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A hyperplane section of a projective variety $X\subset {\mathbb{P}}^n$ is the scheme-theoretic intersection $X\cap H$ with a projective hyperplane $H$. Its associated line bundle is ${\mathcal{O}}_X(1)$.

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

For finite-dimensional Complex projective space ${\mathbb{CP}}^r$, use the filtration by its linear projective subspaces. The difference ${\mathbb{CP}}^j\setminus {\mathbb{CP}}^{j-1}$ is ${\mathbb{C}}^j$, so there is one topological cell in every even dimension $0,2,\dots ,2r$, and none in odd dimensions. Every cellular boundary vanishes for degree reasons. The integral cohomology is therefore $\mathbb{Z}$ in those even degrees and zero elsewhere.

For the cohomology ring, let $x\in H^2({\mathbb{CP}}^r;\mathbb{Z})$ be the Poincare dual of a projective hyperplane, using the complex orientation. Generic hyperplanes meet transversely, and a collection of $j$ of them intersects in ${\mathbb{CP}}^{r-j}$. The geometric interpretation of the cup product identifies $x^j$ with the dual of that intersection. Restricting to a linear ${\mathbb{CP}}^j$, the $j$ hyperplanes meet in one point, with positive sign because the transverse normal directions are complex lines. HenceThus $x^j$ is a primitive generator in degree $2j$, not merely a nonzero multiple. There are no classes beyond degree $2r$, soThe point case $r=0$ has ring $\mathbb{Z}$. Taking the union of the finite projective spaces gives $H^{\ast }({\mathbb{CP}}^{\infty };\mathbb{Z})=\mathbb{Z}[x]$, since each fixed cohomological degree stabilizes. This proves the cohomology ring of complex projective space computation.

For the rank bound for a nonsingular complex bilinear map, write $a={\dim }_{\mathbb{C}}U$, $b={\dim }_{\mathbb{C}}V$, $R=\mathrm{im}\varphi$, and $s={\dim }_{\mathbb{C}}R$. Assume $a,b\ge 1$. Injectivity with either nonzero factor fixed implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Thereforeis well defined and continuous. Scaling either vector only scales the output, and in local projective charts the formula is given by bilinear coordinates with no simultaneous zero. Let $h$ be the target's degree-two generator and $x,y$ the pullbacks of the two factor generators. The Künneth theorem givesFixing $[v]$ or $[u]$ makes $F$ a projective linear embedding induced by an injective linear map. Such an embedding pulls back the hyperplane class to the factor's hyperplane class. Since the two restrictions determine $H^2$ of the product, $F^{\ast }h=x+y$, with the generator of a point factor interpreted as zero.

Put $d=a+b-2$. In the source cohomology ring, every term in $(x+y{)}^d$ vanishes except the one with exponents $a-1,b-1$, andThe binomial coefficient is a positive integer and the final monomial generates the top integral cohomology. If $s\le d$, the class $h^d$ in ${\mathbb{CP}}^{s-1}$ would be zero. This contradiction provesWhen $a=b=1$, the nonzero pure-tensor image already gives $s\ge 1$. A positivity convention is genuinely necessary: if $U=0$, $V={\mathbb{C}}^2$, and $W=0$, all restrictions to the zero tensor subspaces are injective, but the literal bound would read $0\ge 1$. The displayed theorem is therefore the nonzero-factor version. If a nonzero factor is infinite-dimensional, either factor restriction already bounds the image dimension below by that factor dimension; the sum of two positive cardinal dimensions, with at least one infinite, is their maximum, giving the corresponding cardinal bound.

The assertion is true. The standard filtration ${\mathbb{CP}}^0\subset \cdots \subset {\mathbb{CP}}^n$ gives one cell in each dimension $0,2,\dots ,2n$. There are no odd-dimensional cells, so all cellular differentials vanish. The cellular homology theorem and the universal coefficient theorem for cohomology give one copy of $\mathbb{Z}$ in each even degree from zero to $2n$, and zero in every other degree.

