Timeline for why isn't the mobius band an algebraic line bundle?
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22 events
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| Apr 29, 2010 at 16:07 | comment | added | Minhyong Kim | All that you write is correct. To be frank, the analytic triviality has always seemed a bit mysterious to me as well. Regarding the triviality of $H^2$, I guess the easiest way to see this is to note that a compact Riemann surface minus some points is homotopic to a graph. | |
| Apr 29, 2010 at 15:03 | comment | added | Qfwfq | P.S.: ah, I see now why top. H^2 is trivial: a noncompact differentiable $n-$manifold doesn't have a fundamental class. | |
| Apr 29, 2010 at 14:57 | comment | added | Qfwfq | Misprint: in the above comments, of course, I actually meant $H^1(X^{an}-p,\mathcal{O}_{X-p}^{an, \times})\rightarrow H^2(X^{an}-p,\mathbb{Z})$ is an $\cong$. | |
| Apr 29, 2010 at 14:53 | comment | added | Qfwfq | (continued) Thus, if what I've written makes sense, we have shown an easy example in which there are algebraically nontrivial line bundles that are analytically trivial. So the possibility of using holomorphic functions instead of only algebraic would give more "degrees of freedom" to trivialize an alg. nontrivial bundle. Are there any mistakes in my reasoning? thank you. | |
| Apr 29, 2010 at 14:52 | comment | added | Qfwfq | (continued) MinhyongKim: could you please elaborate a bit why every affine curve would have trivial "topological $H^2$" (I suppose you mean the $H^2(X^{an},\mathbb{Z})$, right?)... Anyway, given this fact and the above isomorphism, one would conclude that the analytic Picard group of $X-p$ would be trivial. (to be continued) | |
| Apr 29, 2010 at 14:42 | comment | added | Qfwfq | (continued) On the other hand, workin analytically, we can write down the exponential sequence and, taking into account that $X-p$ is affine hence Stein, we have that the first Chern class is an isomorphism $H^1(X^{an}, \mathcal{O}_X^{an, \times})\rightarrow H^2(X^{an}, \mathbb{Z})$. (to be continued) | |
| Apr 29, 2010 at 14:35 | comment | added | Qfwfq | (continued) If every line bundle on $X-p$ was algebraically trivial, that is, if $Pic(X-p)=0$, then $Pic(X)$ would be isomorphic to $\mathbb{Z}$, which is well known not to be the case. (to be continued) | |
| Apr 29, 2010 at 14:31 | comment | added | Qfwfq | (continued) So, take $X$ an elliptic curve; then $X−p$ is an affine smooth algebraic curve ($p$ is any closed point of $X$ ). Now, applying Proposition 6.5 and Corollary 6.15 from Hartshorne to the present case, we have the exact sequence: $0\rightarrow \mathbb{Z}\rightarrow Pic(X) \rightarrow Pic(X-p) \rightarrow 0$ (to be continued) | |
| Apr 29, 2010 at 14:25 | comment | added | Qfwfq | @MinhyongKim: I'm getting very confused, as your comment seems (superficially) to contradict my example with the elliptic curve minus one point. My goal in that example was to provide an example of an affine alg curve with algebraically nontrivial line bundles on it. (to be continued) | |
| Apr 29, 2010 at 13:41 | comment | added | Minhyong Kim | Yes, an affine algebraic curve has trivial topological $H^2$, so all line bundles are topologically trivial. But then, they are also Stein spaces, so this implies that line bundles are analytically trivial. | |
| Apr 29, 2010 at 12:42 | comment | added | Qfwfq | @MinhyongKim: what do you mean by "this" in "this implies analytic triviality"? You mean that on an affine algebraic variety (e.g. an affine curve) having trivial (topological) first Chern class implies being holomorphically trivial, right? | |
| Apr 29, 2010 at 10:39 | comment | added | David E Speyer |
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 $\mathrm{arg}(z)=2 \mathrm{arg}(w)$.
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| Apr 29, 2010 at 8:56 | comment | added | Minhyong Kim | The professor may have been talking about the analytic category. What's true is that affine varieties over $C$ are Stein manifolds, so that coherent analytic sheaf cohomology vanishes as well. Then the exponential sequence for $O_X$ implies that an analytic line bundle is determined by its topological first Chern class. In the case of an affine algebraic curve, for example, this implies analytic triviality. | |
| Apr 29, 2010 at 8:16 | comment | added | Qfwfq | @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. | |
| Apr 29, 2010 at 8:14 | comment | added | Max Flander | ah that explains it | |
| Apr 29, 2010 at 8:11 | comment | added | Pete L. Clark | @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. | |
| Apr 29, 2010 at 8:10 | comment | added | Qfwfq | Affine algebraic varieties behave trivially w.r.t. cohomology of coherent sheaves, but their geometry can still be nontrivial (even from the purely topological viewpoint). | |
| Apr 29, 2010 at 8:07 | comment | added | Pete L. Clark | 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 $\mathbb{R}$. | |
| Apr 29, 2010 at 8:05 | comment | added | Max Flander | thanks for that, the thing about line bundles being trivial was told to me today by a topology professor, i guess she was confused. | |
| Apr 29, 2010 at 8:03 | vote | accept | Max Flander | ||
| Apr 29, 2010 at 7:58 | history | edited | Qfwfq | CC BY-SA 2.5 |
added 7 characters in body
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| Apr 29, 2010 at 7:47 | history | answered | Qfwfq | CC BY-SA 2.5 |