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Could any one give me hint how to show $SO(n)$ is connected? I understand that it is closed subset, I can prove $O(n)$ is not connected. Edit: suddenly got some idea, any matrix from $SO(n)$ can be written in this from, where first and second row are like below $$\left(\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{array}\right).$$

we can make $f(t)= tA+(1-t)I$ this is a path between $A$ and $I$. can this idea be rigorised?

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$O(n)$ is not connected. In fact $O(n)=[O(n)\cap \det^{-1}(1)]\cup[O(n)\cap\det^{-1}(-1)]$ and $SO(n)=O(n)\cap\det^{-1}(1)$. –  Mercy Aug 5 '12 at 9:49
    
@Mercy just for got to put Not :) –  miosaki Aug 5 '12 at 9:55
    
Do you know the normal form for such matrices? –  mland Aug 5 '12 at 10:01
    
@mland No, I dont know what is Normal form' –  miosaki Aug 5 '12 at 10:05
    
Your updated question more or less is the normal form.. But of course you have to take sums of these kinds of matrices.. Then you do not want to take a liner path between A and Id (else you could always do that) but homotop the angle $\theta$ to zero.. –  mland Aug 5 '12 at 10:16

2 Answers 2

up vote 2 down vote accepted

You can prove connectedness of SO(n) by induction on $n$: Sketch: For any unit vectors $v,w\in {\mathbb R}^n$ there is $1$-parameter family of matrices $A_\theta\in SO(n)$ such that $A_0(w)=w$ and $A_1(w)=v$ -- take $A_\theta$ to be the matrix composed of your $2\times 2$ matrix (acting on the space spanned by $v,w$) and the identity matrix on the orthogonal complement to $Span(v,w).$

Fix $v\in \mathbb R^n.$ Take any $M\in SO(n).$ Suppose $M(v)=w$. Then $M$ is connected through $A_\theta^{-1}M$ to a matrix $N$ which fixes $v$. Take an orthogonal basis of the orthogonal complement of $v$. Then $N$ is represented by an $SO(n-1)$ matrix wrt to this basis. By inductive assumption $N$ is connected to $I$ by a path.

I skipped the base step. Note that the base step fails for $O(n)$ for $n$ odd.

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This should probably be a comment on student's answer, but I cannot preview comments...

One way to restate his/her argument is to notice that there is a smooth action of $SO(n)$ on $S^{n-1}\subseteq\mathbb R^n$ which is transitive. If $s$ is a point in the sphere, the stabilizer of $s$ in $SO(n)$ is isomorphic to $SO(n-1)$, as one easily sees, so we have a map $p:g\in SO(n)\mapsto gs\in S^n$ which is surjective and, in fact, a fibration with fiber $SO(n-1)$. We can apply to it the long exact sequence in homotopy and an induction to show $SO(n)$ is connected. Alterantively (in fact, equivalently, but less technologically, maybe) one can show that the domain of the map $p$ is connected because its codomain and its fibers are, and the map is proper, or locally trivial.

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