Lemma 15.19.4C390030D09AE

Proof. For the differentiated calculation write $U=u_\varepsilon /(2\pi )$ and $P=\rho /(2\pi )$, so that $\mathrm{dd}^{\mathrm{c}}u_\varepsilon =\mathrm{i}\partial \bar\partial U$. Put

The previous lemma and the smooth positive density give $0<d_1\leq D\leq d_2$. Choose a fixed $C_0$ so that $v=U-P+C_0>1$, and set

where $A$ will be fixed below. For each $\varepsilon $, $S$ is smooth on $X$ and $v$ tends to infinity at $Z$. Thus $H\to -\infty $ there, and $H$ attains its maximum on $X\setminus Z$. We bound this maximum independently of $\varepsilon $.

In local holomorphic coordinates, $h$ and $g$ denote the coefficient matrices of $\eta $ and $\omega _\varepsilon $ with respect to $\mathrm{i}\, \mathrm{d}z_ j\wedge \mathrm{d}\bar z_ k$, and commas denote coordinate derivatives, as in $h_{i\bar j,k}=\partial _ kh_{i\bar j}$ and $h_{i\bar j,\bar k}=\partial _{\bar k}h_{i\bar j}$. At a maximum point choose holomorphic coordinates in which

A linear change first normalizes $h$ and diagonalizes $g$. In these coordinates, the quadratic change $z_ i=w_ i+\frac12\sum _{j,k}a^ i_{jk}w_ jw_ k$, with symmetric coefficients chosen so that $a^ i_{ik}=-h_{i\bar i,k}$, then kills $h_{i\bar i,k}$. Its coefficients are uniformly bounded on a fixed finite cover. Write

The lowered torsion coefficients $T_{ij\bar k}$ and their first derivatives are uniformly bounded: the potential Hessian cancels, leaving only derivatives of $\theta +\varepsilon \eta $.

Let $\mu _{\mathrm{loc}}$ be the density of $\mu $ with respect to the coordinate volume form. Differentiating $S=h^{k\bar\ell }g_{k\bar\ell }$ and the equation $\log \det g=\log c_\varepsilon +\log \mu _{\mathrm{loc}}$ gives

The second derivatives of $g$ contribute $\sum _{i,j,k}|g_{i\bar j,k}|^2/(\lambda _ i\lambda _ j)$, up to $-CR-C$, by commuting the potential derivatives. The first derivatives of $h$ contribute $-2\operatorname {Re}\sum _{i,j,k}\lambda _ i^{-1}h_{j\bar k,\bar i}g_{k\bar j,i}$. Its terms with $j=k$ vanish. Replace $g_{k\bar j,i}$ by $g_{i\bar j,k}+T_{ik\bar j}$ and apply weighted Cauchy–Schwarz. This consumes exactly the terms of the preceding square with $j\ne k$ and costs at most $CRS+CR$. The second derivatives of $h^{-1}$ cost $CRS$. Finally the fixed density term $-C$ is absorbed using $SR\geq n^2$.

Since $\partial _ iS=\sum _ jg_{i\bar j,j}+E_ i$ at the maximum point, weighted Cauchy–Schwarz also gives

Indeed expand around $\partial S-E$ and discard the resulting term $-|E|_ g^2$. Combining with (15.19.5) yields

At the maximum, $\partial \log S=\gamma '\partial v$. Moreover (15.19.1) gives $\Delta _ gv\leq n-\delta R$, and $|E|_ g^2\leq CR$. Young's inequality with the positive coefficient $-\gamma ''=v^{-2}$ therefore gives

Since $\gamma '=A+v^{-1}$, choose $A$ with $\delta A-C\geq 1$. The resulting inequality is

All constants are independent of $\varepsilon $.

If the maxima of $H$ were unbounded, then at a sequence of maximizing points

Thus $S/v\to \infty $ and $S\to \infty $. Inequality (15.19.6) would imply $R\leq 2C$ along this sequence. But

a contradiction. Hence $H$ is uniformly bounded. On $K$, both $U$ and $P$ are bounded, so $S\leq C_ K$. The determinant lower bound then gives $\lambda _ i\geq d_1/C_ K^{n-1}$, completing the proof. $\square$