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I'm taking a class in manifolds, and the Hopf fibration recently came up. I'm trying to get a handle on it, so I'm going to try and explain what I think is going on, and hopefully math.stackexchange will either correct me, or have more enlightening/sophisticated ways to explain what's going on.

The Hopf fibration maps circles (these are the 'fibres') of $S^3$ onto points of $S^2$. Identifying $S^3$ with $SU(2)$, is the above description telling us how many copies of $SO(2)$ (that is, $S^1$) fit inside $SU(2)$? I think this makes sense, since $SU(2)/SO(2)$ is isomorphic to $S^2$.

If one takes a 3-dimensional slice of $S^3$, one ends up with a an image that look like this. Are the fibres of $S^3$ that are being mapped to points of $S^2$ the Villarceau circles one gets from slicing these torii?

The fibre bundle map can be written $S^1\hookrightarrow S^3\xrightarrow{\pi}S^2$. So for $p\in S^2$, $\pi^{-1}(p)=S^1$. Wikipedia says that if $U$ is an open neighborhood of $S^2$, then $\pi^{-1}(U)$ should be homeomorphic to $U \times S^1$. How can this homeomorphism be realized? Is this related to the image I linked above, at all?

Sorry if this seems rambling/undirected. Feel free to ask for clarification etc.

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If you consider a disk bounded by the red circle in your image, all of the copies of $S^1$ (the fibers) intersect that disk exactly once (with the exception of the the red outer circle). You can identify this disk with $S^2$ by mapping all the points on that red boundary circle to one point. Since every point in $S^3$ is on a fiber, you can then define $\pi : S^3 \rightarrow S^2$ by sending that point to where its fiber intersects the disk.

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  • $\begingroup$ Could you explain why every fiber has to intersect the disc bounded by that red circle? What makes that red circle special? $\endgroup$ – Sid Raval Feb 9 '12 at 19:39
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    $\begingroup$ In the image you can think $S^3$ being built starting with the two red circles, and then filling the rest of the space with tori around that red circle. The tori themselves are made up of copies of $S^1$ which twist once around the meridian and once around the longitude. The disk intersects each of these tori at a circle, and each of the $s^1$ fibers intersect that circle once. $\endgroup$ – user16124 Feb 9 '12 at 20:15

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