# Living inside a Hopf fibration, seeing the fibers

• I
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## Main Question or Discussion Point

Suppose I am stuck inside a three-sphere, S^3, given by w^2 + x^2 + y^2 + z^2 = 1. Let me be near the surface where w is zero. Let the surface w = 0 be painted white, this surface divides the three-sphere into two parts? If so say I am stuck on one side.

Is it true that at each point of this painted surface, w = 0, I can attach a little vector which would represent each fiber of a Hopf Fibration for my three-sphere above where the painted two-sphere is to be fibered? I would guess this is not unique? Is there a simple function that would give the direction of my little vectors on the surface w = 0, I want to imagine walking on one side of my surface w = 0 and picturing the little arrows? Hope this is clear and makes sense.

Edit, If the fibers are oriented then the vectors above would come in pairs for each fiber, one pointing "up" and the other pointing "down"? A single fiber both enters and exits "my side" of the three-sphere?

Thanks for any help.

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Infrared
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I'm not totally what your question is, but I'll make a couple remarks. I identity $\mathbb{R}^4=\mathbb{C}^2$.

For a point $p=(w_0+ix_0,y_0+iz_0)\in S^3$, the fiber $F_p$ containing $p$ is $\{e^{i\theta}p\}.$ In particular, if $w_0=x_0=0$, then $F_p$ is contained in the plane $w=0$ (and also in the plane $x=0$). So you have many fibers that lie in your surface $w=0$, not transverse to it. [Edit: as pointed out, this is only one fiber]

You can orient each fiber: assign to $p\in S^3$ the tangent vector $\frac{d}{dt}\big\vert_{t=0}e^{it}p=ip$. This defines a nonzero tangent vector field to each fiber. In real coordinates, the tangent vector at $p=(w,x,y,z)$ is $(-x,w,-z,y)$. Does this satisfy your request for "a simple function" for the direction of the tangent vectors?

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I'm not totally what your question is, but I'll make a couple remarks.
So I think it is true that a three-sphere can be equally divided into two equal separated parts by the two-sphere w = 0 ? If that is true I wondered if one could picture how, if they did at all, the fibers intersected the surface w = 0. Thank you for your help, will try and understand your reply a little better.

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Does this satisfy your request for "a simple function" for the direction of the tangent vectors?
That is pretty simple and sweet, thanks.

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For a point $p=(w_0+ix_0,y_0+iz_0)\in S^3$, the fiber $F_p$ containing $p$ is $\{e^{i\theta}p\}.$ In particular, if $w_0=x_0=0$, then $F_p$ is contained in the plane $w=0$ (and also in the plane $x=0$). So you have many fibers that lie in your surface $w=0$, not transverse to it.
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What does theta range over? Thanks.

mathwonk
Homework Helper
I could be wrong, but the nice calculation in post #2 seems to show that exactly one fiber lies in w=0, namely the one through the point w=0=x. the others seem to meet the 2 sphere each in 2 polar oppøsite points, and transversely, where theta = 0 or π. theta of course ranges over [0,2π). was i too careless?

Indeed since a fiber is cut by a complex line, it is hard to see how two distinct complex lines, which meet only in one point in C^2, could lie inside a real 3 space in C^2. Indeed a complex line generally meets the real 3 space w=0 in a real line, so the induced map, restrited to the 3 space w=0, is the one from the 2 sphere to the 2 diml projecttive plane. Hence the fibers of that map are pairs of antipodal points on the 2 sphere, of which exactly one collection of them lie on the equatorial circle cut by the unique complex line contained in w=0.

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lavinia
Gold Member
I could be wrong, but the nice calculation in post #2 seems to show that exactly one fiber lies in w=0, namely the one through the point w=0=x. the others seem to meet the 2 sphere each in 2 polar oppøsite points, and transversely, where theta = 0 or π. theta of course ranges over [0,2π). was i too careless?
If the 3 sphere is the points $(x+iy,z+iw)$ of distance 1 to the origin then the condition $w=0$ says that the second coordinate is purely real. If $z=0$ as well then one has the equator of the sphere and it is preserved by multiplication by $e^{iθ}$. If $z≠0$ then $e^{iθ}z$ is purely real only if $θ$ is a multiple of $π$ and this gives two antipodal points on the 2 sphere $w=0$.

