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Let $S^3$ be a unit cube in $\mathbb{R}^3$ with the boundary points identified. Let $\chi$ be a continuous, surjective map from $S^3$ onto $SO(3)$ where the preimage $\chi^{-1}(x)$ has a cardinality of 2 for every $x \in SO(3)$. From my intuitive understanding this should correspond to a homeomorphism to the double cover of $SO(3)$, $S^3 \rightarrow Spin(3)$ and therefore exist even though a one folded map (homeomorphism) $S^3 \rightarrow SO(3)$ does not exist.

Is this intuitive understanding correct, i.e. does such a continuous map exist? If so, do you know of any visualization of that?

Chris Grossack
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akreuzkamp
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    Yes, your intuitive understanding is right. If you look around for the double cover of $SU(2)$ over $SO(3)$, you'll find lots of people talking about exactly this. Since $SU(2)$ is homeomorphic to $S^3$. Imo the best way to understand this double cover is via quaternions, though this also takes some meditation. See this answer for more. – Chris Grossack Mar 29 '22 at 14:23
  • Maybe I'm misunderstanding your question, but there's another way to see this without quaternions (albeit not so visualizable). First, $SO(3)$ is homeomorphic to $\mathbb{R}P^3$, and the usual quotient map $S^3 \to \mathbb{R}P^3$ is a 2-fold covering. Since $S^3$ is simply connected it's the universal cover of $SO(3)$. Since $Spin(n)$ is defined as the universal cover of $SO(n)$ (which is necessarily 2-to-1 since $\pi_1(SO(n)) \cong \mathbb{Z}/2\mathbb{Z}$), there's a homeomorphism $S^3 \to Spin(3)$ making a commutative diagram with the covering maps just by uniqueness of universal covers. – kamills Mar 29 '22 at 14:57
  • Thanks! :) The most important bit for me was to double check that I didn't miss something important here. Regarding the visualization: I am imagining an image of a 3D grid of arrows (or anything better suited to visualize a rotation not just a direction) that allows to track the two twists of the field and how it continuously turns into no rotation at the border. – akreuzkamp Mar 29 '22 at 15:53
  • @akreuzkamp . How do you indentify the boundary points of the unit cube in $\mathbb R^3$ ? Note that this can easily produce a torus and not $SO(3)$. The latter is the unit ball with antipodal points identified. – Kurt G. Mar 29 '22 at 18:33
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    To be precise : $SO(3)$ is the unit sphere in $\mathbb R^{\color{red}{4}}$ with antipodal points identified. Surely a sphere can be continuously deformed into the surface of a unit cube. However this doesn't make $SO(3)$ a torus. Visualising this in $\mathbb R^2$ (torus and unit circle) should make this clear. – Kurt G. Mar 30 '22 at 05:47
  • @KurtG. I identify all boundary points as a single point, that gives me an $S^3$, not $SO(3)$ and not a torus.

    I don't get the torus thing, but stll your comment gave the crucial hint to get the intuitive understanding: I can imagine how the cube corresponds to $S^3$ and therefore, which points of the 3 cube are antipodal on $S^3$. That tells me which points of the box map onto the same rotation: The center of the box maps onto no rotation, the sphere which marks the halfway from the center to the boundary map onto 180° rotations around different axes and so on.

     – akreuzkamp Mar 30 '22 at 08:52
  • The torus thing was just a warning because I fell into that trap once. So we agree that $S^3=SU(2)$ and when we identify antipodal points we get $SO(3)$. – Kurt G. Mar 30 '22 at 09:03
  • @akreuzkamp . When I think of it. You identify all boundary points as a single point, and that gives you $S^3$ ? Please elaborate. – Kurt G. Mar 30 '22 at 09:10
  • @KurtG.The interior of the cube is just homeomorphic to $\mathbb{R}^3$. The one-point compactification of $\mathbb{R}^3$ is $S^3$ (https://en.wikipedia.org/wiki/Alexandroff_extension#Compactifications_of_continuous_spaces). Identifying the boundary of the box as a single point is the same thing as the one-point compactification of the interior. I bet there is a more direct proof, but that's how I think about it. – akreuzkamp Mar 30 '22 at 13:00

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