# Constructing S^3 from a S^2 and a bunch of S^1's?

Gold Member
Suppose I sit in R^4 with a single 2-sphere, x^2+y^2+z^2=1, and an infinite number of circles of radius one that are free to move in R^4. Is it true that one can arrange these infinite number of circles so that their union is the three-sphere S^3, defined by x^2+y^2+z^2+w^2=1 and each circle is disjoint from all other circles but intersects the 2-sphere at one point? Is this a Hopf fibration?

Thanks for any help!

WWGD
Gold Member
If every circle is disjoint from every other circle, then it is not the Hopf fibration .

Gold Member
If every circle is disjoint from every other circle, then it is not the Hopf fibration .

If not disjoint, how many points does each pair of circles share?

Thanks!

WWGD
Gold Member
The linking number for the Hopf fibration is 1, it is the Hopf invariant of the Hopf fibration. Notice that 1+2 (coming from S^1 and S^2 respectfully) =3, which makes it harder for the subspaces(submanifolds) to miss each other, e.g., a plane and a line in #\mathbb R^3# (actually, if you have codimensions n,m in an ambient k-manifold, with n+m<k, then the submanifolds ca be homotoped to miss each other. ) , and that the linking number is a homotopy invariant. I think if the map were null-homotopic, the linking number would be 0. This shows that #\pi_3(S^2)# is non-trivial

Last edited:
Ben Niehoff
Gold Member
The circles are disjoint. This is a separate issue from their linking number.

WWGD
Gold Member
My bad, if the circles overlapped, they would project down to a common point. I wasn't connecting it to the linking point, I was just mistaken.

WWGD
Gold Member
Still, Spinnor, I am not sure I understand: the Hopf fibration goes from $S^3 \rightarrow S^2$ with fiber $S^1$ over every point in $S^2$ , so what do you mean by the intersection of the circles with the 2-sphere? The circles, as fibers, live in the 3-sphere. I assume, [itex] S^2 [\itex] is embedded in the 3-sphere, but how so? If this were a vector bundle, then there is a natural version of a 0-section, where you embed the base in the top space, but this is not a vector bundle, so you need to determine how the 2-sphere is embedded in the 3-sphere to be able to determine the intersection. Maybe as a big-circle-type embedding?

Last edited:
Gold Member
Sorry for the response delay.

Using the simple objects, a two-sphere, circles, and the four dimensional Euclidian space, R^4, I wondered if I could construct a three-sphere without having to pass the circles through each other or the two-sphere. So it looks like the circles are linked but moving them around in R^4 they can be linked? The extra freedom of R^4 allows us to do this without the circles passing through each other? The other question is are the circles "flat" and of the same radius?

Thanks for the help!

Ben Niehoff
Gold Member
The circles are "flat", yes, and they all have radius 1. In fact, they are each great circles of the 3-sphere.

Spinnor
Gold Member
The circles are "flat", yes, and they all have radius 1. In fact, they are each great circles of the 3-sphere.

That is cool! Thank you!

WWGD
Gold Member
The circles are "flat", yes, and they all have radius 1. In fact, they are each great circles of the 3-sphere.
Do you mean the circles are planar, unknotted and (if planar, so that we can set up a 'standard' xy-coordinate system), are translates of {(x,y): x^2+y^2=1?

WWGD
Gold Member
Ah, never mind, for some reason I cannot delete my previous post.

WWGD
Gold Member
Sorry for the response delay.

Using the simple objects, a two-sphere, circles, and the four dimensional Euclidian space, R^4, I wondered if I could construct a three-sphere without having to pass the circles through each other or the two-sphere. So it looks like the circles are linked but moving them around in R^4 they can be linked? The extra freedom of R^4 allows us to do this without the circles passing through each other? The other question is are the circles "flat" and of the same radius?

Thanks for the help!
And let me continue showing off the two things I know:

To add to Ben's answer, if you see the Wikipedia page, they give you the explicit map for the Hopf fibration and they show you that the fiber over a point is a set of complex numbers with |z|=1 ; this is a "flat circle", or what I think is called a geometric (planar, unknoted) circle, and this holds because i) S^2 is homeomorphic to complex projective 1-space, and 2) multiplication by a complex number of modulus 1 is equivalent to rotation by the argument of the number. And remember that the circles live ,in a sense, in R^4 , but they are also in S^3, which constrains the "unlinking". If the circle fibers in the Hopf fibration could be unlinked (without changing the homotopy class of the fibration), then ## \pi_3(S^2) = \mathbb Z ## would be trivial (it has a single generator). I wish I had a better answer than that. Maybe if I knew 3 things.....

