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...

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8 Answers8


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.

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    could any one explain me in a little detail about the point (ii)? – Marso 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? –  Jul 18 '12 at 16:02
  • yes sir just now I want to know that. – Marso Jul 18 '12 at 16:06
  • @Patience: I asked you about two alternatives. Which one is the one you are asking about? –  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. – Marso Jul 18 '12 at 16:09
  • @Patience I answered the question you posted today. –  Jul 18 '12 at 16:25

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.

Qiaochu Yuan
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  • It would be interesting to see why $GL_n$ also has the stronger property of being path-connected as a subset of $\mathbb{R}^{n^2}$ namely that two nearby (for the Euclidean distance) matrices of positive determinant can always be connected by a path lying close to them. – Mikhail Katz Apr 20 '16 at 10:31
  • @MikhailKatz: $GL_n$ is not path-connected since it is not connected. But any of its connected components are path connected, because of the local path-connectedness. This is in the paragraph before the example in the wikipedia link... the last sentence of the paragraph. – André Caldas Jan 22 '20 at 22:40

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 $i \neq j$.

Alexander Thumm
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    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

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:

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Bruno Stonek
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    @Qiaochu: I don't understand why you deleted the pages. See the discussion (I had already started) in meta: http://meta.math.stackexchange.com/questions/3495/on-the-inclusion-of-pages-of-text-as-images-in-answers – Bruno Stonek Jan 21 '12 at 20:45

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.

