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Right now I'm looking into the 2-fold covering of SO(3) by SU(2). I'm using the Unit-quaternion approach. I think from an algerbraic standpoint I understand the homomorphism quite well by now and I also know that both are in fact Lie-Groups, but I have more or less no clue how to proof that its a covering map. From my research up to this point I concluded that you can throw like 2-4 theorems of Lie-Group theory onto it and be finished, but I have no background in the theory of Lie-groups(I know what a Lie group is, but thats it).

Therefore my question would be if there is a nice "elementary" topological proof of this or some ("short") literature with which I could get the theoretical background I would need.

edit: I think I should go into some more details about the homomorphism. I know how to cover $\mathbb{RP}^3$ with $\mathbb{S}^3$. The motivation for the map is the connection between the quaternions and the Rotations of $\mathbb{R}^3$. one can show that the map $f:S \rightarrow SO(3)$ $x\rightarrow(q\rightarrow xqx^{-1})$ is a surjective 2:1 homomorphism from the Lie-Group $SU(2)\cong S$ onto the lee-group SO(3) with kernel {1,-1}.

I'd like to show that this map is a covering map and if possible without using lie-algebras.

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    $\begingroup$ It's hard to give specific advice, since it's not clear what tools do and do not want to use, but topologically these spaces are $\Bbb S^3$ and $\Bbb R \Bbb P^3 \cong \Bbb S^3 / \Bbb Z_2$, where the nontrivial element of $\Bbb Z_2$ acts antipodally. Once you have these topological identifications, you're basically done. $\endgroup$ – Travis Apr 14 '17 at 13:49
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The comment by Travis says it all. From the algebra we see that topologically the map $SU(2) \to SO(3)$ is the map "Identify each point on $S^3$ with its antipode" from $S^3$ to $\mathbb{RP}^3$. So at that point we can forget the algebra and ask ourselves the purely topological question: is the map 'Identify anti-podal points' from $S^n$ to $\mathbb{RP}^{n}$ a covering map? I believe the book 'Topology' by Munkres covers this type of quotient map really well, but I also guess that if you already know what it means to be a quotient map you can solve this yourself.

Note that on the algebraic side the only thing I added to all the much harder stuff already in your answer is the insight that 'multiplication with -1' maps each point in $SU(2) \cong S^3$ to its antipode, which is of course obvious from the quaternionic picture you already gave.

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  • $\begingroup$ I think the "abstract" conjugacy map was what confused me a little bit, but you are totally right in the end its just the antipodal map and with that I can go on working, thanks mate! $\endgroup$ – Hanspeterwurst Apr 15 '17 at 11:12
  • $\begingroup$ I can see how it is confusing. The antipodal map comes from the multiplication action of the group for the element -1, while our interest in the element -1 comes from the conjugation action. There is a sweet irony in the fact that the reason that -1 is the most interesting element for the one action (in fact the only element for which we even wonder what the multiplication action looks like topologically) is that it does exactly nothing in the other action. (Of course now I am just writing it down in a way that makes it seem more confusing than it really is.) $\endgroup$ – Vincent Apr 16 '17 at 14:33
  • $\begingroup$ I started thinking about the covering again :D. how should/could on argue that the homomorphism is continuous, i don't see a good argument i could try. $\endgroup$ – Hanspeterwurst Apr 30 '17 at 17:22
  • $\begingroup$ It depends on how you define the topology on $SO(3)$. If for instance you view $SO(3)$ as a submanifold of the space of all 3 by 3 matrices, then you can write out the map from unit quaternions to 3 by 3 matrices entirely explicitly as a map from $\mathbb{R}^4$ to $\mathbb{R}^9$ and conclude continuity from the fact that is made up entirely of maps you already knew were continuous: addition, multiplication, composition of maps, projection onto a coordinate... It feels ugly with the involvement of these huge 4- and 9-dimensional ambient spaces for a local homeomorphism of 3D stuff but it works. $\endgroup$ – Vincent May 1 '17 at 9:39
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The facts below may not satisfy you... If I remember correctly

  • if a discrete group $G$ acts properly discontinuously and freely on a locally compact Hausdorff space $X$, then the projection $\pi:X\to X/G$ is a (Galois) cover,
  • and I think that for topological groups, if $H\subset G$ is a closed (?) subgroup of $G$, then the projection onto left (or equivalently right) cosets $G\to G/H$ is a principal $H$-fiber bundle.

Some conditions may be required for this last result to work, but it will certainly work for topological groups that are Lie groups. In the first case, take $X=S^3$ and $G=\mathbb{Z}/2\mathbb{Z}$ acting by the antipode map (as mentioned in the comments), and $G=S^3$ and $H=\lbrace\pm 1\rbrace$.

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