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When I hear the phrase "line bundle" the first thing that pops into my head is a mobius band. But this is a bad picture from an algebraic point of view since any line bundle on an affine variety is trivial. Anyway, my question is: is there a way of seeing more concretely what "goes wrong" when you try to construct the mobius band as an algebraic line bundle over ℝ, and what changes when you move to analytic line bundles?

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Consider the real algebraic line bundle $\mathcal{O}(-1)$ over the real algebraic variety $\mathbb{R}\mathbb{P}^1$. It is nontrivial hence continuously isomorphic to the "Moebius" line bundle (there are only 2 line bundles on the circle, up to continuous isomorphism), so its total space is homeomorphic o the "Moebius strip".

By the way, it is false that line bundles on affine algebraic varieties are trivial. One example is the above universal line bundle over $\mathbb{R}\mathbb{P}^1$. If you want an example over $\mathbb{C}$, consider the complement of a point in an elliptic curve: it's an affine variety but its Picard group is far from being trivial.

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thanks for that, the thing about line bundles being trivial was told to me today by a topology professor, i guess she was confused. –  Max Flander Apr 29 '10 at 8:05
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Your comment about $\mathbb{R} \mathbb{P}^1$ is correct, but may be confusing to some. One of the interesting features of real algebraic geometry is that $\mathbb{R} \mathbb{P}^n$ is (also) the real analytic manifold associated to a smooth affine variety. So there is, in this sense, no distinction between affine and projective varieties over $\mathbb{R}$. –  Pete L. Clark Apr 29 '10 at 8:07
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@maxmoo: There is a related true result here: a coherent sheaf on an affine variety $X$ is acyclic for sheaf cohomology: its higher cohomology groups are all zero. One might be tempted to deduce from this that $\operatorname{Pic}(X) = H^1(X,\mathcal{O}_X^{\times}) = 0$. But this is not valid: $\mathcal{O}_X^{\times}$ is not a coherent sheaf. –  Pete L. Clark Apr 29 '10 at 8:11
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@Pete: yes, O^* is not a coherent sheaf simply because it is not a sheaf of O_X-modules, but just a sheaf of abelian groups. –  Qfwfq Apr 29 '10 at 8:16
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Just to expand a little on Pete's point: $\mathbb{RP}^1$ is isomorphic to $x^2+y^2=1$. Over the complex numbers, the former is a sphere and the latter is a sphere with two punctures, but the punctures are complex conjugates of each other, so you don't see them in the real picture. I think you should be able to write the Mobius band explicitly as $\{(x,y,u,v) : \ x^2+y^2=1 \ (u^2-v^2)y=2uv x \}$. In other words, it is the set of pairs of complex numbers $(z,w)$ such that $z$ has norm $1$ and $\mathrm{arg}(z)=2 \mathrm{arg}(w)$. –  David Speyer Apr 29 '10 at 10:39
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