# Are all Lie groups Matrix Lie groups?

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 ...?

• There is a careful discussion of this question here. – user98602 Nov 30 '15 at 15:16
• You might have heard that all compact Lie groups are matrix groups, or that all finite dimensional Lie algebras are matrix Lie algebras; both are actually true. (A Lie algebra of a finite dimensional Lie group is always finite dimensional and moreover carries "most" of the information about the Lie group.) – Max Nov 30 '15 at 15:32
• @Max: Would you write this as an answer? – John Doe Nov 30 '15 at 16:23

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.

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.

• I recommend this wikipedia page for further reading: en.wikipedia.org/wiki/… – Max Nov 30 '15 at 19:12
• Did you mean to write "a finite-dimensional Lie algebra $\mathfrak{g}$ which is the Lie algebra of a matrix group"? – Vectornaut Dec 1 '15 at 5:16
• ah, I actually meant "which is a matrix algebra." thanks for the correction. – Max Dec 1 '15 at 5:19

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.

...a matrix Lie group is a closed subgroup of $$\mathrm{GL}_n(\mathbb{C})$$.

This is correct!

There is also the notion of a Lie group...

A Lie group is a tuple $$(G,\cdot, \tau,\mathcal{A})$$, where $$G$$ is a set, $$\cdot$$ is a group operation on $$G$$, $$\tau$$ is a topology on $$G$$ which turns $$G$$ into a topological manifold, and $$\mathcal{A}$$ is a maximal smooth atlas, such that the maps $$m\colon G\times G\to G,\quad (g,h)\mapsto g\cdot h$$ $$i\colon G\to G,\quad g\mapsto g^{-1}$$ are smooth.

What, for example, is the isomorphism? Is it of abstract groups, manifolds, or ...?

Two Lie groups are said to be isomorphic if they are isomorphic as sets, groups, topological spaces and smooth manifolds. But conveniently enough, it is actually enough to check that they are isomorphic as topological groups, as one can show that every continuous group homomorphism between Lie groups is automatically smooth.

I have also heard something saying that all Lie groups are in fact isomorphic to a matrix Lie group. Is this correct?

This is not true! A classical counterexample is $$\widetilde{\mathrm{SL}_2(\mathbb{R})}$$ as mentioned in other answers.

Another classical counterexample (discussed in a paper by G. Birkhoff from 1936) is obtained by taking the 3-dimensional Heisenberg group $$G_3=\Bigg\{\begin{pmatrix}1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1\end{pmatrix}:x,y,z\in\mathbb{R}\Bigg\}$$ and quotienting out by the normal subgroup $$N=\Bigg\{\begin{pmatrix}1 & 0 & n\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}:n\in\mathbb{Z}\Bigg\}.$$ The resulting factor group $$G_3^*:=G_3/N$$ can be equipped with the quotient topology. This will turn $$G_3^*$$ into a topological manifold, homeomorphic to the product $$\mathbb{R}\times\mathbb{R}\times S^1$$, which in turn is a smooth manifold whose maximal smooth atlas we can let $$G_3^*$$ inherit. It is easy to verify that the group structure on $$G_3^*$$ is "compatible" with this atlas, in the sense that both multiplication and inversion become smooth maps.

We conclude that $$G_3^*$$ equipped with the group operation, topology and smooth maximal atlas described above is a Lie group. However, it is not isomorphic (as Lie groups) to a matrix Lie group.

To prove this, one can show an equivalent statement, namely that $$G_3^*$$ doesn't admit any faithful finite-dimensional complex representations. This is done in Section 4.8 of Lie Groups, Lie Algebras and Representations by B.C. Hall. The idea behind his proof is to note that every representation $$\Sigma\colon G_3^*\to \mathrm{GL}_n(\mathbb{C})$$ of $$G_3^*$$ gives rise to a representation $$\Pi=\Sigma\circ \Phi\colon G_3\to \mathrm{GL}_n(\mathbb{C})$$ of $$G_3$$, where $$\Phi\colon G_3\to G_3^*=G_3/N$$ is the natural projection. By passing to the Lie algebra level, one can show (this takes some effort) that for every such representation $$\Pi$$, it holds that $$\ker(\Pi)\supsetneq \ker(\Phi)=N\,,$$ which implies that the representation $$\Sigma$$ must have a non-trivial kernel and thus cannot be faithful.

So, no, it's not true that every Lie group is isomorphic to a matrix Lie group. Interestingly, though, it actually is true for compact Lie groups:

Theorem. Every compact Lie group $$G$$ admits a faithful finite-dimensional representation, and is thus isomorphic to a matrix Lie group.

This is typically proved as a corollary of the famous Peter-Weyl theorem (see for instance Section IV.3 of Lie Groups Beyond an Introduction by A.W. Knapp).