# Closed Subgroup of $GL(n,\mathbb{K})$ is Lie group.

I am recently doing self-study for Lie Group. I am using Hall’s Elementary Introduction. In this book, it says the matrix Lie group is any subgroup $G$ of $GL(n,\mathbb{C})$, for any convergent sequence in $G$ it will converge to an element A such that A is invertible or A is not invertible.

I have two questions. 1. Why is the matrix Lie group really a Lie group? I know it is a group, but will it be a manifold whose group operation and inverse are smooth functions?

2.In that book, it also adds that the matrix Lie group is a closed subgroup of $GL(n,\mathbb{C})$.

My understanding is if we use the relative topology on $GL(n,\mathbb{C})$, we know that a set which contains all its limit points is closed. Hence, the matrix Lie group is closed under the relative topology. Am I correct?

• is it clear to you why it is a manifold? – Simonsays Apr 4 '18 at 14:45
• @Simonsays I know $GL(n,\mathbb {C})$ is a Lie group. However, I am not sure whether the matrix Lie group is a Lie group? – Hamio Jiang Apr 4 '18 at 14:47
• why don't you consider some particular cases like $SL(n,\mathbb{C})$ or $SL(2,\mathbb{R})$ . Even if you understand the general result, it gives you a better feeling. – Orest Bucicovschi Apr 4 '18 at 14:56
• @orangeskid I just wondered whether the definition here coincides with the “rigorous definition” of Lie groups. Indeed, in Hall’s book, it presents a lot of examples about the matrix Lie groups. – Hamio Jiang Apr 4 '18 at 14:58

## 1 Answer

1. This is a non-trivial theorem due to John von Neumann: every closed subgroup of $GL(n,\mathbb{C})$ is a Lie group. It was generalized to Lie groups by Elie Cartan. It's called the closed subgroup theorem.
2. It is closed because we're in a metric space and because in a metric space a set $S$ is closed if and only if it contains the limit of every convergent sequence of elements of $S$.