I am trying to understand a piece of proof in Brian Hall's book: Lie Groups, Lie Algebras, and Representations.
The Lie algebra $\mathfrak{g}$ associated with a given matrix Lie group $G$ is defined first as a set:
Definition 3.18. Let $G$ be a matrix Lie group (a closed subgroup of $GL_n(\mathbb{C})$). The Lie algebra of $G$, denoted $\mathfrak{g}$, is the set of all matrices $X$ such that $e^{tX}$ is in $G$ for all real numbers $t$.
And then $\mathfrak{g}$ is given the structure of a Lie algebra.
Theorem 3.20. Let $G$ be a matrix Lie group with Lie algebra $\mathfrak{g}$. If $X$ and $Y$ are elements of $\mathfrak{g}$, the following results hold.
- (1) $sX\in\mathfrak{g}$ for all real numbers $s$;
- (2) $X+Y\in\mathfrak{g}$;
- (3) $XY-YX\in\mathfrak{g}$.
A step proving (3) says the following:
By (1) and (2), $\mathfrak{g}$ is a real subspace of $M_n(\mathbb{C})$, from which it follows that $\mathfrak{g}$ is a (topologically) closed subset of $M_n(\mathbb{C})$.
Question: Could anyone elaborate that why the fact that $\mathfrak{g}$ is a real vector space implies that it is closed in $M_n(\mathbb{C})$?
Best Answer
If $V$ is a vector subspace of some $\mathbb{C}^n$ (and $M_n(\mathbb{C})$ is basically $\mathbb{C}^{n^2}$), then it has a basis $\{e_1,e_2,\ldots,e_k\}$. Extend it to a besis $\{e_1,e_2,\ldots,e_n\}$ of $\mathbb{C}^n$. Then$$V=\{\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n\in\mathbb{C}^n\,|\,\alpha_{k+1}=\cdots=\alpha_n=0\}$$and therefore it is a closed subset of $\mathbb{C}^n$.