= Solution
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> is
$$
\omega(u,v)=g(Ju,v),\qquad
\omega=\frac{i}{2}\sum_{a,b}h_{a\bar b}\,dz^a\wedge d\bar z^b,
$$
where $g=\operatorname{Re}(\sum h_{a\bar b}\,dz^a\otimes d\bar z^b)$ in this convention. It is real and of type $(1,1)$, since $g(Ju,Jv)=g(u,v)$. Choose an adapted real orthonormal frame $E_j,F_j=JE_j$ and its dual coframe $e^j,f^j$. Then $\omega=\sum_j e^j\wedge f^j$, giving the <Riemannian volume form>
$$
\boxed{\operatorname{vol}_g=e^1\wedge f^1\wedge\cdots\wedge e^n\wedge f^n=\frac{\omega^n}{n!}.}
$$
We use the complex-linear <Hodge star operator>, the extension of the real metric star characterized on equal-degree forms by
$$
\alpha\wedge *\overline\beta=\langle\alpha,\beta\rangle_g\operatorname{vol}_g.
$$
It maps $(p,q)$-forms to $(n-q,n-p)$-forms and satisfies $**\alpha=(-1)^{p+q}\alpha$. It is distinct from the conjugate-linear map $\alpha\mapsto*\overline\alpha$. 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> formula
$$
\boxed{*\omega=\frac{\omega^{n-1}}{(n-1)!}.}
$$
This computation uses no closedness assumption on $\omega$.
Use the integrated Hermitian inner product $(\alpha,\beta)=\int_X\alpha\wedge*\overline\beta$. The <formal adjoints> $d^*$ and $\bar\partial^*$ are defined by $(d\alpha,\beta)=(\alpha,d^*\beta)$ and $(\bar\partial\alpha,\beta)=(\alpha,\bar\partial^*\beta)$; compact manifolds here have no boundary. The <Hodge Laplacian> and <Dolbeault Laplacian> are
$$
\Delta_d=dd^*+d^*d,\qquad
\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial.
$$
The <Dolbeault Hodge decomposition on a compact Hermitian manifold> states that
$$
\mathcal A^{p,q}(X)=\mathcal H_{\bar\partial}^{p,q}\oplus
\bar\partial\mathcal A^{p,q-1}(X)\oplus
\bar\partial^*\mathcal A^{p,q+1}(X)
$$
is an orthogonal direct sum, with finite-dimensional $\mathcal H_{\bar\partial}^{p,q}=\ker\Delta_{\bar\partial}$. Each <Dolbeault cohomology> class
$$
H_{\bar\partial}^{p,q}(X)=\ker(\bar\partial:\mathcal A^{p,q}\to\mathcal A^{p,q+1})/\bar\partial\mathcal A^{p,q-1}
$$
has a unique representative in $\mathcal H_{\bar\partial}^{p,q}$. A form is harmonic exactly when both $\bar\partial\alpha$ and $\bar\partial^*\alpha$ 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 give
$$
(L\alpha)\wedge*\overline\beta
=\alpha\wedge\omega\wedge*\overline\beta
=\alpha\wedge*\overline{\Lambda\beta}.
$$
The wedge pairing is nondegenerate, so $*\overline{\Lambda\beta}=L*\overline\beta$. Apply another star to the degree-$(r-2)$ form and use that the star and $L$ commute with conjugation. This yields
$$
\boxed{\Lambda\beta=(-1)^{r-2}*L*\beta=(-1)^{p+q}*L*\beta.}
$$
For 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 $[\Lambda,\partial]=i\bar\partial^*$. Since $\Lambda$ is real, conjugation gives $[\Lambda,\bar\partial]=-i\partial^*$, hence
$$
\bar\partial^*=-i[\Lambda,\partial],\qquad
\partial^*=i[\Lambda,\bar\partial].
$$
The commutators are ordinary commutators because $\Lambda$ has even degree. Using $\partial^2=0$ we obtain, by direct expansion,
$$
\partial\bar\partial^*+\bar\partial^*\partial
=-i\left(\partial\Lambda\partial-\partial^2\Lambda+\Lambda\partial^2-\partial\Lambda\partial\right)=0.
$$
Conjugation gives $\bar\partial\partial^*+\partial^*\bar\partial=0$ as well. Expanding the two unmixed Laplacians and using $\partial\bar\partial=-\bar\partial\partial$ gives the same operator:
$$
\begin{aligned}
\Delta_\partial
&=i\left(\partial\Lambda\bar\partial-\bar\partial\Lambda\partial-\partial\bar\partial\Lambda-\Lambda\partial\bar\partial\right)\\
&=\Delta_{\bar\partial}.
\end{aligned}
$$
Finally expand $d=\partial+\bar\partial$ and $d^*=\partial^*+\bar\partial^*$; the mixed terms just proved zero disappear. This derives the <Kähler Laplacian identity>
$$
\boxed{\Delta_d=\Delta_\partial+\Delta_{\bar\partial}=2\Delta_{\bar\partial}.}
$$
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, gives
$$
b^r(X)=\sum_{p+q=r}h^{p,q},\qquad h^{p,q}=\dim_{\mathbb C}\mathcal H_{\bar\partial}^{p,q}.
$$
The 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 \b[all odd Betti numbers are even].
For $1\leq k\leq n$, the <Kähler form> power $\omega^k$ is closed. If it were exact, say $\omega^k=d\eta$, then
$$
\int_X\omega^n=\int_Xd(\eta\wedge\omega^{n-k})=0
$$
by <Stokes theorem>. But $\omega^n=n!\operatorname{vol}_g$ has strictly positive integral on the nonempty compact manifold. Hence the <powers of a compact Kähler form have nonzero cohomology classes>, and
$$
\boxed{b^{2k-1}(X)\in2\mathbb Z_{\geq0},\qquad b^{2k}(X)\geq1\quad(1\leq k\leq n).}
$$
For an example take the <Hopf surface>
$$
H=(\mathbb C^2\setminus\{0\})/\langle z\mapsto2z\rangle.
$$
The 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 $1\leq\|z\|\leq2$ covers the quotient, proving compactness. The map
$$
[z]\longmapsto\left(z/\|z\|,\ \log\|z\|\bmod\log2\right)
$$
is 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 \b[this compact complex surface admits no Kähler metric].
Back to article page
