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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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    $\begingroup$ 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. $\endgroup$
    – Xiaolei Wu
    Commented Mar 21, 2012 at 4:20
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    $\begingroup$ For reference this was asked (and answered in the same traditional form!) at math.stackexchange.com/questions/122612/non-linear-lie-groups $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 21, 2012 at 4:41
  • $\begingroup$ This is a special case of a question that was already asked. $\endgroup$
    – S. Carnahan ♦
    Commented Mar 21, 2012 at 11:42

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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 PlanetMath.

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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    $\begingroup$ In general, do these Lie groups have their corresponding Lie algebra linearization? $\endgroup$
    – YKY
    Commented May 5, 2017 at 11:29
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    $\begingroup$ planetmath.org/encyclopedia/ExamplesOfNonMatrixLieGroup.html does not exist any more. However this could also be found with details in "B.C. Hall, Lie Groups, Lie Algebras, and Representations, Graduate Texts in Mathematics 222, Springer 2015 DOI 10.1007/978-3-319-13467-3_5" pp 103-105. $\endgroup$
    – Mahmood Al
    Commented Jul 6, 2018 at 19:00
  • $\begingroup$ @YKY: Every Lie group has a Lie algebra. Every finite dimensional real Lie algebra has a faithful finite dimensional representation (Ado's theorem). But the Lie groups with that Lie algebra might not share that representation, and if they do there might be a discrete normal subgroup acting trivially. $\endgroup$
    – Ben McKay
    Commented Jul 2, 2022 at 15:26

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