Relationship between (algebraic) vector bundle and (topological) vector bundle on $\mathbb{RP}^{1}$($\mathbb{P}^{1}$)

Question

We know the one dimensional real projective space $\mathbb{RP}^{1}$ is isomorphic $S^{1}$, so there are only two line bundles (up to isomorphic) on it: the cyclinder and Mobius band. However the Picard group of one dimensional projective space (in algebraic geometry) $\mathbb{P}^{1}$ is $\mathbb{Z}$, the line bundle or locally free sheaves on it are $\mathscr{O}(m), m\in \mathbb{Z}$. I am wondering if there are relationships between the line bundle on $\mathbb{RP}^{1}$ and $\mathbb{P}^{1}$? If there are have relationships, why there are only two line bundles on $\mathbb{RP}^{1}$, but infinitely many on $\mathbb{P}^{1}$?

Answer 1

The reason a topological line bundle has order 2 (under tensor product) is because we can pick a metric on the bundle that allows us to identify the bundle with its dual. Then since a one dimensional vector space tensor its dual is naturally $\mathbb{R}$ we can identify this vector bundle with the trivial one. However, in algebraic geometry this does not work since picking a metric is very arbitrary and will not allow you to make this identification while staying in the realm of varieties. Since we have so much more room to maneuver in topology, often we can end up with less isomorphism classes of objects then in other settings.

You can see a similar result when considering real vector bundles versus complex vector bundles. It isn't true that a complex line bundle has order 2 because we can only identify a line bundle with its conjugate dual. The extra action of $i$ causes there to be more isomorphism classes of bundles (at least for projective space).