Pierre-Yves Gaillard
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  • sequence in manifold? how come? – Marso 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 $C=M$. – Marso 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? – Marso 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
  • thank you :), now please tell me why the fiber is connected and the base is connected and what is base here? why base is connected and that implies total space is connected? fiber of some element $y\in\mathbb{R}^n\setminus\{0\}$ is those $x\in G_n: \pi(x)=y$ right? – Marso Jul 18 '12 at 20:26
  • and what was your $s(x)$ would you tell me why such map exist? – Marso Jul 18 '12 at 20:35
  • and it would be nice if you explain why by constant rank theorem it is a surjective submersion with fiber $G_{n-1}$ – Marso Jul 18 '12 at 20:37
  • Dear @Patience: I'll think about your questions, but I'm busy right now. I'll try to get back to you. Thanks for your interest, and your ... patience. – Pierre-Yves Gaillard Jul 18 '12 at 20:52
  • ok dear sir, I will wait for your explaination :) – Marso Jul 18 '12 at 20:58
  • Dear @Patience: I think it's better to take things one after the other. Why is $\pi$ a surjective submersion with fiber $G_{n-1}$? The fact that it is a surjection with fiber $G_{n-1}$ follows from linear algebra. The definition of a submersion implicitly used here is that it is a map locally given (in appropriate charts) by linear surjections. Then, the Constant Rank Theorem implies that a smooth map of maximal rank from an open subset of $\mathbb R^{m+k}$ to $\mathbb R^m$ is a submersion. – Pierre-Yves Gaillard Jul 19 '12 at 05:37
  • ok I understant this one. – Marso Jul 19 '12 at 08:14
  • @Patience: Do you have any other question? – Pierre-Yves Gaillard Jul 19 '12 at 08:21
  • regarding $s(x)$, please tell me what was that.why such exist? – Marso Jul 19 '12 at 08:22
  • and why the fiber is $G_{n-1}$, please explain sir. – Marso Jul 19 '12 at 08:27
  • @Patience: Existence of $s$: Any submersion admits local sections. This follows again from the Constant Rank Theorem. When I say "the fiber is $G_{n-1}$" I of course mean "any fiber is homeomorphic to $G_{n-1}$". This is clear for the fiber above $e_1$. For a general fiber, this results from the following Lemma: Let $G$ be a group acting transitively on a set $X$. For $a,b\in X$ let $G(b,a)$ be the set of those $g\in G$ such that $ga=b$. [$G(b,a)$ is the fiber above $b$ of $g\mapsto ga$.] Let $g_0$ be in $G(b,a)$. Then $G(b,a)=G(a,a)g_0$ ... – Pierre-Yves Gaillard Jul 19 '12 at 09:05
  • ... In particular if $G$ is a topological groups, then $G(b,a)$ is homeomorphic to $G(a,a)$. – Pierre-Yves Gaillard Jul 19 '12 at 09:05
  • ok could you tell me why the fiber above $e_1$ i.e $\{g\in G_n: \pi(g)=e_1\}$ is connected? – Marso Jul 19 '12 at 10:44
  • @Patience: I've just corrected a mistake (which I wouldn't have noticed without your help!). The fiber is $G_{n-1}\times\mathbb R^{n-1}$, not just $G_{n-1}$. The fiber $\pi^{-1}(e_1)$ is the set of all $n$ by $n$ matrices with first column $e_1$ and positive determinant. Such a matrix is of the form $$\begin{pmatrix}1&b\\ 0&D\end{pmatrix}$$ with $1\in\mathbb R$, $b\in\mathbb R^{n-1}$, $D\in G_{n-1}$. – Pierre-Yves Gaillard Jul 19 '12 at 11:14
  • yes sir, I have just calculated myself for $3\times 3$ :) – Marso Jul 19 '12 at 11:16
  • could you tell me why that is the fiber ? i mean why homeomorphic to $G_{n-1}\times\mathbb{R}^{n-1}$, what is the homeomorphism? :( – Marso Jul 19 '12 at 11:45
  • @Patience: In the above notation, to $$g:=\begin{pmatrix}1&b\\ 0&D\end{pmatrix}\in\pi^{-1}(e_1)$$ attach $(D,b)\in G_{n-1}\times\mathbb R^{n-1}$. – Pierre-Yves Gaillard Jul 19 '12 at 12:23
  • ok so as $G_{n-1}$ and $\mathbb{R}^{n-1}$ is connected so is their cartesian product. – Marso Jul 19 '12 at 12:55
  • why if each fiber is connected implies the whole space is connected?any intuition or proof? – Marso Jul 19 '12 at 12:59
  • and last but not the least why this map is surjective submersion? – Marso Jul 19 '12 at 13:09
  • @Patience: I tried to explain why $\pi$ is a surjective submersion in the comment containing the word "charts" (you can search this word). From an intuitive viewpoint, it may be more suggestive to think in term of path connectedness. Let $f:X\to Y$ be our open surjection (as in Edit B, but with "path connected" instead of "connected" throughout). Take $x,x'\in X$, and take a path $p$ from $f(x)$ to $f(x')$. Then the local sections will enable you to "lift" $p$ to a path from $x$ to $x'$. – Pierre-Yves Gaillard Jul 19 '12 at 13:53
  • @Patience: PS. The fibers being path connected, it suffices to join any given point in the fiber of $x$ to some point in the fiber of $x'$. – Pierre-Yves Gaillard Jul 19 '12 at 14:07

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. $\square$

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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!

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We will show that $G:=Gl_n({\bf R})^+$, set of matrices of positive determinant, is path connected.

If $A=[a_1\cdots a_n]\in G$, consider $H:={\bf R}^{n-1}$ which is spanned by $\{ a_i\}_{i=2}^n$ Here there exists some $v_1=e_i$ or $-e_i$ s.t. $$ \angle (v_1,a_1) < \pi $$ where $e_i$ is a canonical basis. Clearly there exists a path, which do not change sign of determinant, so that we have $$ A_1:=[ v_1\ a_2\cdots a_n] $$

By repeating this process we have $A_n\in SO(n)$. That is if $SO(n)$ is connected, then we are done

Note the fact that $SO(n)/SO(n-1)=S^{n-1}$. Since $SO(2)$ is connected, by induction, we complete the proof.

HK Lee
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