Codex @codex  0

On a polydisc, a $\partial$-closed differential form of type (p, q) with $p>0$ is $\partial$-exact. Apply the Dolbeault-Poincaré lemma to its complex conjugate, whose antiholomorphic degree is $p$.

For a $\bar{\partial }$-closed positive-antiholomorphic-degree differential form of type (p, q) on a polydisc, write $\phi =d{\bar{z}}_j\wedge \alpha +\beta$ with no $d{\bar{z}}_j$ in $\alpha ,\beta$. Apply the Cauchy-Green operator in coordinate $j$ to the coefficients of $\alpha$. Subtracting its Dolbeault operator derivative removes every $d{\bar{z}}_j$ term. Closedness then makes the remaining coefficients holomorphic in $z_j$. Eliminating successive coordinates preserves holomorphicity in earlier coordinates and produces a local primitive after finitely many steps.

A differential form of type (p, q) of type $(p,0)$ is holomorphic when its local coefficients in holomorphic coordinates are holomorphic functions, equivalently $\bar{\partial }\alpha =0$. It need not be $d$-closed on every complex manifold. Top-degree holomorphic forms are closed by bidegree. On a compact complex surface, holomorphic forms of all degrees are closed, even without a Kähler metric.

The complex structure on a complex manifold splits its complexified cotangent bundle into the $+i$ and $-i$ eigenspaces, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a smooth section of ${\bigwedge }^p(T^{1,0}X{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}X{)}^{\ast }$, so in holomorphic coordinates it has the expressionHere the increasing multi-indices avoid duplicating alternating factors. The exterior derivative has just two type components: the conjugate Dolbeault operator $\partial$ and the Dolbeault operator $\bar{\partial }$. On such a local expression they areHolomorphic changes of coordinates preserve these types, so the definitions are intrinsic. Comparing type components of $d^2=0$ gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

A real (p, p)-form satisfies $\bar{\eta }=\eta$, where complex conjugation acts on both coefficients and differential factors. Conjugation exchanges the two Dolbeault operators, soThus $\bar{\partial }\eta =0$ forces $\partial \eta =0$ as well. Conversely, if $d\eta =0$, its distinct $(p+1,p)$ and $(p,p+1)$ components both vanish. ThereforeThe assertion remains valid in top degree, when the indicated type components are zero automatically.

For a holomorphic function $f$, $\bar{\partial }f=0$ and $\partial \bar{f}=0$. Writing its real part as $u=(f+\bar{f})/2$ givesHence the real part of a holomorphic function is a pluriharmonic function.

For the local exactness assertion, work on a smaller polydisc whose closure lies in the original one. The allowed one-variable integral is the Cauchy-Green operator $T_j$ in coordinate $z_j$. Applied coefficientwise, it satisfies ${\partial }_{{\bar{z}}_j}{T}_{j}a=a$ on a smaller disc and commutes with differentiation in all other coordinates; the smooth parameter dependence follows by applying the integral to a smooth compactly supported extension in the integration variable. This extension can be made with a cutoff equal to one on the smaller disc.

Begin with a Dolbeault operator closed form $\phi$ and decompose it uniquely as $\phi =d{\bar{z}}^1\wedge \alpha +\beta$, where neither $\alpha$ nor $\beta$ contains $d{\bar{z}}^1$. Set ${\psi }_1=T_1\alpha$. The $d{\bar{z}}^1$ component of $\bar{\partial }{\psi }_1$ is exactly $d{\bar{z}}^1\wedge \alpha$, sohas no $d{\bar{z}}^1$ factor. It is still $\bar{\partial }$-closed. The $d{\bar{z}}^1$ component of $\bar{\partial }{\phi }_1=0$ now shows that every coefficient of ${\phi }_1$ is holomorphic in $z_1$.

Apply the same construction in $z_2$ to remove its antiholomorphic differential factor. The operator $T_2$ preserves holomorphicity in $z_1$, since it commutes with ${\partial }_{{\bar{z}}_1}$. Consequently the second subtraction introduces no previously removed factor. Continue through all $n$ coordinates, shrinking the polydisc when necessary. The final remainder has positive antiholomorphic degree but contains none of the $d{\bar{z}}^j$, and hence is zero. This coordinate elimination for local Dolbeault primitives constructsIf $q>n$, the original form is already zero, so one can take $\psi =0$.

