Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 118 / 5

This is the Poincaré residue of a meromorphic top form with a simple pole. Write $z=z_1$ and locally $\alpha =(dz/z)\wedge \beta$, with $\beta$ a holomorphic $(n-1)$-form. The proposed residue is the pullback of $\beta$ to $Y$.

First it does not depend on the choice of $\beta$: if $dz\wedge (\beta -{\beta }^{\prime })=0$, then at each point the difference has a factor $dz$, and its pullback to $Y$ is zero. Now change the local defining coordinate to $t=az$, where $a$ is a nowhere-zero holomorphic function. Such a factor exists because both coordinates define the same smooth complex analytic hypersurface. If $\alpha =(dt/t)\wedge \eta$, thenAt points of $Y$, this is an equality in the ambient exterior-power fibre, and the last term is zero. It follows that $dz\wedge (\beta -\eta )=0$ there, so their pullbacks to $Y$ agree. This also covers changes of the tangential coordinates.

The restrictions are holomorphic top forms on $Y$, and their agreement on overlaps makes them a global section of the canonical bundle. Thus the residue is independent of the adapted coordinates:The ambient-fibre comparison is important: pulling an ambient $n$-form directly back to the $(n-1)$-dimensional hypersurface would give zero and would not prove the required independence.

On ${\mathbb{C}}^{n+1}\setminus \{0\}$, let $\Omega =d{z}_0\wedge \cdots \wedge d{z}_n$ and let $E={\sum }_i{z}_i{\partial }_{z_i}$ be the Euler vector field. The numerator is its interior product of a differential form, soBoth numerator and denominator have homogeneous scaling degree $n+1$. Thus this meromorphic $n$-form is invariant under constant nonzero scalings, and ${\iota }_E\alpha =0$ makes it horizontal for the quotient to Complex projective space.

To check descent using the hint, take a local holomorphic section $Z$ of the quotient and another section $Z^{\prime }=\lambda Z$, where $\lambda$ is a nowhere-zero holomorphic function. In the evaluationevery term involving $d\lambda$ has two collinear arguments and vanishes. The remaining term is ${\lambda }^{n+1}{Z}^{\ast }({\iota }_E\Omega )$, while $f(\lambda Z)={\lambda }^{n+1}f(Z)$. Hence $Z^{\prime \ast }\alpha =Z^{\ast }\alpha$. The forms therefore glue to a meromorphic section of the canonical bundle of Complex projective space.

On the affine chart $z_j\ne 0$, use the section $z_j=1$ and list the remaining coordinates as $t_1,\dots ,t_n$ in increasing original-index order. If $F$ is the resulting polynomial, thenAt a point of $Y$, some derivative $F_{t_r}$ is nonzero. Otherwise all the homogeneous derivatives except possibly $f_{z_j}$ would vanish; the Euler homogeneous function theorem gives ${\sum }_i{z}_i{f}_{z_i}=(n+1)f=0$ there, forcing $f_{z_j}=0$ too. This contradicts the permitted smoothness criterion. The holomorphic inverse function theorem now makes $F$ a transverse coordinate, so the pole is simple. On a neighbourhood with $F_{t_r}\ne 0$, rearranging the wedge product of differential forms givesBy part (a), its Poincaré residue is consequentlyThe remaining $t$ coordinates form a holomorphic coordinate system on $Y$, and $F_{t_r}$ is nonzero in this neighbourhood. The displayed top form is therefore nowhere zero. These neighbourhoods cover $Y$ and part (a) guarantees agreement on their overlaps. The residue is a nowhere-vanishing global holomorphic section:This is the explicit residue trivialization for a degree n+1 projective hypersurface, consistent with the adjunction formula $K_Y\cong {\mathcal{O}}_Y((n+1)-n-1)$, but obtained here directly from the indicated form.