Hopf fibration of 3-sphere

[cianfa72]

TL;DR

About the Hopf fibration of the 3-sphere

As shown by this animation, the fibers of the Hopf fibration of the 3-sphere are circles (click on a point on the sphere to visualize the associated fiber). As far as I understand, they never intersect and their union is the 3-sphere itself.

I'd be sure whether the circles in the animation are given by stereographic projection of the 3-sphere from a point, say the "equivalent" of the $S^2$ north-pole. Assuming the viewpoint of 3-sphere defined by its embedding in $\mathbb C^2$ as "ambient" space, the 3-sphere is given as the locus of complex pairs $(z,w)$ such that $|z|^2 + |w|^2 = 1$. Then one can employ, for instance, the stereographic projection from the complex point $(0,i)$ that is $(0,0,0,1)$ using the natural identification $\mathbb C^2 \cong \mathbb R^4$. Such a stereographic projection maps the 3-sphere on the 3D space $v=0$ where $z = x +iy, w = u + iv$ including $\infty$.

Did I understand it correctly ? Thanks.

 

[cianfa72]

BTW, the fibers $S^1$ of the Hopf fibration are supposed to be what R. Penrose calls Clifford parallels in his book "The Road to Reality".

 

[Hornbein]

I don't really understand it but I think stereographic projection introduces much distortion that we don't see in the Hopf fibration. The Hopf fibration is a bijective map from points on the 2-sphere to a special class of great circles on the 3-sphere. The way I possibly understand it is that we fix a more or less arbitrary circle on the 4-sphere. Then any other great circle in that class has two angles relative to that circle. These two angles correspond to longitude and latitude on the 2-sphere. Note that the longitude angle is from -pi to pi, the latitude angle is from -pi/2 to pi/2. In general if we have two 2-planes that intersect at a point then their relative orientation can be characterized by two angles. The point is the center of the 3-sphere and the intersection of each plane containing that point with the 3-sphere is a great circle in that Hopf class.

I don't like their saying "the circles are the same distance apart." What they are trying to say is that if A and B are circles in the same Hopf fibration then each point in circle A has the same distance to the point nearest to it on circle B. This minimum distance varies from 0 to pi/2 depending on the choice of A and B. Relations are pretty much linear which is why I question whether stereographic is involved.

I also often see it said that the great circles are linked, but I don't understand what they mean by this.

I hope I'm not leading you astray.

 

[cianfa72]

I don't like their saying "the circles are the same distance apart." What they are trying to say is that if A and B are circles in the same Hopf fibration then each point in circle A has the same distance to the point nearest to it on circle B. This minimum distance varies from 0 to pi/2 depending on the choice of A and B.

Yes, this is the same for parallel straight lines in Euclidean space. The minimum distance between parallels A and B depends on our choice of which A and B are supposed to be.

 

[Hornbein]

From Wikipedia

Stereographic projection of the Hopf fibration induces a remarkable structure on R3, in which all of 3-dimensional space, except for the z-axis, is filled with nested tori made of linking Villarceau circles. Here each fiber projects to a circle in space (one of which is a line, thought of as a "circle through infinity"). Each torus is the stereographic projection of the inverse image of a circle of latitude of the 2-sphere. (Topologically, a torus is the product of two circles.) These tori are illustrated in the images at right. When R3 is compressed to the boundary of a ball, some geometric structure is lost although the topological structure is retained (see Topology and geometry). The loops are homeomorphic to circles, although they are not geometric circles.

So they are not saying that fibers are linked. They are saying that the images of the fibers in the stereographic projection are linked.

 

[cianfa72]

So they are not saying that fibers are linked. They are saying that the images of the fibers in the stereographic projection are llinked.

No, being pairwise linked (as in the image with keyrings in wikipedia link) is invariant under stereographic projection.

Happy 2026 !

 

[Hornbein]

No, being pairwise linked (as in the image with keyrings in wikipedia link) is invariant under stereographic projection.

Happy 2026 !

I say that circles (1-spheres) can't be linked in 4-space. In 4-space circles can always be separated, which to me means they aren't linked. 2-spheres can be linked in 4-space, (N-2)-spheres in N-space with N>2. An (N-2)-sphere is functionally a ring.

Hmm, can 0-spheres be link in 2-space? No, but they can be linked in 1-space.

 

[cianfa72]

I say that circles (1-spheres) can't be linked in 4-space. In 4-space circles can always be separated, which to me means they aren't linked.

True but not relevant. What matters is the (intrinsic) $\mathbb S^3$ topology. In this topology (homeomorphic to two 3-balls glued along their common spherical boundary) the fibers of the Hopf fibration can never be unlinked/separated.

