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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.