# How can I visualize principal bundles?

I'm trying to learn how to think about principal bundles where the fibre is a lie group with local trivialization $ϕ^{-1}_i:π(U_i)→U_i×G$ . For example $ϕ^{-1}_i:π(S^2)→S^2×U(1)$ (if that makes sense) . But I don't know how to think of this (and other products with lie groups like that) geometrically in 3D space. How does $S^2×U(1)$ look like? I know that U(1) rotates things, but I can't visualize a thing that rotates other things. What does it mean if I put U(1) at every point on $S^2$? It get's even worse if I want to think about a connection on a principal bundle which is defined on the tangent space of the lie group. I guess there is a lot of higher dimensional stuff going on, but isn't there an easy 3D example which captures every concept at once?

• There is a unique non-trivial principal $O(1)$-bundle over $S^1$, namely the unique connected double cover of $S^1$. This is easy enough to visualise. Oct 13, 2014 at 11:27
• @ZhenLin What does this have to do with the question...? Oct 13, 2014 at 12:58
• @OP Do you know that $U(1) \cong S^1$ (as Lie groups), where $S^1$ is the circle, aka the complex numbers of modulus $1$? Oct 13, 2014 at 12:59
• @NajibIdrissi It is the easiest non-trivial example of a principal $G$-bundle that can be embedded in $\mathbb{R}^3$. Oct 13, 2014 at 16:35

There's no doubt you will have to learn to stretch your geometric intuition quite a bit in order to understand principal bundles. The intuition is built up in little steps, each with their own way of visualization, and there might not ever be one big picture in which to visualize everything, except for the very simplest examples.

Regarding principal $U(1)$ bundles, let me start with a few more-or-less obvious things. $U(1)$ is the circle group. Algebraically, $U(1)$ can be represented in a few ways: as unit circle $S^1$ in the complex plane, with the group operation being multiplication of unit complex numbers $z_1,z_2$ producing another unit cmoplex number $z_1 \cdot z_2$; or as angles, i.e. as elements of the quotient group $\mathbb{R}/2\pi \mathbb{Z}$, and the group operation on two angles $\theta_1,\theta_2$ produces an angle $\theta_1+\theta_2$. These two algebraic descriptions are related by the universal covering homomorphism $\mathbb{R}\to S^1$ given by $\theta \mapsto e^{2\pi i \theta}$.

You say you cannot visualize a thing that rotates other things, but perhaps you can visualize that $U(1)$ rotates any oriented round circle $C$: an element of $U(1)$ represented by an angle $\theta$ rotates the circle $C$ through the angle $\theta$.

Here's an example of a principal $U(1)$ bundle you can visualize entirely in $\mathbb{R}^3$: the bi-infinite annulus $\mathbb{R} \times S^1$, with projection map $\mathbb{R} \times S^1 \to \mathbb{R}$ defined by $(x,z) \mapsto x$. It can be embedded into $\mathbb{R}^3$ by the map $(x,z) \mapsto (x,\text{Re}(z),\text{Im(z)}$.

Regarding the principal bundle $U(1) \times S^2$, it cannot be globally embedded into $\mathbb{R}^3$, so you're back to building your intuition up in little steps. In this case, remember that the universal cover of the circle $U(1)=S^1$ is $\mathbb{R}$, and so the universal cover of $U(1) \times S^2$ is $\mathbb{R} \times S^2$. We can think of $\mathbb{R} \times S^2$ as a principal $\mathbb{R}$ bundle over $S^2$, and there is quite a nice embedding $\mathbb{R} \times S^2 \mapsto \mathbb{R}^3$ defined by mapping $(x,p) \in \mathbb{R} \times S^2$ to the scalar multiple $e^x p$. If you fix $p$ and let $x$ vary the image is the "deleted ray" through $p$ based at the origin, but with the base point at the origin deleted.

The final step of visualizing $U(1) \times S^2$ requires you to use your knowledge of quotient topologies: from the universal covering map $\mathbb{R} \to U(1)$ we obtain a universal covering map $\mathbb{R} \times S^2 \to U(1) \times S^2$. You can think of this quotient another way, using the fact that $U(1)$ is the quotient of the interval $[0,2\pi]$ where the points $0$ and $2\pi$ are identified. Thus $U(1) \times S^2$ can be visualized as the quotient of all the points in $\mathbb{R}^3$ between the inner sphere of radius $e^0=1$ and the outer sphere radius $e^{2\pi} \approx 535.49$, then identifying each point $p$ on the inner sphere with the corresponding point $e^{2\pi} p$ on the outer sphere.

• Hi Lee, thx for reply. Do you mean this : mathworld.wolfram.com/images/gifs/circbuns.gif ? Oct 13, 2014 at 20:30
• @JonnyPython: that looks like a different way of viewing things than my answer. Oct 13, 2014 at 22:20