differential of a Lie group representation

Question

Let $G$ be a subgroup of $GL_n(\mathbb{R})$ and $\rho: G \rightarrow GL(\mathbb{R}^n)$ be the standard representation . If look at the differential at the identity: $$ d\rho_e: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathbb{R}^n)$$ Is it correct to say that $\operatorname{im}(d\rho_e)=\mathfrak{g}$ ?

If yes is there condition to say when a representation, which is not the standard representation, $\rho: G \rightarrow GL(\mathbb{R}^n)$ will have its image as a subspace $\operatorname{im}(d\rho_e) \subset \mathfrak{g}$ ?

I am a little bit confused and I hope that this question makes sense. Thanks to everyone who took the time to read this I really appreciate it.

Answer 1

Your first question makes sense and the answer is affirmative. What you call “the standard representation” is simply the inclusion of $G$ in $GL_n(\mathbb{R})$. Therefore, what you get after differentiating is the inclusion of $\mathfrak g$ in $\mathfrak{gl}_n(\mathbb{R})$.

I don't know an answer to your other question. My instinct says that no simple condition exists.

Answer 2

About your second question: Unless $ker(d\rho_e)\ne 0$, there is not much you can say in general. There are way too many, say, solvable Lie algebras and their representations. However, if you assume that, say, ${\mathcal g}$ is simple, $\rho(G)\ne e$ and $G$ is connected, then every such representation $\rho$ will an automorphism of $G$, and there are not that many of those: If you mod out the inner automorphisms (given by conjugations $x\mapsto gxg^{-1}$ for a fixed $g\in G$), you will have only finitely many automorphisms left.