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For a group action of $G$ on $X$, the orbit of a group action through $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $H_x=\{g:gx=x\}$. A transitive group action has one orbit. A simply transitive group action has exactly one element carrying any given point to any other; equivalently it is transitive and every stabilizer subgroup is trivial.

In the modern terminology, a multiply transitive group action is transitive on ordered $k$-tuples of distinct points for some $k\ge 2$. In older geometric usage, the phrase can instead mean transitive but not simply transitive, so that there are several elements carrying one point to another. Under either interpretation the action is transitive, and that alone is what the following argument needs. These meanings should not be confused: a transitive action with nontrivial stabilizers need not be two-transitive.

Fix $x_0\in X$ and put $H=H_{x_0}$. DefineThis is well-defined because every element of $H$ fixes $x_0$. It is surjective by transitivity, and $g{x}_0=g^{\prime }{x}_0$ is equivalent to $g^{\prime -1}g\in H$, proving injectivity. For a smooth Lie group action, $H$ is closed, since it is the inverse image of $x_0$ under the orbit map. The orbit map has constant rank: translations in $G$ and the action diffeomorphisms of $X$ intertwine its differentials. Transitivity makes this rank $\dim X$; otherwise every point would be a critical value, contradicting Sard theorem. Its local submersion charts give smooth local sections and hence a smooth inverse for $F$. ThereforeChoosing $x_1=a{x}_0$ replaces $H$ by $aH{a}^{-1}$. Thus the stabilizer subgroup is determined by the base point, and intrinsically only up to conjugacy. A simply transitive group action gives $H=\{e\}$.

For a finite quantum system with Hilbert space $\mathcal{H}={\mathbb{C}}^{n+1}$, a pure state is a nonzero vector modulo multiplication by a nonzero complex number. Equivalently, it is a unit vector modulo its physically irrelevant phase. Consequently the space of pure states is the space of complex lines,The qualification “pure” is essential: arbitrary quantum states include mixed density matrices and do not form Complex projective space.

The unitary group $U(n+1)$ acts transitively on these lines: choose unit vectors in two given lines, complete each to an orthonormal basis, and map one basis to the other by a unitary transformation. The stabilizer of the line $\mathbb{C}{e}_0$ consists exactly of block matrices $\mathrm{diag}(z,V)$ with $z\in U(1)$ and $V\in U(n)$. Indeed a unitary transformation preserving the line also preserves its orthogonal complement. The homogeneous space description is thereforeIts real dimension is $(n+1{)}^2-1-n^2=2n$. This action is generally not two-transitive: the absolute inner product between unit representatives of two lines is invariant.

The Hopf fibration is the projection $\pi :S^{2n+1}\to {\mathbb{CP}}^n$, $z\mapsto [z]$. The right action $z\cdot \lambda =z\lambda$, $\lambda \in S^1$, is free, and its orbits are exactly the fibres. On the chart $U_i$ where $z_i\ne 0$, write $w_j=z_j/z_i$ for $j\ne i$. A smooth local section isEvery unit vector over this chart has the unique expression $z={\sigma }_i([z])\lambda$, where $\lambda =z_i/|z_i|$. These formulas provide local trivializations ${\pi }^{-1}(U_i)\cong U_i\times S^1$, proving that this is a principal bundle with fibre $S^1$. For $n\ge 1$ it has no global section: a global section would give $S^{2n+1}\cong {\mathbb{CP}}^n\times S^1$, contradicting their fundamental groups. For $n=0$ the base is a point and the bundle is trivial.

On a complex manifold, holomorphic coordinates give the decomposition $d=\partial +\bar{\partial }$, withFor $\alpha ={\sum }_{I,J}{a}_{I\bar{J}}d{z}^I\wedge d{\bar{z}}^J$, the conjugate Dolbeault operator $\partial$ and the Dolbeault operator $\bar{\partial }$ are explicitlyThese are the type components of the intrinsic exterior derivative, so are independent of coordinates. The identity $d^2=0$, separated by type, gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. A holomorphic p-form on a complex manifold is a $(p,0)$-form whose coefficients are holomorphic functions; equivalently it is a $(p,0)$-form killed by the Dolbeault operator.

