# I Hopf fibration, stereoprojected fibers look close, can be?

#### Spinnor

Gold Member
In the stereographic projection of a Hopf fibration of a three-sphere, say at the 30 second point of the video below,

the fibers appear to get arbitrarily close to each other. Are they or can they be close in the three-sphere? If you and I took a walk on two very close fibers of the three-sphere and in the same direction, would our paths remain close through one full circuit of the fibers?

Is there a "left" handed and a "right" handed fibration of the three-sphere?

Are there more complicated fibrations of the three-sphere involving circles as fibers?

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#### Spinnor

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Given that, can you define what you mean by close?
Given some small distance delta, can I always find a pair of fibers that are never more than a distance delta from each other as we go around the fibers.

Thanks.

#### fresh_42

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2018 Award
Given some small distance delta, can I always find a pair of fibers that are never more than a distance delta from each other as we go around the fibers.

Thanks.
In general, yes, although your definition of distance lacks a lot of information. I would not bet that this will be the case in all metrics, but for the usual suspects, yes. And every path, which is a function $\gamma\, : \,[0,1] \longrightarrow F$ can be right and left handed, resp. forwards and backwards: simply use $\gamma (1-t)$ instead.

#### WWGD

Gold Member
What metric are you using on the 3-Sphere?

#### WWGD

Gold Member
Given some small distance delta, can I always find a pair of fibers that are never more than a distance delta from each other as we go around the fibers.

Thanks.
I am not sure I understand. Can't you just take any two close points in the sphere, project them and then lift them back up?

#### Spinnor

Gold Member
What metric are you using on the 3-Sphere?
Is there a "natural" metric, ds^2 = [x2-x1]^2 + [y2-y1]^2 + [z2-z1]^2 + [w2-w1]^2

given by the embedding of the three-sphere in R^4 where (x1,y1,z1,w1) and (x2,y2,z2,w2) are two nearby points of nearby fibers.

I am confused with your question, and that is a good thing. There must be more possibilities then I am thinking of.

I am trying to understand also if two nearby fibers are parallel in some sense, I would like them to not just stay close but to be the same distance apart as we circuit a pair of fibers. I had this thought that if I lived in a three-sphere and had two laser pointers (ideal mathematical ones) that were nearby and nearly parallel and turned them on, eventually the light from each would come back and illuminate the backs of the laser pointers. By careful aiming I could get the beams of light to be "parallel" in the sense that the beams of light were always the same distance apart.

Edit, the following must be wrong. I also wondered if by carefully aiming I could get the beams to swap so to speak, laser pointer a illuminates the back of laser pointer b and vise versa in which case the beams of light would require two circuits to return to the original laser pointers. I would like to know if such things are possible in this interesting space.

Please forgive my sloppy non-mathematical language.

Thank you.

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#### Infrared

Gold Member
I would like them to not just stay close but to be the same distance apart as we circuit a pair of fibers.
Yes, this happens. Viewing a point $(a,b,c,d)\in S^3$ as a point $p=(a+bi,c+di)=(z,w)\in\mathbb{C}^2$, the fiber containing the point $p$ is the set of points $\alpha p=(\alpha z,\alpha w)$ such that $\alpha$ is a complex number of length $1$. Now if $p'=(z',w')$ is another point, on a different fiber, then we see that $d(p,p')=d(e^{i\phi}p,e^{i\phi}p')$ for any complex number $e^{i\phi}$ on the unit circle. So distance is preserved if we travel on both fibers with the same angular velocity.

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#### Spinnor

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So distance is preserved if we travel on both fibers with the same angular velocity.
Thank you for that proof!

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#### lavinia

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The three sphere is the set of points in $C^2$ of the form $(re^{ix},se^{iy})$ with $r^2+s^2 = 1$. The fiber passing through $(re^{ix},se^{iy})$ is the circle $(re^{i(x+θ)},se^{i(y+θ)})$. From this formula one sees that the fiber circles are pairwise disjoint.

While no two fiber circles intersect, they are nevertheless linked like two metal rings in a chain. Each fiber circle passes through every other fiber circle exactly once. It is this linking that distinguishes the Hopf fibration. There are other circle bundles over the 2 sphere where the linking number is higher than 1 but I don't think any of the manifolds are the 3 sphere.

To visualize the link consider the equator of the 3 sphere. This is a 2 sphere for which the second coordinate of $(re^{ix},se^{iy})$ is real or equivalently $y$ is a multiple of $π$. The only fiber circle on this sphere is the circle with $s=0$. Every other fiber circle intersects this 2 sphere in two opposite poles and so links with the circle $s=0$ exactly once. By a rotation of the 3 sphere any fiber circle can be moved onto $s=0$ so every pair of fiber circles links exactly once.

