Timeline for What is the relation between vector bundles on a manifold and grassmanians?
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| Jul 20, 2010 at 18:43 | comment | added | Tom Goodwillie | Oh, I see. For me, the point about maps $X\to \mathbb G(k,n)$ is that they correspond to vector bundles on $X$ expressed as sub (or quotient) bundles of trivial bundles. (I guess I'm thinking like a topologist here.) Your map expresses $L^*$ as a quotient of a trivial rank three bundle, so I never thought of considering it as being associating it with $L\oplus L^*$. | |
| Jul 20, 2010 at 16:38 | comment | added | Francesco Polizzi | @Tom Take $L= \mathcal{O}_{\mathbb{P}^1}(-2)$. Then $L \oplus L^*$ has three global sections, and we have a map $\mathcal{O}^3 \to L \oplus L^*$. Of course this map is not surjective, since the first summand in the right has no sections. However, we still have a map $\mathbb{P}^1 \to \mathbb{G}(1,3)$; the point is that this map is $degenerate$, in fact its image is a smooth conic contained in a $\mathbb{P}^2 \subset \mathbb{G}(1,3)$. Of course this $\mathbb{P}^2$ corresponds to the three sections of $L^*$. | |
| Jul 20, 2010 at 16:00 | comment | added | Tom Goodwillie | I'm still a little mixed up. My $L$ does not have enough sections. Therefore, I think, neither my $L\otimes L^*$ nor its dual has enough sections. (The only reason why I added $L$ to its dual was because I did not know whether I wanted to make a bundle that does not have enough sections or whether I wanted to make a bundle whose dual does not have enough sections.) But of course we agree that you don't get a map to the Grassmannian unless something has enough sections. | |
| Jul 20, 2010 at 14:50 | comment | added | Francesco Polizzi | In your example $L^*$ is ample, since $L$ must have negative degree over $CP^1$. Anyway, your example works taking an elliptic curve $E$ instead of $CP^1$, and a degree $0$ line bundle $L$ on $E$ with no non-zero sections. The point is that in this case $L \oplus L^*$ is not a quotient of $V^* \otimes \mathcal{O}_E$, (I'm adopting the Grothendieck convention here, I hope not to make confusion with the duals), so there is actually no map to the Grassmannian. I was a bit sloppy, in fact I should have said "the answer is yes provided that you have enough sections". Thank you for the remark. | |
| Jul 20, 2010 at 14:31 | comment | added | Tom Goodwillie | Let $L$ be a homolomorphic line bundle on $\mathbb CP^1$ admitting no global section except $0$. Then $L\oplus L^*$ does not give a morphism to a Grassmannian manifold in any obvious sense, does it? I mean, a bundle associated with such a morphism should be such that either it or its dual is spanned by global sections, depending on your conventions. (These are the ampleness considerations mentioned in the other thread.) | |
| Jul 20, 2010 at 10:52 | comment | added | Francesco Polizzi | I edited the answer, thank you. So, the answer to your new question is yes. In fact, for all schemes S the morphisms f : S \to Grass(V,r) are in 1-1 correspondence with the locally free rank r quotients V^* \otimes \mathcal{O}_S \to \mathcal{F} via f <-> f^*Q, where Q is the universal quotient bundle on the Grassmannian. This is the so-called universal property of the Grassmannian. I think that this is the dual construction of the one proposed by Bugs Bunny in his comment. | |
| Jul 20, 2010 at 10:20 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Jul 20, 2010 at 10:15 | comment | added | mihail | I think it must be vector space instead of vector field. my question : does a vector bundle on compact complex manifold give a morphism to a grassmanian manifold | |
| Jul 20, 2010 at 10:05 | history | answered | Francesco Polizzi | CC BY-SA 2.5 |