# If $X$ is a smooth complex projective variety, can the arithmetic genus be computed from its topological cohomology?

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?

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

• Hi, just a curious question : do you have a reference for these kind of computations ? – user171326 Jan 3 '17 at 8:48
• 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 3 '17 at 9:29
• Awesome, thanks a lot for all these precisions ! – user171326 Jan 3 '17 at 9:32
• I'm glad to help, if you want more details let me know @N.H. – user347489 Jan 3 '17 at 9:36
• Thanks, I will surely do (and I will surely have questions in the future !) – user171326 Jan 3 '17 at 9:37