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For a matrix-valued gauge field $A$, let $F=dA+A\wedge A$ be its gauge field strength; products below combine matrix multiplication with the wedge product of differential forms. The five-form can be written $Q_5=\mathrm{Tr}(A(dA{)}^2+\frac{3}{2}{A}^{3}dA+\frac{3}{5}{A}^5)$. Applying the exterior derivative, using its graded product rule and the graded cyclic property of the matrix trace, gives $d{Q}_5=\mathrm{Tr}(F^3)$. The six-form is an invariant polynomial in the gauge curvature, while its local potential depends on the connection.

For positive coordinate charts put $J=\partial y/\partial x$. The metric tensor transforms by $g_x=J^T{g}_{y}J$, giving $\det g_x=(\det J{)}^2\det g_y$. The top wedge product of differential forms transforms by $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. Since $\det J>0$, the two factors match. Thus the local expressions define one global Riemannian volume form.

For a smooth map $\varphi :M^{\prime }\to M$ and a connection on a vector bundle $E\to M$, the curvature form of a connection on the pullback vector bundle satisfiesas an endomorphism of $E_{\varphi (p)}$. In a frame of a vector bundle written as a row, with section components written as columns, the connection one-form is $\Omega$ and its curvature form of a connection is $d\Omega +\Omega \wedge \Omega$. The pullback connection has matrix ${\varphi }^{\ast }\Omega$; commutation of the pullback of a differential form with the exterior derivative and wedge product of differential forms proves the identity. No pushforward to an ambient vector field is required: $d\varphi (X)$ is a vector field along a map.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

An exact symplectic manifold has a symplectic form that is an exact differential form, written here $\omega =-d\lambda$. A positive-dimensional example without boundary cannot be compact, by the Generalized Stokes theorem applied to the top wedge product of differential forms.

The exterior derivative is the unique graded derivation $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$ that agrees with the differential on functions and satisfies $d^2=0$.

On a smooth planar one-form $\alpha =Pdx+Qdy$, the product rule and antisymmetry of the wedge product of differential forms giveFor $\alpha =xdy$, the coefficient is $1$: there is a nonzero integral around the boundary of every positively oriented rectangle. For $\beta =d(xy)$, the coefficient is ${\partial }_{x}x-{\partial }_{y}y=0$. More generally, the identity $d^2=0$ makes every exact differential form a closed differential form. It expresses cancellation between mixed partial derivatives locally; it does not imply that every closed differential form has a global potential.

Exterior multiplication by a fixed differential form is the linear operator taking the wedge product of differential forms on the left. For a one-form $\omega$, it raises degree by one. Anticommutativity gives $e(\omega )e(\eta )+e(\eta )e(\omega )=0$ for one-forms. On the exterior algebra of an inner product space, pairing this operation with contraction produces Clifford multiplication $c(v)=e(v^{♭})-{\iota }_v$.

Composition of endomorphisms and wedge product of differential forms give the class $[{\Theta }^k]\in H^k(M,{\Omega }_M^k\otimes \mathrm{End}E)$. Even total degree of vector-bundle curvature and the noncommutative telescoping identity give ${\Theta }_1^k-{\Theta }_0^k=\bar{\partial }{\sum }_{j=0}^{k-1}{\Theta }_1^{j}a{\Theta }_0^{k-1-j}$ whenever ${\Theta }_1-{\Theta }_0=\bar{\partial }a$. Thus all powers are metric independent; no commutativity of endomorphisms is assumed.

The exterior derivative is an $\mathbb{R}$-linear operator $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ which agrees with the differential of a smooth function on degree zero, satisfies $d^2=0$, and obeys the graded Leibniz rule

For the uniqueness of the exterior derivative from its axioms, first note that these axioms force locality. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ equal to one near $p$ and supported where $\alpha =0$. Then $0=d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $d\alpha (p)=0$. Thus $d\alpha (p)$ depends only on the form near $p$.

