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In the chart $x_j=X_j/X_0$ of Complex projective space, differentiate the quotient coordinates to obtain this formula for $\pi :{\mathbb{C}}^{n+1}\setminus \{0\}\to {\mathbf{P}}^n$. The kernel is the radial line $\mathbb{C}X$. A degree-one homogeneous vector field on the punctured cone descends to a holomorphic vector field, since its scaling cancels the inverse scaling of $d\pi$. This gives the fiberwise construction of the Euler sequence on complex projective space.

At a point $[X]\in {\mathbf{P}}^n$, the complex tautological line bundle has fiber $\mathbb{C}X\subseteq {\mathbb{C}}^{n+1}$. The hyperplane line bundle is its dual, $[H]=\mathcal{O}(1)$, and $\mathcal{O}(m)=\mathcal{O}(1{)}^{\otimes m}$. On $U_i=\{X_i\ne 0\}$ let $t_i=X/X_i$ be the tautological frame and $e_i$ its dual. On overlaps $e_j=(X_j/X_i)e_i$. The sheaf of holomorphic sections of a vector bundle of this dual bundle is the sheaf ${\mathcal{O}}_{{\mathbf{P}}^n}(1)$.

If a nonzero linear form $\ell$ cuts out the projective hyperplane $H$, it gives a holomorphic section $s_H$ of $\mathcal{O}(1)$ with a simple zero along $H$. Dividing a local section $s$ by $s_H$ gives a meromorphic function with no poles away from $H$ and at most a simple pole on $H$. Conversely multiplication by $s_H$ turns such a function into a holomorphic section. These mutually inverse local constructions agree on overlaps and prove the holomorphic line bundle associated to a divisor identification

The nonnegative hyperplane twist cohomology by restriction calculation givesTo derive it from the permitted vanishing, use the hyperplane exact sequence for twisting sheaveswhere $i:H\hookrightarrow {\mathbf{P}}^n$ and $H\cong {\mathbf{P}}^{n-1}$. Local multiplication by a defining equation is injective, and restriction gives exactly its cokernel. Induct on $n$, starting from a point, and then on $m$. For $m=0$, higher sheaf cohomology vanishes by assumption and global holomorphic functions are constant by compactness and the maximum principle. For $m>0$, the long exact sequence in sheaf cohomology, the inductive vanishing for $\mathcal{O}(m-1)$ and the dimension induction on $H$ give vanishing in every positive degree andSumming this recurrence gives $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. Homogeneous degree-$m$ monomials give that many independent sections, so they form a basis. The printed extra equality to zero cannot hold in degree zero, already because $h^0({\mathbf{P}}^n,\mathcal{O})=1$; the boxed calculation gives the intended dimensions.

For the differential of the projective quotient map, the differential of a smooth map is computed by applying a tangent vector to coordinate functions. Since $x_j=X_j/X_0$,ThusUnder replacing $X$ by $\lambda X$, $d\pi$ on a constant ambient tangent vector scales by ${\lambda }^{-1}$, while a homogeneous linear form $L$ scales by $\lambda$. Therefore $d{\pi }_X(L(X){\partial }_{X_j})$ is independent of the representative. Its chart coefficients are holomorphic, so it is a global holomorphic vector field. The radial Euler vector field lies in the kernel of $d\pi$, either from the same formulas or because it differentiates a scaling orbit. Hence

Describe the Euler sequence on complex projective space fiberwise as well as in coordinates. At a line $\ell =\mathbb{C}X$, the middle bundle has fiber $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ and the holomorphic tangent bundle has fiber $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Send a homomorphism to its composition with the quotient. The map from $\mathcal{O}$ sends a scalar to that scalar times the inclusion $\ell \hookrightarrow {\mathbb{C}}^{n+1}$. In homogeneous notation these maps areA local section of $\mathcal{O}(1)$ is a local degree-one homogeneous function on the punctured cone, so the second expression is representative-independent. Quotienting is surjective, and its kernel consists exactly of scalar inclusion maps. This proves the short exact sequence of locally free sheavesSince it is locally split, dualizing and tensoring by $\mathcal{O}(1)$ preserves exactness and gives the cotangent Euler sequence in homogeneous coordinatesOn global sections the last map identifies ${\mathbb{C}}^{n+1}$ with the basis of linear forms in $H^0(\mathcal{O}(1))$. Its kernel is zero. This proves the vanishing of global sections of the once-twisted projective cotangent bundle; left exactness gives $\boxed{h^0({\mathbf{P}}^n,{\Omega }^1(1))=0}$.

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.

A complex-linear endomorphism $A$ of ${\mathbb{C}}^{n+1}$ induces a holomorphic vector field on Complex projective space by differentiating the projective transformations $[z]\mapsto [e^{tA}z]$. Its zeroes are precisely the projective eigenlines of $A$; a diagonalizable $A$ with distinct eigenvalues therefore gives exactly $n+1$ zeroes.