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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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    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](http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation#Conversion_to_and_from_the_matrix_representation) for an explicit formula. See also [this nice blog post](http://qchu.wordpress.com/2011/02/12/su2-and-the-quaternions/) by Qiaochu Yuan.2012-03-07

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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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    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.2012-03-07
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    Wouldn't those change orientation?2012-03-07
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    I don't think so. Its eigenvalues are $\{-1,-1,1\}$2012-03-07
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    @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.2012-03-07
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    How do you find an eingenvector to the eingenvalue 1 ? Can't the case $\{-1, e^{i\theta}, -e^{-i\theta}\} $ occur ?2012-03-08
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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.