Proposition 11.1.16.2A8C104899E8 Suppose that every pair $u,v\in \mathcal H_\omega $ satisfies
F5F2F7988675
11.1.17
Then $\mathrm M_\omega $ is constant on $\mathcal H_\omega $.
B5345EE28748The global comparison principle of Kähler geometry compares the Monge–Ampère measures of two potentials on the set where one lies below the other. Adding a constant to a potential leaves its Monge–Ampère measure unchanged, and after a large constant shift this set becomes the whole manifold. Such an inequality would therefore compare total masses, and the varying mass of example 11.1.12 rules it out.
Proposition 11.1.16.2A8C104899E8 Suppose that every pair $u,v\in \mathcal H_\omega $ satisfies
F5F2F7988675
11.1.17
Then $\mathrm M_\omega $ is constant on $\mathcal H_\omega $.
Proof. Fix $u,v\in \mathcal H_\omega $. By compactness there is a constant $C$ such that $u<v+C$ everywhere. Since $\omega _{v+C}=\omega _ v$, applying (11.1.17) to $u$ and $v+C$ gives $\mathrm M_\omega (v)\leq \mathrm M_\omega (u)$. Exchanging the roles of $u$ and $v$ gives the opposite inequality. $\square$
Corollary 11.1.18.9578EA44189A The inequality (11.1.17) does not hold for every Hermitian form on a compact complex manifold.
Proof. Let $X$, $\omega $, $u_ t$ and $B=\int _ X\beta _1\wedge \beta _2$ be as in example 11.1.12. For $-\pi ^{-2}<t<0$, set $u=0$ and $v=u_ t+C$ with $C>|t|$. Then $\{ u<v\} =X$, but
Thus this pair violates (11.1.17). $\square$
The modified comparison principle of theorem 14.19.6 compensates for the derivatives of the background. Conversely, under the equivalent conditions of theorem 11.1.7, the ordinary comparison principle holds for bounded potentials, by 14.25.3.
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