Lemma 16.15.323CF8D2A3054

Proof. The form is closed if and only if its $(2,1)$-part $\frac{\mathrm i}{2\pi }\partial _ le_{i\bar j}\, \mathrm dz_ l\wedge \mathrm dz_ i\wedge \mathrm d\bar z_ j$ and the conjugate $(1,2)$-part vanish, that is, if and only if $\partial _ le_{i\bar j}=\partial _ ie_{l\bar j}$; this holds in all holomorphic coordinates. If $z=\Phi (w)$ with $\Phi $ holomorphic and $J=D\Phi $ its complex Jacobian matrix, then $\mathrm dz_ i=J_{ia}\, \mathrm dw_ a$, so the new coefficients are $\tilde e(w)=J(w)^{\mathrm T}e(\Phi (w))\overline{J(w)}$.

Let $A=\overline{e(z_0)^{-1/2}}$, where $e(z_0)^{-1/2}$ is the positive square root of the inverse matrix; its operator norm is $\lambda (z_0)^{-1/2}$, where $\lambda (z)=\min _{|\xi |=1}\sum _{i,j}e_{i\bar j}(z)\xi _ i\bar\xi _ j$ is the smallest eigenvalue of $e(z)$, and $A^{\mathrm T}e(z_0)\bar A=e(z_0)^{-1/2}e(z_0)e(z_0)^{-1/2}=I$. In the coordinates $v$ with $z=z_0+Av$ the coefficient matrix $e'$ satisfies $e'(0)=I$. Put $c^ m_{ij}=\partial _{v_ i}e'_{j\bar m}(0)$, which is symmetric in $i$ and $j$ by the closedness, and define $v=w-\frac12c(w,w)$, that is, $v_ m=w_ m-\frac12c^ m_{ij}w_ iw_ j$. Then $J_{ia}=\partial v_ i/\partial w_ a=\delta _{ia}-c^ i_{aj}w_ j$, so $J(0)=I$ and $\partial _{w_ p}J_{ia}(0)=-c^ i_{ap}$. Differentiating $\tilde e_{a\bar b}=J_{ia}e'_{i\bar j}(v(w))\overline{J_{jb}}$ at $w=0$, and using that $\overline{J_{jb}}$ and $\bar v$ are antiholomorphic in $w$, we find

by the closedness, and $\partial _{\bar w_ p}\tilde e_{a\bar b}(0)=\overline{\partial _{w_ p}\tilde e_{b\bar a}(0)}=0$. Put $\Phi _{z_0}(w)=z_0+A(w-\frac12c(w,w))$. Its derivative at $0$ is the invertible matrix $A$, so it is a biholomorphism near $0$ by the holomorphic inverse function theorem (proposition 5.2.4(1)).

By the chain rule, the derivatives of $\tilde e$ at $0$ are polynomials in the derivatives of $e$ at $z_0$ and in the entries of $A$, $\bar A$ and $c$, and the entries of $c$ are polynomials in the first derivatives of $e$ at $z_0$ and in the entries of $A$ and $\bar A$. The derivatives of $e$ are bounded on compact sets, and so are the entries of $A$, because the function $\lambda $ is continuous and positive, as a minimum over the unit sphere of a function that is continuous in $(z,\xi )$; hence these polynomials are bounded for $z_0\in K$. Finally, if $Q$ is unitary, replacing $\Phi _{z_0}(w)$ by $\Phi _{z_0}(Qw)$ replaces $\tilde e(w)$ by $Q^{\mathrm T}\tilde e(Qw)\bar Q$, which equals $Q^{\mathrm T}\bar Q=\overline{Q^*Q}=I$ at $0$, still has vanishing differential at $0$, and has derivatives at $0$ bounded by a constant depending only on $n$ times those of $\tilde e$. $\square$