Components of the linear group

Idea. Components are always closed. They are open as soon as the space is *locally connected*, and $GL_n(\mathbb{R})$ is an open subset of a normed space, so it is.

1. Components are clopen.
- *Closed.* The closure of a connected set is connected, so the closure of a component $K$ inside $GL_n$ is a connected set containing $K$; by maximality it equals $K$.
- *Open.* $GL_n(\mathbb{R})=\det^{-1}(\mathbb{R}\setminus\{0\})$ is open in $M_n(\mathbb{R})$ because $\det$ is continuous (it is a polynomial in the entries). So for $A\in K$ there is an open ball $B(A,r)\subset GL_n(\mathbb{R})$. A ball is convex, hence connected (Problem 1). Therefore $B(A,r)\cup K$ is connected (two connected sets sharing the point $A$), so $B(A,r)\subset K$ by maximality. Thus $K$ is open.

2. Disconnectedness. The map $\det:GL_n(\mathbb{R})\to\mathbb{R}\setminus\{0\}$ is continuous and onto: $\det(\operatorname{diag}(c,1,\dots,1))=c$ for any $c\ne0$. Hence
$$GL_n(\mathbb{R})=\{\det>0\}\sqcup\{\det<0\},$$
a union of two disjoint non-empty open sets. So $GL_n(\mathbb{R})$ is not connected, and each component lies entirely in $\{\det>0\}$ or in $\{\det<0\}$.

Remark. The next problems show that there are *exactly two* components, $\{\det>0\}$ and $\{\det<0\}$. Note the contrast with $GL_n(\mathbb{C})$, which is connected: the determinant takes values in $\mathbb{C}^\times$, which is connected.