What is the usual topology on a vector space?

Question

I do not understand the topology of a Lie group clearly. Let $G$ be a Lie group and $T_eG$ be its tangent space at the identity $e \in G$. Why $\operatorname{Aut}(T_eG)$ is an open subset of the vector space of endomorphisms of $T_eG$ (i.e., $\operatorname{End}(T_eG)$)? What does “open” mean?

Answer 1

First, this has nothing to do with Lie groups or Lie algebras. The only important part is that $T_eG$ is a vector space.

For any vector space $V$ (say, finite dimensional over the reals), the set $\operatorname{End}(V)$ is naturally a finite dimensional vector space. Hence, $\operatorname{End}(V)$ is isomorphic to $\mathbb{R}^N$ for some $N$ (in fact, $N = (\dim V)^2$). Use any choice of isomorphism to topologize $\operatorname{End}(V)$. This choice of isomorphism is equivalent to choosing a basis of $V$.

Now, one has the determinant $\det:\operatorname{End}(V)\rightarrow\mathbb{R}$ which is given as a polynomial in the entries of the matrices in $\operatorname{End}(V)$ (they are matrices after choosing a basis), and hence is continuous.

Since $\det$ is continuous, $\det^{-1}(\mathbb{R}-\{0\})$ is an open subset of $\operatorname{End}(V)$. But this subset is precisely $\operatorname{Aut}(V)$, the invertible transformations from $V$ to $V$.