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