Representatives of $\pi_1(SO(2))=\mathbb{Z}$ may be given by paths$$\theta\mapsto\left(\begin{array}{lr}\cos(n\theta)&-\sin(n\theta)\\\sin(n\theta)&\cos(n\theta)\end{array}\right).$$However, $\pi_1(SO(4))=\mathbb{Z}/2$. Is it true that the path$$\theta\mapsto\left(\begin{array}{lrcc}\cos(n\theta)&-\sin(n\theta)&&\\\sin(n\theta)&\cos(n\theta)&&\\&&1&\\&&&1\end{array}\right)$$can be continuously deformed to $\theta\mapsto\text{id}$ for even $n$? What does the homotopy look like?

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    $\begingroup$ You can already do this in $SO(3)$. This can be done using quaternions. $\endgroup$ – Lord Shark the Unknown Jun 16 '18 at 19:24

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