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The Newlander-Nirenberg theorem says that a smooth almost complex structure is integrable if and only if its Nijenhuis tensor vanishes.

An almost complex structure is a smooth endomorphism $J:TM\to TM$ with $J^2=-I$; in particular the real dimension is even. Complexifying the tangent bundle splits it into the $i$ and $-i$ eigensubbundlesinterchanged by conjugation. On a complex manifold, these are locally spanned by ${\partial }_{z^j}$ and ${\partial }_{{\bar{z}}^j}$. An integrable almost complex structure is one arising from a holomorphic coordinate atlas. Its $T^{0,1}$ sections are closed under the Lie bracket. The obstruction is the Nijenhuis tensorExpanding this expression in the two eigentypes shows that $N_J=0$ is equivalent to involutivity of $T^{0,1}$. Holomorphic coordinates make this condition necessary; the Newlander-Nirenberg theorem gives its sufficiency for smooth $J$. Integrability also gives $d=\partial +\bar{\partial }$, with ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault operator on $T^{1,0}M=T_M^{\prime }$ defines its holomorphic vector bundle structure.

A real connection on a vector bundle $\nabla$ on $TM$ preserves $J$ when $\nabla J=0$. Its complexification then preserves both eigensubbundles. Restricting to $T_M^{\prime }$ gives a complex connection $D$. Conversely any complex connection on $T_M^{\prime }$ gives such a real connection by adjoining its conjugate on $T^{0,1}$ and restricting to the conjugation-fixed real tangent bundle. Equivalently, transfer $D$ through the real isomorphism $v\mapsto (v-iJv)/2$. These operations are inverse. Preserving $J$ alone does not impose the holomorphic compatibility condition $D^{0,1}={\bar{\partial }}_{T^{\prime }}$.

For integrable $J$, the latter condition has a precise real-connection interpretation: the mixed-type torsion tensor vanishes. In holomorphic coordinates it says $D_{{\partial }_{{\bar{z}}^i}}{\partial }_{z^j}=0$. Reality also gives ${\nabla }_{{\partial }_{z^j}}{\partial }_{{\bar{z}}^i}=0$. Since these coordinate fields commute, their torsion is zero. Conversely a $J$-preserving real connection with zero mixed torsion has these two derivatives equal; they belong to opposite eigentypes, so both vanish, proving $D^{0,1}={\bar{\partial }}_{T^{\prime }}$. This is the holomorphic tangent connection and mixed torsion criterion.

Full torsion-freeness imposes the additional symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ in $D_{{\partial }_{z^i}}{\partial }_{z^j}={\Gamma }_{ij}^k{\partial }_{z^k}$, and its conjugate. Such a $J$-preserving torsion-free connection exists on any complex manifold: local holomorphic-coordinate flat connections have these properties, and a real partition of unity combines them globally. Preservation of $J$ and zero torsion persist because both conditions are affine in the connection. Conversely if $J$ is merely almost complex, a torsion-free connection preserving it forces $N_J=0$: replace brackets by ${\nabla }_{X}Y-{\nabla }_{Y}X$ in the displayed tensor and use $\nabla J=0$; all terms cancel. Thus this connection condition detects integrability, without imposing any metric condition.

A Hermitian metric is a real Riemannian metric $g$ with $g(JX,JY)=g(X,Y)$. It induces a Hermitian metric $h$ on $T_M^{\prime }$ and the fundamental two-form $\omega (X,Y)=g(JX,Y)$. The unique Chern connection satisfies both $D^{0,1}={\bar{\partial }}_{T^{\prime }}$ and metric compatibility. In a holomorphic frame with metric matrix $H$ and coefficient-column convention, its matrix is $H^{-1}\partial H$. It need not be torsion-free when transferred to $TM$.