Last edited:
lavinia
Gold Member
Suppose I sit in R^4 with a single 2-sphere, x^2+y^2+z^2=1, and an infinite number of circles of radius one that are free to move in R^4. Is it true that one can arrange these infinite number of circles so that their union is the three-sphere S^3, defined by x^2+y^2+z^2+w^2=1 and each circle is disjoint from all other circles but intersects the 2-sphere at one point? Is this a Hopf fibration?

Thanks for any help!
Your idea is not correct, but it does suggest a way to think of the Hopf fibration.

For this, It will be easier to think of R^4 as C^2, the space of all pairs of complex numbers. These are all points
$$(re^{iθ},se^{iα})$$
In this notation, the unit 3 sphere is all points with r^2 + s^2 =1
If you take the 2 sphere to be the points on the 3 sphere with w coordinate equal to zero then in this notation it is the points on the 3 sphere where α is a multiple of π. That is: the imaginary part of the second complex coordinate is zero.

The circles in the Hopf fibration are the orbits of points under multiplication by the complex numbers of norm 1. So the circle passing through the point ,
$$(re^{ix},se^{iy})$$
on the 3 sphere, is all points $$(re^{ix+θ},se^{iy+θ})$$ where θ varies from 0 to 2π.

Here are some things to check.
- If s ≠ 0 then a circle has exactly two points with real second coordinate and thus intersects the 2 sphere in exactly 2 points. These two points are opposite poles on the two sphere.
- If s = 0, the the circle lies entirely on the 2 sphere and is, in fact, a great circle.
- Every circle intersects this 2 sphere. There are no circles that are disjoint from it.

To summarize: each circle in the Hopf fibration intersects the two sphere in opposite poles except for one circle that lies entirely on the 2 sphere and it is a great circle.

With any fibration, E -> B, the base space ,B, is the quotient space of E obtained by identifying each fiber to a single point. For the Hopf fibration, one identifies each circle to a single point. The quotient space,B, is a topological 2 sphere.

Since every circle has non-empty intersection with the 2 sphere, the quotient space is also equal to the projection of the 2 sphere onto the base B. This projection identifies the opposite poles - since they lie on the same circle - and then crushes the remaining great circle to a single point. So the base,B is a 2 sphere with the opposite poles in the northern and southern hemispheres glued together and the equator crushed to a point.
This is again a 2 sphere. Opposite hemispheres are pasted together to get a single half sphere and then its bounding circle is identified to a point to again get a 2 sphere.

BTW:

- One can also see the linking of the fiber circles from this picture. Each circle that intersects the 2 sphere in two opposite poles is linked to the equator. The linking number is one because it passes through each hemisphere exactly once.

There is nothing special about this particular equator. Given any two great circles on the 3 sphere one of them can be considered as the equator of an embedded 2 sphere.

- Since the Hopf fibration describes the 3 sphere as the total space of a circle bundle over the 2 sphere, one can ask what it looks like over the northern and southern hemispheres. and over the equator. Over each of these three pieces of the 2 sphere, the bundle is trivial. Over the equator it is therefore the Cartesian product of two circles, that is, it is a torus. Over the two hemispheres, it is the Cartesian product of a half sphere with a circle and this is a solid torus. One can see this by observing that a half sphere is really a disk (just flatten it out) and then one has a disk of circles. If you cut a solid torus transversally, the slice is a disk and a circle perpendicular to it passes through each of its points. So a solid torus is a disk of circles.

So one has two solid tori, one above each hemisphere and they are glued together along their boundaries, the torus above the equator. So the Hopf fibration reveals that the 3 sphere is two solid tori glued together along their boundaries.

One may ask, what are all of the possible manifolds that can be obtained from gluing two solid tori along their boundaries. There certainly are others, for instance the tangent circle bundle to the 2 sphere. I am not sure of the answer.

Last edited: