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23.1.67 Positive quadratic measures52E95F708FD4

The mass formula of proposition 23.1.4 gives

\[ M_\omega (\lambda f)-6=\lambda ^2\bigl (M_\omega (f)-6\bigr ) \]

for every smooth real $f$ and real $\lambda $. To use this scalar identity in a weak limit, we also need a positive measure that represents the quadratic term.

Lemma 23.1.68.1E196E409754 Let $f$ be smooth, $0<t<1$, and $\mathrm{dd}^{\mathrm{c}}f\geq -t\omega $. Set $H=\mathrm{dd}^{\mathrm{c}}f+t\omega $, $d=M_\omega (f)-6$, and $\nu =3\omega \wedge H^2$. Then $d\geq 0$ and

\[ \begin{gathered} \int _ X\omega ^2\wedge H=6t,\qquad \int _ X\omega \wedge H^2=d/3+6t^2,\qquad \int _ XH^3=td+6t^3,\\ \bigl \| (\omega +\mathrm{dd}^{\mathrm{c}}f)^3-\omega ^3-\nu \bigr \| _{\mathrm{TV}} \leq 36t+2td+24t^3. \end{gathered} \]

Proof. The Gauduchon identity $\mathrm{dd}^{\mathrm{c}}\omega ^2=0$ gives $\int _ X\omega ^2\wedge \mathrm{dd}^{\mathrm{c}}f=0$, and Stokes' formula gives $\int _ X(\mathrm{dd}^{\mathrm{c}}f)^3=0$. Expanding the three integrals with these identities proves the asserted values, and proposition 23.1.4 gives $d\geq 0$. The identity of measures

\[ (\omega +\mathrm{dd}^{\mathrm{c}}f)^3 =(1-t)^3\omega ^3+3(1-t)^2\omega ^2\wedge H +(1-t)\nu +H^3 \]

has four positive terms. After subtracting $\omega ^3+\nu $, the total masses of the remaining terms bound the total variation by $18t+18t+t(d+18t^2)+td+6t^3$, as asserted. $\square$

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