# Chern classes generating cohomology

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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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". –  Vamsi Nov 16 '10 at 1:35
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. –  Donu Arapura Nov 16 '10 at 2:05
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 $K(X)\otimes Q$. –  Donu Arapura Nov 16 '10 at 2:28
Your question looks fine now, I removed my initial comments. –  Donu Arapura Nov 16 '10 at 12:14