A rank-$r$ holomorphic vector bundle has local trivializations $E{|}_{U_i}\cong U_i\times {\mathbb{C}}^r$ whose changes are $(x,v)\mapsto (x,g_{ij}(x)v)$, with $g_{ij}:U_i\cap U_j\to {\mathrm{GL}}_r(\mathbb{C})$ holomorphic. These transition functions of a vector bundle satisfy $g_{ii}=I$ and $g_{ij}{g}_{jk}=g_{ik}$. Equivalently, for local frames we use the convention $e_j=e_i{g}_{ij}$. A holomorphic section is a section having holomorphic coordinate functions in every such holomorphic local frame; the condition is unchanged by the holomorphic invertible transitions.

The holomorphic cotangent bundle has local frames $d{z}^1,\dots ,d{z}^n$. Under another holomorphic coordinate system $w$, the chain rule givesThe Jacobian matrix is holomorphic and invertible, with holomorphic inverse supplied by the inverse coordinate change. These are precisely holomorphic bundle transitions. A section written as ${\sum }_b{f}_b(z)d{z}^b$ is holomorphic exactly when every $f_b$ is holomorphic, sothe space of holomorphic one-forms. In particular, being a holomorphic section is not the same as being a $d$-closed one-form on an arbitrary complex manifold.

The complex tautological line bundle isOn $U_i=\{Z_i\ne 0\}$ use the nowhere-zero frame $s_i([Z])=Z/Z_i$. Its components are holomorphic affine coordinates and the constant one in position $i$. On overlapsso the frame changes are nowhere-zero holomorphic functions. This proves that $\mathcal{O}(-1)$ is a holomorphic line bundle, and its inclusion into the trivial bundle is a holomorphic bundle map.

A global holomorphic section composed with that inclusion has $n+1$ component holomorphic functions on compact connected Complex projective space. Each is constant: its absolute value attains a maximum; in a coordinate ball the one-variable maximum modulus principle along complex lines makes it locally constant, and connectedness propagates that constant globally. Thus the section is a fixed vector $v\in {\mathbb{C}}^{n+1}$ lying in every fibre, namely every line $\mathbb{C}Z$. For $n\ge 1$ the intersection of all these lines is zero. HenceThe positive-dimension convention is necessary: ${\mathbb{CP}}^0$ is one point and this section space is $\mathbb{C}$.

A holomorphic line bundle $L\to X$ is a rank-one holomorphic vector bundle: it has local trivializations $L{|}_{U_i}\cong U_i\times \mathbb{C}$ whose fiberwise linear transition functions of a vector bundle are nowhere-zero holomorphic functions. Equivalently, choose local holomorphic frames $e_i$ and write $e_j=g_{ij}{e}_i$ on overlaps, with $g_{ij}\in {\mathcal{O}}^{\ast }(U_i\cap U_j)$ and the cocycle relation on triple overlaps. A holomorphic section over $U$ has coefficients $s_i$ holomorphic on $U\cap U_i$ in $s=s_i{e}_i$; consistency means $s_i=g_{ij}{s}_j$.

The dual bundle has fiber $L_x^{\ast }={\mathrm{Hom}}_{\mathbb{C}}(L_x,\mathbb{C})$. Its dual frames $e_i^{\ast }$ are characterized by $e_i^{\ast }(e_i)=1$, and the frame relation becomes $e_j^{\ast }=g_{ij}^{-1}{e}_i^{\ast }$. Reciprocals of nowhere-zero holomorphic functions are holomorphic, and their cocycle relations persist. This defines the holomorphic dual vector bundle structure on $L^{\ast }$, with holomorphic evaluation pairing $L^{\ast }\otimes L\to {\mathcal{O}}_X$.

Assume now that $X$ is compact and connected, and let $s$ and $t$ be nonzero global holomorphic sections of $L$ and $L^{\ast }$, respectively. The function $t(s)$ is a global holomorphic function. The maximum modulus principle, applied in local complex coordinate lines near a maximum and then propagated on the connected manifold, makes it constant. Moreover it is not identically zero: the zero locus of a nonzero holomorphic section has empty interior by the identity theorem, so the two nonvanishing loci are dense open sets and intersect. Thus $t(s)=c\ne 0$ everywhere.