In a coordinate chart write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Since $d(d{x}^i)=d^2{x}^i=0$, the axioms forceThis also proves existence. Define $d$ by that formula in each chart. The product rule for partial derivatives gives the graded Leibniz rule, and symmetry of mixed partial derivatives against antisymmetry of the wedge product of differential forms gives $d^2=0$. On overlaps it still agrees with the differential of every function, by the chain rule, so the preceding uniqueness argument in the second coordinate system makes the formulas agree. They therefore glue to a global exterior derivative.

For a differential one-form $\omega ={\sum }_i{a}_{i}d{x}^i$, the exterior derivative of a one-form evaluated on vector fields follows directly:In $X(\omega (Y))-Y(\omega (X))$, the additional terms are ${\sum }_i{a}_i[X(Y^i)-Y(X^i)]=\omega ([X,Y])$, where the bracket is the Lie bracket of vector fields. Hence

The de Rham cohomology iswith zero denominator in degree zero. The Poincaré lemma states that a closed differential form of positive degree on a star-shaped open subset of Euclidean space is an exact differential form; consequently every closed positive-degree form is locally exact on a smooth manifold.

For the suspension isomorphism for top de Rham cohomology of spheres, put $U=S^n\setminus \{\text{south pole}\}$ and $V=S^n\setminus \{\text{north pole}\}$. Both are diffeomorphic to ${\mathbb{R}}^n$, while $W=U\cap V$ deformation retracts onto the equatorial $S^{n-1}$. The Mayer--Vietoris sequence for de Rham cohomology comes from the short exact sequence of cochain complexesThe final map is surjective: a partition of unity ${\chi }_U+{\chi }_V=1$ lets a form $\eta$ on $W$ be lifted by extending ${\chi }_V\eta$ to $U$ and $-{\chi }_U\eta$ to $V$ by zero. The Poincaré lemma makes $H^{n-1}(U),H^{n-1}(V),H^n(U),H^n(V)$ vanish when $n>1$. Exactness givesIf $j:S^{n-1}\hookrightarrow W$ is the equatorial inclusion, homotopy invariance of de Rham cohomology makes $j^{\ast }$ an isomorphism. Thereforeis injective, indeed an isomorphism. Concretely, choose local primitives $d{\alpha }_U=\omega {|}_U$, $d{\alpha }_V=\omega {|}_V$, and send $[\omega ]$ to $[j^{\ast }({\alpha }_U-{\alpha }_V)]$. Their difference is closed; changing either primitive changes it by an exact form, so this agrees with the cohomological construction.

A symplectic manifold is a smooth manifold $P$ with a closed, nondegenerate two-form $\omega$. Nondegeneracy forces $\dim P=2n$. Examples are ${\mathbb{R}}^{2n}$ with $\omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, an oriented surface with a nowhere-zero area form, and every cotangent bundle with its canonical form.

A Poisson manifold has a bivector $\Pi$ such that $\{f,g\}=\Pi (df,dg)$ is an antisymmetric bilinear Poisson bracket, is a derivation in each argument, and satisfies the Jacobi identity. Its rank need not be full or constant. The zero bivector gives a Poisson manifold on any smooth manifold, including an odd-dimensional or nonorientable one. A nonzero example on ${\mathbb{R}}^3$ is the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$: it has rank two off the origin and rank zero at the origin, so it does not arise from a symplectic form on the whole three-dimensional space. Even on ${\mathbb{R}}^2$, $\Pi =x{\partial }_x\wedge {\partial }_y$ is Poisson, since its Jacobi condition is a three-vector and hence vanishes, but its degeneracy on $x=0$ prevents it from being the inverse of a global symplectic form. A manifold carrying a degenerate Poisson structure might separately admit some other symplectic structure; these are distinct claims.

At each point of a symplectic manifold, linear nondegeneracy gives a basis with $\omega ={\sum }_i{e}^i\wedge f^i$. Its top wedge product of differential forms isThus ${\omega }^n/n!$ is a global nowhere-zero volume form and selects a consistent symplectic orientation. This proves every symplectic manifold is orientable without choosing a global frame.

