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How do you prove that the universal covering of $SO(3, \mathbb{R})$ is $S^3$ ? Or equivalently, that it is diffeomorphic to $P_3\mathbb{R}$ ?

Thank you for your answers.

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    $\begingroup$ The group $SU(2) \cong S^3$ of unit quaternions acts by conjugation on the imaginary quaternions. Check that this gives a surjective homomorphism $SU(2) \twoheadrightarrow SO(3)$ with kernel $\pm \operatorname{id}$. See here for an explicit formula. See also this nice blog post by Qiaochu Yuan. $\endgroup$
    – t.b.
    Mar 7, 2012 at 16:00

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Any element of $SO(3)$ is rotation about an axis in $\mathbb{R}^3$ - that is each element can be represented by an axis of rotation and an angle of rotation.

$\mathbb{RP}^3$ is $\mathbb{D}^3$ with antipodal points identified. Given a point in $\mathbb{D}^3$ it is some distance (between -1 and 1) on a vector from the origin. This vector gives you the axis of rotation for a point in $\mathbb{RP}^3$. We still need the angles of rotation, but these will be given by the distance between -1 to 1. Scale those values to be $-\pi$ to $\pi$, Note that then antipodal points of $\mathbb{D}^3$ are mapped to the same point in $\mathbb{RP}^3$

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  • $\begingroup$ Thank you for your answer. Your first statement seems false : i think you forgot the elements which are a relexion on an axis and a rotation-reflexion on the orthogonal plane. $\endgroup$ Mar 7, 2012 at 16:16
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    $\begingroup$ Wouldn't those change orientation? $\endgroup$
    – Aru Ray
    Mar 7, 2012 at 16:18
  • $\begingroup$ I don't think so. Its eigenvalues are $\{-1,-1,1\}$ $\endgroup$ Mar 7, 2012 at 16:40
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    $\begingroup$ @SelimGhazouani: Note that this is simply a rotation by $180°$ around the third eigenvector, e.g. it can be written $$\begin{pmatrix} \cos\theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ for $\theta = \pi$. aru is right. It is a standard theorem in linear algebra that any matrix $A\in SO(3)$ can be brought into the above form with some $\theta$ by a orthogonal change of coordinates. You can also look at it this way: Given $A$, we can find an eigenvector $v$ to the eigenvalue $1$ and the restriction of $A$ to the 2-dim subspace $v^\perp$ must be a rotation. $\endgroup$
    – Sam
    Mar 7, 2012 at 18:03
  • $\begingroup$ How do you find an eingenvector to the eingenvalue 1 ? Can't the case $\{-1, e^{i\theta}, -e^{-i\theta}\} $ occur ? $\endgroup$ Mar 8, 2012 at 7:32
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One can also show that the adjoint representation of su(2) is so(3) and find the covering map quite explicitly.

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