Let $X$ be a smooth projective complex algebraic variety. The Hodge decomposition tells us that $H^n(X, \mathbf C) = \oplus H^{p,q}$.


Here is my question:



> 
> For what kind of $X$ is $H^{2n}(X) = H^{n,n}$ for all $n$ (I don't
>  care about the odd cohomology groups $H^{2n+1}$)? Has this condition been studied in the literature?
> 
> 
> 


Or a slightly more relaxed question: given some $X$, can one construct a related variety $X'$ so that the property holds for $X'$? Or, maybe, give bounds on $dim H^{2n} - dim H^{n,n}$ etc.


Thank you!



When is the Hodge diamond concentrated in $H^{n,n}$'s?
A more restrictive condition, which probably has been studied, is to look at varieties with Hodge level $\le 1$. This means that $h^{pq}=0$ unless $|p-q|\le 1$. If you assume the general Hodge conjecture, then this is equivalent to the motive being a summand of $\bigoplus C\_i(n\_i)$ for some smooth projective curves $C\_i$. If just want examples, they are easy enough to find. Take your favourite variety with cohomology generated by algebraic cycles (e.g. a toric variety, a flag variety…) and then optionally take the product with a curve.


I am a bit skeptical about the last two questions. If $X$ is a degree $d$ surface, then $\dim H^2 - h^{11}$ grows with $d$. I am not sure what you would take for $X'$ in such a case.


