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For each of the operators $D=d,\partial ,\bar{\partial },d^c$, define its nonnegative Laplacian by ${\Delta }_D=D{D}^{\ast }+D^{\ast }D$ using the $L^2$ formal adjoint. In particular, $\Delta =d{d}^{\ast }+d^{\ast }d$, ${\Delta }_{\partial }=\partial {\partial }^{\ast }+{\partial }^{\ast }\partial$, ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$, and ${\Delta }^c=d^c(d^c{)}^{\ast }+(d^c{)}^{\ast }{d}^c$. Use the d c operator conventionThis normalization, without an extra factor of one-half, is the one for which the requested equality with ${\Delta }^c$ holds.

The Lefschetz operator of a Kähler manifold is $L\alpha =\omega \wedge \alpha$, and the adjoint Lefschetz operator is its pointwise metric adjoint $\Lambda =L^{\ast }$. It lowers type by $(1,1)$ and is a real operator because $\omega$ and the metric are real. Write $P=\partial$ and $B=\bar{\partial }$, use $[A,D]=AD-DA$, and use $\{A,D\}=AD+DA$ for the anticommutators of the odd-degree operators. The allowed relation among the Kähler identities and its complex conjugate giveThese identities alone suffice for the calculation. Since $P^2=B^2=0$,Replacing $P$ by $B$ in the same calculation gives $\{B,P^{\ast }\}=0$. Thus both mixed terms vanish when expanding the Hodge Laplacian of $d=P+B$:The anticommutation $PB=-BP$ now proves equality of the two summands. Indeed,which are the same expression. Finally, $(d^c{)}^{\ast }=i(P^{\ast }-B^{\ast })$, so its Laplacian expands asThis proves the Kähler Laplacian identity, with its real-operator version included:

The action of the almost complex structure on forms is $J\alpha (v_1,\dots ,v_k)=\alpha (J{v}_1,\dots ,J{v}_k)$. It multiplies a form of type $(p,q)$ by $i^{p-q}$. Since $\Delta =2{\Delta }_{\bar{\partial }}$ preserves each bidegree, it commutes with this action of $J$. Thus $\Delta (J\alpha )=J(\Delta \alpha )$; the invertibility of $J$ givesThis argument applies to arbitrary sums of types, not only to pure-type forms.

The ddbar lemma for a compact Kähler manifold states that a $d$-closed form of pure type is $\partial \bar{\partial }$-exact if it is $d$-exact, $\partial$-exact, or $\bar{\partial }$-exact. Conversely, a $\partial \bar{\partial }$-exact form is $d$-closed and exact for all three operators. In particular a $d$-exact, $d$-closed $(p,q)$-form equals $\partial \bar{\partial }\beta$ for a form $\beta$ of type $(p-1,q-1)$.

The difference $\gamma =\tilde{\omega }-\omega$ of the two Kähler forms is a real $(1,1)$-form, is $d$-closed, and is $d$-exact by equality of their de Rham cohomology classes. The ddbar lemma therefore gives a smooth complex-valued function $h$ with $\gamma =\partial \bar{\partial }h$. Reality and anticommutation implySet $f=(h-\bar{h})/(2i)$, which is real. It follows directly that $i\partial \bar{\partial }f=\gamma$, giving the global potential for cohomologous Kähler forms

If $f_1,f_2$ are two real potentials, $u=f_1-f_2$ satisfies $\partial \bar{\partial }u=0$, so it is a pluriharmonic function. The allowed relation among the Kähler identities gives ${\bar{\partial }}^{\ast }\bar{\partial }u=-i[\Lambda ,\partial ]\bar{\partial }u=-i\Lambda \partial \bar{\partial }u=0$, because $\Lambda \bar{\partial }u=0$ by degree. The Kähler Laplacian identity then gives $\Delta u=2{\bar{\partial }}^{\ast }\bar{\partial }u=0$. Integration by parts on compact $X$ givesso $u$ is constant on each connected component. Consequently the potential is unique up to one additive constant when $X$ is connected; on a disconnected $X$, one constant per component is the exact conclusion. Without the usual connectedness convention the printed uniqueness assertion needs this qualification: distinct constants on the components of a disjoint union leave the same pair of Kähler forms unchanged.

On a complex manifold, the exterior derivative splits by type as $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator $\partial$ raises holomorphic degree by one and the Dolbeault operator $\bar{\partial }$ raises antiholomorphic degree by one. Their identities areUse the convention $d^c=i(\bar{\partial }-\partial )$ for the d c operator; then $d{d}^c=2i\partial \bar{\partial }$. The alternative convention with a factor $1/2$ changes this last factor only. Complex conjugation of differential-form type interchanges $\partial$ and $\bar{\partial }$ without reversing the wedge order. Thus, for a real form $\eta$,The resulting form is real, independently of the degree or type components of $\eta$.

