Timeline for The correspondence between affine vector bundles and f.g. projective modules
Current License: CC BY-SA 2.5
26 events
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| Nov 9, 2010 at 18:30 | comment | added | roger123 | Ok, then I do understand, finally and we have produced the most comments to a question :-). | |
| Nov 9, 2010 at 17:53 | comment | added | t3suji | Thank you, makes better sense now. The usage of `linear' is very confusing, by the way. For the second point: but $\phi_p$ is not part of the data in topological picture! One should require a structure of a vector space on the fiber, but you don't fix an isomorphism with the standard vector space. If you want to imitate this, you should require an isomorphism with the symmetric algebra of a vector space, not with the polynomial algebra (which is the symmetric algebra of the standard vector space). For the third point: yes, this would solve it. | |
| Nov 9, 2010 at 17:30 | comment | added | roger123 | Thank you for the comment, t3suji. With 'linear' I mean that every $a\in k$ is mapped to $a$ and $X_i$ is mapped to $\sum a_{i,j} X_j$. I want $\phi_p$ to be part of the data, like in the topological definition. The third point is very helpful, thank you! It is 'solved' if I replace each $k(p)$ by $R_p$, and demand the automorphism on $R_p[X_1,...,X_n]$ induced by the restriction to be linear, right? Again, thank you for the very helpful comments. | |
| Nov 9, 2010 at 17:08 | comment | added | t3suji |
First of all, if you use linear' in the sense compatible with grading', please say so. (E.g.: is the map $k[x]\to k[x]:p(x)\mapsto p(x+1)$ linear in your sense?) Secondly, why would you want $\phi_p$ to be part of the data? Are you trying to define the notion of `vector bundle with fixed basis in every fiber'? Thirdly, you do not impose any compatibility between isomorphisms on different sets $D(a_i)$, only between these isomorphisms and isomorphisms on fibers. This is going to be trouble if A has nilpotents.
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| Nov 9, 2010 at 11:19 | vote | accept | roger123 | ||
| Nov 8, 2010 at 22:49 | comment | added | roger123 | I still do not understand. Since the isomorphism of the fiber $A_{k(p)}$ to the polynomial ring $k(p)[X_1,...,X_n]$ belongs to the data, $A_{k(p)}$ gets the grading from this isomorphism. Then everything should follow from the linearity of the restriction isomorphism, right? | |
| Nov 8, 2010 at 14:06 | comment | added | t3suji | @roger123: Same problem remains:As Willberd van der Kallen says, you need a grading on A and isomorphisms must respect it. | |
| Nov 8, 2010 at 12:46 | comment | added | roger123 | Oh, I beg your pardon! I forgot the "linear" in the definition above and made an edit. Now, everything should be ok, right? | |
| Nov 8, 2010 at 12:44 | history | edited | roger123 | CC BY-SA 2.5 |
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| Nov 7, 2010 at 14:28 | comment | added | Wilberd van der Kallen | Yes, the isomorphisms should all be graded. Notice also that the answer to your problem involves taking the degree one part of each algebra and then using the patching data to patch these degree one parts as modules. That way you get a projective module, possibly dual to the one you were after. | |
| Nov 5, 2010 at 20:46 | comment | added | roger123 | Thank you very much for the correction! I am trying to understand what you mean and I need a little more time to do that. Maybe, you could help me with this: Have I only to assume that $A$ is graded or should any of the maps in the definition also be a graded? | |
| Nov 5, 2010 at 15:33 | comment | added | Wilberd van der Kallen | It is hidden in the italicized word `linear'. For polynomial rings an automorphism of algebras respects the grading exactly when it is given by a linear map in the degree one part. When thinking about a projective module $P$ over a base ring $R$, the algebra of functions is the symmetric algebra over $R$ on the dual module, but as a graded algebra, so as not to forget the $R$-module structure on $P$. Your gluing maps should respect the module structure on $P$. For instance, try how you would describe a translation by one on the affine line over a field. You get a map that is not graded. | |
| Nov 5, 2010 at 14:01 | comment | added | roger123 | Sorry but I did not get your point. The above definition is the affine translation of the scheme theoretic one of Hartshorne. Where is the grading hidden in Hartshorne's definition? | |
| Nov 5, 2010 at 8:40 | comment | added | Wilberd van der Kallen | No, I am not taking Proj. You want scalar multiplication on your bundle or what? | |
| Nov 4, 2010 at 6:45 | comment | added | roger123 | Why should $A$ be graded? (vgl. Hartshorne Ex.5.18.) Do you perhaps mix this up with projective bundles? | |
| Nov 3, 2010 at 15:05 | comment | added | Wilberd van der Kallen | I am missing the grading on $A$ in the above. One needs a grading to capture the multiplication with scalars. Recall that a grading amounts to having $\mathbb G_m$ act. | |
| Nov 3, 2010 at 6:15 | comment | added | Harry Gindi | This is also discussed in EGA 2 in the section on the relative proj construction. | |
| Nov 3, 2010 at 6:02 | answer | added | Sándor Kovács | timeline score: 2 | |
| Nov 2, 2010 at 23:08 | answer | added | Greg Muller | timeline score: 4 | |
| Nov 2, 2010 at 21:57 | comment | added | roger123 | Ok, I understand that. But why does the above correspond to a locally free module of finite rank? | |
| Nov 2, 2010 at 21:15 | comment | added | Martin Brandenburg | It is a standard commutative algebra fact that a module is locally free of finite rank if and only if it is finitely generated and projective. You can find it in every good textbook (and of course, in Bourbaki). | |
| Nov 2, 2010 at 19:10 | answer | added | Andrew Parker | timeline score: 1 | |
| Nov 2, 2010 at 18:51 | comment | added | roger123 | Yes, it is a special case. Unfortunately I do not understand the general case and what happens there with the symmetric algebra. This is why I would like to see how this works in the affine case. | |
| Nov 2, 2010 at 18:47 | comment | added | Mike Skirvin | Isn't this just a special case of the correspondence between vector bundles and locally free $\mathcal{O}_X$-modules? In which case, take your vector bundle and look at its sheaf of global sections (which will just be an $R$-module in your affine setting). | |
| Nov 2, 2010 at 18:44 | history | edited | roger123 | CC BY-SA 2.5 |
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| Nov 2, 2010 at 18:10 | history | asked | roger123 | CC BY-SA 2.5 |