Why is the image of a faithful representation of a Lie group a Lie subgroup?

Question

Let $\varphi:G\to\operatorname{GL}_n(\mathbb{R})$ be a faithful Lie group representation, i.e. an injective Lie group homomorphism. Then is $\varphi(G)$ a Lie subgroup of $\operatorname{GL}_n(\mathbb{R})$? Somehow I think this shouldn't be to difficult to see, but I don't quite get it. I know that $\varphi(G)$ is an immersed submanifold because $\varphi$ is injective and of constant rank, but how to show that the multiplication $m:\varphi(G)\times\varphi(G)\to\varphi(G)$ is smooth?

Answer 1

You are perhaps confused about the definitions, specifically the definition of Lie subgroup (which you have not specified in your question; I will do so below following the most common convention).

An immersion is a $C^\infty$ map $\phi:N \to M$ between manifolds such that the differential $d \phi_n$ is injective at each $n \in N$. A submanifold of a manifold $M$ is a pair $(N,\phi)$ consisting of a manifold $N$ and a one-to-one immersion $\phi:N \to M$. If $G$ is a Lie group, a Lie subgroup of $G$ is a pair consisting of a Lie group $H$ and a group homomorphism $\phi: H \to G$ such that the pair $(H,\phi)$ is a submanifold of the manifold $G$.

With these definitions it is clear that checking that a faithful representation $\phi:G \to \mathrm{GL}_n(\mathbf{R})$ is a Lie subgroup amounts to checking that it is an immersion (which it is, as you have written above---though I would rather see it by observing that it suffices to prove that it is an immersion at the identity $1 \in G$, which follows from the fact that all vectors in the kernel exponentiate to elements in the kernel).