In particular, $s$ never vanishes. The map $(x,a)\mapsto as(x)$ is a holomorphic bundle isomorphism $X\times \mathbb{C}\to L$; in a local frame its inverse divides by the nowhere-zero holomorphic coefficient of $s$. Conversely, a holomorphic trivialization supplies the constant unit sections of both $L$ and $L^{\ast }$. We have proved the two-section criterion for holomorphic triviality:

Write points of Complex projective space as lines $[Z]\subset {\mathbb{C}}^{n+1}$. The complex tautological line bundle iswith fiber that very line. Its dual is the hyperplane line bundle $\mathcal{O}(1)$. On the affine chart $U_i=\{Z_i\ne 0\}$, the vector $e_i([Z])=Z/Z_i$ is a nowhere-zero holomorphic frame of $\mathcal{O}(-1)$. On $U_i\cap U_j$,Both ratios are nowhere-zero holomorphic functions of the affine coordinates. They exhibit the holomorphic transition functions of a vector bundle for the two bundles explicitly.

Choose a nonzero linear functional $\ell :{\mathbb{C}}^{n+1}\to \mathbb{C}$ whose zero hyperplane is $H$. At $[Z]$, its restriction to $\mathbb{C}Z$ is an element of that line's dual and therefore defines a holomorphic section $s_{\ell }$ of $\mathcal{O}(1)$. In the dual frame above its coefficient isThis expression is holomorphic even along $H$, and its zero locus is exactly $H$. Consequently $s_{\ell }{|}_{{\mathbb{P}}^n\setminus H}$ is the requested nowhere-zero section; its global holomorphic extension vanishes precisely on $H$. The extension need not, and does not, stay nowhere zero on the hyperplane.

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.

An almost complex structure is a smooth real bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular the real dimension is even, say $2m$. After complexifying, the type decomposition of the complexified tangent bundle isThe corresponding differential forms of type (p, q) are smooth sections of ${\Lambda }^p(T^{1,0}M{)}^{\ast }\otimes {\Lambda }^q(T^{0,1}M{)}^{\ast }$, regarded as alternating forms by the wedge product of differential forms. A form of type $(p,q)$ vanishes unless its arguments include $p$ vectors of type $(1,0)$ and $q$ of type $(0,1)$.

Besides vanishing of the Nijenhuis tensor, two equivalent descriptions of an integrable almost complex structure are: the smooth sections of $T^{0,1}M$ are closed under Lie bracket; and every point has complex coordinates in which $J$ is multiplication by $i$, with holomorphic changes of coordinates. The equivalence with existence of these coordinates is the Newlander-Nirenberg theorem for smooth $J$. A further equivalent condition is that the exterior derivative has only the two type components $d=\partial +\bar{\partial }$, with no components of bidegree $(2,-1)$ or $(-1,2)$. Indeed evaluation of $d$ of a $(1,0)$-form on two $(0,1)$ vector fields detects the $(1,0)$ component of their bracket; complex conjugation detects the other component.

For the almost complex structure induced by a complex atlas, write $z^j=x^j+i{y}^j$ and put $J{\partial }_{x^j}={\partial }_{y^j}$, $J{\partial }_{y^j}=-{\partial }_{x^j}$. The differentials of holomorphic transition maps are complex linear, so these local definitions agree. In these coordinates $J$ has constant coefficients and all coordinate vector fields commute. The expression defining the Nijenhuis tensor is tensorial, so evaluating it on this frame proves it is zero everywhere. Thus the induced structure is integrable.

Here is a direct proof for the smooth submanifolds with invariant complex tangent spaces are complex assertion. At a point of $Y$, its tangent space is a complex-linear subspace, of real dimension $2k$. Make a complex-linear change of ambient holomorphic coordinates so that this tangent space is ${\mathbb{C}}^k\times 0$. Projection onto the first $k$ coordinates has invertible real differential on $Y$. The real inverse function theorem states that a smooth map between equal-dimensional manifolds with invertible differential is a diffeomorphism on sufficiently small neighbourhoods. It therefore writes $Y$ locally as a smooth graph $w=F(z)$ over an open set in ${\mathbb{C}}^k$.

A tangent vector to this graph is $(v,dF(v))$. Its image under $J$ is $(iv,idF(v))$. Tangent-space invariance impliesso $F$ satisfies the Cauchy-Riemann equations. Smooth functions satisfying these equations are holomorphic: restrict to each coordinate complex line, use the one-variable criterion, and apply the iterated Cauchy integral formula on a small polydisc to get jointly holomorphic coefficients. Hence the graph parametrization is holomorphic. Restrictions of ambient holomorphic coordinates supply holomorphic transition maps between these graph charts. The restricted structure on $Y$ is an integrable complex structure. This also proves the induced structure agrees with the original restriction, rather than merely producing some complex structure on $Y$.