 

[Hornbein]

True but not relevant. What matters is the (intrinsic) $\mathbb S^3$ topology. In this topology (homeomorphic to two 3-balls glued along their common spherical boundary) the fibers of the Hopf fibration can never be unlinked/separated.

I confess that this is completely over my head. I also know from experience that I am never able to find explanations of such things that I am able to understand.

When I was in math graduate school I came across a survey book on research on the Riemann hypothesis. I couldn't understand even the first page.

 

[cianfa72]

I confess that this is completely over my head. I also know from experience that I am never able to find explanations of such things that I am able to understand.

Try thinking of $\mathbb S^3$ in terms of the space obtained by gluing the spherical boundary of two 3-balls in 3D space. This space is homeomorphic to $\mathbb S^3$ and a fiber of the Hopf fibration is basically a circle passing through two antipodal points on the common/glued sphere.

 

[lavinia]

I also often see it said that the great circles are linked, but I don't understand what they mean by this.

@Hornbein

Intuitively, in Euclidean 3 space, two non-intersecting closed loops are linked if they cannot be separated without crossing over each other even if they can be bent or stretched in trying to do so. Similarly, on the 3 sphere two fibers are linked if they cannot be separated on the 3 sphere without crossing over each other. (They can be separated if you allow them to leave the three sphere and wander around freely in R^4).

Another way which works for smooth curves is to fill one curve in to be the boundary of a disklike region. If the other curve is transverse to this region then one can count the number of times it intersects it after adding a sign that accounts for the direction it is coming from. The sign would be determined by the direction of its tangent vector at the points of intersection. If the signed number isn't zero then the two curves are linked.

 

[cianfa72]

Intuitively, in Euclidean 3 space, two non-intersecting closed loops are linked if they cannot be separated without crossing over each other even if they can be bent or stretched in trying to do so. Similarly, on the 3 sphere two fibers are linked if they cannot be separated on the 3 sphere without crossing over each other. (They can be separated if you allow them to leave the three sphere and wander around freely in R^4).

Can you explain why they can always be separated in the "ambient" $\mathbb R^4$ ?

 

[lavinia]

Can you explain why they can always be separated in the "ambient" $\mathbb R^4$ ?

What are your thoughts on this?

 

[cianfa72]

What are your thoughts on this?

Sorry, I've not idea for this....

 

[lavinia]

Another way which works for smooth curves is to fill one curve in to be the boundary of a disklike region. If the other curve is transverse to this region then one can count the number of times it intersects it after adding a sign that accounts for the direction it is coming from. The sign would be determined by the direction of its tangent vector at the points of intersection. If the signed number isn't zero then the two curves are linked.

As an example, consider the equator of the three sphere, the circle in S^3 whose third and fourth coordinates are both zero. It divides the equatorial 2 sphere, the sphere whose fourth coordinate is zero, into two hemispheres and it is the boundary of both. Any fiber of the Hopf fibration except the equator itself intersects each these hemispheres transversally in exactly one point so all fibers are linked with the equator.

 

[lavinia]

Sorry, I've not idea for this....

OK. So how would you move two linked loops in R^3 off of each other into R^4 without them crossing over each other?

How would you move a point in the northern hemisphere of a sphere in R^3 to the southern hemisphere without crossing the equator if the point is allowed to move anywhere in R^3?

If you know the picture of the Klein bottle as a surface in R^3 that intersects itself, how would you avoid the intersection if it was allowed to be stretched and bent in R^4?

 

[cianfa72]

OK. So how would you move two linked loops in R^3 off of each other into R^4 without them crossing over each other?

The idea could be "lift" one the two linked loops in R^3 along the fourth dimension to separate it from the other.

How would you move a point in the northern hemisphere of a sphere in R^3 to the southern hemisphere without crossing the equator if the point is allowed to move anywhere in R^3?

Yes, move it along the third dimension "outside" of the 2-sphere inside R^3.

If you know the picture of the Klein bottle as a surface in R^3 that intersects itself, how would you avoid the intersection if it was allowed to be stretched and bent in R^4?

It should be analogous to two linked circles in R^2: they intersect. To separate them, lift one of them in R^3 (same argument applies to the Klein bottle: to avoid the intersection move "that part of the surface" along the fourth dimension in R^4 (Klein bottle, indeed, can be smoothly embedded into R^4).

 

[lavinia]

The idea could be "lift" one the two linked loops in R^3 along the fourth dimension to separate it from the other.


Yes, move it along the third dimension "outside" of the 2-sphere inside R^3.


