Pome

14.26.2 Three rank-one directions0CFD28034339

On $U=\{ A\ne 0,B\ne 0\} $ put

\[ r=\log \frac{|A|^2}{|B|^2},\qquad P=\mathrm{dd}^{\mathrm{c}}\log |A|^2,\quad Q=\mathrm{dd}^{\mathrm{c}}\log |B|^2,\quad R=\mathrm{d}r\wedge \mathrm d^{\mathrm c}r. \]

The logarithms are computed in a local homogeneous lift, but their Hessians do not depend on the lift. The forms $P,Q$ are pullbacks of Fubini–Study forms from the two projective lines $\mathbb P(\mathbb C^2)$. All three forms are semipositive of rank one, and

6970E9123EB5 14.26.3
\begin{equation} P^2=Q^2=R^2=0,\qquad P\wedge Q\wedge R>0. \end{equation}

Indeed, in local coordinates $A=(1,a)$ and $B=\tau (1,b)$, $P$ and $Q$ are positive multiples of $\mathrm{i}\, \mathrm{d}a\wedge \mathrm{d}\bar a$ and $\mathrm{i}\, \mathrm{d}b\wedge \mathrm{d}\bar b$, while

\[ \partial r=\frac{\bar a\, \mathrm{d}a}{1+|a|^2} -\frac{\mathrm{d}\tau }{\tau } -\frac{\bar b\, \mathrm{d}b}{1+|b|^2}. \]

Its $\mathrm{d}\tau $ component is nonzero, so the three rank-one forms correspond to independent directions. In particular, for nonnegative $x,y,z$,

71DFC7D0861C 14.26.4
\begin{equation} (xP+yQ+zR)^3=6xyz\, P\wedge Q\wedge R, \end{equation}

and the form is positive exactly when all three coefficients are positive.

Since $\mathrm{dd}^{\mathrm{c}}r=P-Q$, the chain rule gives, for a smooth function $G:\mathbb R\to \mathbb R$,

1C52D8C53893 14.26.5
\begin{equation} \mathrm{dd}^{\mathrm{c}}G(r)=G'(r)(P-Q)+G''(r)R. \end{equation}

Let $\sigma (r)=\mathrm{e}^ r/(1+\mathrm{e}^ r)$. Since $\log (|A|^2+|B|^2)=\log |B|^2+\log (1+\mathrm{e}^ r)$, we have

DE71ED12948B 14.26.6
\begin{equation} \omega _ X=\sigma (r)P+\sigma (-r)Q+\sigma '(r)R. \end{equation}

A function of $r$ that is constant near each end of $\mathbb R$ extends smoothly across $X\setminus U$. A form $xP+yQ+zR$ whose coefficients equal those of (14.26.6) for $|r|\geq 2$ extends smoothly by $\omega _ X$ across $X\setminus U$.

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