Let $X$ be a smooth and proper complex analytic space, or a Kahler complex manifold. Is the image $\Lambda$ of the finitely generated abelian group $H^{2i}(X,\mathbf{Z})$ into $J := H^{2i}(X,\mathbf{C})/F^iH^{2i}(X,\mathbf{C})$ a full lattice?, where $F^{\bullet}$ is the Hodge filtration. In other words, is $J/\Lambda$ compact?



Full lattice images and Hodge decomposition
No, except in some exceptional cases. Take for instance $i=1$. Then the kernel of the composition $H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathbb{C})\rightarrow J$ is the Neron-Severi group $NS(X)\subset H^2(X,\mathbb{Z})$; its dimension is the Picard number $\rho $, and $\operatorname{rk}\Lambda=b\_2-\rho $. 
If $\Lambda $ is a lattice in $J$, we have $\operatorname{rk}\Lambda=2\dim J=2h^{2,0}$, which implies $\rho =h^{1,1}$; in other words, $X$ has maximal Picard number. For K3 surfaces, for instance, $\rho $ can take all values from 1 to $h^{1,1}=20$; the general case is $\rho =1$, while the K3 with $\rho =20$ form a discrete locus inside the moduli space.


