Timeline for What functor does Grassmannian represent?
Current License: CC BY-SA 2.5
9 events
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| Nov 27, 2009 at 14:50 | comment | added | Andrea Ferretti | This is exactly what I meant (both comments). In any case if you think that adding the remark that restriction that the subbundle has constant rank clarifies the matters, then it is good to keep it. | |
| Nov 26, 2009 at 16:37 | comment | added | Steven Sam | Andrea: If I am interpreting your comment correctly, are you thinking of a vector bundle as the fibers of a locally free sheaf? If so, then what you mentioned is fine. In my comment I was implicitly thinking of vector bundles and locally free sheaves as the same thing (since the question was tagged with "algebraic geometry" I thought this seemed reasonable). | |
| Nov 25, 2009 at 22:43 | comment | added | Kevin H. Lin | Andrea: So in summary, are you saying that what I wrote is correct, and that I can remove the statements that I had added in the square brackets? And you are saying that if my statements had involved locally free sheaves rather than vector bundles, then I would have to keep the statements in the square brackets? | |
| Nov 25, 2009 at 10:18 | comment | added | Andrea Ferretti | This is not true. There may exist a an injection of locally free sheaves such that the quotient is not torsion free. Still a sub-vector-bundle is contained in the vector bundle fiberwise, so the quotient is always a vector bundle. Said otherwise let E and F be vector bundles. One can have an injective morphism of sheaves between (the sheaf sections of) E and F without E being a sub-vector-bundle of F. This is because a morphism of sheaves is injective iff it is injective on stalks, not on fibers. | |
| Nov 24, 2009 at 22:09 | history | edited | Kevin H. Lin | CC BY-SA 2.5 |
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| Nov 24, 2009 at 21:44 | comment | added | Steven Sam | Kevin: if we're working in the algebraic category, this isn't right. You need that the quotient of the subbundle is torsionfree. So it is better to say that the functor represents quotient bundles (because the kernel is automatically locally free). | |
| Nov 24, 2009 at 21:39 | history | edited | Kevin H. Lin | CC BY-SA 2.5 |
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| Nov 24, 2009 at 21:31 | history | edited | Kevin H. Lin | CC BY-SA 2.5 |
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| Nov 24, 2009 at 21:25 | history | answered | Kevin H. Lin | CC BY-SA 2.5 |