Definition of Lie groups representation I'm new to Lie groups, and I have some problems with the definition of Lie groups representation.
Let $G$ be a real or complex Lie group, the definition of (finite dimensional) Lie group representatin is exactly the definition of group representation, up to some conventions:
Let $G$ be a group, let $V$ be a real or complex finite dimensional vector space, and let $\pi:G\to GL(V)$ be a morphism of groups. Then $(V,\pi)$ is a finite dimensionsal representation of $G$
If $G$ is a real or complex Lie group, we want $\pi$ to be a morphism of Lie groups, and $GL(V)$ to be a Lie group.
The only way I can think of $GL(V)$ as a Lie group is $V\cong\mathbb{C}^n$ (and respectively $\mathbb{R}^m$) where $\dim(V)=n$, then, $GL(V)$ has a structure of a Lie group.
Am I right?
Thanks in advance.
 A: You are correct. To be precise:
Let $V$ be a finite dimensional vector space over $K$ (here $K=\mathbb{R}$ or $K=\mathbb{C}$). Fix some basis of $V$. Then any linear map $f:V\to V$ can be represented as $n\times n$ matrix over $K$, where $n=\dim(V)$ by looking how basis vectors are represented after passing through $f$. This is the classical linear algebra. Moreover the ring of all linear maps $End(V)$ is naturally isomorphic to the matrix ring $M_n(K)$ of all $n\times n$ matrices over $K$.
In that situation the group of all invertible linear maps $GL(V)\subseteq End(V)$ is naturally isomorphic to the group of all invertible matrices $GL_n(K)\subseteq M_n(K)$.
Finally any $n\times n$ matrix over $K$ can actually be interpreted as an element of $K^{n\times n}$. Now $M_n(K)$ is actually isomorphic (as linear space) to $K^{n\times n}$. Therefore both $M_n(K)$ and $GL_n(K)$ inherit topological and metrical structure of $K^{n\times n}$. Moreover $M_n(K)\simeq K^{n\times n}$ inherits differential structure of $K^{n\times n}$, it is a manifold. But since the determinant is a continuous function and $GL_n(K)=\det^{-1}(K\backslash\{0\})$ then $GL_n(K)$ is an open subset of $K^{n\times n}$. Thus it is a manifold as well. The only thing that is left is to show that the matrix multiplication and taking inverse is smooth, which requires some work, but it is not that hard.