Let $x$ be the Poincare dual of a projective hyperplane, with the complex orientation. It has degree two and evaluates to $+1$ on a complex projective line, so it is the positive generator of $H^2$. We use the intersection interpretation of the cup product: the product of duals of oriented submanifolds in transverse position is the dual of their oriented intersection. Distinct transverse complex hyperplanes intersect in ${\mathbb{CP}}^{n-j}$ after $j$ intersections, with positive complex orientation. Thus $x^j$ is the dual of that linear subspace.

Pairing $x^j$ with a transverse linear ${\mathbb{CP}}^j$ gives one positively oriented intersection point. Therefore $x^j$ is a primitive generator of $H^{2j}$, for every $0\le j\le n$. There is no cohomology above dimension $2n$, so $x^{n+1}=0$. These facts show that the surjective graded ring map from $\mathbb{Z}[x]$ has exactly the indicated kernel:This proves the cohomology ring of complex projective space, rather than only its additive groups. For $n=0$ it is $\mathbb{Z}$, with $x=0$.

Use ${\mathbb{F}}_2$ coefficients throughout. Write $h=c_1(\mathcal{O}(1))\mathrm{mod}2$ on Complex projective space and $a=w_1({\gamma }_d)$ on Real projective space. We use the standard cohomology ring of complex projective space ${\mathbb{F}}_2[h]/(h^{d+1})$, with $|h|=2$, and the real projective calculation from the preceding solution. The complex projective calculation follows, for example, from its one cell in each even dimension together with the fact that transverse projective hyperplanes represent the powers of $h$; $d$ such hyperplanes have one intersection point.

For the standard inclusion $i:{\mathbb{RP}}^d\hookrightarrow {\mathbb{CP}}^d$, the restricted complex tautological line is ${\gamma }_d{\otimes }_{\mathbb{R}}\mathbb{C}$. As a real vector bundle it is ${\gamma }_d\oplus {\gamma }_d$, whose total Stiefel–Whitney class is $(1+a{)}^2=1+a^2$. The Stiefel–Whitney class of the underlying real bundle of a complex line identifies its degree-two class with its First Chern class reduced modulo two. Passing to the dual line defining $h$ changes only a sign, which disappears modulo two. Thus mod-two restriction from complex to real projective space gives $i^{\ast }h=a^2$. In particular the answer for dimension two is the ring mapIt sends $1$ to $1$, is an isomorphism in degree two, and sends $h^2$ to zero.

Now assume $k\ge 1$ and put $M={\mathbb{CP}}^{2k}$, $N={\mathbb{RP}}^{2k}$, $U=M\setminus N$, with inclusions $i:N\hookrightarrow M$ and $j:U\hookrightarrow M$. The real dimensions are $4k$ and $2k$, respectively. A tubular neighborhood and excision identify the cohomology of $(M,U)$ with that of the normal disk bundle relative to its sphere bundle. Every real normal bundle is oriented over ${\mathbb{F}}_2$, so the Thom isomorphism theorem givesUnder this identification the relative-to-absolute map is the cohomological Gysin map of an embedding $i_{!}$. The exact sequence of the pair is thereforeHere $\delta$ denotes the ordinary connecting homomorphism followed by the inverse of the isomorphism supplied by the Thom isomorphism theorem.

To compute $i_{!}$, use mod-two Poincare duality and its evaluation formulaFor $0\le r\le k$, take $b=a^{2r}$ and $z=h^{k-r}$. The right side is $⟨a^{2k},[N{]}_2⟩=1$. Since the target group is generated by $h^{k+r}$ and $⟨h^{2k},[M{]}_2⟩=1$, this provesOdd-degree classes have zero image because $M$ has no odd-degree cohomology. Exactness now givesFor precision, below degree $2k-1$ the restriction map is an isomorphism. In degree $2k-1$ the next map is $i_{!}:H^0(N)\to H^{2k}(M)$, an isomorphism, so the complement group is zero. In even degrees $2k+2r$, $0\le r<k$, the preceding $i_{!}$ is an isomorphism and the following odd-degree $i_{!}$ is zero; consequently $\delta :H^{2k+2r}(U)\to H^{2r+1}(N)$ is an isomorphism. The intervening odd groups and the top group $H^{4k}(U)$ vanish.

