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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 $\boxed{h^0({\mathbf{P}}^n,{\Omega }^1(1))=0}$.

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.