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14.26.9 Two forms with the same cubeDBEDC79E97CD

The cancellation identities now turn the scalar coefficients into two semipositive forms with the same cube. Their average is strictly positive except at the two ends of the central interval.

By the extensions described above, the form and the function

\[ \omega _0=x(r)P+y(r)Q+z(r)R,\qquad w=F(r) \]

are smooth on $X$. The form is semipositive, and positive except on the hypersurfaces $M_- =\{ r=-1\} $ and $M_+=\{ r=1\} $. Put

\[ H=\tfrac 12\mathrm{dd}^{\mathrm{c}}w=p(P-Q)+qR,\qquad A_0=\omega _0-H,\quad B_0=\omega _0+H. \]

For $|r|\geq 1$ we have $H=0$. On $|r|<1$,

\[ A_0=aP+(2p+b)Q+(z_0-q)R,\qquad B_0=(2p+a)P+bQ+(z_0+q)R. \]

All coefficients are nonnegative, and by (14.26.4) and (14.26.8), $A_0^3=B_0^3=0$ on this band. Outside it both cubes equal $\omega _0^3$. Thus $A_0,B_0$ are smooth semipositive forms on $X$ with equal cubes. This common top-degree form vanishes on the closed band $|r|\leq 1$ and is positive on its complement, including near $X\setminus U$.

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