Show that the special orthogonal group $SO_n(\mathbb{R})$ is arcwise connected and compact

Question

I'm doing the following exercise:

Using that we know that if $M\in SO_n(\mathbb{R})$ there exists a $P\in GL_n(\mathbb{R})$ such that $M=PM'P^{-1}$, where $M'$ has zeros on everywhere except on its diagonal, formed by blocks of the form:

$$\begin{bmatrix} \cos(\theta_i) & \sin(\theta_i) \\ -\sin(\theta_i) & \cos(\theta_i) \\ \end{bmatrix}$$

or by $1\times 1$ blocks of the form $$\begin{bmatrix} 1 \\ \end{bmatrix}$$

Prove that $SO_n(\mathbb R)$ is arcwise connected and compact.

I have the compact part because previously I had to prove that $O_n(\mathbb{R})$ is connected, so I showed that the special orthogonal group is a closed subgroup of the orthogonal. But I'm not able to prove the arcwise connection of the special orthogonal group.

Thanks for your time.

Answer 1

You can get an arc connecting each such block matrix to $\operatorname{Id}_n$ using the map $t\mapsto\left(\begin{smallmatrix}\cos(t\theta_i)&-\sin(t\theta_i)\\\sin(t\theta_i)&\cos(t\theta_i)\end{smallmatrix}\right)$ ($t\in[0,1]$). More precisely: if you have a matrix $M$ in, say, $SO_5(\mathbb{R})$ and if you know that there is an invertible matrix $P$ sauch that$$PMP^{-1}=\begin{pmatrix}\cos(\theta_1)&-\sin(\theta_1)&0&0&0\\\sin(\theta_1)&\cos(\theta_1)&0&0&0\\0&0&1&0&0\\0&0&0&\cos(\theta_2)&-\sin(\theta_2)\\0&0&0&\sin(\theta_2)&\cos(\theta_2)\end{pmatrix},$$then you consider the path$$\begin{array}{ccc}[0,1]&\longrightarrow&SO_5(\mathbb{R})\\t&\mapsto&P^{-1}\begin{pmatrix}\cos(t\theta_1)&-\sin(t\theta_1)&0&0&0\\\sin(t\theta_1)&\cos(t\theta_1)&0&0&0\\0&0&1&0&0\\0&0&0&\cos(t\theta_2)&-\sin(t\theta_2)\\0&0&0&\sin(t\theta_2)&\cos(t\theta_2)\end{pmatrix}P,\end{array}$$which connects $M$ to $\operatorname{Id}_5$.

Answer 2

If you could show $M'$ had the form $\exp(tA)$ for some skew-symmetric matrix $A$, then $M'$ is joined to $I$ by the path $t\mapsto\exp(tA)$. Then also $M$ will be joined to $I$ by a path.