Let $\pi :T^{\ast }M\to M$ and let a point of the cotangent bundle be $(x,p)$ with $p\in T_x^{\ast }M$. Define the canonical Liouville one-form intrinsically by ${\theta }_{(x,p)}(V)=p(d\pi (V))$. In local coordinates it is $\theta =p_{i}d{q}^i$, soThis definition is global, $d\omega =0$ by $d^2=0$, and its coordinate matrix has invertible blocks. Hence $T^{\ast }M$ is symplectic for every smooth $M$, even if $M$ itself is not orientable.

The coordinate Jacobi condition for a Poisson bivector isNecessity follows by applying the Jacobi identity to three coordinate functions. For arbitrary functions, the second-derivative terms in the Jacobiator cancel by antisymmetry of $\Pi$; the terms left are exactly this coefficient multiplying $f_i{g}_j{h}_k$, proving sufficiency.

For a symplectic form, take ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. With these signs the matrix of $\Pi$ is $-{\omega }^{-1}$. Differentiation gives ${\partial }_{\ell }\Pi =\Pi ({\partial }_{\ell }\omega )\Pi$. The cyclic expression above consequently equalswhich vanishes because $d\omega =0$. This directly proves that a symplectic manifold defines a Poisson manifold.

Let $X$ be a connected compact Riemann surface of geometric genus $g$, equivalently a smooth projective curve over $\mathbb{C}$, and put $V=H^0(X,K_X)$. Choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$ with $a_i\cdot b_j={\delta }_{ij}$. For a closed differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The underlying bilinear identity isIt holds for any two smooth closed one-forms. In particular, the Riemann bilinear relations for holomorphic differential forms arefor $\omega ,\eta \in V$ and $\omega \ne 0$ in the inequality. The inequality fixes the orientation and the sign convention.

To prove the identity, cut $X$ along the symplectic basis to a fundamental polygon $P$. On $P$, a closed one-form $\alpha$ has a primitive $F$. The Stokes theorem givesOpposite copies of each cut have opposite orientations, while their values of $F$ differ by the corresponding period of $\alpha$. Pairing those edges therefore leaves $A_i(\alpha ){\int }_{b_i}\beta -B_i(\alpha ){\int }_{a_i}\beta$ for the $i$th handle. Summing proves the identity. The same calculation can be made with a small disk about the polygon's vertices removed; its extra boundary terms tend to zero, so the vertices cause no additional term.

For holomorphic differential forms $\omega ,\eta$, their wedge product of differential forms vanishes: locally both are multiples of $dz$. This proves the first relation. If $\omega =f(z)dz$ and $z=x+iy$, thenA nonzero holomorphic differential form is nonzero on an open set, so the integral is strictly positive. Applying the identity to $\omega ,\overline{\omega }$ proves the second relation.

These relations first prove that the map $V\to {\mathbb{C}}^g$ taking $a$-periods is injective: if every $A_i(\omega )$ vanishes, the positive integral just computed would vanish. Since $\dim V=g$ by the Riemann-Roch theorem, it is an isomorphism. Consequently there is a unique basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$. Set ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. The first relation gives ${\tau }_{ij}={\tau }_{ji}$, and the second gives, for any nonzero column $c$,Thus the normalized period matrix of a complex torus is symmetric and has positive-definite imaginary part:

The integration map sends a cycle $\gamma$ to the functional $\omega \mapsto {\int }_{\gamma }\omega$ in $V^{\ast }$. In the normalized basis its image is the period latticeIndeed, the real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: its imaginary part is $Yy$, and $Y$ is invertible. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and has compact quotient. This proves thatis a well-defined $g$-dimensional complex torus, rather than merely a quotient by an arbitrary subgroup.

