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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

For $U_{i_0\dots i_q}=U_{i_0}\cap \cdots \cap U_{i_q}$, the Čech cochain complex has groupsTerms in ${\delta }^2$ cancel in pairs, and Čech cohomology is $\ker \delta /\mathrm{im}\delta$. A degree-zero cocycle is exactly a family of compatible local sections. The sheaf gluing axiom gives their unique global section, proving ${\check{H}}^0(\mathcal{U},\mathcal{F})\cong \mathcal{F}(X)$. If the cover has affine finite intersections, a quasi-coherent sheaf has no higher cohomology on those intersections by vanishing of quasi-coherent cohomology on an affine scheme. The acyclic cover theorem then identifies all Čech groups with sheaf cohomology. In particular, a finite affine open cover of a separated variety has this property.

For the sheaf of units of the structure sheaf, a multiplicative degree-one cocycle is a family $g_{ij}\in {\mathcal{O}}_X^{\ast }(U_i\cap U_j)$ satisfying $g_{ij}{g}_{jk}=g_{ik}$, with $g_{ji}=g_{ij}^{-1}$. It glues trivial rank-one free modules into an invertible sheaf. Changing the local frames multiplies $g_{ij}$ by a coboundary, and two sets of transition data give isomorphic line bundles precisely when their cocycles differ this way. Tensoring line bundles multiplies their cocycles. Thus line bundles trivialized by an open cover give the group isomorphismFor the remaining arguments, work over the algebraically closed ground field. On the irreducible variety $V$, put $\mathcal{Q}={\mathcal{K}}^{\ast }/{\mathcal{O}}_V^{\ast }$, a quotient of sheaves of abelian groups. A global section of $\mathcal{Q}$ is locally represented by rational functions $f_i$ whose ratios are regular units. The corresponding unit cocycle $f_j/f_i$ defines an invertible sheaf. A single global rational function has trivial cocycle. Conversely, every line bundle has a nonzero rational section: choose a nonzero vector in its one-dimensional fibre at the generic point and express it in local frames. This supplies such local $f_i$. If the associated line bundle is trivial, changing frames makes all $f_i$ restrictions of one rational function. Thereforeis exact. This is the Cartier-divisor description of the Picard group. The sheaf of nonzero rational functions on an irreducible variety is flasque, since all restrictions between nonempty open sets are the identity on $k(V{)}^{\ast }$. Apply the long exact sequence in sheaf cohomology to $1\to {\mathcal{O}}_V^{\ast }\to {\mathcal{K}}^{\ast }\to \mathcal{Q}\to 1$. Since $H^1(V,{\mathcal{K}}^{\ast })=0$, its connecting map has exactly the cokernel just computed, provingFinally the Segre description of a smooth quadric surface identifies $V$ with ${\mathbb{P}}^1\times {\mathbb{P}}^1$. The two rulings of a smooth quadric surface have classes $F_1,F_2$ generating $\mathrm{Pic}(V)\cong {\mathbb{Z}}^2$. For completeness, remove one line in each ruling: the remaining chart is the affine plane, with factorial coordinate ring $k[s,t]$ and trivial divisor class group. The localization sequence for the divisor class group makes $F_1,F_2$ generators, and their degrees on the two ruling lines prove independence. The hyperplane class, and hence the conic $C$, has bidegree $(1,1)$. By Picard-group localization on a smooth variety, the Picard group of a smooth affine quadric surface isExplicitly, the restriction of ${\mathcal{O}}_V(1,0)$ is nontrivial: if it were trivial on $U$, its rational trivialization would have divisor supported on $C$, forcing $(1,0)$ to be an integer multiple of $(1,1)$, which is impossible.

The hyperplane sections through $P$ are spanned by the three linear forms $x_1,x_2,x_0-x_3$. Their common zero on $X$ is precisely $P$, so away from $P$ the linear system of divisors is base-point-free and defines the projection from a point on a smooth quadricIn particular, the chart-ratio construction in part (a) makes this a morphism of schemes on $X\setminus \{P\}$, not merely a rational map there.

Write a target point as $[u:v:w]$. The corresponding line through $P$ consists of pointsSubstituting in the quadric equation givesRemoving $P$ means $\mu \ne 0$, so set $\mu =1$. Over the residue field $K$ of the target point, its scheme-theoretic fibre is consequentlyIf $u\ne v$, this is one reduced point, with $\lambda =-uw/(u-v)$. If $u=v$ and $uw\ne 0$, it is empty: the corresponding line meets the quadric only at the removed point, with intersection multiplicity two. If $u=v$ and $uw=0$, the line lies entirely on the quadric, and the scheme-theoretic fibre is ${\mathbb{A}}_K^1$, the line with $P$ removed.

The last case occurs at exactly $[0:0:1]$ and $[1:1:0]$. Their lines are respectivelythe two rulings of a smooth quadric surface through $P$. Thus the image consists of the complement of the line $u=v$, together with those two points. The line $u=v$ parametrizes directions in the tangent plane $x_1=x_2$ at $P$. The calculation describes the scheme-theoretic fibres over arbitrary target points and works in every characteristic.

Over an algebraically closed field of characteristic different from two, a projective change of coordinates puts every smooth quadric surface in the formThe Segre embedding identifies this surface with ${\mathbb{P}}^1\times {\mathbb{P}}^1$ byTwo distinct fibres of either coordinate projection are disjoint lines in one of the rulings of a smooth quadric surface. Explicitly,both lie on $V$ and have no common projective point. Thus $V$ contains a pair of disjoint lines.

Over an algebraically closed field, remove a smooth hyperplane conic $C$ from a smooth quadric surface $Q$. The rulings of a smooth quadric surface generate $\mathrm{Pic}(Q)={\mathbb{Z}}^2$, and $C$ has class $(1,1)$. Picard-group localization on a smooth variety gives the quotient by $\mathbb{Z}(1,1)$. The restricted ruling bundle ${\mathcal{O}}_Q(1,0)$ is a generator, exhibiting a nontrivial line bundle on an affine surface.

For $X=\{x_0{x}_1=x_2{x}_3\}$ and $P=[1:0:0:1]$, the displayed map is defined on $X\setminus \{P\}$. Its fibre over $[u:v:w]$ has equation $(u-v)\lambda +uw=0$ on an affine line. Away from $u=v$ the fibre is one reduced point. On that line it is empty except at $[0:0:1]$ and $[1:1:0]$, where it is a punctured ruling line, hence an affine line. The two exceptional lines are the rulings of a smooth quadric surface through $P$.