the 2-fold covering $SU(2)\rightarrow SO(3)$, a problem with the topological viewpoint

Question

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.

Answer 1

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.

Answer 2

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 $H=\lbrace\pm 1\rbrace$.