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In Voinsin's book [1], Theorem 11.32 (page 280) says:

"If X is an algebraic variety, these subgroups of $Hdg^{2k}(X) coincide."

However, the proof did not show that the subgroup generated by cycle classes (denoted by $A$) is containded in the subgroup generated by Chern classes of vector bundles (denoted by $B$) when $k>1$. In fact it just claims that $B\subseteq A$.

There are two questions:

(1) How to show $A\subseteq B$?

(2) Why is the condition that $X$ is an algebraic variety necessary?

[1] C. Voisin, Hodge theory and complex algebraic geometry, Vol. I, Cambrige Univ Press, 2002

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    $\begingroup$ If $X$ is a smooth projective variety, then for any subvariety $V$ of codimension $r$, we can resolve $O_V$ by a finite complex of vector bundles $E_.$. But then $c_r(E_.)=[V]$ as an algebraic cycle. This kind of thing is discussed extensively in Fulton's book on intersection theory. I forget when this works in the non-projective case. In the projective case, the construction of this resolution is definitely very algebraic, and unlikely to carry over to complex manifolds. $\endgroup$
    – Minhyong Kim
    Commented Jul 12, 2011 at 11:07
  • $\begingroup$ Given the standard comparison isomorphism between $K$-theory and $K'$-theory, I guess the argument above must work on regular schemes (say Noetherian). $\endgroup$
    – Minhyong Kim
    Commented Jul 12, 2011 at 13:44

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To expand slightly on Minhyong's comment, the key facts can be found in Fulton's Intersection Theory. If you look at the comment following corollary 18.3.2, you'll see an isomorphism (in slightly different notation) $$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$ where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence, and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly the statement you want.

Note that this is false if

  1. you omit the $\mathbb{Q}$, or
  2. if you work on a general (compact) complex manifold because there may not be enough vector bundles or subvarieties. To be clear, I mean that conclusion in cohomology is false: In Zucker, "Hodge conjecture for cubic fourfolds" Compositio 1977, you can find an example of a torus with a nonzero integral $(1,1)$ class which is not a divisor, but it would necessarily lie in the image of $c_1$.
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To complete Donu's answer, the Hodge classes and the group of classes generated by Chern classes of coherent sheaves can be very different even in the Kähler world. Claire Voisin gave an example of a 4-dimensional torus $X$ (hence Kähler) with:

(1) $\mathrm{Hdg}^4(X,\mathbb{Q})\neq0$

(2) $c_2(\mathcal{F})=0$ for any coherent sheave $\mathcal{F}$ on $X$.

The corresponding article is "A counterexample to the Hodge conjecture extended to Kähler varieties" (in IMRN n.20, 2002).

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