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?

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    $\begingroup$ I'm guessing you're taking your Calabi-Yau 3-fold to be simply-connected (otherwise a torus can be given a complex structure not satisfying these conditions). In that case, note that $h^{1,0}, h^{0,1}$ are 0 since they have to add up to $b_1 = 0$ (by Kahler conditions). Then use $c_1 = 0$ and Noether's formula/Hirzebruch-Riemann-Roch to conclude that $h^{2,0} = 0$. By the symmetry $h^{p,q} = h^{q,p}$ we have on any Kahler manifold, and the symmetry $h^{p,q} = h^{n-p,n-q}$ by Serre duality (valid for any compact complex manifold), the other identities follow. $\endgroup$ – Aleksandar Milivojevic Feb 8 at 0:24
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    $\begingroup$ In general dimension $n$, for a Calabi-Yau $X$ we have $h^{n,0} = 1$ (and hence $h^{0,n}$ by symmetry) since $H^0(X, \Omega^n)$ is one-dimensional; namely, we have a holomorphic volume form (by definition of Calabi-Yau) which is a nowhere-zero section of the canonical bundle. Then any other section is a holomorphic function (which is constant, since $X$ is compact) times this section, i.e. $dim H^0(X, \Omega^n) = 1$. $\endgroup$ – Aleksandar Milivojevic Feb 8 at 0:27
  • $\begingroup$ @AleksandarMilivojevic thanks for your answer. :) $\endgroup$ – Bruce Lee Feb 9 at 1:14

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