The relations also provide its canonical principal polarization of a complex torus. With the convention that a Hermitian form is conjugate-linear in its first argument, put $H(z,w)={\overline{z}}^{,t}{Y}^{-1}w$ and $E=\mathrm{Im}H$. Symmetry of $\tau$ givesIn particular $E$ is integral and unimodular on $\Lambda$, and it is compatible with the complex structure. These are exactly the positivity and integrality conditions for a Riemann form on a complex torus; they make the complex torus a Jacobian variety with a principal polarization of a complex torus. If instead one uses a first-argument-linear Hermitian form, the same alternating form is $-\mathrm{Im}H$. Changing the symplectic basis changes the coordinates but not this polarized quotient. For $g=0$, $V=0$ and the Jacobian variety is a point.

An almost complex structure is a smooth real vector bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular, $M$ has even real dimension $2n$. Extend $J$ complex-linearly to $T_{\mathbb{C}}M=TM{\otimes }_{\mathbb{R}}\mathbb{C}$. Its minimal polynomial divides $(u-i)(u+i)$, whose roots are distinct, so the type decomposition of the complexified tangent bundle isThe smooth projections are $(I-iJ)/2$ and $(I+iJ)/2$. Complex conjugation exchanges the two rank-$n$ eigenbundles. A $(1,0)$-covector annihilates $T^{0,1}M$ and a $(0,1)$-covector annihilates $T^{1,0}M$. Thus the differential forms of type (p, q) are the smooth sections ofHere the tensor product is identified with its image under the wedge product of differential forms. For a real tangent vector $v$, the vectors $v-iJv$ and $v+iJv$ exhibit the two types directly.

We now construct the almost complex structure from a decomposable volume form. Locally write $\Omega ={\theta }_1\wedge \cdots \wedge {\theta }_n$. The nonvanishing of $\Omega \wedge \overline{\Omega }$ says that the $2n$ covectors ${\theta }_1,\dots ,{\theta }_n,{\overline{\theta }}_1,\dots ,{\overline{\theta }}_n$ form a complex coframe. In particular, the span of the ${\theta }_j$ has the intrinsic descriptionIndeed, expansion in that coframe shows that every coefficient of a conjugate factor must vanish if $\alpha \wedge {\Omega }_x=0$. Therefore this span is independent of the chosen local factorization. The local coframes show that $W$ is a smooth rank-$n$ vector subbundle, andDefine $J_{\Omega }^{\ast }$ to be multiplication by $i$ on $W$ and by $-i$ on $\overline{W}$. It commutes with complex conjugation, so is the complexification of a real smooth bundle endomorphism, whose dual defines $J_{\Omega }$ on $TM$. It has square $-I$, and its $(1,0)$-covectors are exactly $W$. Conversely, any such almost complex structure must have this same action on the entire coframe. Thus $J_{\Omega }$ exists globally and is unique, and every factor in every allowed decomposition is a $(1,0)$-form.

Finally suppose $d\Omega =0$. Since ${\theta }_j\wedge \Omega =0$, the exterior derivative product rule givesDecompose $d{\theta }_j$ into types $(2,0)$, $(1,1)$ and $(0,2)$ using $J_{\Omega }$. The first two wedge products with the $(n,0)$-form $\Omega$ vanish by dimension. On $(0,2)$-forms, wedging with $\Omega$ is injective: the monomials $\Omega \wedge {\overline{\theta }}_a\wedge {\overline{\theta }}_b$ are linearly independent. For $n=1$ the $(0,2)$ space is zero, so the statement still holds. Hence every $d{\theta }_j$ has no $(0,2)$ component. For an arbitrary $(1,0)$-form $\alpha ={\sum }_j{a}_j{\theta }_j$, the extra terms $d{a}_j\wedge {\theta }_j$ likewise have only types $(2,0)$ and $(1,1)$. The bracket and differential-form criteria for integrability proved below now giveLocal decomposability has been used essentially; the nonvanishing top-degree product alone does not provide the required coframe.

