Corollary 18.20.447BA03C63283

Proof. For $\alpha =0$, the background is pulled back from $Y\times Z$, and by proposition 5.43.5 every potential in it descends to $Y\times Z$. The corresponding currents have no component in the $T$-directions, so they cannot dominate a positive multiple of a Hermitian form on $M$. Thus the background is not big.

If $B_ i=0$, choose any Hermitian form $\omega _ T$ on $T$ and write $s:M\to T$ for the projection. The positive $(m+\ell +k-1,m+\ell +k-1)$-form

is closed, because $\omega _ T^\ell $ has top degree on $T$. Its pairing with $\Theta _\alpha $ is zero, since the only possible contribution contains the integral of $\sigma _ i$, whereas its pairing with a positive product Hermitian form is strictly positive. By Stokes' theorem, no current in the background is bounded below by a positive multiple of a Hermitian form.

When all areas are positive, write $\sigma _ i=B_ i\eta _ i/A_ i+\mathrm{dd}^{\mathrm{c}}\chi _ i$. Subtracting the smooth pullback of $\sum _ i\chi _ i\circ \pi _ i$ gives the current $p^*\sum _ i(B_ i/A_ i)\pi _ i^*\eta _ i+\varpi ^*\alpha +r^*\kappa $. Positivity of $\alpha $ in the $T$-directions makes this form strictly positive, which proves bigness. For the stated choice of $\alpha $, the exterior derivative of $\Theta _\alpha $ has the component $\mathrm dh\wedge \omega _ T$, with one direction in $Y$ and two in $T$. It is nonzero for nonconstant $h$ and cannot be canceled by the component with three directions in $T$. Hence $\mathrm d\Theta _\alpha \ne 0$. $\square$