Prove that the manifold $SO(n)$ is connected

Solution 1:

We have that $S^n \cong SO(n+1)/SO(n)$.

Using the fact that:

If $H <G $ is a closed subgroup and both $H$ and $G/H$ are connected, then $G$ is connected,

the claim follows by induction.

Solution 2:

1. Given two vectors $v,w$ of same length (say of length $1$) in $\mathbb{R}^n$, let's first verify that there is indeed a path in $\gamma(t) \in SO(n)$ so that $\gamma(t)v$ "starts off" at $v$ and "arrives" at $w$. More precisely, we look for $\gamma: [0,1] \rightarrow SO(n)$ s.th. $$\gamma(0)v = v,$$ and $$\gamma(1)v=w$$ hold. Therefore we choose a basis of $\mathbb{R}^n$: Take $v$ as first basis vector and a second normalized basis vector $u$ orthogonal to $v$ s.th. $u$ lies in the span of $v$ and $w$ (if $v$ and $w$ are linearly independent, $u$ can be obtained by applying Gram-Schmidt, if not take any $u$ orthogonal to $v$). Complete the basis arbitrarily.
   The idea is now to apply basic knowledge of rotations in two dimensions to the subspace $V:=\text{span}(v,u)$ and construct a rotation that leaves the complement of $V$ invariant. Since $w\in V$ and is normalized, there is an angle $\varphi$ s.th. (in the above constructed basis): $$w = \begin{bmatrix} cos\varphi & sin\varphi & 0 \\ -sin\varphi & cos\varphi & 0 \\ 0 & 0 & I_{n-2} \end{bmatrix} v$$ Now, we can define our path: $$\gamma(t):= \begin{bmatrix} cos(t\varphi) & sin(t\varphi) & 0 \\ -sin(t\varphi) & cos(t\varphi) & 0 \\ 0 & 0 & I_{n-2} \end{bmatrix}$$ that is clearly in $SO(n)$ and takes $v$ to $w$ when applied to $v$.
   This seems to correspond to your "spinning process".
2. For a given orthonormal basis $(a_1,...a_n)$ apply this recursively: there is a path $\gamma_1(t) \in SO(n)$ s.th. $a_1$ is taken to $e_1$, i.e. $(\gamma_1(1)a_1,...,\gamma_1(1)a_1)= (e_1, \gamma_1(1)a_2,...,\gamma_1(1)a_n)$. We then choose $\gamma_2$ as taking $\gamma_1(1)a_2$ to $e_2$.
   But here we have to watch out:
   $\gamma_2$ has to leave $e_1$ invariant, otherwise we destroy our previous work. But this is indeed the case, since: $e_1 \perp e_2$ and $e_1 \perp \gamma_1(1)a_2$ (the second being true because applying $\gamma_1$ to an orthonormal basis results in an orthonormal basis) and so $e_1$ is in the complement of the subspace in which the rotation happens (see first step) and is thus left invariant. Continuing this yields a composed path $ \gamma := \gamma_1 \star \gamma_2 \star ... \star \gamma_{n-1}$ s.th. $\gamma(0) = I_n$ and $(\gamma(1)a_1,...,\gamma(1)a_n)=(e_1,...,e_n)$.
   $SO(n)$ is therefore path-connected.