How to visualise $\mathcal{O}(1)$ on $\mathbb{P}^1$

Question

I'm always told that the geometry behind going from $\mathcal{O}$ to $\mathcal{O}(1)$ is like going from the cylinder to the Mobius strip. $\mathcal{O}$ here is the structure sheaf on $\mathbb{P}^1=\text{Proj} \mathbb{C}[x,y]$, and $\mathcal{O}(1)$ is obtained from it by the standard twisting construction.

However, I can see how the Mobius strip is obtained from the cylinder by cutting it open and gluing it back together with a twist. However, I have no idea how to take the trivial line bundle (as a "geometric object") on the sphere (Let's think about $\mathbb{P}^1$ as the Riemann sphere), cut it open, and twist it? Has someone ever attempted to make some sort of visualisation of how this would work?

Answer 1

For the record, here is is an expansion of my comment and a summary of the answers to the question I linked to.

First, let's see how $\mathbb{P}^2 \setminus \{x\}$ is a line bundle on $\mathbb{P}^1$. For concreteness, give $\mathbb{P}^2$ coordinates and take $x = [0:0:1]$. The $\mathbb{P}^2 \setminus \{x\} \to \mathbb{P}^1$ is a line bundle, where the fiber above $[z_0,z_1] \in \mathbb{P}^1$ is the line $\{[z_0:z_1:\lambda] : \lambda \in \mathbb{A}^1\}$.

To see that this line bundle is $\mathcal{O}(1)$, consider the global section $$[z_0:z_1] \mapsto [z_0:z_1:z_0].$$ It has exactly one zero at $[0:1]$, so the degree of this bundle is $1$, hence it is $\mathcal{O}(1)$.

A more geometric way to describe this is as follows. Remove a point $x$ from $\mathbb{P}^2$ and consider a line $L \subset \mathbb{P}^2$ disjoint from $x$. Then projection away from the point $x$ onto $L$ is a map $\mathbb{P}^2 \setminus \{x\} \to L \cong \mathbb{P}^1$ which realizes the line bundle $\mathcal{O}(1)$.

Answer 2

The stuff about "twisting" has to do with the euler class. Over $\mathbb{C}$, you can replace your line bundle with an oriented $R^2$ bundle. Then you can pose the question of finding consistent polar coordinates - if there is no twisting this is doable (because the nontwisted case is just $P^1 \times \mathbb{R}^2$, so you can pull back the coordinates). The osbtruction to twisting can be thought of as twisting of the polar angle, and it can be measured by a class in $H^2(\mathbb{P^1}, \mathbb{Z}) \cong \mathbb{Z}$.

There are a few ways to build this class -- you can use the general cohomological machinery described on the wikipedia page, and it is also possible to construct a differential form. I think the differential form gives a better geometric feeling -- the form is built by trivializing your bundle on some cover, putting polar coordinates on each trivialization, and then studying discrepancies of the angle on the overlaps. The result of this construction is a 2-form that will pullback to give the differential of the angle form (the 1 form on the total space that restricts to $d \theta$ on each fiber) along the map $E \to X$, where $E$ is the total space.

I learned this from this essay (section 1.2), where there is more detail and explicit computation: http://w3.impa.br/~massaren/files/CCCAG.pdf

On $\mathbb{CP}^1$, by Poincare duality $H^2$ is isomorphic to $H_0$, and from $H_0$ we have a degree map. Following these maps, $O(n)$ gets sent to $n$, so $O$ is not twisted, but $O(1)$ is twisted in some way. $O(2)$ is twisted by twice as much as $O(1)$, $O(-1)$ twists in the "opposite" direction.

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Another (more concrete) approach to this feeling of twistedness are the transition maps for the trivializations using the $S^2 - N$ and $S^2 - S$ (remove north and south poles, respectively) cover of $\mathbb{CP}^2$. Every line bundle on $P^1$ will trivialize on this cover. Try to write down the transition map and think about what that map looks like on $\mathbb{C}^*$. This is related to the "clutching" construction in algebraic topology.

For varieties where the trivializations are not so easy, the euler class still works for intuition.

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Is this visualization? It's more of a justification of a feeling of "twistedness." However, I think that is also "geometric."

Answer 3

As the saying goes, comparison is no reason. Do not take seriously the Möbius strip comparison. It is better to understand what a fiber bundle is: a projection map which looks like a product; then to see the example of$$\mathbb{C}^n - \{0\} \to \mathbb{P}^{n - 1},$$which gives a fiber bundle with fiber $\mathbb{C}^\times$; then add a $0$ on each fiber in order to get a line bundle which is $\mathcal{O}(-1)$. Etc.