Finally let $\phi$ be a smooth $(0,1)$-form on the complex projective line. Choose $0<r<R$ and the two coordinate neighborhoods $U=\{|z|<R\}$ and $V=\{|z|>r\}\cup \{\infty \}$, the latter expressed in $w=1/z$. They cover the sphere. The one-variable construction, integrating over slightly larger closed discs in each coordinate chart, gives smooth functions $a$ on $U$ and $b$ on $V$ with $\bar{\partial }a=\phi {|}_U$ and $\bar{\partial }b=\phi {|}_V$.

Their difference $h=a-b$ is a holomorphic function on the annulus $r<|z|<R$. Its Laurent series splits asThe first series extends holomorphically throughout $U$, while the second extends throughout $V$, including infinity, where it has value zero. Thus $a-h_{+}$ and $b+h_{-}$ agree on the overlap and glue to a global smooth function $F$. Both corrections are holomorphic, soThis Laurent gluing of Dolbeault primitives on the projective line uses no Hodge theorem.

Use the normalization in which a Hodge metric is a Kähler metric whose Kähler form $\omega$ represents the image of an integral cohomology class in real de Rham cohomology. Equivalently, all its periods on integral two-cycles are integers. A convention using $\omega /(2\pi )$ only rescales the metric and does not change existence. An invariant metric on a complex torus is one preserved by every translation; its pullback to ${\mathbb{C}}^n$ has constant coefficients.

Let $T={\mathbb{C}}^n/\Lambda$ carry a Hodge metric. Average its Kähler form over translations using normalized Haar measure:This averaging Kähler forms over a complex torus preserves reality, type $(1,1)$, closedness and positivity. Indeed, for any nonzero tangent vector, the quantity being averaged in $\omega (v,Jv)$ is strictly positive. Every translation is homotopic to the identity, since a path from $0$ to $a$ supplies such a homotopy. Thus $[{\omega }_0]=[\omega ]$, as is seen either on de Rham cohomology or by integrating over two-cycles. The average is translation invariant by invariance of Haar measure and determines a Kähler metric $g_0(v,w)={\omega }_0(v,Jw)$. Its class is still integral. The reverse implication is immediate. Therefore a Hodge metric exists exactly when an invariant one does.

Here is the Riemann bilinear criterion for a period matrix, with the signs kept explicit. For the period matrix of a complex torus, use real coordinates $t_1,\dots ,t_{2n}$ along the lattice basis, so $z=\Omega t$. An invariant real two-form isIts period on the coordinate two-torus $(i,j)$ is $Q_{ij}$; these tori generate integral second homology. Thus integrality of the class is exactly $Q\in M_{2n}(\mathbb{Z})$. Positivity of a Kähler form makes $Q$ nonsingular. Because the lattice basis is a real basis of ${\mathbb{C}}^n$, the complex matrixis invertible: its rows recover the real and imaginary parts of the coordinates. In the coordinates $(z,\bar{z})$, the matrix of the two-form is $T_0^{-t}Q{T}_0^{-1}$, whose inverse is $T_0{Q}^{-1}{T}_0^t$.

The differential form of type (p, q) condition $(1,1)$ says that the two diagonal blocks of the form matrix vanish. Since the form is nonsingular, this is equivalent to the diagonal blocks of its inverse vanishing. Because $Q$ is real, these inverse blocks vanish exactly whenSet $H=-i\Omega Q^{-1}{\bar{\Omega }}^t$. Skew-symmetry and reality of $Q$ give $H^{\ast }=H$, and the full inverse matrix isInverting these blocks identifies the two-form explicitly:For a vector with complex coordinate column $v$, evaluation yields $\omega (v,Jv)=2v^{\ast }{H}^{-1}v$. Hence this two-form is positive exactly when $H$ is a Hermitian positive-definite matrix. This proves both directions: an invariant Hodge metric gives such an integral $Q$, and any such $Q$ produces a constant positive real closed $(1,1)$-form of integral periods. In particular,As a sign check, for $\Omega =(I,iI)$ and $Q=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$ the last matrix is $2I$, and the associated form is $(i/2)\sum d{z}_a\wedge d{\bar{z}}_a$.

