In this post, I want to give an example of a Lie group which is not isomorphic to a subgroup of a matrix group. This contrasts with the algebraic picture, where every affine algebraic group can be realized as a subgroup of a matrix group. I’ll just sketch the details.
The group we’ll use is the universal cover G of . One can realize as a fiber bundle with fiber over via the map which sends a matrix to its first row, so its fundamental group is . Similarly, we can use this map to show that is a fiber bundle with fiber over and hence is simply connected.
Now let be any smooth map. It induces a map on Lie algebras . We can use this to define a map by . It’s straightforward to check that f preserves the Lie bracket. Since is simply connected, f is the derivative of some smooth map .
To show that r cannot be injective, we will show that the composition is not injective, where the second map is the usual inclusion. I claim that this composition is the same map as the composition , where is the covering space map. By construction, these two compositions induce the same differentials on their Lie algebras. Now using the fact that both G and are connected, this implies that the map on the level of Lie groups must be equal as well. But is not injective: we saw above that the fundamental group for is nontrivial, so we are done.