11.47.13 The full envelope on a Hermitian productC6C2D9DCA88F

Proof. Every competitor restricts to a projective slice as a $\kappa $-plurisubharmonic function or as $-\infty $ identically. To check the exact lower bound when $H$ is nonclosed, choose a smooth quadratic correction $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq H$ on a coordinate polydisk and let $v_\epsilon =(v+q)*\eta _\epsilon -q$ on a smaller polydisk, where $\eta _\epsilon $ is the rescaling of a fixed smooth nonnegative radial probability kernel. Radial plurisubharmonic convolutions decrease to the canonical $v+q$, and

Restrict this smooth inequality to a fixed slice. The corrected restrictions decrease either to $-\infty $ or to a plurisubharmonic function in local $L^1$; in the latter case the background error tends to zero distributionally. Subtracting $q$ gives the exact $\kappa $ bound, and these canonical restrictions patch on the connected slice. The everywhere obstacle constraint now gives $v\leq \operatorname {pr}_1^*E_ F$ on every slice. Upper regularization preserves this bound. Conversely every base competitor lifts, since $h_0>0$, proving equality of the full envelopes.

The bounded truncations $\max (E_ F^ H,-j)$ are pullbacks. By proposition 5.28.2, lemma 5.28.4, dimensional multilinearity leaves only the term with $k$ projective factors and $m$ factor entries. Restrict to strict superlevel sets and let $j\to \infty $ to obtain the stated non-pluripolar measure and its finite mass. The residual regularity and uniform estimates pull back in a fixed smooth product atlas. Finally $E_ F^ H\geq u_\alpha -K_ F$. At $t=1$ take the stretching witness $v_1=E_ F^ H$; for $t>1$ take $v_ t=u_\alpha -tK_ F$, so that

Together with ordinary domination this is definition 16.28.61, including the canonical values at the common pole. $\square$

Proof. Take $Y=(\mathbb C^2\setminus \{ 0\} )/\langle \zeta \mapsto \mathrm e\zeta \rangle $. With $s=\log |\zeta |^2$, the form $h_0=\mathrm{dd}^{\mathrm{c}}s+\mathrm ds\wedge \mathrm d^{\mathrm c}s$ is invariant under the quotient. Its two summands are positive on the tangential and radial complex directions, respectively, so it is Hermitian. Moreover $\mathrm dh_0=-\mathrm ds\wedge \mathrm{dd}^{\mathrm{c}}s\ne 0$. The Hopf surface is diffeomorphic to $S^3\times S^1$ and has odd first Betti number. Thus $X=\mathbb P^2\times Y$ is a non-Kähler fourfold with $\mathrm dH\ne 0$, and proposition 11.47.14 applies. $\square$