A nonsingular analytic subvariety is a closed analytic subset which near each of its points is the zero set of a holomorphic submersion of constant rank; equivalently, it is locally a coordinate subspace in holomorphic coordinates. The holomorphic implicit function theorem says that if a holomorphic map to ${\mathbb{C}}^r$ has differential of rank $r$ at a zero, its nearby zero set is a complex submanifold of codimension $r$.

The homogeneous equation defines a subset of Complex projective space because multiplying a vector by $\lambda \ne 0$ multiplies its quadratic value by ${\lambda }^2$. Its gradient on ${\mathbb{C}}^4$ is $2Qz$, which is nonzero for $z\ne 0$. More explicitly, in an affine chart set $z_i=1$. If all derivatives of the restricted equation vanished at one of its zeros, all components of $2Qz$ except possibly the $i$th would vanish. The Euler homogeneous function theorem gives ${\sum }_j{z}_j{\partial }_{j}p=2p=0$, forcing the remaining component to vanish as well. This contradicts invertibility of $Q$. The holomorphic implicit function theorem thus proves $S$ is a smooth complex surface.

To see its geometry, reduce the nondegenerate symmetric bilinear form to diagonal form over $\mathbb{C}$. There is a vector of nonzero self-pairing, since otherwise polarization would make the entire form zero. Normalize such a vector using a square root of its self-pairing. Its orthogonal complement remains nondegenerate, so induction gives an orthonormal basis. Apply the same argument to the nondegenerate form $w_0{w}_3-w_1{w}_2$: the two forms are equivalent under an invertible linear change of variables. The allowed projective linear transformation therefore reduces $S$ toView its coordinates as the nonzero rank-one matrix $A=\left(\begin{array}{cc}{w}_0& w_1\\ w_2& w_3\end{array}\right)=u{v}^T$. At $[A]$, holding $[u]$ fixed and varying $[v]$ gives a projective line, and holding $[v]$ fixed and varying $[u]$ gives another. They are distinct and meet at $[A]$. In fact these are the only lines through it: if a second rank-one matrix is $a{b}^T$, the entire pencil has determinant zero only if $\det (u,a)\det (v,b)=0$, so one projective factor is fixed. These are the two rulings of a smooth quadric surface.

The Segre embedding is the holomorphic mapEvery nonzero rank-one matrix has such a factorization, unique up to reciprocal rescaling, so $\Phi$ is bijective. Its inverse is holomorphic on each open set where a matrix entry $a_{ij}\ne 0$: the projective column $[a_{0j}:a_{1j}]$ recovers $[u]$, and the projective row $[a_{i0}:a_{i1}]$ recovers $[v]$. These formulas agree on overlaps. Thus the Segre description of a smooth quadric surface gives the promised stronger conclusion

An irreducible complex analytic hypersurface is a closed analytic subset of pure complex codimension one which cannot be expressed as the union of two proper closed analytic subsets. Here a local defining function of a complex analytic hypersurface means a reduced local equation: a holomorphic germ generating its vanishing ideal, with each local branch occurring once. The word reduced matters. Merely prescribing the zero set would allow both $f$ and $f^2$, and the requested uniqueness would then be false.

The local facts used are these: in holomorphic coordinates the local ring ${\mathcal{O}}_{X,p}$ is $\mathbb{C}\{z_1,\dots ,z_n\}$, a regular local ring and a unique factorization domain; its units are exactly the germs nonzero at $p$; and the analytic Nullstellensatz identifies the zero-set ideal of a germ $f$ with $\sqrt{(f)}$. A reduced germ is a product of distinct irreducible factors. Unique factorization makes its principal ideal radical: if $f\mid h^r$, every irreducible factor of $f$ divides $h$, so $f\mid h$. Thus two reduced germs with the same hypersurface germ have $(f)=(g)$. Write $g=uf$, $f=vg$. The local ring is a domain, so $uv=1$. Hence $u$ is a holomorphic unit and $u(p)\ne 0$. This proves reduced local defining functions differ by a holomorphic unit. Global irreducibility does not imply local irreducibility at a singular point; the same argument uses the product of all local branches and therefore covers that case too.