Choose holomorphic coordinates centered at the point $p$ and a polydisc $U\subset {\mathbb{C}}^n$ about the origin. The local blowup of a complex manifold at a point is the incidence manifoldAway from $z=0$ the direction is forced to be $[\ell ]=[z]$, giving a biholomorphism with $U\setminus \{0\}$. Glue this punctured part to $X\setminus \{p\}$. The charts below make the result a complex manifold. The projection is proper near $p$, since it is the restriction of projection from a closed subset of $U\times {\mathbb{CP}}^{n-1}$; globally the modification is proper and is an isomorphism off $p$. Its exceptional divisor is

On ${\ell }_i\ne 0$, let $t=z_i$ and $u_j={\ell }_j/{\ell }_i$ for $j\ne i$. ThenThese are coordinates on the open domain where $t$ and all $t{u}_j$ lie in the original polydisc; at $t=0$ all finite $u_j$ are allowed. Their inverse reads the direction ratios and $z_i$ from the incidence point. In this chart $E$ is the nonsingular hypersurface $t=0$. On overlap with chart $k\ne i$, the holomorphic coordinate changes areThey are holomorphic with holomorphic inverses where $u_k\ne 0$, and the $n$ charts cover every point of $E$. Intrinsically, changing the original coordinates by $F$ lifts via $[\ell ]\mapsto [A(z)\ell ]$, where $F(z)=A(z)z$, $A(0)=d{F}_0$, and $A$ is holomorphic and invertible near zero. One can take $A(z)={\int }_0^{1}d{F}_{sz}ds$ on a sufficiently small polydisc. Away from the origin the lift is uniquely forced; this proves coordinate-independence of the gluing near $E$.

The canonical bundle is $K_X={\Lambda }^n{\Omega }_X^1$. In the $i$th blowup chart the top-form Jacobian iswhere the $u_j$ occur in increasing order. Every term with a second $dt$ vanishes in the wedge product of differential forms. Thus the differential gives a map ${\sigma }^{\ast }{K}_X\to K_{\tilde{X}}$ vanishing to order $n-1$ along $E$ and nowhere else. To express this through the supplied meromorphic-section hypothesis, take a nonzero meromorphic section $s$ of $K_X$ and put $D=\mathrm{div}(s)$. Locally $s=f(z)d{z}_1\wedge \cdots \wedge d{z}_n$, soHere ${\sigma }^{\ast }D$ is the total divisor pullback, including any additional exceptional multiplicity from $f$; it is not merely the strict transform. A line bundle with a nonzero meromorphic section is isomorphic to the line bundle of that section's divisor, as its local coefficient ratios are its frame transitions. Pullback of divisor line bundles and additivity of analytic divisor line bundles therefore give the canonical bundle formula for a point blowup:For $n=1$ the blowup is already an isomorphism, consistently with the zero exponent.

For the projective-plane construction, an invertible linear change of homogeneous coordinates puts $x=[1:0:0]$. Lines through $x$ are parametrized by $[a:b]\in {\mathbb{CP}}^1$, with equation $b{Z}_1-a{Z}_2=0$. Thus the required map on the punctured plane isIt is well-defined under rescaling and holomorphic on the covering sets $Z_1\ne 0$ and $Z_2\ne 0$, where its target coordinates are $Z_2/Z_1$ and $Z_1/Z_2$. These sets cover the complement of $x$.

The graph closure isOver the complement of $x$ it is the graph of $f$. In the affine chart $Z_0\ne 0$, its equation is exactly the two-dimensional local incidence model for the blowup of a complex manifold at a point, so its first projection identifies it with the required blowup globally. The second projection is a holomorphic map $\tilde{f}:S\to {\mathbb{CP}}^1$ extending $f\circ \sigma$. Explicitly, the two blowup charts giveThese formulas agree when $v=1/u$ and remain holomorphic at $t=0$. Hence the pencil extends over the exceptional divisor, where it records the projective tangent direction. This is the projective pencil resolved by a point blowup.

