Fundamental group of $SO(3)$

How can I show that the universal cover of $SO(n)$, for $n\ge 3$, is a double cover? And how does that reflect the fact that the fundamental group of $SO(n)$ has two elements? What is the relation between the fundamental group of a topological space and its universal cover? How come that $SU(2)$ is simply connected but $SO(3)$ is not? Thank you!

Solutions Collecting From Web of "Fundamental group of $SO(3)$"

Adapted from this answer.

The fact that the sphere $SU(2)$ is a twofold cover of $SO(3)$ can be seen by viewing $SU(2)$ as the group of unit quaternions, which acts by conjugation on the real $3$-dimensional space of purely imaginary quaternions as explained here; the action can be seen to be by elements of $SO(3)$, and two unit quaternions that have the same action differ by a factor $-1$ (call these antipodes of each other).

This fact, together with the fact that $SU(2)$ is connected, shows that that $SO(3)$ is not simply connected. Indeed, one can take a path from a unit quaternion to its antipode, and map this path to $SO(3)$ (take the rotation action defined by each unit quaterion on the path), where it becomes a loop (in $SO(3)$ its starting and ending point are identified). This loop cannot be contracted in $SO(3)$: if it could, we could perform the corresponding deformation to the path in $SU(2)$ as well, contracting it to a point while keeping the endpoints antipodes of each other all the time, which is absurd.

And $SU(2)$ is simply connected because the set of unit quaternions is homeomorphic to the $3$-sphere $\{\,(a,b,c,d)\in\mathbf R^4\mid a^2+b^2+c^2+d^2=1\,\}$ (the $n$-sphere is simply connected for all $n>1$). Therefore forming a new loop in $SO(3)$ by going around the one indicated above twice, so that the result lifts to a loop in $SU(2)$, the new loop can be contracted in $SO(3)$ (just contract the loop “covering” it in $SU(2)$ to a point, and project that deformation back to $SO(3)$). One can conclude from this that the fundamental group of $SO(3)$ has two elements.

For $SO(n)$ things are more complicated to describe explicitly.

There are advanced techniques that I am not familiar with (e.g. using fiber bundles, differential geometry) to show that $\pi_1(SO(n)) = \pi_1(SO(3))$ for all $n > 3$. Now $SO(3)$ is homeomorphic to $\Bbb{R}P^3$ which by the Van – Kampen Theorem has $\pi_1$ isomorphic to that for $\Bbb{R}P^2$. This of course is known to have fundamental group isomorphic to $\Bbb{Z}/2\Bbb{Z}$.