A divisor on a complex manifold is a locally finite integer combination of irreducible analytic hypersurfaces. On a compact manifold only finitely many have nonzero coefficients. Choose local meromorphic equations $f_i$ for a divisor $D$, by products of powers of reduced local equations. Their ratios are holomorphic units on overlaps. For the holomorphic line bundle associated to a divisor, choose frames satisfyingThe ratios obey the cocycle identity and so define a holomorphic line bundle $[D]$. This convention makes the local meromorphic sections $f_i{e}_i$ agree, with divisor $D$. Changing the equations by units changes frames and gives an isomorphic bundle.

For hyperplanes $H_1,H_2$ in Complex projective space, choose nonzero linear equations ${\ell }_1,{\ell }_2$. The degree-zero quotient $r={\ell }_1/{\ell }_2$ is a global meromorphic function and has divisor $H_1-H_2$. A principal divisor on a complex manifold gives a trivial associated bundle: using the same global meromorphic equation on every chart makes all frame transitions equal to one. The allowed tensor-product rule now givesIf the hyperplanes coincide the same conclusion is immediate. Equivalently, their common bundle is the hyperplane bundle $\mathcal{O}(1)$.

Locally at the centre of a blowup of a complex manifold at a point, replace a coordinate neighbourhood in ${\mathbb{C}}^n$ byAway from zero the direction is uniquely $[z]$, so this map is an isomorphism there; glue it to $X\setminus \{p\}$. In the chart ${\ell }_j\ne 0$, write $z_j=u$ and $z_i=u{t}_i$ for $i\ne j$. These are smooth holomorphic coordinates on the blowup. The exceptional divisor is $u=0$, globally $E={\sigma }^{-1}(p)\cong \mathbb{P}(T_{p}X)$.

The canonical bundle formula for a point blowup isThe local chart also explains the exponent: pulling back a coordinate volume form gives $\pm u^{n-1}du\wedge {\bigwedge }_{i\ne j}d{t}_i$, so the Jacobian section has divisor $(n-1)E$.

Since $Y$ is smooth at the centre, choose coordinates with $Y=\{z_1=0\}$. Its strict transform is a genuine analytic hypersurface. In a blowup chart $j\ne 1$, it is the closed set $t_1=0$: off $E$, the original equation $u{t}_1=0$ reduces to $t_1=0$, and taking the closure restores the points with $u=0$. In chart $j=1$, there are no such off-exceptional points, and the strict transform is empty there. These descriptions agree on overlaps and show analyticity, with $\tilde{Y}\cap E\cong {\mathbb{CP}}^{n-2}$. The total transform has the extra factor $u$ exactly once, soUnder the given anticanonical assumption, this yieldsTo prove the only-if direction, one must rule out an accidental trivial power of $[E]$. The local blowup is the tautological line bundle over ${\mathbb{CP}}^{n-1}$, with $E$ its zero section. Its normal bundle is therefore ${\mathcal{O}}_E(-1)$. The normal bundle of a smooth analytic hypersurface identifies $[E]{|}_E$ with this bundle. If $n\ge 2$, restrict to a projective line inside $E$; the restriction of $[E{]}^{\otimes (2-n)}$ has degree $n-2$. A trivial line bundle has degree zero, so it is trivial only if $n=2$. Conversely the displayed tensor factor is literally trivial when $n=2$. This is the anticanonical strict transform under a point blowup criterion:For completeness, dimension one cannot satisfy the premise in this compact setting: an irreducible analytic hypersurface is one point, and $[-Y]$ has degree $-1$ on its curve component, whereas $K_X$ has degree $2g-2$. Thus no exceptional dimension-one case invalidates the equivalence.

In the compact twistor correspondence, a point $p$ of complexified compactified Minkowski space is a two-dimensional subspace $W_p\subset \mathbb{T}$. The correspondence fiber consists of flags $(\ell ,W_p)$ with $\ell$ a one-dimensional subspace of $W_p$. Its image in projective twistor space is precisely $\mathbb{P}(W_p)$, soThis is the twistor line of $p$. In the affine chart the twistor incidence relation gives $W_p=\{(i{p}^{A{A}^{\prime }}{\pi }_{A^{\prime }},{\pi }_{A^{\prime }}):\pi \in {\mathbb{C}}^2\}$. The linear map $\pi \mapsto (ip\pi ,\pi )$ is injective, and projectivizing it gives the same Complex projective space ${\mathbb{CP}}^1$. The subspace argument also covers points at conformal infinity, where that graph chart fails.

