Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 118 / 1 / Solution

Let $(z^1,\dots ,z^n)$ be a complex manifold chart, with $z^j=x^j+i{y}^j$. The almost complex structure induced by a complex atlas isThus $J^2=-I$. On an overlap, the derivative of a holomorphic map is complex linear and hence commutes with multiplication by $i$. The two coordinate definitions of $J$ therefore agree, so $J$ is globally well-defined.

The complexified cotangent bundle splits into the $i$- and $-i$-eigenspaces of $J^{\ast }$, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a sumSplitting the exterior derivative according to type definesComplex conjugation sends $d{z}^j$ to $d{\bar{z}}^j$ and conjugates the coefficient derivatives. Term by term this gives the complex conjugation of differential-form type identity

Let $J$ act on a $k$-form byIt acts on a $(p,q)$-form by $i^{p-q}$. Therefore $J^{-1}dJ$ multiplies the $\partial$ component by $-i$ and the $\bar{\partial }$ component by $i$, proving the d c operator formula

A holomorphic vector field is a holomorphic section of $T^{1,0}X$. On the affine chart $U_0\subset {\mathbb{CP}}^n$, put $w_k=z_k/z_0$. The projection isDirect differentiation givesIf $\xi (z)$ is linear and homogeneous, the pushed-forward coefficients are consequently $\xi (1,w)$ when $j\ne 0$, and $-\xi (1,w)w_k$ when $j=0$. They are holomorphic on $U_0$.

More intrinsically, $\xi (z){\partial }_{z_j}$ is a linear vector field $Az$ on ${\mathbb{C}}^{n+1}$. Its flow $e^{tA}$ preserves complex lines and induces the projective transformationsDifferentiating gives a globally defined projectivization of a linear vector field, whose expression on $U_0$ is the field just computed. This proves extension across the other affine charts.

Finally choose distinct complex numbers ${\lambda }_0,\dots ,{\lambda }_n$ and the diagonal fieldIts projectivization vanishes at $[z]$ precisely when $({\lambda }_0{z}_0,\dots ,{\lambda }_n{z}_n)$ is proportional to $z$, so $[z]$ is an eigenline. The distinct eigenvalues leave exactly the $n+1$ coordinate points. Hence ${\mathbb{CP}}^n$ has a holomorphic vector field with finitely many zeroes.