Timeline for Algebraic vector bundles on affine punctured plane
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15 events
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| Nov 24 at 2:58 | comment | added | naahiv | @MahdiMajidi-Zolbanin Sorry to revive this old question, I don't understand the deduction from Corollary 4.1.1, which implies that the restriction of $M_{(X,Y)}$ to the punctured spectrum of $\mathbb{C}[X,Y]_{(X,Y)}$ is a trivial bundle. How do we know $M_{(X,Y)}$ is trivial on the entire spectrum? | |
| Feb 5, 2013 at 23:01 | comment | added | Mahdi Majidi-Zolbanin | Yes, that is correct. | |
| Feb 5, 2013 at 22:44 | comment | added | anonymous | Just to make sure I understand right: the corollary 4.1.1 is used in order to show that $\mathcal F$ has a free stalk above the origin - and this gives a trivialization in some neighbourhood of the origin. The conclusion is that $\mathcal F$ is a vector bundle on the affine plane - and as such it has to be free. Thanks again everybody for great answers and comments! | |
| Feb 5, 2013 at 22:41 | comment | added | anonymous | Thanks Mahdi Majidi-Zolbanin for the great answer! I think I finally understand now. Thanks also Hailong Dao for giving the short version of this answer already from the beginning (sorry I didn't know about the extension result from EGA I, p. 174 so that I could follow what you meant). Thanks also Martin Brandenburg for your input - I'll see if I manage to come up with a simple algebraic proof sometime soon. | |
| Feb 5, 2013 at 18:32 | comment | added | Mahdi Majidi-Zolbanin | We already know that $M$ is free when localized at any prime ideal of $\mathbb{C}[X,Y]$ other than $(X,Y)$ because $\tilde{M}$ was the extension of $\mathcal{F}^\prime$. Now localize both $M$ and $\mathbb{C}[X,Y]$ at the ideal $(X,Y)$. You get a regular two-dimensional local ring and a vector bundle $M_{(X,Y)}$ on its punctured spectrum and Corollary 4.1.1 says it's free. This shows $M$ is a vector bundle on $\mathbb{C}[X,Y]$. By Serre's Problem $M$ must be free then. | |
| Feb 5, 2013 at 18:32 | comment | added | Mahdi Majidi-Zolbanin | Even better! Suppose $\mathcal{F}^\prime$ is a vector bundle on the punctured affine plane. Then you can extend $\mathcal{F}^\prime$ to a coherent sheaf $\mathcal{F}$ on the affine plane itself (cf. EGA I, p. 174 here: archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1960__4_/…) Now since we are on the affine space, $\mathcal{F}=\tilde{M}$, where $M$ is a finitely generated module over $\mathbb{C}[X,Y]$. | |
| Feb 5, 2013 at 18:23 | comment | added | anonymous | Another approach would perhaps be to ''projectivize the base'' somehow, and use the fact that vector bundles on $\mathbb P^1$ decomposes into line bundles. Line bundles on $\mathbb A^2_*$ are trivial. Can this be used to see that there is only the trivial vector bundle on $\mathbb A^2_*$? | |
| Feb 5, 2013 at 17:42 | comment | added | anonymous | Dear Mahdi Majidi-Zolbanin, Corollary 4.1.1 states that ''If $A$ is 2-dimensional and regular then every bundle on $Y$ is free.'' Here, as far as I understand, $A$ is a local ring with maximal ideal $\mathfrak m$, $X=Spec(A)$ and $Y$ is the punctured spectrum $X\setminus\mathfrak m$. A bundle is a locally free coherent sheaf. | |
| Feb 5, 2013 at 17:31 | comment | added | Mahdi Majidi-Zolbanin | Dear Anonymous: I don't have Horrock's paper in front of me right now, but I would guess Corollary 4.1.1 states that a vector bundle on the punctured affine plane can be extended to a vector bundle on the affine plane. The triviality of a vector bundle on the affine space is called Serre's Problem, which was solved independently by Quillen and Suslin (cf. en.wikipedia.org/wiki/Quillen–Suslin_theorem). | |
| Feb 5, 2013 at 17:13 | comment | added | anonymous | I would very much appreciate some help in reading Horrocks' article: plms.oxfordjournals.org/content/s3-14/4/689.full.pdf. First of all I wonder if $\mathcal O_X$ -- ''the sheaf of local rings'' -- is just the sheaf of regular functions on $X$? I can't think of a sheaf for which $\mathcal O_X(U)$ is a local ring for every open $U$. My main question is whether Corollary 4.1.1 really can be used to conclude that every vector bundle on $\mathbb A^2_*$ is trivial? It seems that there's a condition on $A$, from the very first page, that $A$ be local. But $\mathbb C[x,y]$ is not local. | |
| Feb 5, 2013 at 14:05 | comment | added | anonymous | Thanks Martin - the reason I posted it as an answer was that I couldn't post comments after I had left the page and come back again without creating a user account. I think the cookies disappeared somehow. I'm sorry for that. Is it correct that the actual way of extending the vector bundle to $\mathbb A^2$ would be to use algebraic Harthog's lemma on 1-cocycles with values in $GL_n(\mathcal O_X)$? I.e. ''extend the transition funtions to the origin'' | |
| Feb 5, 2013 at 9:41 | comment | added | Martin Brandenburg | PS: Please read mathoverflow.net/faq, your "answer" should have been a comment. | |
| Feb 5, 2013 at 9:35 | comment | added | Martin Brandenburg | Yes this was my idea (except that I don't want to extend the vector bundle to $\mathbb{A}^2$ as in Hailong's comment (which is an answer)). For every $n$ there is a bijection between the set of isomorphism classes of rank $n$ vector bundles on $X$ and the Cech cohomology set $\check{H}^1(X,\mathrm{GL}_n(\mathcal{O}_X))$, even for an arbitrary ringed space $X$ (so this is just a fancy way of stating that vector bundles can be equivalently described by transition functions). | |
| Feb 4, 2013 at 16:30 | comment | added | anonymous | Thanks for your comment Martin! All I need is to understand the result, but of course it would be a very nice bonus if it could be understood by means of a short and elementary algebraic proof! | |
| Feb 4, 2013 at 16:23 | history | answered | anonymous | CC BY-SA 3.0 |