I was going through this, where on page 6, it is mentioned that on a $2n$ dimensional Calabi-Yau manifold, $h_{(n,0)} = h_{(0,n)} = 1$. What is the reason for this?
One way to prove this is by introducing a metric on the Calabi-Yau and to compute the set of all solutions to the equation $\Delta \omega^{(n,0)} = 0$ and consequently $\Delta \omega^{(0,n)} = 0$ by Hodge duality, where $\omega^{(m,n)}$ is a $(m,n)$ form. Is it apparent that this gives only one solution?
Also further, on page 7, it is mentioned that for a Calabi Yau 3-fold, the following equations hold.
$h_{(1,0)} = h_{(2,0)}= h_{(0,1)} = h_{(0,2)} = h_{(2,3)} = h_{(3,2)} = h_{(1,3)} = h_{(3,1)} = 0$
How do I see this?