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!

  • $\begingroup$ You are asking a whole lot of questions at the same time. This makes it difficult to give an answer, which is on the one hand comprehensible and on the other hand exhaustive. Maybe you should first ask how the fundamental group of a topological space relates to its universal cover, illustrated with the example of $SO(n)$. $\endgroup$ – Nils Matthes Sep 21 '12 at 10:59

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.

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    $\begingroup$ Not clear at all for a beginer $\endgroup$ – miosaki Sep 21 '12 at 8:39
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    $\begingroup$ I disagree, I think it is clear enough. $\endgroup$ – Lierre Sep 21 '12 at 11:38

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 $\Bbb{Z}/2\Bbb{Z}$.

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    $\begingroup$ The intuition for the fiber bundle technique, on the other hand, is not so bad: Any map from $S^1$ to $SO(n)$ is homotopic to one which lands in the top left $2\times 2$ $SO(2)$ block. Any homotopy of such a map is homotopic to another homotopy which lands in the $SO(3)$ block, so once you get to $SO(3)$, you've seen all there is to see about the fundamental group. $\endgroup$ – Jason DeVito Sep 21 '12 at 12:27
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    $\begingroup$ How I need to use Van-Kampen theorem for proof that $\pi_1$ of $\mathbb{R}P^3$ is homeomorphic to $\mathbb{R}P^2$? $\endgroup$ – kp9r4d Aug 4 '17 at 16:52

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