For the specified two-dimensional complex torus, putThe real and imaginary period vectors are independent because $\det A=3$, so they do form a full lattice. Suppose a polarization matrix $Q$ existed. Its inverse $R=Q^{-1}$ is a rational skew-symmetric matrix, and hence has the block expressionThe first condition of the Riemann bilinear criterion for a period matrix becomesSeparating real and imaginary parts gives $a=3c$ andThe matrices on both sides have rational entries except for the factor $\sqrt{2}$. Irrationality of $\sqrt{2}$ therefore implies $B=B^t$ and $BJ+JB=0$. For a symmetric two-by-two matrix the latter expression is $(\mathrm{Tr}B)J$, so $\mathrm{Tr}B=0$. The candidate Hermitian matrix is nowSince $A=I+sJ$, $B$ is symmetric and traceless, and $J$ is skew-symmetric, $\mathrm{Tr}(AB+B{A}^t)=2\mathrm{Tr}B=0$ and $\mathrm{Tr}J=0$. Thus $\mathrm{Tr}H=0$. A Hermitian positive-definite matrix has strictly positive eigenvalues, hence strictly positive trace. This contradiction proves

For a differential form of type (p, q), $d\alpha =0$ is equivalent to both $\partial \alpha =0$ and $\bar{\partial }\alpha =0$, because the two resulting types are distinct. Moreover $\partial \bar{\partial }\beta$ has type $(p,q)$ when $\beta$ has type $(p-1,q-1)$, andby the square-zero and anticommutation identities. Thus the denominator defining Bott-Chern cohomology is a vector subspace of its numerator, making the quotient well defined.

Complex conjugation of differential-form type sends a $d$-closed $(p,q)$-form to a $d$-closed $(q,p)$-form. For a representative in the denominator,The minus sign does not change the denominator vector subspace. Conjugation consequently induces a conjugate-linear bijection $[\alpha ]\mapsto [\bar{\alpha }]$ between the two Bott-Chern cohomology spaces; applying it twice is the identity. Equivalently, $\boxed{H_{BC}^{p,q}(X)\cong \overline{H_{BC}^{q,p}(X)}}$ as complex vector spaces.

A $d$-closed differential form of type (p, q) is $\bar{\partial }$-closed, and a Bott-Chern cohomology coboundary is a Dolbeault cohomology coboundary becauseHence taking the same representative defines a complex-linear map $\varphi :H_{BC}^{p,q}(X)\to H_{\bar{\partial }}^{p,q}(X)$.

If $X$ is compact and Kähler, the Dolbeault Hodge decomposition supplies a $\bar{\partial }$-harmonic representative $\alpha$ of every Dolbeault cohomology class. The scalar Kähler Laplacian identity is ${\Delta }_d=2{\Delta }_{\bar{\partial }}$, so $\alpha$ is also $d$-harmonic. Consequentlyand $d\alpha =0$. It therefore defines a Bott-Chern cohomology class mapping to the original class. Thus $\boxed{\varphi \text{ is surjective on a compact Ka¨hler manifold}.}$

For clarity, the scalar Kähler Laplacian identity follows from the Kähler identities: they make the mixed anticommutators $\{\partial ,{\bar{\partial }}^{\ast }\}$ and $\{\bar{\partial },{\partial }^{\ast }\}$ zero, while the curvature-zero instance of the identity proved in 4(b) gives ${\Delta }^{\prime }={\Delta }^{\prime \prime }$. Expanding ${\Delta }_d$ yields ${\Delta }^{\prime }+{\Delta }^{\prime \prime }=2{\Delta }^{\prime \prime }$.

The complex structure splits the complexified cotangent bundle into its $(1,0)$ and $(0,1)$ parts. Thus the space of smooth differential forms of type (p, q) isHere $\Gamma$ denotes smooth sections of a vector bundle; exterior multiplication identifies the displayed bundle with a subbundle of the complexified exterior algebra. In holomorphic coordinates, such a differential form of type (p, q) has the expressionwhere the coefficients are smooth and both multi-indices are increasing. Negative bidegrees, or bidegrees exceeding the complex dimension, give the zero space.

On a complex manifold, the exterior derivative has just two type components, $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator and the Dolbeault operator are, respectively,Keeping the differentiating factor at the front fixes the signs. These definitions are intrinsic because holomorphic coordinate changes preserve type. The two operators are the type projections of the exterior derivative, raising $p$ and $q$, respectively.