A point $q\in \mathbb{PT}$ specifies a one-dimensional subspace ${\ell }_q\subset \mathbb{T}$. Its correspondence fiber consists of two-planes $W$ containing ${\ell }_q$. Passing to the quotient gives a bijection, with inverse taking the inverse image of a line,Since ${\dim }_{\mathbb{C}}(\mathbb{T}/{\ell }_q)=3$, this is an isomorphism of Complex projective spaces. ConsequentlyIt is the compact alpha-plane associated with $q$. In an affine chart with $\pi \ne 0$, the solutions of the twistor incidence relation are $x^{A{A}^{\prime }}=x_0^{A{A}^{\prime }}+u^A{\pi }^{A^{\prime }}$, where $u^A$ supplies two complex coordinates. That affine ${\mathbb{C}}^2$ is only an open subset of the compact ${\mathbb{CP}}^2$; distinguishing them is essential to the printed compact correspondence.

An alpha-plane is a totally null complex two-plane with tangent vectors $u^A{\pi }^{A^{\prime }}$ for a fixed primed spinor. A beta plane has tangent vectors ${\lambda }^A{u}^{A^{\prime }}$ for a fixed unprimed spinor. The metric tensor pairs two vectors of either family to zero, since the repeated fixed spinor contracts with an symplectic form. These are the two chiral families of totally null planes; labels for their self-duality depend on the orientation convention.

Globally in the twistor correspondence, writewhere $\ell$ is a line and $H$ a three-dimensional hyperplane in twistor space. Both compact families are Complex projective spaces ${\mathbb{CP}}^2$.

For distinct ${\ell }_1,{\ell }_2$, any $W\in A_{{\ell }_1}\cap A_{{\ell }_2}$ must contain their two-dimensional span. Since $W$ itself has dimension two, it must equal that span. Conversely the span belongs to both planes. Thus two distinct compact alpha-planes meet in exactly one point:In affine coordinates, independent ${\pi }_1,{\pi }_2$ determine a unique matrix $x$ from the two incidence equations. If the primed spinors are proportional while the projective twistors are distinct, no finite affine solution exists. The compact intersection still exists, at conformal infinity; saying the planes are disjoint without this affine qualification would be wrong for the printed space $M$.

Use the standard orientations on the three spheres and the complex orientation on Complex projective space. Let $h\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ satisfy $⟨h^3,[{\mathbb{CP}}^3]⟩=1$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^4)$. The Künneth theorem givesWrite the induced map on cohomology as $f^{\ast }h=pa+qb+rc$ with integers $p,q,r$. Naturality and graded commutativity of the cup product giveThe mapping degree is therefore $d=6pqr$. This is the factorial degree obstruction for products of two-spheres. Every positive value is a positive multiple of six, so the least possible positive degree is at least $6$.

To attain it, identify each oriented sphere with ${\mathbb{CP}}^1$. Send three projective linear forms to the coefficients of their product:whereNo product of nonzero linear forms is the zero polynomial. Rescaling any factor rescales all coefficients by the same nonzero scalar. The map is therefore well-defined and is a holomorphic map on the products of projective charts.

Choose a homogeneous cubic with three distinct roots, all in an affine chart. Its preimages are exactly the $3!=6$ orderings of its three projective linear factors. In root coordinates, the derivative of the elementary-symmetric coefficient map has determinant equal, up to sign, to the Vandermonde determinant, the product of the three pairwise root differences. This is nonzero for distinct roots, so all six preimages are regular. A nonsingular complex-linear derivative preserves the underlying real orientation, since its real determinant is the squared absolute value of its complex determinant. Each preimage consequently contributes $+1$ to the mapping degree. HenceThis polynomial multiplication map to complex projective space provides the requested example without assuming that a prescribed cohomology homomorphism is realizable by a continuous map.

The holomorphic normal bundle is the quotient $N_V=T^{1,0}M{|}_V/T^{1,0}V$, of rank $r$. The tangent sequence is a short exact sequence of holomorphic vector bundles:Taking determinant line bundles gives $\det (T^{1,0}M){|}_V\cong \det (T^{1,0}V)\otimes \det N_V$. Dualizing and rearranging proves adjunction for a smooth submanifold:The determinant identity follows locally by adjoining lifts of a quotient frame to a subbundle frame; changing the lifts adds only off-diagonal blocks, so does not change the determinant.

