TITLE: Prove that the manifold $SO(n)$ is connected
QUESTION [12 upvotes]: The question really is that simple:
Prove that the manifold $SO(n) \subset GL(n, \mathbb{R})$ is connected.
it is very easy to see that the elements of $SO(n)$ are in one-to-one correspondence with the set of orthonormal basis of $\mathbb{R}^n$ (the set of rows of the matrix of an element of $SO(n)$ is such a basis).
My idea was to show that given any orthonormal basis $(a_i)_1^n$ in $\mathbb{R}^n$ there's a continuous deformation from $(a_i)_1^n$ to $(e_i)_1^n$ the usual basis passing only through orthonormal basis. Such a deformation would yield a path between any element of $SO(n)$ and $I$, and the theorem would follow.
Also, a geometric picture is also "simple", one would simply spin the first basis until $a_1$ agrees with $e_1$ and proceed from there. However i'm having a lot of trouble coming up with this spinning process.
Any help would be appreciated.

REPLY [5 votes]: Here's a Riemannian geometry-flavored proof.
It suffices to show that there is a continuous path from any $M_1 \in SO(n)$ to the $n\times n$ identity matrix $I_n$. Write $M_1 = [v,a_2,\ldots,a_n]$ and $I_n = [e_1,\ldots,e_n]$. We have $v \in S^n \subset \mathbb{R}^{n+1}$, and thus $a_2,\ldots,a_n \in v^\perp \cong T_vS^n.$
Let $g$ be the round metric on $S^n$: i.e., $g$ is the restriction of the Euclidean inner product to $T S^n$. Given any path $c:[0,1]\to S^n$ and $0 < t_0 < t_1 < 1$, let $P_{c,t_0,t_1}:T_{c(t_0)}S^n\to T_{c(t_1)}S^n$ be parallel transport along $c$ from $c(t_0)$ to $c(t_1)$, with parallel transport corresponding to the Levi-Civita connection determined by $g$. For any path $c$ and times $t_0,t_1 \in [0,1]$, $P_{c,t_0,t_1}$ is an orientation-preserving isometry.
Now let $c:[0,1]\to S^n$ be any path from $v$ to $e_1$. Define a smooth path $A_1:[0,1]\to GL(n,\mathbb{R})$ via $$A_1(t):=[c(t),P_{c,0,t}(a_2),\ldots,P_{c,0,t}(a_n)].$$
We have $A_1(1) = [e_1,P_{c,0,1}(a_2),\ldots,P_{c,0,1}(a_n)].$ Since parallel transport is an orientation-preserving isometry, im$(A_1)\subset SO(n)$, and 
$$A_1(1) =
\left[
\begin{array}{c|c}
I_1 & 0 \\
\hline
0 & M_2
\end{array}
\right]
$$
for some $M_2 \in SO(n-1)$. Using the standard embedding $S^{n-1}\subset S^n$ and repeating the above process to the submatrix $M_2$, we get a path $A_2:[0,1]\to SO(n)$ from $A_2(0) = A_1(1)$ to 
$$A_2(1) = 
\left[
\begin{array}{c|c}
I_2 & 0
 \\
\hline
0 & M_3
\end{array}
\right]$$
for some $M_3 \in SO(n-2)$. Continuing by induction and concatenating the paths $A_1,\ldots,A_{n-1}$, we obtain a piecewise-smooth path from $M_1$ to $I_n$.