Homeomorphism class of GL_n? For example, it is easy to see that $GL_1(\mathbb{C})$ is a plane minus a point, and $GL_2(\mathbb{R})$ is $\mathbb{R}^4$ with a topological (half-open) cube removed (since the matrices of determinant zero can be described uniquely by an element of $[0,\pi)\times \mathbb{R}\times\mathbb{R}$ with polar coordinates). (I'm not fully content with the latter description though, since it does not make clear for example how many components the space has. This could be fixed by being more specific about how the cube sits in $\mathbb{R}^4$.)
My question is, is there a systematic way to describe the homeomorphism class of $GL_n$ (say over the real or complex numbers)?
 A: This is not a complete answer, but rather a long comment. First, polar decomposition gives a homeomorphism $$\left\{ \begin{array}{ccc} S^+ \times O(n) & \to & GL(n,\mathbb{R}) \\ (S,O) & \mapsto & SO \end{array} \right.,$$ where $S^+$ is the space of positive-semidefinite matrices which is homeomorphic to the space of symmetric matrices: $$\exp : S \overset{\sim}{\longrightarrow} S^+.$$ Then, $O(n)$ has two connected components homeomorphic to $SO(n)$. Thus, $$GL(n,\mathbb{R}) \simeq  \left( S \times SO(n) \right) \coprod \left( S \times SO(n) \right),$$ so $GL(n, \mathbb{R})$ has two connected components homeomorphic to $SO(n) \times \mathbb{R}^{n(n+1)/2}$. Unfortunately, it seems that $SO(n)$ cannot be expressed nicely with respect to the usual spaces. 
In his book, Hatcher gives some simple examples:


*

*$SO(1)$ is a point,

*$SO(2)$ is homeomorphic to the circle $\mathbb{S}^1$,

*$SO(3)$ is homeomorphic to the projective space $\mathbb{R}P^3$,

*$SO(4)$ is homeomorphic to $\mathbb{S}^3 \times \mathbb{R}P^3$,


Nevertheless, you can find a cell structure of $SO(n)$ in Hatcher's book (chapter 3.D).
