Corollary 12.18.1.40D45A12B192 Suppose $g$ is a Hermitian form with $\mathrm{dd}^{\mathrm{c}}g=\mathrm{dd}^{\mathrm{c}}(g^2)=0$, and put $V_ g=\int _ Xg^ n$. For every Hermitian form $h\geq cg$, $c>0$,
\[ v_-(h)\geq c^ nV_ g,\qquad \sup _{T\in \mathscr G_ h}\| k_ T\| _{L^ q(\mathrm{d}V)} \leq c^{-1}V_ g^{-1/n}. \]
Equality holds in the determinant bound when $h=g$ and $c=1$. In particular, on a compact Kähler manifold or a compact complex surface, the critical norm is uniformly bounded over the normalized Gauduchon currents of each fixed Hermitian form.
Proof.
The identity $\mathrm{dd}^{\mathrm{c}}(g^2)=2g\wedge \mathrm{dd}^{\mathrm{c}}g+2\mathrm{d}g\wedge \mathrm d^{\mathrm c}g$ shows that $\mathrm{d}g\wedge \mathrm d^{\mathrm c}g=0$, so theorem 11.1.7 gives $v_-(g)=V_ g$. The quotient in (12.15.5) is monotone and homogeneous in its numerator, so $h\geq cg$ implies $v_-(h)^{1/n}\geq cV_ g^{1/n}$. Theorem 12.9.2 turns these lower-mass bounds into the asserted determinant bounds, with equality when $h=g$.
A Kähler form satisfies both differential identities. On a surface, theorem 11.4.2 gives a metric $g$ with $\mathrm{dd}^{\mathrm{c}}g=0$, and $\mathrm{dd}^{\mathrm{c}}(g^2)=0$ for degree reasons. By compactness, each fixed $h$ satisfies $h\geq cg$ for some $c>0$.
$\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.