How many connected components does $\mathrm{GL}_n(\mathbb R)$ have?

I've noticed that $\mathrm{GL}_n(\mathbb R)$ is not a connected space, because if it were $\det(\mathrm{GL}_n(\mathbb R))$ (where $\det$ is the function ascribing to each $n\times n$ matrix its determinant) would be a connected space too, since $\det$ is a continuous function. But $\det(\mathrm{GL}_n(\mathbb R))=\mathbb R\setminus\{0\},$ so not connected.

I started thinking if I could prove that $\det^{-1}((-\infty,0))$ and $\det^{-1}((0,\infty))$ are connected. But I don't know how to prove that. I'm reading my notes from the topology course I took last year and I see nothing about proving connectedness...

Solution 1:

Your suspicion is correct, $GL_n$ has two components, and $\det$ may be used to show there are at least two of them. The other direction is slightly more involved and requires linear algebra rather than topology. Here is a sketch of how to do this:

i) If $b$ is any vector, let $R_b$ denote the reflection through the hyperplane perpendicular to $b$. These are all reflections. Any two reflections $R_a, R_b$ with $a, b$ linear independent can be joined by a path consisting of reflections, namely $R_{ta+ (1-t)b}, t\in[0,1]$.

ii) Any $X\in O^+(n)$ (orthogonal matrices with positive determinant) is the product of an even number of reflections. Since matrix multiplication is continuous $O(n)\times O(n) \rightarrow O(n)$ and by i) you can join any product $R_a R_b$ with $R_a R_a = Id$ it follows that $O^+(n)$ is connected.

iii) $\det$ shows $O(n)$ is not connected.

iv) $O^-(n) = R O^+ (n)$ for any reflection $R$. Hence $O^-(n)$ is connected.

v) Any $ X\in GL_n$ is the product $AO$ of a positive matrix $A$ and $O \in O(n)$ (polar decomposition). Now you only need to show that the positive matrices are connected, which can be shown again using convex combination with $Id$. This proves the claim.

Solution 2:

Here's another proof. First, by Gram-Schmidt, any element of $\text{GL}_n(\mathbb{R})$ may be connected by a path to an element of $\text{O}(n)$. Second, by the spectral theorem, any element of $\text{SO}(n)$ is connected to the identity by a one-parameter group. Multiplying by an element of $\text{O}(n)$ not in $\text{SO}(n)$, the conclusion follows.

The first part of the proof can actually be augmented to say much stronger: it turns out that Gram-Schmidt shows that $\text{GL}_n(\mathbb{R})$ deformation retracts onto $\text{O}(n)$, so not only do they have the same number of connected components, but they are homotopy equivalent.

Note that $\text{GL}_n(\mathbb{R})$ is a manifold, hence locally path-connected, so its components and path components coincide.

Solution 3:

Yes $GL(\mathbb R^n)$ has exactly two components. An easy proof can be obtained in the following way: Have a look at which elementary operations of the Gauss-algorithm can be presented as paths in $GL(\mathbb R^n)$. Conclude, that any point in $GL(\mathbb R^n)$ can be connected to either $\text{diag}_n(1,1,\dots, 1)$ or $\text{diag}_n(1,1,\dots, -1)$ by a path, where $D = \text{diag}_n(a_1,a_2,\dots, a_n)$ is the diagonal matrix with entries $D_{i,i} = a_i$ and $D_{i,j} = 0$ for $i \neq j$.