Proposition 23.1.4.DDEBFFAFCD1F For every $V\in C^\infty (X,\mathbb R)$,
C46F0B405DC2
23.1.5
\begin{equation} \int _ X(\omega +\mathrm{dd}^{\mathrm{c}}V)^3 =6+6\int _{\mathbb P^2} \left(\int _{\pi ^{-1}(y)}|V_ w|^2\, \mathrm{d}A_ E\right) \omega _{\mathrm{FS}}^2(y). \end{equation}
In particular, positivity of $\omega +\mathrm{dd}^{\mathrm{c}}V$ is not required.
Proof.
Since $\beta $ restricts to $2\mathrm{d}A_ E$ on each fiber,
\[ \int _ X\omega ^3=3\int _ X\alpha ^2\wedge \beta =6. \]
By (11.1.6) and (23.1.3), followed by an integration by parts against the closed form $\alpha ^2$,
\[ \begin{split} \int _ X(\omega +\mathrm{dd}^{\mathrm{c}}V)^3-\int _ X\omega ^3 & =3\int _ X V\mathrm{dd}^{\mathrm{c}}\omega \wedge \mathrm{dd}^{\mathrm{c}}V\\ & =-3\int _ X V\alpha ^2\wedge \mathrm{dd}^{\mathrm{c}}V =3\int _ X\mathrm{d}V\wedge \mathrm d^{\mathrm c}V\wedge \alpha ^2. \end{split} \]
Wedging with $\alpha ^2$ removes every component of $\mathrm{d}V\wedge \mathrm d^{\mathrm c}V$ except the vertical one, which is $\mathrm i|V_ w|^2\mathrm{d}w\wedge \mathrm{d}\bar w/(2\pi )=2|V_ w|^2\mathrm{d}A_ E$. Integration over the fibers proves the formula.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.