# Visualizing Hopf fibration $S^3\to S^2$ as a based map $S^1\to \mathrm{Map}(S^2,S^2)$

A fiber bundle $F\to E\to B$ may be interpreted as $E$ being a bunch of $F$s arranged in the shape of a $B$. For instance, a Mobius band $M$ is a bunch of line segments $[0,1]$ arranged in the shape of a circle $S^1$, so we may write $[0,1]\to M\to S^1$.

The Hopf fibration $S^1\to S^3\to S^2$, then, says that $S^3$ may be interpreted as a bunch of circles $S^1$ arranged in the shape of a sphere $S^2$. Via stereographic projection, $S^3$ may be interpreted as the one-point compactification of $\mathbb{R}^3$, or in other words $\mathbb{R}^3$ with a "point at infinity."

(Artist depiction by Lun-Yi Tsai from here.)

This can be visualized. A vertical line with the point at infninity is a generalized circle, which corresponds to the north pole of the base space $S^2$, while a specific circle around that line corresponds to the south pole of $S^2$. Everywhere in between is a torus which is a circle's worth of circles, just as in between the north and south pole of $S^2$ there are circles' worths of points.

However, another type of visualization should be possible.

Recall smashing pointed spheres is like adding whole numbers; the property $S^n\wedge S^m\simeq S^{n+m}$ implies that homotopy classes of pointed spheres form a free monoid on one generator with respect to the smash product operation. Moreover, there is an adjunction $$\mathrm{Map}(S^n\wedge X,Y)\simeq \mathrm{Map}(X,\Omega^nY),$$ where $\mathrm{Map}(X,Y)$ is the space of basepoint-preserving continuous maps $X\to Y$ equipped with the compact open topology and $\Omega^k=\Omega\cdots\Omega$ where $\Omega Y$ is the loop space $\mathrm{Map}(S^1,Y)$.

Therefore, we should be able to interpret the Hopf fibration $S^3\to S^2$ as a basepoint-preserving continuous map $S^1\to\mathrm{Map}(S^2,S^2)$, which should be mentally visible as a looped animation of a sphere immersed into another sphere.

Question. How do we visualize this?

One idea is to decompose $\mathbb{R}^3$ into (generalized) spheres all tangent to the $xy$-plane at the origin, which is essentially a based map $S^1\to\mathrm{Map}(S^2,S^3)$, and then project down onto $S^2$. Perhaps this can be done by somebody mentally by appropriately intersecting these spheres with the tori of the artist depiction above, but I cannot do that in my head to see the effect in the base space $S^2$.

Here is an analogy to illustrate what I mean some more. The Hopf fibration may be interpreted as a representative of the generator of $\pi_3(S^2)$. Well, $\pi_2(S^2)$ also has a generator, the identity map. It may be interpreted as a based map $S^1\to\mathrm{Map}(S^1,S^2)$ as follows: crumple a rubber band to a single point on the sphere, then wrap the rubber band all the way around the sphere over one side and back down the other until it gets crumpled again down to that single point.

More precisely, if we decompose $\mathbb{R}^2$ into (generalized) circles tangent to the $x$-axis at the origin and then stereographically project up to $S^2$, the result is the same as if we took a tangent plane at the sphere's basepoint, rotated it around a tangent line and intersected with the sphere to get smaller then bigger then smaller again circles.

Presumably something analogous can be done with the Hopf fibration and a "crumpled up balloon" instead of a rubber band.

• It might be more convenient to try to find a (geometrically defined) Z-valued invariant of $\pi_1 \text{Map}(S^2, S^2)$. Once you've done this, it's a matter of fiddling until you find the one of invariant one. – user98602 Jun 8 '17 at 16:48
• The oriented diffeomorphism group of $S^2$ retracts onto $SO(3)$ (reference mathoverflow.net/questions/141794/… ) and $\pi_1(SO(3))$ is cyclic of order $2$. Does a noncontractible loop in $\pi_1(SO(3))$ do the job? Leaving this as a comment rather than an answer, because I'm not sure about the distinction between $\mathrm{Diff}(S^2)$ and $\mathrm{Maps}(S^2, S^2)$. – David E Speyer Feb 10 at 16:05
• To be more precise, degree $1$ diffeomorphisms $S^2 \to S^2$ embeds into degree $1$ maps $S^2 \to S^2$ , but I don't know if this embedding is a homotopy equivalence. – David E Speyer Feb 10 at 16:13
• @DavidESpeyer I am treating my spheres as based spaces, so they have distinguished points, and ${\rm Map}(S^2,S^2)$ is a based space with the constant map $S^2\to\ast$ as its distinguished point. All of the other maps this is homotopic to should likewise then have degree $0$, unlike the elements of ${\rm Diff}(S^2)$ and ${\rm SO}(3)$. I gave a rubber band analogy to illustrate the nontrivial loop in ${\rm Map}(S^1,S^2)$, but can't visualize how it should be generalized. – arctic tern Feb 17 at 16:00