The fact that Chern classes are Hodge classes (and are rational combinations of algebraic cycles) is a part of the proof of the "Gauss Bonnet theorem" (as given in Griffiths and Harris). So my question is why is the fundamental class of every algebraic variety a rational combination of them?
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$\begingroup$ I am sorry (I phrased the question wrongly). I edited my question. I mean why is the Hodge conjecture equivalent to "Rational cohomology is generated by Chern classes of holomorphic vector bundles". $\endgroup$– VamsiNov 16, 2010 at 1:35
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$\begingroup$ Yes, the last statement is equivalent to the Hodge conjecture. Sorry if my joke seemed at your expense. It's interesting that you arrived at the conjecture in this way. You might still want to edit the first sentence. It sounds almost like you are asserting something which isn't true. Non-Hodge classes won't be spanned by Chern classes. $\endgroup$– Donu ArapuraNov 16, 2010 at 2:05
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$\begingroup$ I unfortunately don't have time to write a complete answer to your edited last question, hopefully someone else will. The key point is that the cycle map from the Chow group $CH(X)\otimes Q\to H^(X)$ can be identified with the Chern character of $K(X)\otimes Q$. $\endgroup$– Donu ArapuraNov 16, 2010 at 2:28
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$\begingroup$ Your question looks fine now, I removed my initial comments. $\endgroup$– Donu ArapuraNov 16, 2010 at 12:14
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I am a novice, but here is what I gather from perusing the literature. The Hodge conjecture asserts that all Hodge classes are spanned by algebraic classes. The fact that all algebraic classes are spanned by the chern classes of (holomorphic or algebraic) vector bundles, is proved by resolving the structure sheaf of a subvariety by a finite complex of such vector bundles. I.e. one defines formal "K groups" generated either by isomorphism classes of locally free algebraic sheaves (vector bundles), or more generally by coherent algebraic sheaves, with an equivalence relation defined by formal alternating sums of sheaves occurring in exact sequences. The fundamental result that all coherent algebraic sheaves have resolutions by locally free algebraic sheaves shows these two K groups are isomorphic. Hence the subgroup of algebraic cohomology classes, which is the image of the chern character map on the K group of algebraic coherent sheaves, equals the image of this map on locally free sheaves as well. Since the chern character is generated by chern classes, the result follows. The 1974 book Topics in algebraic and analytic geometry, Princeton mathematical notes #13, by Phillip Griffiths and John Adams discusses this in detail.
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$\begingroup$ Thanks for the reply. But, why is it that the group of algebraic cohomology classes is the image of the Chern character map of coherent sheaves? Anyway, I will try to look it up. Thanks for the reference. $\endgroup$– VamsiNov 16, 2010 at 5:11
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$\begingroup$ I'm a little over my head here, but apparently all algebraic subvarieties define coherent algebraic sheaves. Then the distinction between algebraic classes and classes of vector bundles is that those algebraic subvarieties occurring as zeroes of sections of vector bundles apparently define locally free coherent sheaves. So showing that all algebraic classes are generated by classes of vector bundles, means showing that the sheaves associated to arbitrary subvarieties can be expressed in terms of the locally free sheaves associated to subvarieties cut out by sections of vector bundles. $\endgroup$ Nov 16, 2010 at 6:36
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2$\begingroup$ Vamsi: The ref that Roy gives is a good one and it probably answers all your questions. Another, is Grothendieck, Theorie de classes de Chern. On p 151 eq. (16) you'll see a formula for the cycle class of a subvariety in terms of the Chern classes of its structure sheaf. $\endgroup$ Nov 16, 2010 at 12:11
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$\begingroup$ The argument that the dth homogeneous part of the chern character of the structure sheaf of a subvariety of codimension d, corresponds to the cohomology class of the subvariety, is sketched on page 187 of Griffiths and Adams. I.e. as you realized better than I, it is not tautological. $\endgroup$ Nov 16, 2010 at 15:48