Pome

Lemma 12.17.23BBBC535BFA2

Lemma 12.17.2.3BBBC535BFA2 Let $g$ be a Hermitian form with $\mathrm{dd}^{\mathrm{c}}(g^2)=0$, and let $u\in \mathrm{PSH}(X,g)$ be bounded. The current

\[ E_ u=\tfrac 12\mathrm{dd}^{\mathrm{c}}(u^2)-u\mathrm{dd}^{\mathrm{c}}u \]

is positive, and

649C88380858 12.17.3
\begin{equation} \int _ X(g+\mathrm{dd}^{\mathrm{c}}u)^3 =\int _ Xg^3-3\int _ XE_ u\wedge \mathrm{dd}^{\mathrm{c}}g. \end{equation}

If $-\mathrm{dd}^{\mathrm{c}}g\geq 0$, then $v_-(g)=\int _ Xg^3$.

Proof. Locally write $u=p-a$, with $p$ bounded plurisubharmonic and $a$ smooth. Expanding in $p$ and $a$ and using multilinearity, we obtain closed currents $C_ r=(\mathrm{dd}^{\mathrm{c}}u)^ r$, not necessarily positive, with $C_0=1$. The recursive definition of bounded products, expanded in the same way, gives

\[ \mathrm{dd}^{\mathrm{c}}(uC_ r)=C_{r+1},\qquad r=0,1,2. \]

By lemma 5.27.2 and invariance under smooth corrections, the weighted product $u\mathrm{dd}^{\mathrm{c}}u$ does not depend on the local decomposition. On a smaller ball choose smooth bounded plurisubharmonic $p_ j$ decreasing to $p$, and put $u_ j=p_ j-a$. The same lemma gives $u_ j\mathrm{dd}^{\mathrm{c}}u_ j\to u\mathrm{dd}^{\mathrm{c}}u$, while boundedness gives $u_ j^2\to u^2$ in $L^1$. Hence

\[ \mathrm du_ j\wedge \mathrm d^{\mathrm c}u_ j =\tfrac 12\mathrm{dd}^{\mathrm{c}}(u_ j^2)-u_ j\mathrm{dd}^{\mathrm{c}}u_ j\longrightarrow E_ u. \]

The smooth currents on the left are positive, so their weak limit $E_ u$ is positive and has order zero.

Now expand the cube. Repeated use of distributional Stokes gives

\[ \int _ XC_3=0,\qquad \int _ Xg^2\wedge C_1=\int _ Xu\mathrm{dd}^{\mathrm{c}}(g^2)=0, \qquad \int _ Xg\wedge C_2=\int _ XuC_1\wedge \mathrm{dd}^{\mathrm{c}}g. \]

Since $\mathrm{dd}^{\mathrm{c}}g$ is closed, $\int _ X\mathrm{dd}^{\mathrm{c}}(u^2)\wedge \mathrm{dd}^{\mathrm{c}}g=0$, so the last integral equals $-\int _ XE_ u\wedge \mathrm{dd}^{\mathrm{c}}g$. This proves (12.17.3). If $-\mathrm{dd}^{\mathrm{c}}g\geq 0$, the correction term is nonnegative, and $u=0$ attains the lower bound $\int _ Xg^3$. $\square$

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.