What is the Geometric Interpretation of Principal Bundles with Lie Group Fibres?

[JonnyMaddox]

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?

 

[homeomorphic]

I don't really try to picture most of those things directly. You can easily visualize a thing that rotates other things in two dimensions because all you need to know is the angle of rotation. What do you get if you take all possible angles? A circle. My mental image for S^2 cross U(1) is just S^2, with a movable point on it that has a circle glued onto it. Since it's a 3-manifold, it should be visualizable that way. Take S^2 cross an interval, which is a ball with the middle hollowed out. And then glue the inside of the ball to the outside. You can think of it sort of like a pacman square, so that if you go outside the ball, it will wrap around and you'll come out on the other end. That would be one way to think of it. But those sorts of pictures can be pretty specific to the nature of the bundle you are dealing with.

In general, I just picture the fiber as a sort of abstract blob, the same way I think of any generic topological space. So, it's generally just a blob over every point in the base space, where all the blobs over different points are supposed to be the "same".

I'm too lazy to get into connections right now, but we did a thread on that a little while ago.

 

[JonnyMaddox]

Hi homeomorphic, thanks for reply. You mean you visualize $S^2 \times U(1)$ like this (base $S^2$ fiber $U(1)$)

But that looks absurd. Does that mean I can visualize the rotation group $SO(3)$ in $R^{3}$ just as a sphere which rotates??
Then ok take again take $S^2 \times U(1)$, now what is the tangent space of the $U(1)$ group? Is it just the tangent vectors on this circle on the sphere? I also found this picture of the torus and $U(1)$.

Are there any other similar examples similar to these ones?

 

[homeomorphic]

Something like that, except you have to keep in find that each circle is a fiber over one point in the sphere, but that kind of picture doesn't capture the twisting of the bundle if it's not a product. Also, I tend to visualize one fiber at a time and then to get the whole bundle, I can vary the base point however I want, so that I only see a little slice of it at a time, rather than the whole thing. Trying to visualize the whole thing is like trying to visualize a labyrinth all at once. Too complicated. You'd be better off visualizing one room at a time and if you want to visualize the whole thing, just walk through the labyrinth and picture only one room at a time. The analogy isn't perfect, of course.

SO(3) is more complicated. To specify a 3-d rotation, you have to have the angle of rotation, as well as an axis. If you just have a 2-sphere, that only gives you the axis. The full space is homeomorphic to RP^3, which you'd have to understand as a 3-manifold in order to picture. One description is to take a ball and identify antipodal points. But you might not be so concerned with this sort of thing, if you just wanted to deal with an SO(3) bundle. I'd just picture an SO(3) bundle as the base space with a little blob over each point, with each blob being homeomorphic to SO(3).

 

[lavinia]

- A torus is a trivial circle bundle over the circle. The fiber circles can be rotated to give an action of SO(2).

- The 2 sphere is a Z/2Z bundle over the projective plane. The fibers are pairs of antipodal points. The action of Z/2Z is to map each fiber point to its antipode.

This is an example of a principal bundle with discrete structure group.
Another example is the torus as a Z/2Z bundle over the Klein bottle.

- Look up visualizations of the Hopf fibration. This is another circle bundle over the sphere but the total space is not SO(3). It is the 3 sphere.

What is the action of U(1) on the 3 sphere for the Hopf fibration?

- If you slice the 2 sphere along the equator, it splits into two topological disks. The tangent circle bundle splits into two solid tori. Their common boundaries are the torus at the equator. So the tangent circle bundle is two solid tori glued together along their boundaries.

See if you can figure out what the gluing map is.

- Here is an example of a connection on a principal bundle that can be visualized.

The bundle is the torus as a U(1) bundle over the circle. Take any smooth section of this bundle. This is a smooth closed curve in the torus that intersects each fiber in a single point. At each intersection point define the line through the tangent vector to the curve to be horizontal. Rotate this line along the fiber circle to get the horizontal space at every fiber point. This is a connection.

The same construction works for a curve that intersects the fibers in more than one point as long as the tangent directions are parallel.

 

[homeomorphic]

Your problem might not be one of visualization, but also of motivation. I would recommend becoming acquainted with vector bundles before principal bundles. You could try reading the intro to this:
http://www.math.cornell.edu/~hatcher/VBKT/VBpage.html

He discusses the tangent bundle of a sphere there, in a nice geometrical way, although I think normal bundles are a little more natural in terms of motivation because they come up naturally if you want to describe a little neighborhood of a lower-dimensional submanifold, sitting in a higher-dimensional manifold. The tubular neighborhood is diffeomorphic to the normal bundle.

As Hatcher shows in that book he is working on, you can then arrive at the general concept of a fiber bundle by various constructions, like considering the unit sphere in each fiber of a vector bundle, or an orthogonal frame in each fiber. Frame bundles are the canonical example of a principal bundle.

 

[homeomorphic]

Another way to picture an SO(3) bundle is to picture what a point in the bundle looks like. It's an orthonormal frame (i.e. a basis for the fiber of some vector bundle). So, if you want an SO(3) bundle over S^2, you can think of it as orthonormal frames parametrized by S^2. A sphere's worth of frames, possibly distinct from the product space.

 

[homeomorphic]

A sphere's worth of frames, possibly distinct from the product space.

I think this got a little bit messed up in translation from my mental image. It's a spheres worth of SO(3)s, so there are lots of frames at each point, not just one, which is what I made it sound like.

 

[JonnyMaddox]

Hi homeomorphic, lavinia, thank you for your answers, I have to think about them a bit. And @homeomorphic: Thx for the link. I already own hatcher's algebraic topology book, I'll read the chapters on both books. I reply back when I have a question again.