Kähler commutators and preservation of harmonic types

For $n\ge 1$, fix $x_0\in X$ and choose holomorphic coordinates with $h_{a\bar{b}}(0)=\frac{1}{2}{\delta }_{ab}$, where $h_{a\bar{b}}=g_{\mathbb{C}}({\mathrm{\partial }}_{z_a},{\mathrm{\partial }}_{{\bar{z}}_b})$ for the complex-bilinear extension of $g$. Then $\omega =i{\sum }_{a,b}{h}_{a\bar{b}}d{z}_a\wedge d{\bar{z}}_b$, and $d\omega =0$ implies ${\mathrm{\partial }}_c{h}_{a\bar{b}}={\mathrm{\partial }}_a{h}_{c\bar{b}}$. Put ${\mathrm{\Gamma }}_{ac}^b=2{\mathrm{\partial }}_a{h}_{c\bar{b}}(0)$ and make the symmetric quadratic change $z_b=w_b-\frac{1}{2}{\sum }_{a,c}{\mathrm{\Gamma }}_{ac}^b{w}_a{w}_c$. Its derivative at zero is the identity, so [F4] makes it a holomorphic coordinate change near zero. In the transformed matrix $h_{a\bar{b}}^{\mathrm{\prime }}={\sum }_{i,j}{h}_{i\bar{j}}(z(w))(\mathrm{\partial }{z}_i\mathrm{/}\mathrm{\partial }{w}_a)\overline{(\mathrm{\partial }{z}_j\mathrm{/}\mathrm{\partial }{w}_b)}$, one has ${\mathrm{\partial }}_c{h}_{a\bar{b}}^{\mathrm{\prime }}(0)={\mathrm{\partial }}_c{h}_{a\bar{b}}(0)-\frac{1}{2}{\mathrm{\Gamma }}_{ac}^b=0$. The antiholomorphic derivatives vanish by the Hermitian symmetry. Thus $h^{\mathrm{\prime }}=\frac{1}{2}I+O(\mathrm{\mid }w{\mathrm{\mid }}^2)$, and the induced form metric, its inverse and the volume density have vanishing first derivatives at zero. Formula [F2] therefore gives the flat adjoint formulas at this point. When $n=0$ the claimed identities are immediate from [F1]; assume $n\ge 1$ below.

Work at such a normal point with the constant-coefficient operators on the coframe $d{z}_j,d{\bar{z}}_j$: write $a_j=d{z}_j\wedge$, $b_j=d{\bar{z}}_j\wedge$ and let $c_j=a_j^{\ast }$, $d_j=b_j^{\ast }$ be the pointwise adjoints of [F1], so that $\{c_j,a_k\}=2{\delta }_{jk}$, $\{d_j,b_k\}=2{\delta }_{jk}$ and the mixed anticommutators $\{c_j,b_k\}$, $\{d_j,a_k\}$ vanish (the factor two is $h(d{z}_j,d{z}_j)$). Then $\mathrm{\partial }={\sum }_j{a}_j{\mathrm{\partial }}_{z_j}$, $\bar{\mathrm{\partial }}={\sum }_j{b}_j{\mathrm{\partial }}_{{\bar{z}}_j}$, ${\mathrm{\partial }}^{\ast }=-{\sum }_j{c}_j{\mathrm{\partial }}_{{\bar{z}}_j}$, ${\bar{\mathrm{\partial }}}^{\ast }=-{\sum }_j{d}_j{\mathrm{\partial }}_{z_j}$, $L=\frac{i}{2}{\sum }_j{a}_j{b}_j$ and $\mathrm{\Lambda }=-\frac{i}{2}{\sum }_j{d}_j{c}_j$, and direct expansion using the anticommutation relations gives $\mathrm{\Lambda }\mathrm{\partial }-\mathrm{\partial }\mathrm{\Lambda }=-i{\sum }_j{d}_j{\mathrm{\partial }}_{z_j}=i{\bar{\mathrm{\partial }}}^{\ast }$ (the mixed pairs $[d_j{c}_j,a_k]=2{\delta }_{jk}{d}_j$), $\mathrm{\Lambda }\bar{\mathrm{\partial }}-\bar{\mathrm{\partial }}\mathrm{\Lambda }=i{\sum }_j{c}_j{\mathrm{\partial }}_{{\bar{z}}_j}=-i{\mathrm{\partial }}^{\ast }$, ${\bar{\mathrm{\partial }}}^{\ast }L-L{\bar{\mathrm{\partial }}}^{\ast }=i{\sum }_j{a}_j{\mathrm{\partial }}_{z_j}=i\mathrm{\partial }$ and ${\mathrm{\partial }}^{\ast }L-L{\mathrm{\partial }}^{\ast }=-i{\sum }_j{b}_j{\mathrm{\partial }}_{{\bar{z}}_j}=-i\bar{\mathrm{\partial }}$. Every one of these operators is a differential operator of order at most two whose coefficients involve at most first derivatives of the metric, so equality at every normal point proves the four identities globally; the normalization enters only through the factor $h(d{z}_j,d{z}_j)=2$, and both sides scale the same way if the metric is rescaled by a constant, so no normalization is mixed inside the computation.

