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is there a relation between vector bundles on a manifold and plücker embeddings

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Yes, there is. Could you be more specific about what kind of relations you had in mind? –  Ben Webster Jul 20 '10 at 9:52
    
I mean is it related to plücker embedding of our manifold to 'some' projective space –  mihail Jul 20 '10 at 10:03
    
Mihail, your statement is a bit unclear: Plucker embedding is from Grassmannian into a projective space, not a random manifold into a projective space. Do you really mean to use (suitable) vector bundles to make maps to a Grassmannian (pulling back its universal quotient bundle to the given $V$), and then consider how the Plucker embedding composes back to the original manifold? –  BCnrd Jul 20 '10 at 14:53
    
yes this is what I mean. thanks BCnrd! –  mihail Jul 20 '10 at 19:47
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Mihail: ok, but what's the actual question? So when the bundle is generated by finitely many global sections you get the map to a Grassmannian expressing that setup on the manifold as a pullback of the universal one over the Grassmannian, and also then to a projective space by the Plucker map, and then what? Do you want a criterion for when the latter is a closed embedding, expressed in terms of the original sections of the bundle, or something else...?? –  BCnrd Jul 20 '10 at 20:31
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3 Answers

It is fun to give an answer when there is no question :-)) LOL...

Anyway, pick your vector bundle $V$ on a manifold $X$ and consider its top exterior power $\Lambda^m V$. Now you have Plucker map from $X$ to the projective space $P(\Gamma (X,\Lambda^m V)^\ast)$ of the dual space of the global sections. It is given by mapping of $p\in X$ to the restriction map of the global sections $\Gamma (X,\Lambda^m V)\rightarrow \Lambda^m V_p$. Notice that $\Lambda^m V_p$ is a one-dimensional vector space without a natural basis - different choices of basis give different functionals, so the map naturally goes to projective space.

You'd better have your global sections finite-dimensional, which you get by adding holomorphicity + compactness. If you want embedding, you have to play with ampleness, etc.

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Probably replace $V_p$ with $\wedge^m(V)_p$ in both places. But then this is all about $\wedge^m(V)$, and $V$ has sort of dropped out of the picture. So one doesn't see a Plucker map in this picture. But if $V$ is generated by $n$ global sections (no need for holomorphic structure or finite-dim. hypotheses), the quotient map $O^n \rightarrow V$ gives a map to a Grassmannian, and then composing with a Plucker embedding gives a map to a projective space, resting on the rank-1 quotient $\wedge^m(O^n) \rightarrow \wedge^m(V)$ (and then one can bring in ampleness considerations, etc.). –  BCnrd Jul 20 '10 at 13:59
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Yeah, thanks, doc! –  Bugs Bunny Jul 20 '10 at 21:00
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Thaaat's all, folks. –  BCnrd Jul 21 '10 at 0:00
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If $V$ is any finite dimensional vector space over a field $k$ and $r$ is any integer between 0 and $\dim(V)$, you can see the Grasmannian $Grass(V, r)$ as the functor $(Sch/k) \to (Sets)$ which associates to any k-scheme of finite type $S$ the set of all subsheaves $K \subset \mathcal{O}_S \otimes_k V$ such that the quotient $F:=\mathcal{O}_S \otimes_k V / K$ is locally free of rank $r$. In this way the Plucker embedding is given by

$Grass(V, r) \to \mathbb{P}(\Lambda^r V)$

$[\mathcal{O}_S \otimes_k V \to F] \to [\mathcal{O}_S \otimes_k \Lambda^rV \to \det(F)]$.

This description is slightly less intuitive than the "classical" Grasmannian (but of course it is just its relative version); on the other hand it can be naturally generalized to the case where $V$ is replaced by a coherent sheaf $\mathcal{V}$ on $S$, in particular by a vector bundle. See the book of Huybrechts and Lehn "The geometry of moduli spaces of sheaves", p. 41 for further details.

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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 –  mihail Jul 20 '10 at 10:15
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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. –  Francesco Polizzi Jul 20 '10 at 10:52
    
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.) –  Tom Goodwillie Jul 20 '10 at 14:31
    
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. –  Francesco Polizzi Jul 20 '10 at 14:50
    
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. –  Tom Goodwillie Jul 20 '10 at 16:00
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I note that it is possible to move over from the manifold point of view, to use the more refined approach of algebraic geometry. This is because the Plucker embedding is actually "algebraic".

In that setting, Grothendieck's Quot scheme would classify quotients(or, looking at the kernel, subbundles) of a free vector bundle. Actually, we use locally free sheaves instead of vector bundles. Indeed, this Quot scheme is stratified for each polynomial, classifying the quotients with a certain "Hilbert polynomial".

And then the Grassmannian is just a particular component of the Hilbert scheme, corresponding to just one particular Hilbert polynomial. The construction of this particular case is precisely via the Plucker embedding, realizing the Grassmannian as a projective variety. On the other hand, in the construction of Quot scheme I have read, it is constructed as some subscheme of the Grassmannian. In this sense, the Plucker embedding is very intertwined with the study of vector bundles.

(Here one might note that this answer is somewhat similar to Francesco Polizzi's).

Added note(prompted by BCnrd's comments below): If you are not into algebraic geometry, please forget completely about this post. If you ever get into it, then you might come back and read it again.

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Anweshi, why do you say that a Grassmannian is a component of a Hilbert scheme, when to the contrary (as you note later) Hilbert schemes with a fixed Hilbert polynomial are naturally found inside of Grassmannians? Anyway, I disagree with the advice to leave the manifold setting, since the basics of Grassmannians work equally well in all geometric settings (smooth, real-analytic, complex-analytic, rigid-analytic, algebraic, etc.), so one can study the question in whatever geometric setting is most familiar. Bringing in Hilbert schemes here seems like a long detour for no purpose. –  BCnrd Jul 20 '10 at 20:29
    
@BCnrd: When I wrote "component of the Hilbert scheme", what I had in mind was the following. If $E$ is a locally free sheaf of rank $n$ and if the polynomial $P = n -m$, then $Quot_{E, P}$ is the same as $Gr(n,m)$. About your other objections: I just wanted to enlighten the OP about the Hilbert scheme, leading him on to moduli problems. If you(or others) feel that this answer is too uninformative to be kept, please leave one more comment and I would be happy to delete it. –  Anweshi Jul 20 '10 at 21:22
    
Anweshi, it's fine; I just thought it might freak out the OP to be told that one should learn about Quot schemes etc. to understand Grassmannians (which of course isn't really necessary). –  BCnrd Jul 21 '10 at 0:03
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