# The Hopf Fibration

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The Hopf fibration is a mapping from the three-dimensional sphere, $$S^3$$, to the two-dimensional sphere, $$S^2$$, such that the pre-image of each point in $$S^2$$ is a circle. In the visualization above, the sphere in the lower left displays the selected point of $$S^2$$. In the central canvas, you see the pre-image of that point, that is all the points that map to the selected point under the Hopf fibration.

The word fibration refers to maps from one space to a parameter space, where the pre-image of points in the parameter space correspond to fibers in the parametrized space, and we can take a neighborhood around each point for which the pre-image is a simple Cartesian product of a fiber with the neighborhood. In this case the parameter space is $$S^2$$, the parametrized space is $$S^3$$, and the fibers are all circles ($$S^1$$).

You can use the parameter controls to select all of the different points of $$S^2$$ and view the corresponding fibers in $$S^3$$. You'll notice that the fiber corresponding to the south pole is the unit circle in the $$xy$$-plane, and the fiber corresponding to the north pole is an infinite vertical line coinciding with the $$z$$-axis.

The formula for the Hopf fibration is given by $(2 x_1 x_3 + 2 x_2 x_4,\;2 x_2 x_3 - 2 x_1 x_4,\; x_1^2 + x_2^2 - x_3^2 - x_4^2)$

## Wait, why does the formula depend on four coordinates?

In the visualization above, it appears as though the Hopf fibration has the standard three-dimensional space as its domain, so how is it possible to have four coordinates in its formula?

Up until now, I've left out a very important detail. I said that the Hopf fibration maps the three-dimensional sphere to the two-dimensional sphere. We can see the two-dimensional sphere very clearly in the bottom-left corner. However, the domain just looks like a standard three-dimensional space extending infinitely in all directions. There's nothing particularly spherical looking about it.

To resolve this confusion, let's think more carefully about what the space $$S^3$$ actually is. By looking at the two-dimensional sphere $$S^2$$ in the corner, we can see that it consists of all points that are distance one away from the origin in three-dimensional space. By analogy, that means that $$S^3$$ consists of all points that are distance one away from the origin in four-dimensional space.

Unfortunately, we can't directly visualize this because we only perceive three spatial dimensions. However, we do have a pretty good workaround, which moves this object in four-dimensional space into the three dimensions that we are familiar with.

In Discovering Stereographic Projection, we saw how, with the exception of a single point, we can identify $$S^2$$ with the infinite plane. To do so, points in $$S^2$$ are mapped to the plane via lines emanating from the north pole.

On $$S^3$$, we can't display the lines emanating from the "top" of the sphere, but we can still perform this same procedure algebraically, projecting lines from the point $$(0,0,0,1)$$ through the sphere and onto a point $$(x,y,z,0)$$, where the fourth coordinate is 0. We use the three coordinates $$x,y,z$$, to visualize the Hopf fibration above, essentially replacing $$S^3$$ with standard three-dimensional space. We just need to keep in mind that there is a single point at infinity that we have left out.

This translation from $$S^3$$ to three-dimensional space explains why the fiber for the north pole is a line, and not a circle. In its proper home, $$S^3$$, the fiber is actually a circle through the point $$(0,0,0,1)$$

## Fiber formulas

In the visualizations, the parameter controls actually select point(s) on $$S^2$$, so to create the visualization, it's actually more useful to have the formulas for the fiber of a given point on $$S^2$$. For a point on $$S^2$$ with spherical coordinates $$\theta, \phi_1$$, the formula for the fiber in $$S^3$$ is:

\begin{aligned} x_{1} &=& \cos \left({\frac {\phi _{1}+\phi _{2}}{2}}\right)\sin \frac{\theta}{2} \\ x_{2} &=& \sin \left({\frac {\phi _{1}+\phi _{2}}{2}}\right)\sin \frac{\theta}{2} \\ x_{3} &=& \cos \left({\frac {\phi _{2}-\phi _{1}}{2}}\right)\cos \frac{\theta}{2} \\ x_{4} &=& \sin \left({\frac {\phi _{2}-\phi _{1}}{2}}\right)\cos \frac{\theta}{2} \end{aligned}

where $$\phi_2$$ ranges from $$0$$ to $$4\pi$$ as one moves along the fiber in $$S^3$$.

## Tori within $$S^3$$

In the visualization below, I've visualized the family of fibers corresponding to a circle of points in $$S^2$$. I also turned on camera controls, so you can drag and pinch the canvas to view the fibers from different angles.

If you look at the fibers for circles parallel to the equator, you can see how the space is decomposed into a family of tori. At the south pole, the tori degenerates into the unit circle along the $$xy$$ plane. As the circle of points moves up toward the north pole, the inner radius of the torus moves from 1 to 0, the outer radius of the torus moves from 1 to infinity and the height of the torus goes from 0 to infinity. For any given value of $$\theta$$ and $$\phi$$, as $$r$$ goes from -1 to 1, an infinite family of tori partition the entire space, and each point in the space belongs to exactly one of the tori.

If you look at a circle of points at the equator and then vary $$\theta$$ between 0 and $$\pi$$, you can watch as the torus flips itself inside out!

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The Hopf fibration wraps $$S^3$$ around $$S^2$$ in a non-trivial way. It can't be continuously molded and deformed to be a constant map. In some sense, the Hopf fibration is fundamental among such non-trivial maps, in the sense that all non-trivial maps can be deformed to either be the Hopf fibration itself, or the Hopf fibration composed with itself an integer number of times.