A smooth differential form of degree $p$ is a smooth section of ${\Lambda }^p{T}^{\ast }M$, soIt is equivalently a smooth alternating covariant $p$-tensor field. In a coordinate chart $x^1,\dots ,x^n$ it has a unique expressionwith smooth coefficients. At $p=0$ these are the smooth functions, and for $p>n$ the space is zero.

Define the exterior derivative in a chart byTo check that it is well-defined, let $y^1,\dots ,y^n$ be other smooth coordinates. The expression $d{y}^a={\sum }_i({\partial }_i{y}^a)d{x}^i$ satisfiesbecause the coefficients are symmetric and the wedge products antisymmetric. The coordinate formula also satisfies the graded Leibniz rule, and for a function $b$ the chain rule gives $d_{x}b={\sum }_a(\partial b/\partial y^a)d{y}^a=d_{y}b$. Applying these two facts to $\omega ={\sum }_I{b}_{I}d{y}^I$ yields $d_x\omega ={\sum }_I{d}_y{b}_I\wedge d{y}^I=d_y\omega$. Thus the local operators agree on overlaps and define a global linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$.

A second application in any chart givesagain by symmetry of mixed derivatives and antisymmetry of the wedge product of differential forms. Consequently every exact differential form is closed. Put ${\Omega }^{-1}(M)=0$ and define the de Rham cohomology byThe identity $d^2=0$ is exactly what puts the denominator inside the numerator, making this quotient meaningful. Since ${\Omega }^p(M)=0$ for $p>\dim M$, $H_{\mathrm{dR}}^p(M)=0$ in those degrees.

Fix the coefficient-column convention: write the local frame as a row $e=(e_1,\dots ,e_r)$ and a section as $s=eu$ for a column $u$ of functions. Define the connection matrix $\theta =({\theta }_{ij})$ byEach entry is a smooth one-form. Define the curvature form of a connection byThe same Leibniz cancellations as for tangent-bundle curvature show that this is linear over smooth functions in $X,Y,s$. Thus it is a two-form with values in the endomorphisms of $E$. Its curvature matrix $\Theta =({\Theta }_{ij})$ is specified by $R^E(X,Y)e_j={\sum }_i{e}_i{\Theta }_{ij}(X,Y)$.

The covariant exterior derivative on coefficient forms is $d+\theta \wedge$. Squaring it on $u$ givesTherefore the Cartan curvature matrix equation isMatrix-valued wedge products here combine matrix multiplication with the wedge product of differential forms, in the stated order.

Using $d^2=0$ and the graded Leibniz rule,On the other hand, associativity of the endomorphism-valued exterior product givesThe cubic terms cancel, proving the requested Bianchi identity

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

Use matrix multiplication together with the wedge product of differential forms, suppressing wedge symbols in this calculation. Set $F=dA+A^2$ and $a=\delta A$. The gauge field strength varies as $\delta F=da+Aa+aA$, the covariant exterior derivative of $a$. The matrix trace has graded cyclicity, $\mathrm{Tr}(\alpha \beta )=(-1{)}^{pq}\mathrm{Tr}(\beta \alpha )$ for forms of degrees $p,q$. Thus the Chern-Simons 5-form can be expanded asFor example, the two cross terms in $F^{2}A$ both become $\mathrm{Tr}(A^{3}dA)$; combining the remaining two terms gives the displayed coefficients. This expansion keeps track of the noncommuting products rather than treating $A$ as a scalar one-form.

Put $B=dA$. Vary the first term and use the graded exterior derivative to move derivatives off $a$:Here $dB=0$. For the next term use $d(A^3)=B{A}^2-ABA+A^{2}B$. Graded cyclicity then givesFinally all five contributions to $\delta \mathrm{Tr}(A^5)$ are equal under the matrix trace, giving $5\mathrm{Tr}(a{A}^4)$. Combining these results proves the variation of the Chern-Simons 5-form:Consequently, on an oriented five-manifold $N$,The boundary term vanishes for a closed manifold, for variations supported in its interior, or when the tangential boundary connection is held fixed. A different boundary variational problem would require boundary conditions or a boundary action as well.