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On the real $\textbf P^1$ we have these algebraic line bundles: $\mathscr O(1)$ and $\mathscr O(-1)$.

Which one corresponds to the Möbius strip? (Both are $1$-twists of $\textbf P^1\times\textbf A^1$, so how to distinguish them? Yes, by means of their transition functions, but how do they tell me which one has the Möbius strip as total space?) And what is the total space of the other one?

I can only imagine that non-orientability should correspond to the absence of global sections, so I would bet on $-1$, but with no real reason.

Also, I can't figure if to every $\mathscr O(d)$ there corresponds a different total space, or there are repetitions. Of course, by viewing those bundles as holomorphic bundles, there are only two surfaces up to diffeomorphism. But what about the algebraic category?

Thank you for any help!

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1 Answer 1

up vote 5 down vote accepted

Edit
This is a corrected answer. Many thanks to Ben who pointed out the idiocy of my original version.

Given an arbitrary field $k$, the Picard group of $\mathbb P^1_k$ (consisting of the isomorphism classes of algebraic line bundles) is isomorphic to $\mathbb Z$ via the degree map $$\text {deg} : \text {Pic} (\mathbb P^1_k) \stackrel {\cong }{\to} \mathbb Z ,$$ the inverse isomorphism being $$ \mathbb Z \stackrel {\cong }{\to} \text {Pic} (\mathbb P^1_k) :n\mapsto \mathscr O(n) $$ In the case $k=\mathbb R$ things become quite interesting because the real points $\mathbb P^1_\mathbb R (\mathbb R)$ of the projective line $\mathbb P^1_\mathbb R$ are endowed with the structure of a real differentiable manifold diffeomorphic to the circle $S^1$.
And that manifold has a differentiable Picard group $\text {Pic}^{\text {diff}} (\mathbb P^1(\mathbb R)) $ of order two generated by the Möbius bundle $M$, so that we have a group isomorphism $$ \text {Pic}^{\text {diff}} (\mathbb P^1(\mathbb R)) \stackrel {\cong }{\to} \mathbb Z/2\mathbb Z: M\mapsto \bar 1 $$
We then have a forgetful group homomorphism $\text {Pic} (\mathbb P^1_\mathbb R)\to \text{Pic}^{\text {diff}} (\mathbb P^1(\mathbb R)) $ forgetting the algebraic structure of a line bundle and retaining only its differentiable structure.
In the above identification this morphism is just reduction modulo $2$: $$\text {Pic} (\mathbb P^1_\mathbb R)\to \text{Pic}^{\text {diff}} (\mathbb P^1(\mathbb R))\cong \mathbb Z: \mathscr O(n) \mapsto M^n\cong \bar n$$
In particular both $\mathscr O(1)$ and $\mathscr O(-1)$ are sent to the Möbius bundle, which answers your question (I hope!)

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Dear Georges, thank you! (in fact I'm quite surprised.) –  Brenin Oct 25 '12 at 9:38
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Dear Georges - I don't think this is right! In particular, I don't believe that $\mathbb{P}_\mathbb{R}^1$ can be isomorphic to $\operatorname{Spec} \mathbb{R}[x,y]/(x^2+y^2-1)$. The latter is affine! The former doesn't have nonconstant global sections! –  Ben Blum-Smith Apr 11 '13 at 18:38
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Also, isn't it the case that $\mathcal{O}(-1)$ has no nonzero global sections while $\mathcal{O}(1)$ has a 2-dimensional space of global sections? –  Ben Blum-Smith Apr 12 '13 at 17:56
    
Dear @Ben, you are absolutely right and what makes things worse: I've known this for quite some time! I had of course intended to correct my answer but somehow forgot to do that. Thanks for reminding me: I have now modified my answer . Leaving a wrong answer for so long certainly deserves my downvote and does not dispense me from apologizing to all users: I am very sorry for my delay in correcting this answer. –  Georges Elencwajg Apr 12 '13 at 19:16
    
@GeorgesElencwajg: thank you very much for correcting your answer. I really appreciate it. –  Brenin Apr 13 '13 at 18:36

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