why isn't the mobius band an algebraic line bundle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:29:35Z http://mathoverflow.net/feeds/question/22950 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22950/why-isnt-the-mobius-band-an-algebraic-line-bundle why isn't the mobius band an algebraic line bundle? Max Flander 2010-04-29T07:18:42Z 2010-04-29T07:58:17Z <p>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 &#8477;, and what changes when you move to analytic line bundles?</p> http://mathoverflow.net/questions/22950/why-isnt-the-mobius-band-an-algebraic-line-bundle/22954#22954 Answer by Qfwfq for why isn't the mobius band an algebraic line bundle? Qfwfq 2010-04-29T07:47:02Z 2010-04-29T07:58:17Z <p>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".</p> <p>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.</p>