For a smooth divisor, choose reduced local defining functions $f_i$, with $f_i=u_{ij}{f}_j$ and $u_{ij}$ holomorphic and nowhere zero. The holomorphic line bundle associated to a divisor $[V]={\mathcal{O}}_M(V)$ has local frames $e_i=f_i^{-1}$, with $e_j=u_{ij}{e}_i$. The products $f_i{e}_i$ define its canonical section vanishing on $V$. Along $V$,Thus $v\mapsto d{f}_i(v)e_i$ patches to a map $T^{1,0}M{|}_V\to [V]{|}_V$, whose kernel is $T^{1,0}V$ and which is surjective because each reduced defining function has nonzero differential there. Hence $N_V\cong [V]{|}_V$. This is the normal bundle of a smooth analytic hypersurface. The divisor construction uses a closed hypersurface: an arbitrary nonclosed embedded hypersurface need not be a divisor. For example, a punctured line in ${\mathbb{C}}^2$ cannot be the support of a divisor, since its missing limit point would also lie in every local holomorphic zero set containing that line.

Let $s$ be the section of $E$ defining $V$. The intended hypothesis is that it is a regular zero locus: $ds$ has rank $r$ on $V$. In a local trivialization, differentiation of the component functions definesThis is intrinsic: differentiating a change-of-frame matrix introduces terms multiplied by $s$, which vanish on $V$. Its kernel is $T^{1,0}V$, so it induces an isomorphism of holomorphic normal bundles. ConsequentlyThis is the normal bundle of a regular zero locus. Here $\det E=L_1\otimes \cdots \otimes L_r$.

Regularity is essential if “vanishing” is read only as equality of underlying sets. On ${\mathbb{P}}^2$, the section $Z_0^2$ of $\mathcal{O}(2)$ vanishes set-theoretically on a smooth line $V$. Its derivative vanishes on $V$, andThus smoothness of the underlying zero set alone does not establish the claimed normal-bundle isomorphism. The preceding proof applies when the equations define the reduced smooth submanifold, equivalently are transverse to zero.

For Complex projective space, put $\mathcal{O}(a)=[aH]$. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Quotienting $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ by its scalar maps into $\ell$ yields the Euler sequenceIts determinant gives $K_{{\mathbb{P}}^n}\cong \mathcal{O}(-n-1)=[-(n+1)H]$. For a smooth complete intersection defined regularly by the homogeneous equations, apply the normal-bundle result to $E={⨁}_i\mathcal{O}(d_i)$:This is the canonical bundle of a regular projective complete intersection. The same regularity qualification is needed here: the squared-line example would otherwise predict ${\mathcal{O}}_{{\mathbb{P}}^1}(-1)$ for its canonical bundle, whereas the line has $K_V={\mathcal{O}}_{{\mathbb{P}}^1}(-2)$.

Let $q:{\mathbb{CP}}^8\to Q$ collapse the indicated Complex projective space. Write $x\in H^2({\mathbb{CP}}^8;\mathbb{Z})$ for the positive generator, so the cohomology ring of complex projective space is $\mathbb{Z}[x]/(x^9)$. The inclusion of ${\mathbb{CP}}^3$ sends $x$ to its corresponding generator and is an isomorphism in degrees $0,2,4,6$.

The CW complex structure makes the inclusion a cofibration, so the positive-degree cohomology of $Q$ identifies with the relative cohomology of the pair. Its long exact sequence shows that $q^{\ast }$ is injective in positive degrees, with image the additive subgroup generated by $x^4,x^5,x^6,x^7,x^8$. Equivalently, the surviving cellular cochain complex has one copy of $\mathbb{Z}$ in each of degrees $8,10,12,14,16$ and zero differential. Let $u_i$ denote the class with $q^{\ast }{u}_i=x^i$, for $4\le i\le 8$.

Naturality of the cup product, together with the injectivity of $q^{\ast }$, now determines all multiplication:Every other product of positive-degree basis elements is zero. Indeed, their degrees exceed $16$. In particular, the surviving classes do not form a polynomial algebra on a degree-eight generator: the independent classes in degrees $10,12,14$ must also be retained. This is an instance of the cohomology ring of a collapsed projective subspace.

