8

I realised yesterday (to my slight embarrassment) that I don't understand how the simplest Hodge numbers $h^{0,q}(X)$ and $h^{q,0}(X)$ behave when $X$ is singular. But I am sure the M.SE community can set me straight!

Let's stick to normal projective varieties over $\mathbf{C}$.

Question 1: for such a variety $X$, is the "conjugation symmetry" $h^{0,q}=h^{q,0}$ still true? The standard explanation for this (or rather the more general symmetry $h^{p,q}=h^{q,p}$) in the nonsingular case uses the complex of $(p,q)$-forms; it isn't clear to me why it should be true in the singular case, although I admit I haven't thought about it much.

Question 2: are the Hodge numbers $h^{0,q}(X)= \mathrm{dim}_\mathbf{C} \, H^q(X,\mathcal{O}_X)$ birational invariants of normal projective varieties? That is, if $X$ is normal and $Y \rightarrow X$ is a birational morphism from a nonsingular variety, is it true that $h^{0,q}(X)=h^{0,q}(Y)$? This is true for $X$ nonsingular, but the proof I know uses the symmetry from Question 1. I think there is a proof via the composite functor spectral sequence if $X$ has rational singularities (although I didn't check carefully). Again, I don't see a reason it should be true in general, but I don't know a counterexample.

  • I think I got the superscripts the right way round... –  Oct 08 '13 at 14:08
  • 1
    I just observe that in (2), if $X$ is a normal surface with $H^2(X, O_X)=0$, then $h^{0,1}(Y)=h^{0,1}(X)$ forces the singularities be rational by Leray spectral sequence. – Cantlog Oct 08 '13 at 20:01
  • 3
    I would be glad to know why my question was downvoted. –  Feb 21 '14 at 14:24

1 Answers1

1

Just a comment about rational singularities. If $f\colon Y \to X$ and $g\colon X \to Spec k$ then the cohomology of $O_Y$ is given by taking cohomology of the complex $Rg_* Rf_* O_Y$. If $f$ has rational fibers, then $Rf_* = O_Y$ so that $H^p(Y,O_Y) = H^p(Rg_* Rf_* O_Y) = H^p(Rg_* O_X) = H^p(X,O_X).$

More generally, one can use the projection formula for any vector bundle (or perfect complex) on $X$, $Rf_* Lf^*E = E$, so that $H^p(Y,f^*E) = H^p(X,E)$.

John Salvatierrez
  • 487
  • 3
  • 11
  • Dear John, thanks for that comment. Indeed, this was what I had in mind when I said above that I thought it could prove it for rational singularities. –  Oct 31 '13 at 08:44