How to Draw Complex Line Bundles

Question

I am giving a presentation soon on the Classification of Complex Line Bundles and I would like to have some very "basic" visualizations to use as examples.

Background and Context

I am considering the Cech-cohomology of a principal $ \mathbb{C}^{*} $ bundle, where my sheaf $\underline{\mathbb{C}}_M^{*}$ is the sheaf of smooth $\mathbb{C}^{*}$ valued functions on the manifold $M$. Using the exponential sequence of sheaves $$ 0 \to \mathbb{Z}(1) \to \underline{\mathbb{C}}_M \to \underline{\mathbb{C}}_M^{*} \to 0$$ we get an isomorphism (via properties of cohomology and the connecting homomorphism) $$H^1(M, \underline{\mathbb{C}}_M^{*}) \cong H^2(M, \mathbb{Z}(1)) $$

It turns out that $H^1(M, \underline{\mathbb{C}}_M^{*}) $ is also isomorphic to the group of isomorphism classes of principal-$\mathbb{C}^{*}$ bundles over $M$. Since the principal- $\mathbb{C}^{*}$ bundles are in one-to-one correspondence with the complex line bundles, it should be evident how this all relates to my title.

My Questions

(1) Given the above information, and some knowledge of cohomology, there should be only a trivial principal- $\mathbb{C}^{*}$ bundle on the circle $S^1$. How can we see this visually?

*See my example/analogue below.

(2) Similarly, how can we visualize a non-trivial principal- $\mathbb{C}^{*}$ bundle on the standard 2-dimensional torus?

*Example/Analogue:

So consider a circle bundle on $S^1$, then we can consider a section of the bundle like so:

alt text http://www.cheynemiller.com/Math/Figures_files/SectionOnU.png

Now, given two sections on adjacent trivializations,

alt text http://www.cheynemiller.com/Math/Figures_files/TransitionUaUb.png

We can imagine deforming one section into another, to get our transition functions. Now, I can also believe that any such family of sections can be deformed into a global section, so again I want to know why this necessarily doesn't work on the Hopf bundle via pictures.

Answer 1

To see that the Hopf bundle is not trivial, one considers its restrictions to the subspaces $N,S\subset S^2 = \mathbf{CP}^1$, with $N = \mathbb{C}^*\cup\{\infty\}$ and $S = \mathbb{C}$ where it is trivial. One should be able to write down sections given this description. Then the transition function $\mathbb{C}^* \to S^1$ can be written down. You can think of this as a map $\mathbb{C}^* \to \mathbb{C}^\ast$, and hence calculate the integral of it around $S^1 \subset \mathbb{C}^*$. This gives you the index of the transition function, which is non-zero.

You can then calculate the index of any transition function for the Hopf bundle, using the fact that it will be a Cech cocycle equivalent to the one you've written down.

Lastly, you can calculate the index of the transition function for the trivial $S^1$-bundle on $S^2$, and find this is not equal to that for the Hopf bundle.

In fancier language, because $\mathbb{C}^*$ is not simply connected, one finds that you cannot deform the Hopf bundle's transition function, which is not null-homotopic, to the transition function for the trivial bundle, which is null-homotopic.

In pictures, one has that the transition function for the Hopf bundle, restricted to the circle, loops once around the origin, but the trivial bundle's transition function is constant. If people are happy with believing that continuously deforming the transition function gives equivalent bundles (one could motivate this by writing down Cech coboundaries on $\mathbb{CP}^1$ that give the equivalence), and that transition functions which cannot be deformed to each other give inequivalent bundles (this is the important point), then this is pretty much the best picture you'll get.

Answer 2

To answer (1), the idea would be to try and construct a trivialization "by hand". For principal bundles, that means a section. Think of a circle as an interval with the endpoints identified. Constructing the section on an interval is easy (path-lifting property of a bundle) but then you have to check that you can ensure the endpoints meet up. This follows from the fact that you've chosen your bundle to be one for a connected group.

To answer (2) it might make sense to get the basic idea of why bundles are classified by cohomology classes. The idea is very simple and beautiful and comes from what's known as Whitney's Embedding Theorem in manifold theory. This is the theorem that says that every manifold can be realized as a submanifold of euclidean space. Moreover, if the euclidean space has large-enough dimension, your embedding is unique up to isotopy. The classification of vector bundles is highly analogous -- every vector bundle is the pull-back of a map to a classifying bundle. What this means is that up to a fibrewise isomorphism, you can think of all the fibers of your vector bundle as sitting as vector subspaces of some very large Euclidean space. Provided this Euclidean space is large enough, this gives a bijective correspondence between bundles over your space and homotopy-classes of maps from the space to the appropriate Grassmann manifold. In the case of complex line bundles (what you're interested in), the corresponding Grassmann manifold is a $K(\mathbb Z,2)$-space, and such maps correspond to elements of $H^2$ of your space, this is a classical obstruction-theoretic argument of Serre's.

The complex line bundles over a torus are just the trivial line bundle over a torus connect-sum with the line bundle over $S^2$ having euler class $1$ (a punctured $\mathbb CP^2$). I suppose you could view this as the normal bundle of an embedded torus in a connect-sum of $\mathbb CP^2$'s.