Are all Lie groups Matrix Lie groups?

Question

I have beard a bit about so-called matrix Lie groups. From what I understand (and I don't understand it well) a matrix Lie group is a closed subgroup of $GL_n(\mathbb{C})$.

There is also the notion of a Lie group. It is something about a smooth manifold of the manifold $M_n(\mathbb{C})$.

I have also hear something saying that all Lie groups are in fact isomorphic to a matrix Lie group. Is this correct? Could someone give me a bit more detail about this? What, for example, is the isomorphism? Is it of abstract groups, manifolds, or ...?

Answer 1

Not all Lie groups are matrix groups. Consider the metaplectic group. From wikipedia:

The metaplectic group $M_{p_2}(\mathbb{R})$ is not a matrix group: it has no faithful finite-dimensional representations. Therefore, the question of its explicit realization is nontrivial. It has faithful irreducible infinite-dimensional representations, such as the Weil representation described below.

Answer 2

A Lie group is a group $(G,m,i)$ where $m \colon G \times G \rightarrow G$ is the multiplication and $i \colon G \rightarrow G$ is the inverse map that is also a smooth manifold such that $m$ and $i$ are smooth maps.

Many Lie groups are subgroups of $\mathrm{GL}_n(\mathbb{R})$ but it is not true that any Lie group is isomorphic to a subgroup of $\mathrm{GL}_n(\mathbb{R})$. For example, the universal cover of $\mathrm{SL}_2(\mathbb{R})$ is not a matrix group.

Answer 3

As other answers mention, it is not true that any Lie group is a matrix group; counterexamples include the universal cover of $SL_2(\mathbb{R})$ and the metaplectic group.

However it is true that all compact Lie groups are matrix groups, as a consequence of the Peter-Weyl theorem.

It is also true that every finite-dimensional Lie group has a finite-dimensional Lie algebra $\mathfrak{g}$ which is a matrix algebra. (This is Ado's theorem.)

In some sense, the Lie algebra of a Lie group captures "most" of the information about the Lie group. Finite-dimensional Lie algebras are in bijective correspondence with finite-dimensional simply-connected Lie groups. So given an arbitrary Lie group $G$, passing to its Lie algebra amounts to passing to the universal cover of the connected component of the identity $\widetilde{G_1}$. Note though that simply-connected Lie groups are not in general matrix groups; $\widetilde{SL_2(\mathbb{R})}$ is a counterexample.