It should be analogous to two linked circles in R^2: they intersect. To separate them, lift one of them in R^3 (same argument applies to the Klein bottle: to avoid the intersection move "that part of the surface" along the fourth dimension in R^4 (Klein bottle, indeed, can be smoothly embedded into R^4).

Right. BTW: Your answer proves that the Klein bottle can be embedded in R^4.

So for two linked fibers in 3 the 3 sphere, is the worry that they would have to crossover each other just to get outnof the 3 sphere into R^4?

 

[cianfa72]

So for two linked fibers in the 3 sphere, is the worry that they would have to crossover each other just to get out of the 3 sphere into R^4?

No. By lifting one of them in the fourth dimension out of the 3-sphere in R^4, they can be separated.

 

[lavinia]

No. By lifting one of them in the fourth dimension out of the 3-sphere in R^4, they can be separated.

Here is a way of doing it using the Hopf fibration. Slide the two linked fibers along the rays emananting from the north pole of the 3 sphere until they reach R^3. Then raise one circle up into R^4 so that its fourth cooridinate is not zero. (This all occurs in R^4)

 

[cianfa72]

Here is a way of doing it using the Hopf fibration. Slide the two linked fibers along the rays emananting from the north pole of the 3 sphere until they reach R^3. Then raise one circle up into R^4 so that its fourth coordinate is not zero. (This all occurs in R^4)

Ok, so basically the first step employs the stereographic projection from the north pole of the 3-sphere to R^3 (i.e. $x_4 = 0$) -- all this occurs within R^4.

 

[lavinia]

Ok, so basically the first step employs the stereographic projection from the north pole of the 3-sphere to R^3 (i.e. $x_4 = 0$) -- all this occurs within R^4.

yes

I just wanted to emphasiize that by sliding the linked pair of fibers along the rays emanating from the north pole of the three sphere the linked fibers are continuously moved into R^3. Stereograaphic projection by itself without this sliding is just a mapping and is not a movement of the fibers.

 

[cianfa72]

I just wanted to emphasize that by sliding the linked pair of fibers along the rays emanating from the north pole of the three sphere the linked fibers are continuously moved into R^3. Stereographic projection by itself without this sliding is just a mapping and is not a movement of the fibers.

Yes, let me say the "process" of sliding the linked pair of fibers occurs along the rays of the stereographic projection from the north pole of the 3-sphere to the hyperplane $x_4 = 0$ in R^4.

 

[lavinia]

@Hornbein

Intuitively, in Euclidean 3 space, two non-intersecting closed loops are linked if they cannot be separated without crossing over each other even if they can be bent or stretched in trying to do so. Similarly, on the 3 sphere two fibers are linked if they cannot be separated on the 3 sphere without crossing over each other. (They can be separated if you allow them to leave the three sphere and wander around freely in R^4).

Another way which works for smooth curves is to fill one curve in to be the boundary of a disklike region. If the other curve is transverse to this region then one can count the number of times it intersects it after adding a sign that accounts for the direction it is coming from. The sign would be determined by the direction of its tangent vector at the points of intersection. If the signed number isn't zero then the two curves are linked.

Another way to think about the linking of two fibers in the Hopf fibration is to look at the topology of their complement in S^3. Take for instance, the complement of the equator(the unit circle in the xy-plane) and the fiber that passes through the north pole of S^3. Since the north pole is removed, the complement is mapped homeomorphically into R^3 under stereographic projection. The image is R^3 minus both the unit circle in the xy-plane and the z-axis.

Problem: Show that R^3 minus the z-axis and the unit circle in the xy-plane can be continuously deformed onto a torus centered around the unit circle. Convince yourself that this cannot be done if the z-axis is replaced by a line that lies outside of the unit circle.

This shows the difference between two linked and two unlinked closed loops,

 

[cianfa72]

Another way to think about the linking of two fibers in the Hopf fibration is to look at the topology of their complement in S^3. Take for instance, the complement of the equator(the unit circle in the xy-plane) and the fiber that passes through the north pole of S^3. Since the north pole is removed, the complement is mapped homeomorphically into R^3 under stereographic projection. The image is R^3 minus both the unit circle in the xy-plane and the z-axis.

Here the equator (i.e. $x^2 + y^2 = 1, z=w=0$) is a fiber of S^3. So you're saying to take the stereographic projection on $w=0$ of both its complement in S^3 and the complement of fiber passing through the north pole of S^3 (let's call such a fiber F). The image under the stereographic projection of F is the z-axis while the image of the equator is the unit circle in xy-plane. Hence, as projection of the complement, one gets what you said.

