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

I had this simple question when formulating the Todd class question.

Does there exist an example of proper morphism $f:X\to Y$ together with nontrivial homology class $t\in H^*(X)$ such that for all coherent sheaves on $X$ the equality $f_*(\mathop{\text{ch}}(u)\cdot t) = 0$ holds?

share|cite|improve this question
up vote 8 down vote accepted

Why not just take a class that is orthogonal to all the algebraic classes on $X$? For instance you can take $Y$ to be a point, and take $X$ to be a generic abelian surface over $\mathbb{C}$, i.e. and abelian surface with $NS(X) = \mathbb{Z}$. The Chern character of any coherent sheaf on $X$ is contained in $H^{0}(X)\oplus NS(X)\oplus H^{4}(X)$. Take now any $t \in NS(X)^{\perp} \subset H^{2}(X)$, i.e. a transcendental cohomology class of degree two.

share|cite|improve this answer
Yes indeed, I forgot ch must be algebraic :) – Ilya Nikokoshev Jan 4 '10 at 11:06

As Tony Pantev points out, it is easy to make an example by taking non-algebraic classes.

If you impose that $t$ is algebraic, and take $X=Y$, you are very close to stating Grothendieck's Conjecture D. Let $X$ be smooth and projective. The conjecture is that any algebraic class in $H^*(X)$, which is orthogonal to all algebraic classes, is zero torsion.

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.