MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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. – Minhyong Kim Jul 12 '11 at 11:07
Given the standard comparison isomorphism between $K$-theory and $K'$-theory, I guess the argument above must work on regular schemes (say Noetherian). – Minhyong Kim Jul 12 '11 at 13:44
up vote 10 down vote accepted

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 $c_1$.
share|cite|improve this answer

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.