Let $Y$ be a "nice" scheme. I am thinking projective varieties over an algebraically closed field, for now, but I am open to more general results.

In terms of singular homology (coefficients in $\mathbb{Z}$), one can define the Euler characteristic $\chi(Y)$.

My question is:

Can I express $\chi(Y)$ in terms of the Euler characteristic of certain coherent sheaves on $Y$, in terms of sheaf cohomology?

Most preferably, I would like $$\chi(Y)=\chi(Y,\mathcal{F})$$ for some particular sheaf $\mathcal{F}$.

I am sorry if this is really trivial or widely known, my searching and asking (in the real world) has led me nowhere so far.

  • 4
    $\begingroup$ Jesko -- if you are willing to accept complexes of sheaves rather than just sheaves, then in characteristic 0 one may take the sum over all smooth strata of the Euler characteristics of the complexes of differential forms. $\endgroup$
    – algori
    Jul 14, 2011 at 8:42
  • 3
    $\begingroup$ Well, in a smooth flat family the fibres are diffeomorphic by Ehresmann theorem, so the topological Euler number is constant. But if the family is not smooth, it can definitely vary (think of a smooth plane cubic degenerating to a nodal one). $\endgroup$ Jul 14, 2011 at 9:22
  • 2
    $\begingroup$ Over $\mathbb{C}$ one can use hodge theory to write $H^k(X,\mathbb{C})=\oplus_{p+q=k}H^q(X,\Omega_X^p)$, where $\Omega_X^p$ is the sheaf of $p$-forms on $X$. $\endgroup$ Jul 14, 2011 at 12:25
  • 1
    $\begingroup$ Of course, you would want the choice of $\mathcal{F}$ to be natural in some sense, and also you would want $Y$ to be proper so that the euler characteristic is finite. And following algori comment, $\F$ may as well be a virtual sheaf (i.e. a formal linear combination). When your variety is smooth and projective, Daniel's suggestion of $\sum \pm \Omega^p_X$ is the most reasonable. Other choices are possible in general, but I won't go into it, unless I have a clearer idea of what you need it for. $\endgroup$ Jul 14, 2011 at 13:23
  • 1
    $\begingroup$ Also the top Chern class of the tangent sheaf gives the topological Euler characteristic (of a smooth, projective, complex variety). $\endgroup$ Jul 14, 2011 at 15:14

1 Answer 1


First, the answer/reference here might be exactly what you are looking for.

On the other hand, perhaps you just want to relate natural algebro-geometric structure to the classical Euler characteristic. Here is one way to do that:

A pure Hodge structure of weight $k$ is a finite dimensional complex vector space $V$ such that $V=\bigoplus_{k=p+q} H^{p,q}$ where $H^{q,p}=\overline{H^{p,q}}$. This gives rise to a descending filtration $F^{p}=\bigoplus_{s\ge p}H^{s,k-s}$. Define $\mathrm{Gr}^{p}_{F}(V)=F^{p}/ F^{p+1}=H^{p,k-p}$.

A mixed Hodge structure is a finite dimensional complex vector space $V$ with a real ascending weight filtration $\cdots \subset W_{k-1}\subset W_k \subset \cdots \subset V$ and a descending Hodge filtration $F$ such that $F$ induces a pure Hodge structure of weight $k$ on each $\mathrm{Gr}^{W}_{k}(V)=W_{k}/W_{k-1}$. Then define $H^{p,q}= \mathrm{Gr}^{p}_{F}\mathrm{Gr}^{W}_{p+q}(V)$ and $h^{p,q}(V) =\dim H^{p,q}$.

Let $Z$ be any quasi-projective algebraic variety. The cohomology groups with compact support $H^k_c(Z)$ are endowed with mixed Hodge structures by seminal work of Pierre Deligne.

The Hodge numbers of $Z$ are $h^{k,p,q}_{c}(Z)= h^{p,q}(H_{c}^k(Z))$, and the $E$-polynomial is defined as $$ E(Z; u,v)=\sum _{p,q,k} (-1)^{k}h^{k,p,q}_{c}(Z) u^{p}v^{q}. $$

From this, one gets the classical Euler characteristic $\chi(Z)=E(Z;1,1)$.

Note: if the counting function of $Z$ over finite fields is a polynomial in the order of the finite field, then $E(Z)$ is exactly the counting polynomial. From this point-of-view, in this case, the Euler characteristic is the number of $\mathbb{F}_1$-points.

  • 1
    $\begingroup$ Indeed, the latter is what I want(ed)! Thanks for answering this old question. $\endgroup$ Apr 30, 2016 at 7:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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