Problem: Show that R^3 minus the z-axis and the unit circle can be continuously deformed onto a torus centered around the unit circle. Convince yourself that this cannot be done if the z-axis is replaced by a line that lies outside of the unit circle.

The torus you are talking about should be a 3D torus since that image is three dimensional. What do you mean with

a torus centered around the unit circle ?

 

[lavinia]

Here the equator (i.e. $x^2 + y^2 = 1, z=w=0$) is a fiber of S^3. So you're saying to take the stereographic projection on $w=0$ of both its complement in S^3 and the complement of fiber passing through the north pole of S^3 (let's call such a fiber F). The image under the stereographic projection of F is the z-axis while the image of the equator is the unit circle in xy-plane. Hence, as projection of the complement, one gets what you said.


The torus you are talking about should be a 3D torus since that image is three dimensional. What do you mean with

After continuous deformation it is a regular 2d torus. I had trouble describing the situation in a clear way even though the picture is clear, so I left it as a problem. If I knew how to post drawings this would be easier to see but I don't. My apologies. Here is another stab at it.

Given a symmetrical 2d torus in R^3, there is a circle in its interior which is equidistant from all of its points. Such a torus can be thought of as being centered on that circle although I admit that the language is vague. The torus I meant was any torus that is centered on the unit circle in the xy-plane and which is small enough that it does not intersect with the z-axis.

The continuous deformation would be a mapping that over a finite time interval moves all of the points onto the torus. Here are some things to consider.

- Each point in R^3-(z-axis,unit circle) lies on a unique line segment of shortest length that intersects the unit circle in the xy-plane at right angles. Notice BTW that this is not true of the z-axis or of the unit circle itself since each lie on infinitely many line such segments.

- Each such line segment intersects the torus at a point that minimizes the distance of the point to the torus.

- This picture helped me. Consider a plane that contains the z-axis and which intersects the torus at right angles to create a cross sectional circle. There are two such circles. Any point in the plane except the z-axis and the centers of the two circles can be connected to one of the circles by a unique straight line segment that is closest to the circle. This segment will intersect the circle at right angles.

 

[cianfa72]

After continuous deformation it is a regular 2d torus. I had trouble describing the situation in a clear way even though the picture is clear, so I left it as a problem. If I knew how to post drawings this would be easier to see but I don't. My apologies. Here is another stab at it.

Sorry, R^3 minus (z-axis, unit circle on xy-plane) is an open set of R^3, so it is a (topological) manifold of dim 3 while a regular 2d torus is a manifold of dim 2. How can they be homeomorphic?

 

[lavinia]

Sorry, R^3 minus (z-axis, unit circle on xy-plane) is an open set of R^3, so it is a (topological) manifold of dim 3 while a regular 2d torus is a manifold of dim 2. How can they be homeomorphic?

They are not homeomorphic. The three dimensional space deforms continuously onto a 2 dimensional torus. The idea of continuous deformation is important in topology. A space that can be continuously deformed onto another has many of the same topological properties

Here is the first example that I ever saw. This will be a continuous deformation of R^n onto a point.

The map H(x,t) will be a continuous function from R^n x [0,1] into R^n so that at time zero it is the identity map, H(x,0)=x, and at time 1 it maps all of R^n to the origin, H(x,1) = the origin in R^n.

Define H(x,t) = (1-t)x .

So all of R^n in a finite time is continuously shrunk to a single point. Each point in R^n slides along the line segment connecting it to the origin until it stops at the origin. A space that can be continuously deformed to a point is called contractible. R^n is contractible. Amazing. Notice that there is a big difference between this and just mapping a space to a point. That can always be done for any space. Mapping to a point is always continuous. But deforming to a point is quite different. Most spaces are not contractible. For instance no closed manifold is contractible.

Here is a related example. Consider the tangent bundle to an n-manifold. The map H(x,t) = (1-t)x shrinks each tangent vector,x, down to the origin of its tangent vector space. ( here the manifold is considered to be the zero vectors in each tangent space.) So the tangent bundle of a manifold can be continuously deformed onto the manifold. For that matter, any vector bundle over a topological space can be continuously deformed onto the base space.

Here is a third related example: Consider R^n minus the origin. The idea is the same. Points in R^n-the origin are moved along line segments through the origin until they hit the unit sphere. Problem: Write down the deformation function H(x,t).

Just as a reminder. The whole point of all of this is to show that the linking of two fibers in the Hopf fibration can be detected by showing that their complement in S^3 can be continuously deformed onto a torus whereas the complement of two unlinked circles cannot be.