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?


1 Answer 1


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 $b_r=\sum_{p+q=r}h^{p,q}$.

  • $\begingroup$ Hi, just a curious question : do you have a reference for these kind of computations ? $\endgroup$
    – user171326
    Jan 3, 2017 at 8:48
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    $\begingroup$ 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. $\endgroup$
    – user347489
    Jan 3, 2017 at 9:29
  • $\begingroup$ Awesome, thanks a lot for all these precisions ! $\endgroup$
    – user171326
    Jan 3, 2017 at 9:32
  • $\begingroup$ I'm glad to help, if you want more details let me know @N.H. $\endgroup$
    – user347489
    Jan 3, 2017 at 9:36
  • $\begingroup$ Thanks, I will surely do (and I will surely have questions in the future !) $\endgroup$
    – user171326
    Jan 3, 2017 at 9:37

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