It remains to establish multiplication. Put $x=j^{\ast }h$. Since $h^k=i_{!}1$, exactness gives $x^k=0$. Choose $y\in H^{2k}(U)$ with $\delta y=a$, which is possible by the just-computed isomorphism. Naturality of the relative cup product, and the module property of the Thom isomorphism theorem, giveThus $\delta (x^{r}y)=a^{2r+1}$ for $0\le r<k$. Each class $x^{r}y$ is nonzero and generates its upper-half group, while $1,x,\dots ,x^{k-1}$ generate the lower-half groups. Also $y^2=0$ because $H^{4k}(U)=0$. These classes account for every group, so there are no further relations:By the field-coefficient Künneth theorem this is precisely the cohomology ring of ${\mathbb{CP}}^{k-1}\times S^{2k}$. For $k=1$ the relation $x=0$ leaves the sphere's ring. The calculation identifies rings and does not claim a homotopy equivalence of the spaces; $k=0$ is outside the expression involving ${\mathbb{CP}}^{k-1}$.

The tautological bundle over Complex projective space isOn $U_i=\{z_i\ne 0\}$, the vector $e_i([z])=z/z_i$ is a holomorphic frame. On $U_i\cap U_j$,so the transition functions are holomorphic. Define $\mathcal{O}(1)=\mathcal{O}(-1{)}^{\ast }$ and, for $k\in \mathbb{Z}$, define $\mathcal{O}(k)=\mathcal{O}(1{)}^{\otimes k}$ when $k\ge 0$ and $\mathcal{O}(k)=\mathcal{O}(-1{)}^{\otimes (-k)}$ when $k<0$. Its transition functions are the corresponding $k$th powers of those of $\mathcal{O}(1)$, equivalently the $(-k)$th powers of those of $\mathcal{O}(-1)$.

A nonzero linear functional $\ell \in ({\mathbb{C}}^{n+1}{)}^{\ast }$ restricts on each projective line to define a global holomorphic section $s_{\ell }$ of $\mathcal{O}(1)$. Its zero divisor is the projective hyperplane $H_{\ell }=\{[z]:\ell (z)=0\}$, so every hyperplane is an effective divisor of $\mathcal{O}(1)$.

Conversely, let $D$ be the divisor of a nonzero section of $\mathcal{O}(1)$. In the affine chart $z_0=1$, the section is represented by an entire function $f$ on ${\mathbb{C}}^n$. Compatibility with the other projective charts gives the linear growth bound in the question along every complex affine line. The supplied Liouville-type result makes $f$ affine-linear, and homogenizing it gives a linear functional $\ell$ on ${\mathbb{C}}^{n+1}$. Thus $D=H_{\ell }$.

In local coordinates centred at $x$, the blowup of a complex manifold at a point is modeled byOn the chart $v_i\ne 0$, write $u=z_i$ and $t_j=v_j/v_i$; then $z_j=u{t}_j$, giving holomorphic coordinates $(u,(t_j{)}_{j\ne i})$. These charts show that the blowup is a complex manifold, that $\sigma$ is holomorphic, and that its exceptional fiber is ${\mathbb{CP}}^{n-1}$. A biholomorphic coordinate change at the centre lifts by sending a punctured point together with its limiting tangent direction to its image; the chart formulas extend across the exceptional divisor. Hence the construction is independent of coordinates up to biholomorphism.

For $a(z)=-z$ on ${\mathbb{C}}^2$, defineThis is holomorphic and satisfies $\sigma \circ \tilde{a}=a\circ \sigma$. The blowup $\widetilde{mathbb{C}^2}$ is the total space of $\mathcal{O}(-1)\to {\mathbb{CP}}^1$, and $\tilde{a}$ acts as multiplication by $-1$ in every fiber. The invariant fiber coordinate is $w=u^2$. If $u_j=g_{ji}{u}_i$ are the transition laws for $\mathcal{O}(-1)$, then $w_j=g_{ji}^2{w}_i$, which are the transition laws for $\mathcal{O}(-2)$. Thus the local quotient maps $(t,u)\mapsto (t,u^2)$ glue, give the quotient a complex-manifold atlas even along the fixed zero section, and yield