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

Question

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...

Answer 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.

Answer 2

It is as you say: $Gl_n(\mathbb{R})$ has two components, $Gl_n(\mathbb{R})^+$ and $Gl_n(\mathbb{R})^-$. This is theorem 3.68, p.131 in Warner's "Foundations of differentiable Manifolds and Lie Groups". The preview in Google Books contains the relevant pages.

Well, I've added it for comfort:

Answer 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 $i \neq j$.

Answer 4

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.

Answer 5

Let $G_n$ be the subgroup formed by the elements of $GL_n(\mathbb R)$ whose determinant is positive.

It suffices to prove that $G_n$ is connected.

We prove this by induction on $n$.

The case $n=1$ is trivial.

Assume that $n$ is at least $2$, and that $G_{n-1}$ is connected. Let $e_1$ be the first vector of the canonical basis of $\mathbb R^n$.

By the Constant Rank Theorem, the map $$ \pi:G_n\to\mathbb R^n\setminus\{0\},\quad g\mapsto ge_1 $$ is a surjective submersion with fiber $G_{n-1}\times\mathbb R^{n-1}$.

The fiber and the base being connected, so is the total space.

EDIT. To make the answer more complete, let's prove:

If $f:M\to N$ is a surjective submersion in the category of smooth manifolds, if $N$ is connected, and if $f^{-1}(y)$ is connected for all $y$ in $N$, then $M$ is connected.

Indeed, let $C\subset M$ be a connected component. It suffices to prove that $f(C)$ is closed.

Let $(c_i)$ be a sequence in $C$ such that $f(c_i)$ tends to some point $f(a)\in N$.

It is enough to find a sequence $(d_i)$ in $C$ satisfying

$\bullet$ $f(d_i)=f(c_i)$ for all $i$,

$\bullet$ $d_i$ tends to $a$.

There exist an open neighborhood $U$ of $a$ in $M$ and a smooth map $s:f(U)\to U$ such that

$\bullet$ $f(s(x))=x$ for all $x$ in $f(U)$,

$\bullet$ $s(f(a))=a$.

We can assume that each $f(c_i)$ is in $f(U)$. Then it suffices to set $d_i:=f(c_i)$.

EDIT B. Here is a mild generalization of the previous edit.

Let $f:X\to Y$ be an open continuous surjection between topological spaces. Assume that $X$ is locally connected, that $Y$ is connected, that $f^{-1}(y)$ is connected for all $y$ in $Y$, and that there is, for each $x$ in $X$, an open neighborhood $U_x$ of $x$ in $X$ and a continuous map $s_x$ from $f(U_x)$ to $U_x$ such that $f\circ s_x$ is the identity of $f(U_x)$. Then $X$ is connected.

Let $C$ be a connected component of $X$. It suffices to show that $f(C)$ is closed.

Let $x$ in $X$ be such that $f(x)$ is in the closure of $f(C)$. It suffices to show that $x$ is in $C$.

Let $U$ be an open neighborhood of $x$ in $X$. It suffices to show that $U$ intersects $C$.

We can suppose $U=U_x$.

As $f(U)\cap f(C)$ is nonempty, we can pick an element $y$ in this subset.

Then $s_x(y)$ is in $U$ by construction, and $s_x(y)$ is in $C$ because

$\bullet\ $ $y$ is in $f(C)$,

$\bullet\ $ $s_x(y)$ is in the connected subspace $f^{-1}(y)$,

$\bullet\ $ $C$ is a connected component.