So the picture is that the Hopf fibration pinches the sphere $w=0$ at the equator to make two 2 spheres joined at a point then indentifies these two spheres to make a single sphere.

Or it forms a projective plane out of the 2 sphere $w=0$ by identifying antipodal points then crushes a great circle in the projective plane to a point.

If one first identifies all antipodal points on the 3 sphere the quotient is the real projective space $RP^3$ and the sphere $w=0$ becomes a real projective plane. The fiber circles are wrapped around themselves twice so are still circles and they intersect the projective plane in exactly one point except for the equator of the sphere which becomes a "great circle" in the projective plane.

- This also illustrates that the fiber circles are linked. Circles in 3 space that intersect a sphere in different pairs of antipodal points have linking number ±1 depending on relative orientation. They also have linking number ±1 with any great circle on the sphere that they do not intersect.

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Gold Member
If you do a Google image search of the Hopf fibration you will get many pictures where we see the fibers by sterographic projection but I wanted to see these fibrations as a person stuck inside a three-sphere might see them examining some small region where space was roughly flat. So with your help I think I have a better picture of what is going on, no paint needed. Please correct me where I go wrong.

Consider the three-sphere, S^3, given by w^2 + x^2 + y^2 + z^2 = 1. Divide S^3 into three parts by points with w < 0, w > 0, and w = 0. Topologically we get two open balls and a two-sphere, S^2? Further divide our two-sphere, the points where w = 0, into three parts, points where z < 0, z > 0, and z = 0, respectively the southern hemisphere, the norther hemisphere, and the equator. Let the equator represent a single fiber of our fibration and give it an orientation if we may. On the surface of this two-sphere let us attach little unit vectors in the following way to represent how one possible fibration intersects the surface w = 0. On the northern hemisphere attach a unit vector pointing outward perpendicular to the surface for each point of the surface, each vector represents a single oriented fiber as it passes through the surface w = 0 (or must the vectors be skew and not perpendicular to the surface, skewness depending on latitude?). On the southern hemisphere do the same but with the vectors pointing inward. Does this describe the look of a possible fibration very close to the surface w = 0? Unfortunately the above seems too simple to be correct, I probably still don't get it.

I read that fibrations in some sense come in pairs, if my fibration above works what is its partner?

lavinia
Gold Member
If you do a Google image search of the Hopf fibration you will get many pictures where we see the fibers by sterographic projection but I wanted to see these fibrations as a person stuck inside a three-sphere might see them examining some small region where space was roughly flat. So with your help I think I have a better picture of what is going on, no paint needed. Please correct me where I go wrong.

Consider the three-sphere, S^3, given by w^2 + x^2 + y^2 + z^2 = 1. Divide S^3 into three parts by points with w < 0, w > 0, and w = 0. Topologically we get two open balls and a two-sphere, S^2? Further divide our two-sphere, the points where w = 0, into three parts, points where z < 0, z > 0, and z = 0, respectively the southern hemisphere, the norther hemisphere, and the equator. Let the equator represent a single fiber of our fibration and give it an orientation if we may. On the surface of this two-sphere let us attach little unit vectors in the following way to represent how one possible fibration intersects the surface w = 0. On the northern hemisphere attach a unit vector pointing outward perpendicular to the surface for each point of the surface, each vector represents a single oriented fiber as it passes through the surface w = 0 (or must the vectors be skew and not perpendicular to the surface, skewness depending on latitude?). On the southern hemisphere do the same but with the vectors pointing inward. Does this describe the look of a possible fibration very close to the surface w = 0? Unfortunately the above seems too simple to be correct, I probably still don't get it.

I read that fibrations in some sense come in pairs, if my fibration above works what is its partner?

