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
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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
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> (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?
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\*See my example/analogue below.



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



How to Draw Complex Line Bundles
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. 