The Dolbeault cohomology isIn degree $(0,0)$ there is no incoming image, and the kernel consists of global holomorphic functions. On a compact connected $X$, $|f|$ attains a maximum. In a coordinate polydisc around a maximizing point, the one-variable maximum modulus principle applied successively along the coordinate directions makes $f$ constant on a smaller polydisc. Alternatively the same argument shows the maximum set is open; it is closed and nonempty, so connectedness makes it all of $X$, and the local constancy then makes $f$ globally constant. Consequently

A holomorphic p-form on a complex manifold is a form of type $(p,0)$ with holomorphic coordinate coefficients, equivalently one killed by $\bar{\partial }$. For degree $n={\dim }_{\mathbb{C}}X$, it is $d$-closed: its $\bar{\partial }$ vanishes by holomorphicity and its $\partial$ has impossible bidegree $(n+1,0)$. Its conjugate is also closed.

The key positivity behind nonzero holomorphic top forms are not exact is explicit. If $\alpha =fd{z}_1\wedge \cdots \wedge d{z}_n$, complex orientation givesThis is nonnegative everywhere and positive on some open set when $\alpha \not\equiv 0$, so its integral on the compact manifold is strictly positive. If $\alpha =d\gamma$, closedness of $\bar{\alpha }$ and the Stokes theorem would giveHere Stokes is used on a compact oriented manifold without boundary, and complex-valued forms are handled by linearity. The contradiction proves a nonzero holomorphic top form is not $d$-exact. For dimension zero the same nonexactness follows directly from the absence of degree $-1$ forms.

Now take a compact complex surface, with no Kähler metric assumed. For a holomorphic one-form $\alpha$, the form $\beta =d\alpha =\partial \alpha$ has type $(2,0)$ and is holomorphic, since $\bar{\partial }\partial \alpha =-\partial \bar{\partial }\alpha =0$. It is exact. The top-form result forces $\beta =0$, proving closedness. This establishes holomorphic forms on a compact complex surface are closed by an argument which does not require a Kähler identity.

If a holomorphic one-form were exact, write $\alpha =df$ for a global smooth complex function. Comparing types gives $\bar{\partial }f=0$. The preceding compactness argument makes $f$ constant on every connected component, hence $\alpha =0$. Holomorphic two-forms are closed and nonexact unless zero by the top-form result. Holomorphic zero-forms are constant on each component, and a nonzero zero-form is never exact because there are no negative-degree forms. Degrees above two contain only zero forms. Therefore every holomorphic form on a compact complex surface is closed, and a nonzero one is never exact.

Let $(z^1,\dots ,z^n)$ be a complex manifold chart, with $z^j=x^j+i{y}^j$. The almost complex structure induced by a complex atlas isThus $J^2=-I$. On an overlap, the derivative of a holomorphic map is complex linear and hence commutes with multiplication by $i$. The two coordinate definitions of $J$ therefore agree, so $J$ is globally well-defined.

The complexified cotangent bundle splits into the $i$- and $-i$-eigenspaces of $J^{\ast }$, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a sumSplitting the exterior derivative according to type definesComplex conjugation sends $d{z}^j$ to $d{\bar{z}}^j$ and conjugates the coefficient derivatives. Term by term this gives the complex conjugation of differential-form type identity

Let $J$ act on a $k$-form byIt acts on a $(p,q)$-form by $i^{p-q}$. Therefore $J^{-1}dJ$ multiplies the $\partial$ component by $-i$ and the $\bar{\partial }$ component by $i$, proving the d c operator formula

A holomorphic vector field is a holomorphic section of $T^{1,0}X$. On the affine chart $U_0\subset {\mathbb{CP}}^n$, put $w_k=z_k/z_0$. The projection isDirect differentiation givesIf $\xi (z)$ is linear and homogeneous, the pushed-forward coefficients are consequently $\xi (1,w)$ when $j\ne 0$, and $-\xi (1,w)w_k$ when $j=0$. They are holomorphic on $U_0$.

More intrinsically, $\xi (z){\partial }_{z_j}$ is a linear vector field $Az$ on ${\mathbb{C}}^{n+1}$. Its flow $e^{tA}$ preserves complex lines and induces the projective transformationsDifferentiating gives a globally defined projectivization of a linear vector field, whose expression on $U_0$ is the field just computed. This proves extension across the other affine charts.

Finally choose distinct complex numbers ${\lambda }_0,\dots ,{\lambda }_n$ and the diagonal fieldIts projectivization vanishes at $[z]$ precisely when $({\lambda }_0{z}_0,\dots ,{\lambda }_n{z}_n)$ is proportional to $z$, so $[z]$ is an eigenline. The distinct eigenvalues leave exactly the $n+1$ coordinate points. Hence ${\mathbb{CP}}^n$ has a holomorphic vector field with finitely many zeroes.