Assume that $M$ is a closed connected Ricci-flat Kaehler manifold $M$ of complex dimension $n\geq 3$ with $h^{2,0}(M)=0$. Is is possible that


* $h^{n, 0}(M)\neq 1$
* $h^{p, 0}(M)\neq 0$ for some $0< p< n$?



Hodge numbers of compact Ricci-flat Kaehler manifold
After replacing by a finite covering space, you must have $h^{n0}=1$, because you have a parallel volume form, so nowhere zero, and any other holomorphic volume form is a holomorphic multiple, so a constant multiple (as holomorphic functions are constant).


Kobayashi, *First Chern class and holomorphic tensor fields*, 1980, proved that any holomorphic tensor on a compact Ricci-flat Kaehler manifold is parallel. Hence it induces a reduction of holonomy group. The holonomy group is a product of groups arising as irreducible holonomy groups, i.e. copies of the special unitary group $SU(n\_i)$ and the compact symplectic group $Sp(n\_j)$. But since you have $h^{20}=0$, you have no $Sp(n\_j)$ factors, so a product of $SU(n\_i)$ factors, i.e. locally a product of Calabi-Yau and flat Kaeher metrics. Unless there is a flat factor in the product, you don't get any parallel tensors except for the holomorphic $n\_j$-forms. By a theorem of Beauville (J.D.G. 1983), after a finite covering, you have a product of simply connected Calabi Yaus with irreducible $SU(n\_i)$ holonomy, $n\_i\ge 3$, and a flat complex torus. But to have $h^{20}=0$, that torus is an elliptic curve.


As Martin de Borbon points out, there are examples of closed complex manifolds with Ricci flat Kaehler metrics with torsion canonical bundle, and then you will get $h^{n0}=0$, but after taking a finite covering space you get $h^{n0}=1$.


