Connectedness of a Lie group 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?
 A: 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. 
A: 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.