As @Infrared pointed out, one can choose your arrows as the unit length tangent vectors to the fiber circles at their points of intersection with the 2 sphere $w=0$. One can consistently choose the direction of these tangents because the Hopf fibration viewed as a circle bundle is orientable. This fails at the equator since it is itself a fiber.

The point of several of the responses is that the 3 sphere does not fiber over the 2 sphere $w=0$. It fibers over a quotient space of this 2 sphere.

In my opinion these tangent vectors do not describe the Hopf fibration because they do not show how the fiber circles are linked, They merely show the direction of the fiber circles at points of intersection with the 2 sphere $w=0$. On the other hand, the picture one gets from stereographic projection does show this linking.

The Hopf fibration shows that the third homotopy group of the 2 sphere $π_3(S^2)$ is non-trivial. Hopf demonstrated this by showing that the fiber circles are linked. Hopf used this linking number argument to show more generally that two mapping of the 3 sphere into the 2 sphere are homotopic if and only if the linking number of circles in the preimages of pairs of points on the 2 sphere are the same. This implies that $π_3(S^2)$ is infinite cyclic since there are mappings with arbitrary linking number. The linking number for the Hopf fibration is 1.

- The Hopf fibration is not the only example of a circle bundle over the 2 sphere. For example the space of tangent unit vectors to the 2 sphere is a circle bundle in which the projection mapping sends each tangent vector to its point of tangency. The total space of this bundle is not the 3 sphere but is the real projective space $RP^3$.

- I have never heard that fibrations come in pairs. Perhaps what was meant is that every orientable sphere bundle has two orientations.

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I still think there is some unit vector field on S^2 (except for say the equator) that gives the direction the fibers as they intersect the two-sphere W = 0, it is not the one I wrote down because, ...

I was told that if two fibers were very close at one point then they were close at all points, nearby fibers remain nearby? Therefore my fibers that are near the equator need to be pointing nearly parallel to the equator so that they have a chance of staying always near the equator fiber. Maybe the fiber that passes through the north and south pole is perpendicular to w = 0 and as we move from north to south pole the fibers become less and less perpendicular to the surface, maybe pointing more eastward or westward (maybe that is the pair of fibrations, eastward verses westward?) as they near the equator they are both nearly tangent to the surface w = 0 and nearly parallel to the equator? I am sure I have been given enough help I just need to work it out.

Thank you!

lavinia
Gold Member
I still think there is some unit vector field on S^2 (except for say the equator) that gives the direction the fibers as they intersect the two-sphere W = 0,
This was shown in post #2.

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This was shown in post #2.
I am likely wrong but I don't think I am looking for tangent vectors given in post 2?

Well I should at least check the simple case where w=x=y=0, z=1 and see what I get.

I get (0,0,-1,0) ?

Lets say we examine the small region around some point p of a fibration. The fiber that goes through that point can point in any direction? I can choose so that at the point w=x=y=0, z=1 the fiber that goes through that point is perpendicular to the surface w=0 at that point, if not rotate our coordinate system so the fiber is perpendicular? If we look at the fibers as they pass through the surface w=0 and rotate our coordinate system the fibers will point in different directions?

Thanks.

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lavinia
Gold Member
I am likely wrong but I don't think I am looking for tangent vectors given in post 2?

Well I should at least check the simple case where w=x=y=0, z=1 and see what I get.

I get (0,0,-1,0) ?

Lets say we examine the small region around some point p of a fibration. The fiber that goes through that point can point in any direction? I can choose so that at the point w=x=y=0, z=1 the fiber that goes through that point is perpendicular to the surface w=0 at that point, if not rotate our coordinate system so the fiber is perpendicular? If we look at the fibers as they pass through the surface w=0 and rotate our coordinate system the fibers will point in different directions?

Thanks.
I think if you want to do this then the tangent vectors best reflect the direction that each fiber points in relative to the tangent plane of the 2 sphere $w=0$. I suggest that you calculate these angles directly.

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I am likely wrong but I don't think I am looking for tangent vectors given in post 2?
I think I am right in claiming I was wrong above.

So with some thought and the tangent vector field given in post 2 I think I have what I was looking for, the tangent vector field of the fibration at the surface w=o. See the sketch below.

