Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question

7 Answers 7

up vote 26 down vote accepted

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.

share|improve this answer
    
could any one explain me in a little detail about the point (ii)? –  El Angel Exterminador Jul 18 '12 at 12:46
    
@Patience what exactly is it you want to know about ii). That any $X\in O^+$ is the product of reflections or the continuity statement? –  user20266 Jul 18 '12 at 16:02
    
yes sir just now I want to know that. –  El Angel Exterminador Jul 18 '12 at 16:06
    
@Patience: I asked you about two alternatives. Which one is the one you are asking about? –  user20266 Jul 18 '12 at 16:07
    
> Any $X\in O^+(n)$ (orthogonal matrices with positive determinant) is the product of an even number of reflection? I am not able to prove this. Please help. –  El Angel Exterminador Jul 18 '12 at 16:09

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.

share|improve this answer

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:

enter image description here enter image description here enter image description here enter image description here enter image description here

share|improve this answer
3  
@Qiaochu: I don't understand why you deleted the pages. See the discussion (I had already started) in meta: meta.math.stackexchange.com/questions/3495/… –  lentic catachresis Jan 21 '12 at 20:45

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 $i \neq j$.

share|improve this answer
3  
This shows there are two components. The map $\det:GL_n(\mathbb{R})\rightarrow \mathbb{R} - \{0\}$ supplies the two-component disconnection. –  ncmathsadist Jan 21 '12 at 13:37

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.

share|improve this answer
    
sequence in manifold? how come? –  El Angel Exterminador Jul 18 '12 at 19:25
    
Dear @Patience: I interpret your comment as being the question "what is a sequence in a manifold?" My answer is: "a sequence in a set $X$ is a map from $\mathbb N$ to $X$". If this interpretation is incorrect, or if you are unsatisfied with my answer, please let me know. –  Pierre-Yves Gaillard Jul 18 '12 at 19:53
    
I agree with your definition, would you tell me how you define convergence in manifold? for example you said $f(c_i)$ tends to some point in $N$? one more thing , could you explain me why did you say enough to prove $f(C)$ is closed? so far I understand we need to prove $M$ is connected, you have taken one component C which is open(am I right?) so enough to prove $C$ is closed then $C=M$. –  El Angel Exterminador Jul 18 '12 at 19:58
    
you applied constant rank theorem which says:If $F : M^m\rightarrow N^n$ is smooth and F has constant rank $r$ on a neighbourhood of $a$ for every $a\in F^{-1}(b)$ then $ F^{-1}(b)$ is a submanifold of $M$ with dimension $m-r$, so would you tell me what is the dimension of $G_n$ and $\mathbb{R}^n\setminus\{0\}$ as a smooth manifold? –  El Angel Exterminador Jul 18 '12 at 20:13
    
@Patience: (a) A sequence $(x_n)$ in a topological space tends to a point $a$ if every neighborhood of $a$ contains $x_n$ for $n$ large enough. A sequentially closed subspace of a metrizable space is closed. (b) Submersions are open. (c) The dimension of a nonempty open subset of $\mathbb R^d$ is $d$. Thus the dimension of $G_n$ is $n^2$, and the dimension of $\mathbb R^n\setminus\{0\}$ is $n$. –  Pierre-Yves Gaillard Jul 18 '12 at 20:19

A proof using action of groups:

Let $GL_n(\mathbb{R})_+= \{ M \in GL_n(\mathbb{R}) \mid \det(M)>0 \}$ act on $\mathbb{R}^n \backslash \{0\}$ in the canonical way; notice that the action is transitive. Let $e_1=(1,0,...,0)$.

Introduce the subgroups $H$ and $K$ defined by $H= \left\{ \left( \begin{array}{cc} 1 & 0 \dots 0 \\ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} & A \end{array} \right) \mid A \in GL_{n-1}(\mathbb{R})_+ \right\}$ and $G= \left\{ \left( \begin{array}{cc} 1 & a_1 \dots a_{n-1} \\ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} & I_{n-1} \end{array} \right) \mid (a_1,...,a_{n-1}) \in \mathbb{R}^{n-1} \right\}$. Then the stabilizer of $e_1$ is $HG$, homeomorphic to $G \times H \simeq \mathbb{R}^{n-1} \times GL_{n-1}(\mathbb{R})_+$.

You deduce that $\mathbb{R}^{n}\backslash \{0\}$ is homeomorphic to $GL_n(\mathbb{R})_+ /( \mathbb{R}^{n-1} \times GL_{n-1}(\mathbb{R}_+))$. Finally, you can conclude by induction using the following lemma:

Lemma: Let $G$ be a topological group and $H$ be a subgroup of $G$. If $H$ and $G/H$ are connected, then $G$ is connected.

Proof: Let $f : G \to \{0,1\}$ be a continuous function. Since $H$ is connected, $f$ is constant on the classes of $G$ modulo $H$, hence a continuous function $\tilde{f} : G/H \to \{0,1\}$. But $G/H$ is connected, so $\tilde{f}$ is constant. You deduce that $f$ is constant, hence $G$ is connected. $\square$

share|improve this answer

You can prove directly that $GL^{+}(n,\mathbb{R}):=\{A \in GL(n,\mathbb{R}): \, det(A)>0 \}$ is connected by induction:

The group $GL^{+}(1,\mathbb{R})$ coincides with the interval $]0,\infty[$. Let $n > 1$ and assume that $GL^{+}(n-1,\mathbb{R})$ is connected and let $p: M(n,n,\mathbb{R}) \mapsto \mathbb{R}^{n}$ the projection of a matrix on its first column. Clearly $p$ is continuous and open, so its restriction to the open set $GL^{+}(n,\mathbb{R})$ is an open map and moreover $p(GL^{+}(n,\mathbb{R}))=\mathbb{R}^{n} \setminus \{0\}$, which is connected. If we prove that the fibers by $p$ are all connected, we finish. (Because we can apply a theorem according to which if $Y$ is a topological connected space and $f: X \mapsto Y$ is connected and surjective such that $f^{-1}(y)$ is connected for all $y \in Y$ and $f$ is an open or closed map, then $X$ is connected.)

But we see that $p^{-1}(1,0,\dots,0)=\mathbb{R}^{n-1} \times GL^{+}(n-1,\mathbb{R})$, so this fiber is connected. Fix a $y \in \mathbb{R}^{n} \setminus \{0\}$. By surjectivity, $p^{-1}(y) \ne \emptyset$; so let $A \in p^{-1}(y)$. Remember that the application $L_A(B)=AB$ is a homeomorphism of $GL^{+}(n,\mathbb{R})$ to itself. Noticing that $p(AB)=Ap(B)$, then follows that $L_A (p^{-1}(1,0,\dots,0))=p^{-1}(y)$ and, therefore, all the fibers are homeomorphic each other... As we shall prove!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.