6

If $X$ is a smooth, projective variety over $\mathbb{C}$, can we deduce the arithmetic genus from the betti numbers? For curves this is possible, but what about higher dimensions?

Elle Najt
  • 21,422

1 Answers1

8

The answer is no. The reason is that the arithmetic genus is defined in terms of finer invariants than the Betti numbers (for cohomology with coefficients in $\mathbb{C}$), namely the Hodge numbers, which depend on the complex/algebraic structure of the space. For this reason we can find spaces whose Hodge numbers differ even when their Betti numbers don't.

The arithmetic genus is defined as: $$ p_a(X)=(-1)^n(\chi(\mathcal{O}_X)-1)=h^{n,0}-h^{n-1,0}+\cdots+(-1)^nh^{1,0}, $$ where the $h^{p,q}$ are a finer invariant than the Betti numbers known a the Hodge numbers. These are defined in terms of the Dolbeault cohomology of $X$ so that: $$ h^{p,q}:=\dim H^{p,q}(X)= \dim H^q(X,\Omega_X^p). $$ Counterexample to the claim:

The Hodge diamond of a K3 surface is given by: $$ h^{p,q}(X)=\begin{pmatrix} 1&0&1\\ 0&20&0\\ 1&0&1 \end{pmatrix} $$ so it's Betti numbers are $b_0=b_4=1$, $b_1,b_3=0$ and $b_2=22$ and its arithmetic genus is $1$.

On the other hand consider $\mathbb{P}^2$ blown-up at $21$ point, this surface has Hodge diamond $$ h^{p,q}(X)=\begin{pmatrix} 0&0&1\\ 0&22&0\\ 1&0&0 \end{pmatrix} $$ and has the same Betti numbers as a K3 surface, but its arithmetic genus is $0$.

Edit: Hodge numbers are finer since by Hodge theorem we have the following decomposition: $$ H^r(X,\mathbb{C})=\bigoplus_{p+q=r}H^{p,q}(X), $$ so that $b_r=\sum_{p+q=r}h^{p,q}$.

user347489
  • 1,920
  • Hi, just a curious question : do you have a reference for these kind of computations ? –  Jan 03 '17 at 08:48
  • 1
    Hi @N.H., the computations depend on each case. For example, by definition a K3 surface has $h^{1,0}=0$ and $h^{2,0}=1$, and this determines all the numbers except $h^{1,1}$. To find the $20$ you use Noether's formula. A more general approach involves the tangent bundle exact sequence together with Hirzebruch-Riemann-Roch (HRR); in this case you can obtain a lot of information from knowing explicitly the tangent and normal bundles of your projective variety. I learned this from Huybrecht's book 'Complex Geometry', specifically the chapter on HRR and the previous one. – user347489 Jan 03 '17 at 09:29
  • Awesome, thanks a lot for all these precisions ! –  Jan 03 '17 at 09:32
  • I'm glad to help, if you want more details let me know @N.H. – user347489 Jan 03 '17 at 09:36
  • Thanks, I will surely do (and I will surely have questions in the future !) –  Jan 03 '17 at 09:37
  • @user347489: Something is wrong with your second example. A hypersurface of bidegree $(2,0)$ in $P^1 \times P^2$ is just the union of two copies of $P^2$. – Sasha Jan 03 '17 at 13:23
  • Hi @Sasha, thanks for bringing this up. I actually thought about it before editing my answer and adding it, but somehow convinced myself that it was something else besides 2 copies of $P^2$. My reasoning was that this is a surface in $P^5$ given by the Segre embedding equations + a set of quadratic equations, from which a priori it is not clear the variety is 2 copies of $P^2$. – user347489 Jan 03 '17 at 19:43
  • What is the additional structure that determines the hodge numbers? Do you need to go all the way to the algebraic structure? – Elle Najt Jan 03 '17 at 19:47
  • @AreaMan I actually don't know exactly, since I studied this from the complex geometry point of view, where the Dolbeault cohomology is defined in terms of a differential that is directly dependent on the complex structure of the space. I believe this translates to algebraic structure by using the isomorphism $H^{p,q}(X)\cong H^q(X,\Omega_X^p)$, but I don't know for sure. This thread may shed some light over the issue: http://math.stackexchange.com/questions/1112438/intuitive-aproach-to-dolbeault-cohomology – user347489 Jan 03 '17 at 19:53
  • good answer but noob notation.. – Rüdiger Jan 06 '17 at 01:36
  • To make the answer a bit less confusing, I think one should define $p_a(X):= (-1)^n(\chi(\mathcal{O}_X)-1)=h^{0,n}-h^{0,n-1}+\cdots+(-1)^{n+1} h^{0,0} - 1$ instead. The current formula, for example, asks for $h^{n-1, 0}(X) = h^0(\Omega_X^{n-1})$ and requires that $h^{0, 0}(X) = h^0(\mathcal{O}_X) = 1$, which relies on other theorems about smooth complex projective varieties (e.g. Serre duality) and the parity of $n$. – D. Zack Garza Feb 10 '24 at 21:23