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

Question

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.

Answer 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.

Answer 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.

Answer 3

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 $I_n$.

Answer 4

For a formal proof, see http://www.jstor.org/discover/10.2307/2315278?uid=378469401&uid=2&uid=3&uid=60&sid=21103743430923. This paper provides a proof of connectedness for all classical groups.

Answer 5

The problem is solved in this MSE post, answer by user20266. (The question was about $GL(n,R)$, but user20266 first gives a proof for the special orthogonal group.)

Answer 6

Let us write $a \sim b$ if and only if $a$ is path-connected to $b$. Note that $\sim$ is an equivalence relation.

We have$$O(n) = \{M \in \text{GL}(n, \mathbb{R}) : MM^\text{T} = I_n\},$$where $I_n$ denotes the identity matrix. Let $M \in O(n)$. Since$$1 = \det(I_n) = \det(MM^\text{T}) = \det(M)\det(M^\text{T}) = \det(M)\det(M),$$then $\det(M) \in \{1, -1\}$. Let$$O_+(n) = \{M \in O(n) : \det(M) = 1\} = GL_+(n, \mathbb{R}) \cap O(n)$$and let$$O_-(n) = \{M \in O(n) : \det(M) = -1\} = GL_-(n, \mathbb{R}) \cap O(n).$$Then $O_+(n)$ and $O_-(n)$ are open in the subspace topology and they disconnect $O(n)$. Check this. Without loss of generality,$$\lambda_1 \le \lambda_2 \le \dots \le \lambda_n.$$Since $\det(M) = 1$, then there must be an even number of $\lambda_i$'s that are negative, say$$\lambda_1, \dots,\,\lambda_{2k}.$$Since $\mathbb{R}^+$ and $\mathbb{R}^-$ are path-connected, then$$D \sim \text{diag}(-1, \dots,\,-1,\,1,\dots,\,1).$$It can be show that $I_2 \sim -I_2$ by showing that $I_2 \sim N$ and $N \sim I_2$, where$$N = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}.$$It follows that $D \sim I_n$ and so$$M \sim UI_n U^\text{T} = I_n.$$Thus, $SO(n)$ is path-connected and therefore connected.