Theorem 8.13.4.53F870DE5980 The following conditions are equivalent:
$\mathrm{dd}^{\mathrm{c}}(\omega ^{n-q})=0$ for every $1\leq q\leq m$;
$\int _ XH_ m(u)$ is independent of the smooth admissible potential $u$;
$\mathcal M_ m(t_1,\ldots ,t_ m)=\int _ X\omega ^ n$ for every bounded admissible list;
every pair of bounded admissible $u,v$ satisfies
\[ \int _{\{ u<v\} }H_ m(v)\leq \int _{\{ u<v\} }H_ m(u). \]
Under these conditions, for every $1\leq r\leq m$ and every fixed bounded admissible list $z_1,\ldots ,z_{m-r}$, one also has
\[ \int _{\{ u<v\} }\mathrm d\mu _ r(v) \leq \int _{\{ u<v\} }\mathrm d\mu _ r(u). \]
Proof.
The first condition implies the third by (8.13.2). The third implies the second by taking all inputs equal to a smooth potential.
To prove that the second implies the first, fix any smooth real $f$. For all sufficiently small real $t$ of either sign, $\omega +t\mathrm{dd}^{\mathrm{c}}f$ is positive, so $tf$ is strictly admissible. The polynomial
\[ \int _ X(\omega +t\mathrm{dd}^{\mathrm{c}}f)^ m\wedge \omega ^{n-m} =\sum _{q=0}^ m\binom mq t^ q \int _ X(\mathrm{dd}^{\mathrm{c}}f)^ q\wedge \omega ^{n-q} \]
is constant on an interval. Consequently every coefficient with $q\geq 1$ vanishes. Polarizing that coefficient by replacing $f$ with $\sum _{a=1}^ q s_ af_ a$ gives
\[ \int _ X\mathrm{dd}^{\mathrm{c}}f_1\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}f_ q \wedge \omega ^{n-q}=0 \]
for all smooth real $f_ a$. Integration by parts in the first factor, using lemma 11.1.2, gives
\[ \int _ Xf_1\mathrm{dd}^{\mathrm{c}}f_2\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}f_ q \wedge \mathrm{dd}^{\mathrm{c}}(\omega ^{n-q})=0. \]
Since $f_1$ is arbitrary, the smooth top-degree form paired with it vanishes pointwise. At each point the other complex Hessians can be prescribed by real quadratic functions times a cutoff. Complexifying this multilinear identity gives all coordinate $(q-1,q-1)$-forms, as in the proof of theorem 11.1.7. They detect every coefficient of $\mathrm{dd}^{\mathrm{c}}(\omega ^{n-q})$, so that form vanishes. When $q=1$, there are no other Hessians and the same conclusion follows directly from the arbitrariness of $f_1$.
Under the third condition, $\mu _ r(w)$ and $\mu _ r(u)$ have the same total mass. 8.13.3 proves the mixed comparison assertion, and $r=m$ gives the fourth condition. Finally, suppose the fourth condition holds and fix smooth admissible $u,v$. Choose $C$ so large that $u<v+C$ everywhere. Constant invariance gives $H_ m(v+C)=H_ m(v)$, so comparison gives $\int _ XH_ m(v)\leq \int _ XH_ m(u)$. Exchanging $u,v$ gives equality, which is the second condition.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.