A hard part was understanding that the unit vector in the w direction was perpendicular the surface w=o as graphed above. Mistakes pointed out appreciated. Thank you for everyone's help.

Edit, If T = (-x,w,-z,y) is a valid tangent field then T = (-x,w,z,-y) is also valid with opposite skew?

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lavinia
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@Spinnor Using your notation in post #15

If $w=0$ then T=(-cosθ,0, -sinθsinφ,sinθcosφ)

Note that as cosθ approaches zero T becomes tangent to the equator (0,0,y,z)

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Note that as cosθ approaches zero T becomes tangent to the equator (0,0,y,z)
As it must?

lavinia
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As it must?
Yes. The tangents to the fibers vary continuously and the equator is a fiber.

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Edit, If T = (-x,w,-z,y) is a valid tangent field then T = (-x,w,z,-y) is also valid with opposite skew?
If the above is true, which I think it is, If I travel along a fiber, say along the w axis and view both fibrations above then is there some sense I can say the fibrations appear to have opposite helicities?

If we look at the above image in a mirror we can see the other fibration?

Thanks.

lavinia
Gold Member
If the above is true, which I think it is, If I travel along a fiber, say along the w axis and view both fibrations above then is there some sense I can say the fibrations appear to have opposite helicities?

View attachment 257437

If we look at the above image in a mirror we can see the other fibration?

Thanks.
What do you mean by helicity and skew? What is the other fibration?

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What do you mean by helicity and skew? What is the other fibration?
Claim, if T = (-x,w,-z,y) is a valid tangent field then T' = (-x,w,z,-y) is also valid tangent field that can not be rotated into T? The tangent fields above give two different fibrations? I think that the image in post #19 when looked at in the mirror gives the other fibration?

With reference to my graph above move around the circle w=x=0, the equator, and consider two fibers that start at a point near the circle w=x=0 given by T and T' above. We are told the fibers will remain near the circle w=x=0 as we go once around that circle. As we go around the circle w=x=0 we will see the fibers "spiral" around the circle w=x=0 in opposite directions and after we go once around the circle the fibers will return to their starting point each fiber linking the equator once. If that is so then as I move along the circle w=x=0 I can say that the fibers "spiral" around the circle w=x=0 with opposite handedness?

I was using the word skew improperly. Thank you for clearing up any misunderstanding on my part.

Infrared
Gold Member
Claim, if T = (-x,w,-z,y) is a valid tangent field then T' = (-x,w,z,-y) is also valid tangent field that can not be rotated into T? The tangent fields above give two different fibrations? I think that the image in post #19 when looked at in the mirror gives the other fibration?
You can let $S^1$ act on $S^3$ instead by $e^{i\theta}\cdot (\zeta_1,\zeta_2)=(e^{i\theta}\zeta_1,e^{-i\theta}\zeta_2)$. Then the vector field you gave is a tangent vector field along each fiber of this new fibration:

$\frac{d}{dt}\vert_{t=0}e^{it}\cdot (\zeta_1,\zeta_2)=(i\zeta_1,-i\zeta_2)$. In real coordinates $\zeta_1=w+ix$ and $\zeta_2=y+iz$, this is exactly $(-x,w,z,-y)$.

lavinia
Gold Member
Thinking about this other fibration it seems that the mapping $(x+iy,z+iw)→(x+iy,z-iw)$ defines an isomorphism between the two fibrations since $e^{iθ}(x+iy, z+iw) →(e^{iθ}(x+iy),e^{-iθ}( z-iw))$.

Fiber circles are mapped to fiber circles.

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Fiber circles are mapped to fiber circles.
I have graphed some of the tangent vectors near the equator graphed above in post #15 and the vectors given by T and T' above seem like they suggest that the fibers "wrap around" the equator with opposite handedness? This to me seemed to give rise to a handedness for each fibration given by T and T'.

My hunch is that if I sit in S^3 and hold a left handed helix in my left hand and hold a right handed helix in my right hand no set of rotations of S^3 sitting in R^4 can change the handedness of one helix into another?