In holomorphic coordinates its coefficients are ${\Gamma }_{ij}^k=h^{k\bar{l}}{\partial }_i{h}_{j\bar{l}}$. Their symmetry in $i,j$ is equivalent to ${\partial }_i{h}_{j\bar{l}}={\partial }_j{h}_{i\bar{l}}$, which says $\partial \omega =0$. The conjugate gives $\bar{\partial }\omega =0$. The result that a torsion-free Chern tangent connection characterizes a Kähler metric follows: the transferred Chern connection is torsion-free exactly when $\boxed{d\omega =0}$, the Kähler metric condition. In that case it equals the Levi-Civita connection, by uniqueness of the metric-compatible torsion-free connection. Equivalently a Hermitian metric is Kähler exactly when its Levi-Civita connection preserves $J$. In the reverse direction $\nabla J=0$ implies $\nabla \omega =0$, and zero torsion gives $d\omega =0$.

This explains the role of Kähler metrics: they make the real Riemannian and holomorphic connection theories coincide. A general complex manifold has torsion-free $J$-preserving connections and has Hermitian metrics, but need not have a connection satisfying both requirements simultaneously. On a Kähler manifold, the Kähler identities then link the Dolbeault Laplacian and Hodge Laplacian, giving ${\Delta }_d=2{\Delta }_{\bar{\partial }}=2{\Delta }_{\partial }$ and the resulting harmonic Hodge decomposition theorem for compact Kähler manifolds.

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

With the normalization used here, $N=2N_J$, where $N_J$ is the standard Nijenhuis tensor. First check that it is a tensor: the Lie bracket identity $[fU,V]=f[U,V]-V(f)U$ givesSkew symmetry gives the same $C^{\infty }$-linearity in the second input. In coordinates from a holomorphic atlas, the real coordinate vector fields ${\partial }_{x_j},{\partial }_{y_j}$ commute, and $J$ sends each to another such field with a constant sign. Every Lie bracket in $N_J$ therefore vanishes on these coordinate fields. Tensoriality then gives $\boxed{N=0}$ on all smooth vector fields. The factor two in this paper does not change this vanishing conclusion.

Use the complex-bilinear extension of the Nijenhuis tensor expression to complex vector fields. If $U,V\in T^{1,0}X$, then $JU=iU$ and $JV=iV$, soFor $U,V\in T^{0,1}X$, the same substitution with $-i$ gives $N(U,V)=-8[U,V{]}^{1,0}$. For inputs of opposite types, the two leading Lie brackets cancel, as do the two terms involving $J$, so $N(U,V)=0$. Applying the type decomposition of the complexified tangent bundle to both inputs consequently givesThus $N=0$ precisely when the Lie bracket of any two sections of either eigenbundle remains in that same eigenbundle. In other words, $N=0$ if and only if both tangent-type distributions are involutive distributions. This proves the equivalence directly and does not assume that an arbitrary almost complex structure already has holomorphic coordinates.

The restriction of $J$ to the $J$-invariant bundle $TY$ is an almost complex structure. Its Nijenhuis tensor is the restriction of the ambient Nijenhuis tensor because vector fields tangent to an embedded submanifold have tangent Lie bracket. The ambient tensor vanishes since $X$ is a complex manifold, so the Newlander-Nirenberg theorem makes the induced structure on $Y$ integrable. The inclusion has complex-linear differential and is therefore holomorphic; hence $Y$ is a complex submanifold.

For a complex submanifold, the holomorphic normal bundle isIf $Y$ is a smooth hypersurface, taking top exterior powers in the holomorphic conormal sequencegivesThe normal bundle of a hypersurface is $N_{Y/X}\cong {\mathcal{O}}_X(Y){|}_Y$, so the adjunction formula is

On $X={\mathbb{CP}}^n\times {\mathbb{CP}}^m$, a bihomogeneous polynomial of bidegree $(d_1,d_2)$ is a section of the holomorphic line bundle $\mathcal{O}(d_1,d_2)$. Its zero locus is smooth precisely when the section is transverse to the zero section, equivalently when $F$ and all of its homogeneous first partial derivatives have no common projective zero. Sincethe adjunction formula yieldsThus a smooth $F$ with $d_1=n+1$ and $d_2=m+1$ defines a complex submanifold with trivial canonical bundle.