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Let $n={\dim }_{\mathbb{C}}X$, and use the complex orientation. Write $g$ for the real Riemannian metric underlying the Hermitian metric. Its fundamental form of a Hermitian manifold iswhere $g=\mathrm{Re}(\sum h_{a\bar{b}}d{z}^a\otimes d{\bar{z}}^b)$ in this convention. It is real and of type $(1,1)$, since $g(Ju,Jv)=g(u,v)$. Choose an adapted real orthonormal frame $E_j,F_j=J{E}_j$ and its dual coframe $e^j,f^j$. Then $\omega ={\sum }_j{e}^j\wedge f^j$, giving the Riemannian volume form

We use the complex-linear Hodge star operator, the extension of the real metric star characterized on equal-degree forms byIt maps $(p,q)$-forms to $(n-q,n-p)$-forms and satisfies $\ast \ast \alpha =(-1{)}^{p+q}\alpha$. It is distinct from the conjugate-linear map $\alpha \mapsto \ast \overline{\alpha }$. In the adapted coframe the star of $e^j\wedge f^j$ is the wedge product of all the other two-dimensional factors, with positive sign. Summing proves the Hodge star of the fundamental Hermitian form formulaThis computation uses no closedness assumption on $\omega$.

Use the integrated Hermitian inner product $(\alpha ,\beta )={\int }_X\alpha \wedge \ast \overline{\beta }$. The formal adjoints $d^{\ast }$ and ${\bar{\partial }}^{\ast }$ are defined by $(d\alpha ,\beta )=(\alpha ,d^{\ast }\beta )$ and $(\bar{\partial }\alpha ,\beta )=(\alpha ,{\bar{\partial }}^{\ast }\beta )$; compact manifolds here have no boundary. The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold states thatis an orthogonal direct sum, with finite-dimensional ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$. Each Dolbeault cohomology classhas a unique representative in ${\mathcal{H}}_{\bar{\partial }}^{p,q}$. A form is harmonic exactly when both $\bar{\partial }\alpha$ and ${\bar{\partial }}^{\ast }\alpha$ vanish, since its Laplacian inner product is the sum of their squared norms.

Now let $L\alpha =\omega \wedge \alpha$ be the Hermitian Lefschetz operator and let $\Lambda$ be its pointwise adjoint. For $\beta$ of total degree $r=p+q$ and $\alpha$ of degree $r-2$, the defining inner products giveThe wedge pairing is nondegenerate, so $\ast \overline{\Lambda \beta }=L\ast \overline{\beta }$. Apply another star to the degree-$(r-2)$ form and use that the star and $L$ commute with conjugation. This yieldsFor degrees below two both sides vanish. The argument also identifies its bidegree as $(-1,-1)$ and does not require the Kähler condition.

Assume now that $d\omega =0$, so the metric is a Kähler metric. Start with the supplied Kähler identities relation $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$. Since $\Lambda$ is real, conjugation gives $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$, henceThe commutators are ordinary commutators because $\Lambda$ has even degree. Using ${\partial }^2=0$ we obtain, by direct expansion,Conjugation gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ as well. Expanding the two unmixed Laplacians and using $\partial \bar{\partial }=-\bar{\partial }\partial$ gives the same operator:Finally expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$; the mixed terms just proved zero disappear. This derives the Kähler Laplacian identity

The Dolbeault Laplacian preserves bidegree. Thus the complex $d$-harmonic forms of total degree $r$ split into their harmonic $(p,q)$ parts. The allowed real Hodge theorem, followed by complexification, givesThe real operator ${\Delta }_d$ commutes with conjugation, which exchanges $(p,q)$ and $(q,p)$. Therefore $h^{p,q}=h^{q,p}$. For odd $r$, no summand has $p=q$, so they pair off in equal dimensions. This proves that all odd Betti numbers are even.

For $1\le k\le n$, the Kähler form power ${\omega }^k$ is closed. If it were exact, say ${\omega }^k=d\eta$, thenby Stokes theorem. But ${\omega }^n=n!{\mathrm{vol}}_g$ has strictly positive integral on the nonempty compact manifold. Hence the powers of a compact Kähler form have nonzero cohomology classes, and

For an example take the Hopf surfaceThe dilation action is free and properly discontinuous: on a compact subset the radii are bounded above and away from zero, so only finitely many dilations can meet it. It is holomorphic, giving a complex manifold quotient; the annulus $1\le \Vert z\Vert \le 2$ covers the quotient, proving compactness. The mapis a diffeomorphism $H\cong S^3\times S^1$. The Künneth theorem gives $b^1(H)=1$. This violates the necessary odd-degree parity, so this compact complex surface admits no Kähler metric.