Use the graded bracket, for which the bracket of two odd operators is their anticommutator and $[\mathrm{\Lambda },\mathrm{\partial }]=i{\bar{\mathrm{\partial }}}^{\ast }$ reads ${\bar{\mathrm{\partial }}}^{\ast }=-i[\mathrm{\Lambda },\mathrm{\partial }]$. By [F3] the operators $\mathrm{\partial },\bar{\mathrm{\partial }}$ are odd and satisfy ${\mathrm{\partial }}^2={\bar{\mathrm{\partial }}}^2=\mathrm{\partial }\bar{\mathrm{\partial }}+\bar{\mathrm{\partial }}\mathrm{\partial }=0$, while $\mathrm{\Lambda }$ is even. Graded Jacobi applied to $\mathrm{\partial },\mathrm{\Lambda },\mathrm{\partial }$ gives $[\mathrm{\partial },[\mathrm{\Lambda },\mathrm{\partial }]]=0$ because the two middle terms cancel and ${\mathrm{\partial }}^2=0$, hence $[\mathrm{\partial },{\bar{\mathrm{\partial }}}^{\ast }]=-i[\mathrm{\partial },[\mathrm{\Lambda },\mathrm{\partial }]]=0$, and its adjoint gives $[\bar{\mathrm{\partial }},{\mathrm{\partial }}^{\ast }]=0$; these are exactly the two vanishing mixed anticommutators. Graded Jacobi applied to $\bar{\mathrm{\partial }},\mathrm{\Lambda },\mathrm{\partial }$ gives $[\bar{\mathrm{\partial }},[\mathrm{\Lambda },\mathrm{\partial }]]=[\mathrm{\partial },[\bar{\mathrm{\partial }},\mathrm{\Lambda }]]$, hence $[\bar{\mathrm{\partial }},{\bar{\mathrm{\partial }}}^{\ast }]=-i[\bar{\mathrm{\partial }},[\mathrm{\Lambda },\mathrm{\partial }]]=-i[\mathrm{\partial },[\bar{\mathrm{\partial }},\mathrm{\Lambda }]]=i[\mathrm{\partial },[\mathrm{\Lambda },\bar{\mathrm{\partial }}]]=[\mathrm{\partial },{\mathrm{\partial }}^{\ast }]$, using the second identity of step 2.1; that is ${\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}={\mathrm{\Delta }}_{\mathrm{\partial }}$. Expanding $d=\mathrm{\partial }+\bar{\mathrm{\partial }}$ and $d^{\ast }={\mathrm{\partial }}^{\ast }+{\bar{\mathrm{\partial }}}^{\ast }$ with the two mixed anticommutators gives ${\mathrm{\Delta }}_d={\mathrm{\Delta }}_{\mathrm{\partial }}+{\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}=2{\mathrm{\Delta }}_{\mathrm{\partial }}=2{\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}$.

For the remaining commutation relations: $[\mathrm{\partial },L]=\mathrm{\partial }\omega \wedge =0$ and $[\bar{\mathrm{\partial }},L]=0$ by [F3]; then $[{\mathrm{\Delta }}_{\mathrm{\partial }},L]=\mathrm{\partial }[{\mathrm{\partial }}^{\ast },L]+[\mathrm{\partial },L]{\mathrm{\partial }}^{\ast }+[{\mathrm{\partial }}^{\ast },L]\mathrm{\partial }+{\mathrm{\partial }}^{\ast }[\mathrm{\partial },L]=-i(\mathrm{\partial }\bar{\mathrm{\partial }}+\bar{\mathrm{\partial }}\mathrm{\partial })=0$, where the first equality uses ${\mathrm{\Delta }}_{\mathrm{\partial }}=\mathrm{\partial }{\mathrm{\partial }}^{\ast }+{\mathrm{\partial }}^{\ast }\mathrm{\partial }$ and $[{\mathrm{\partial }}^{\ast },L]=-i\bar{\mathrm{\partial }}$, and adjunction gives $[{\mathrm{\Delta }}_{\mathrm{\partial }},\mathrm{\Lambda }]=0$; since ${\mathrm{\Delta }}_d=2{\mathrm{\Delta }}_{\mathrm{\partial }}$, ${\mathrm{\Delta }}_d$ commutes with $L$ and $\mathrm{\Lambda }$. Commutation with $\mathrm{\partial }$ and $\bar{\mathrm{\partial }}$ follows from ${\mathrm{\partial }}^2={\bar{\mathrm{\partial }}}^2=0$ and ${\mathrm{\Delta }}_d=2{\mathrm{\Delta }}_{\mathrm{\partial }}=2{\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}$. For the star: for $\alpha$ of degree $k-1$ and $\beta$ of degree $k$ the formal-adjoint identity of [F2] for $d$ and the characterizing property ${\alpha }^{\mathrm{\prime }}\wedge \ast \bar{\beta }=⟨{\alpha }^{\mathrm{\prime }},\beta ⟩dV$ of the star give $\int \alpha \wedge \ast \overline{d^{\ast }\beta }=(-1{)}^k\int \alpha \wedge d\ast \bar{\beta }$, hence, since this holds for all $\alpha$, the operator identity $\ast d^{\ast }=(-1{)}^{k}d\ast$ on degree $k$; substituting it and ${\ast }^2=(-1{)}^{k(m-k)}$ with $m=2n$ into ${\mathrm{\Delta }}_d=d{d}^{\ast }+d^{\ast }d$ shows $\ast {\mathrm{\Delta }}_d={\mathrm{\Delta }}_d\ast$ on every degree, because the adjacent-degree relations are $\ast d^{\ast }=(-1{)}^{k}d\ast$ and $\ast d=(-1{)}^{k+1}{d}^{\ast }\ast$ on degree $k$, and hence $\ast d{d}^{\ast }=d^{\ast }d\ast$ and $\ast d^{\ast }d=d{d}^{\ast }\ast$. Finally ${\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}={\mathrm{\Delta }}_d\mathrm{/}2$ has bidegree $(0,0)$ because $\bar{\mathrm{\partial }}$ and ${\bar{\mathrm{\partial }}}^{\ast }$ shift bidegree by $(0,\pm 1)$, so ${\mathrm{\Delta }}_d$ preserves each ${\mathrm{\Omega }}^{p,q}(X)$. For $n=0$ every operator but the star is zero and all assertions are immediate.