Note that identifying each fiber circle to a point first identifies all of the opposite poles not lying on $s=0$ to make a hemisphere with boundary $s=0$ then crushes all of $s=0$ to a point to make a 2 sphere.

- From the formula $(re^{i(x+θ)},se^{i(y+θ)})$ for a fiber circle one can calculate the distance between points on two fiber circles. The standard stereographic projection in which the n-sphere touches $R^{n}$ at the South pole always increases distances.

- I saw on on line that there are other decompositions of the 3 sphere into circles called Seifert fibrations.

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#### lavinia

Gold Member
A couple side thoughts on the Hopf fibration:

- If $r$ and $s$ are both not zero and are kept fixed while $x$ and $y$ are allowed to vary arbitrarily, then $(re^{ix},se^{iy})$ spans a torus inside the 3 sphere. This torus is completely covered by fiber circles that are pairwise linked with linking number equal to 1. I think this torus when projected into $R^3$ is the torus in the animation the OP @Spinnor posted. I would guess that $r$ and $s$ are chosen to be equal so that the projected image looks completely symmetrical. If the they were unequal the projected torus would look elongated.

- Interestingly all of these tori are geometrically flat. Their Gauss curvature is identically zero. This is because the map

$T(x,y)= (re^{ix},se^{iy})$

preserves the Euclidean metric on the plane. One can literally think of this as bending a strip of paper into a double cylinder. This is not possible in $R^3$ but is possible in $R^4$. The stereographic projections of these tori are not flat but they are conformal images of flat tori.

- If $s$ is allowed to approach zero, one gets a continuum of nested tori which converge to the circle $(e^{i(x+θ)},0)$. These tori together with this circle form a solid torus ( a doughnut filled with jelly). Similarly if one allows $r$ to approach zero one gets a second solid torus with equatorial circle $(0,e^{i(y+θ)})$. These two solid tori share the original torus as their common boundary. And since every point on the 3 sphere lies on one of these solid tori, the 3 sphere can be thought of as two solid tori glued together along their boundaries.

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#### Spinnor

Gold Member
And since every point on the 3 sphere lies on one of these solid tori, the 3 sphere can be thought of as two solid tori glued together along their boundaries.
A three-sphere is also the same (topologically?) as two solid 3-balls identified at their surfaces? Can these two pictures then morph into one another?

Thanks.

#### lavinia

Gold Member
A three-sphere is also the same (topologically?) as two solid 3-balls identified at their surfaces? Can these two pictures then morph into one another?

Thanks.
Yes. Any n-sphere can be obtained as two n-balls glued together along their boundary. Slice the sphere along its equator. One gets two n-balls with the equator sphere as their boundary.

The morphing:

Slice the 3-sphere along the equator to get two hemispheres and project them into $R^3$ to get two ordinary 3 dimensional balls. In the 3-sphere these balls are glued back together along the equator.

From each ball remove a solid cylindrical core so that the two are identified in the 3-sphere at their ends. Glue them together to make a solid torus. What remains are two 3-balls with solid cores removed and in the 3 sphere they are glued together along what remains of their bounday 2-spheres which are each 2-spheres with the two polar ice caps removed. The gluing creates a surface whose boundary is a torus - the two boundaries left by the removed cores glued together at their ends. So already one knows this must be a second solid torus.

To see this directly consider the exterior surface of the cored ball - what remains of its bounding sphere. It is a topological cylinder and is thus homeomorphic to an annulus in the plane. Warp the cored ball so that this cylinder becomes an annulus. In the process the interior core is extruded and becomes the outer boundary of a half bagel excepting the annulus itself. Remembering the identifications, these two half bagels are glued together along their annuli and this makes a second solid torus.

The two solid tori are glued together along their boundaries. For the first solid torus this is the surface of the attached cores. For the second it is the extruded surfaces left by the removed cores.

Circle bundles over the 2 sphere:

Another way to see that the 3-sphere is two solid tori glued together is to consider the Hopf fibration. This interprets the 3-sphere as a circle bundle over the 2-sphere. This bundle is trivial above the two hemispheres of the 2 sphere and is therefore homeomorphic to $D^2× S^1$ a disk Cartesian product a circle and this is a solid torus. Its boundary is the equator $× S^1$ and this is a torus.

This way of looking at the 3-sphere generalizes to any circle bundle over the 2 sphere. For instance the tangent circle bundle of the 2 sphere is not the 3-sphere but rather it is diffeomorphic to the rotation group $SO(3)$. Still by the same argument it is two solid tori glued together along their boundaries. There are infinitely many of these bundles that are not 3 spheres. I am not sure but they may all be quotient spaces of the 3 sphere by that actions of finite cyclic groups and I think - not sure - that these are called lens spaces.

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"Hopf fibration, stereoprojected fibers look close, can be?"

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