The divisor determines the bundle through its local equations. Choose an open cover on which $V$ has a reduced holomorphic defining equation $f_i$, using $f_i=1$ on sets disjoint from $V$. Because $V$ is a complex submanifold of codimension of a submanifold one, its ideal is locally generated by a coordinate; consequently $f_i/f_j$ is a nowhere-zero holomorphic function on each overlap. Inside the sheaf of meromorphic functions on a complex manifold, consider the locally free sheaf of rank oneThese local descriptions agree on overlaps. With $e_i=1/f_i$, its transition functions of a vector bundle are specified byThe ratios satisfy the Čech cocycle condition, and the associated holomorphic line bundle associated to a divisor is $[V]$. The section $1=f_i{e}_i$ is holomorphic and vanishes to order one along $V$. Replacing $f_i$ by $u_i{f}_i$, with $u_i$ a holomorphic unit, replaces the frame by $u_i^{-1}{e}_i$ without changing the subsheaf of meromorphic functions. Refining the cover changes no local sections either. This proves choice independence up to isomorphism rather than merely producing a bundle for one cover.

For a complex manifold of dimension $n$, the canonical bundle isIt is the holomorphic line bundle of holomorphic top-degree differential forms. To compute it on Complex projective space, let $S=\mathcal{O}(-1)$ be the complex tautological line bundle, so that $S^{\ast }=[H]=\mathcal{O}(1)$ is the hyperplane line bundle. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Tensoring the tautological quotient sequence by $S^{\ast }$ gives the Euler sequence on complex projective spaceTaking the top exterior power gives $\det T^{1,0}{\mathbb{P}}^n=\mathcal{O}(n+1)$; taking its dual bundle yieldsHere a negative tensor power means the corresponding positive tensor power of the dual bundle.

The proposed sections over $U$ form a sheaf: compatible maps into $L$ glue uniquely, and their projections remain $\varphi$. Addition and multiplication by holomorphic functions are defined in the fibres of $L$, making this a sheaf of ${\mathcal{O}}_X$-modules. If $e_i$ is a holomorphic local frame of $L$ on $W_i\subset Y$, every section over $U\subset {\varphi }^{-1}(W_i)$ has the unique expressionThus the sheaf is a locally free sheaf of rank one. Its transition functions of a vector bundle are those of $L$ composed with $\varphi$, so the corresponding pullback vector bundle is precisely ${\varphi }^{\ast }L$. This proves invertibility locally and compatibility globally.

For the local blowup of a complex manifold at a point, work on the projective chart $Y_i\ne 0$ and writeAll the incidence equations reduce to $x_j=u{t}_j$. Therefore $(u,(t_j{)}_{j\ne i})$ gives a holomorphic coordinate chart isomorphic to ${\mathbb{C}}^n$. On an overlap with $Y_{\ell }\ne 0$,which is a biholomorphism where $t_{\ell }\ne 0$. The incidence subset is closed in the product of two Hausdorff spaces and is covered by these charts; hence it is a complex manifold of dimension $n$. In each chart the exceptional fibre is $u=0$, so it is a smooth complex submanifold, with its induced projective coordinates givingAway from the origin the inverse of the projection is $x\mapsto (x,[x])$, which is holomorphic; the projection is therefore a biholomorphism outside $E$.

To find the canonical bundle, pull back the nowhere-zero top form on ${\mathbb{C}}^n$. In the chart above, its Jacobian determinant givesThe sign depends only on the ordering of the coordinates. This is a section of $K_X$ with divisor on a complex manifold exactly $(n-1)E$: it has that vanishing order in every exceptional chart and no zeros elsewhere. A nonzero meromorphic section of a holomorphic line bundle with divisor $D$ identifies its holomorphic line bundle with $[D]$, by sending the local generator $1/f_i$ of $\mathcal{O}(D)$ to the nowhere-zero frame obtained by dividing the section by $f_i$. ConsequentlyThis also includes $n=1$, when the local map is an isomorphism and the exponent is zero.

For a general complex manifold, choose a coordinate neighbourhood $U$ of $P$, with $P$ sent to $0$, and replace $U$ by the inverse image of this coordinate neighbourhood in the local blowup of a complex manifold at a point. Glue this space to $M\setminus \{P\}$ along $U\setminus \{P\}$ using the preceding biholomorphism. The resulting charts give a complex manifold $\tilde{M}$ and a holomorphic map $\varphi$; the exceptional fibre is ${\mathbb{P}}^{n-1}$ and the map is a biholomorphism off that fibre. For completeness, the local projection is proper because its inverse image over a compact subset is closed in that subset times compact Complex projective space. This ensures the gluing is Hausdorff: separate points with distinct images downstairs, and points over $P$ inside the local blowup. The atlas is second countable as well.