# Geometric perspective to see that the Hopf-fibration is non-trivial.

I'm holding a seminar in geometry on the Hopf-fibration $$S^3\rightarrow S^2$$ using Quaternions/Rotations of $$3$$-Space to introduce the map. I've found a nice geometric way of presenting everything, except for the fact that the Hopf-fibration is not trivial.

I'd like to present that too geometrically (under stereographic projection). For instance, in the case of the Mobius strip, you can see that it is non-trivial by holding it next to a cylinder and noting that you need to tear it in at least one place to deform it into the said cylinder with one map, but you don't need to tear it if you allow two maps.

Following that analogy, I think the key is properly understanding the implications of every Hopf fiber being linked with the unit circle in the plane, but I can't visualize this properly.

Is there a nice way to see that we need at least two trivializing maps geometrically?

EDIT: It is common to refer to fiber bundles as twisted products, is there some visible "twisting" here?

• If your audience is familiar with fundamental groups, the following answer by @Tsemo Aristide is the simplest that I can find so far. math.stackexchange.com/a/1793990/481770 Commented Jun 24, 2021 at 17:18
• @TeebroProkash sadly they are not, it sure would make life easier! Perhaps you could help with my comment under Andrew's comment though? Commented Jun 24, 2021 at 17:20

Thinking of pairs of complex numbers rather than quaternions, let's write $$(z_{1}, z_{2}) = (x_{1} + iy_{1}, x_{2} + iy_{2})$$. One approach is to consider the hemisphere $$|z_{1}|^{2} + (\operatorname{Re} z_{2})^{2} = x_{1}^{2} + y_{1}^{2} + x_{2}^{2} = 1,\quad y_{2} = 0,\quad x_{2} \geq 0.$$ If $$z_{2} = |z_{2}|e^{i\theta}$$ is non-zero, then the Hopf circle through a point $$(z_{1}, z_{2})$$ of the $$3$$-sphere hits this hemisphere at $$(e^{-i\theta}z_{1}, |z_{2}|)$$. This shows the Hopf fibration is trivial on the open hemisphere. On the other hand, the boundary $$\{z_{2} = 0\}$$ of the hemisphere is a Hopf circle, so the action by scalar multiplication does not extend trivially to the boundary, but instead collapses the boundary to a point.
Another approach is to split the $$3$$-sphere into two solid tori by the Clifford torus $$|z_{1}| = |z_{2}| = \frac{1}{2}$$, and show that each is trivial as a circle bundle over a disk but the attaching map on the boundary glues latitudes of one solid torus to meridians of the other, so the union is not a trivial bundle.
• Yes, this is absolutely how I visualize this, too. For the first argument, please note that the Hopf fibration is just $(z_1,z_2)\mapsto [z_1:z_2]$, thereafter identifying $\mathbb{C}P^1$ with $S^2$. I feel like you see this already. Commented Jun 24, 2021 at 17:02
• I have been playing around with my visualization to see the effect you describe. You can see the Hopf-circles intersecting the hemisphere of $S^2$. I don't understand why this makes it locally trivial though, as the Hopf-map of the great circles is clearly not this intersection in general. @TeebroProkash Commented Jun 24, 2021 at 17:19
• FWIW, if $C$ denotes the boundary circle of the closed hemisphere, the fact that every Hopf circle except $C$ intersects the open hemisphere $H^+$ in one point tells us there is a mapping from $H^+ \times S^1$ into the $3$-sphere whose image is the complement of $C$, and the action is trivial on the first factor, so the restriction of the Hopf bundle to the complement of $C$ is trivial. Commented Jun 24, 2021 at 20:12