote 6 down vote accepted

The traditional example is the universal cover of $SL(2,\mathbb{R})$. You can look e.g. at the wikipedia article on $SL(2,\mathbb{R})$.

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Copied from

While most well-known Lie groups are matrix groups, there do in fact exist Lie groups which are not matrix groups. That is, they have no faithful finite dimensional representations.

For example, let $H$ be the real Heisenberg group

$$H=\{\begin{pmatrix} 1 & a & b\newline 0&1&c\newline 0 &0 &1\end{pmatrix}\mid a,b,c\in\mathbb{R} \},$$

and $\Gamma$ the discrete subgroup

$$\Gamma=\{\begin{pmatrix} 1 & 0 & n\newline0&1&0\newline 0 &0 &1\end{pmatrix}\mid n\in\mathbb{Z}\}.$$

The subgroup $\Gamma$ is central, and thus normal. The Lie group $H/\Gamma$ has no faithful finite dimensional representations over $\mathbb{R}$ or $\mathbb{C}$.

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