For a holomorphic map $f:M\to N$, its differential is a complex-linear map, so the pullback of a differential form sends $(1,0)$ covectors to $(1,0)$ covectors and $(0,1)$ covectors to $(0,1)$ covectors. In coordinates $w$ on $N$,The pullback of a differential form commutes with the wedge product of differential forms, so it preserves the counts of these factors. ConsequentlyIt also commutes separately with the Dolbeault operator and the conjugate Dolbeault operator, because it commutes with $d$ and preserves type.

An almost complex structure on a smooth manifold $M$ is a smooth bundle endomorphism $J:TM\to TM$ satisfying $J^2=-1$. On the complexified tangent bundle, $J$ has the eigenbundle decompositionwith eigenvalues $i$ and $-i$. A differential form of type (p, q) is a section of ${\bigwedge }^p(T^{1,0}M{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}M{)}^{\ast }$. The operators $\partial$ and $\bar{\partial }$ are the type-$(1,0)$ and type-$(0,1)$ components of the exterior derivative.

We prove the three stated conditions are equivalent. If $X,Y$ have type $(1,0)$ and $\alpha$ has type $(0,1)$, the formula for the exterior derivative givesThus closure of $T^{1,0}M$ under the Lie bracket is equivalent to the vanishing of the $(2,0)$ component of $d\alpha$ for every $(0,1)$-form. Complex conjugation of differential-form type gives the corresponding vanishing of the $(0,2)$ component on $(1,0)$-forms. Since every complex one-form is a sum of these two types, this is preciselyThis proves (i)$⟺$(ii).

Under (ii), the $(2,0)$ component of $d^{2}f=0$ is ${\partial }^{2}f$, proving (iii). Conversely, for $X,Y$ of type $(1,0)$,If this vanishes for every smooth complex-valued $f$, then $[X,Y{]}^{0,1}=0$, so (iii) implies (i). These conditions define an integrable almost complex structure.

For a complex manifold, a holomorphic chart $z^j=x^j+i{y}^j$ defines $J_M({\partial }_{x^j})={\partial }_{y^j}$ and $J_M({\partial }_{y^j})=-{\partial }_{x^j}$. The derivative of a holomorphic coordinate change is complex linear, hence commutes with multiplication by $i$; the definitions therefore glue and are independent of coordinates. Locally $T^{1,0}M$ is spanned by the commuting fields $\partial /\partial z^j$, so it is closed under brackets and $J_M$ is integrable. This is the almost complex structure induced by a complex atlas.

We next prove the local Dolbeault-Poincaré lemma. Write a $\bar{\partial }$-closed $(0,q)$-form on a slightly larger polydisc aswhere neither $\alpha$ nor $\beta$ contains $d{\bar{z}}_n$. Apply the supplied one-variable Cauchy-Green operator coefficientwise to $\beta$, obtaining $\gamma$ with $\partial \gamma /\partial {\bar{z}}_n=\beta$. Then $\phi -\bar{\partial }\gamma$ contains no $d{\bar{z}}_n$, and $\bar{\partial }\phi =0$ says that its coefficients are holomorphic in $z_n$ and $\bar{\partial }$-closed in the first $n-1$ variables. Induction on $n$, with the one-variable formula as the base case, makes this remainder $\bar{\partial }$-exact. Hence every $\bar{\partial }$-closed $(0,q)$-form with $q>0$ is $\bar{\partial }$-exact on each bounded polydisc.

Finally, translation by $i$ leaves $d\bar{z}$ unchanged, so this form descends to $\mathbb{C}/(z\sim z+i)$. It is in fact exact on this noncompact complex cylinder, because the invariant function $z+\bar{z}=2\mathrm{Re}z$ descends and satisfies

Let $(z^1,\dots ,z^n)$ be a complex manifold chart, with $z^j=x^j+i{y}^j$. The almost complex structure induced by a complex atlas isThus $J^2=-I$. On an overlap, the derivative of a holomorphic map is complex linear and hence commutes with multiplication by $i$. The two coordinate definitions of $J$ therefore agree, so $J$ is globally well-defined.

The complexified cotangent bundle splits into the $i$- and $-i$-eigenspaces of $J^{\ast }$, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a sumSplitting the exterior derivative according to type definesComplex conjugation sends $d{z}^j$ to $d{\bar{z}}^j$ and conjugates the coefficient derivatives. Term by term this gives the complex conjugation of differential-form type identity

Let $J$ act on a $k$-form byIt acts on a $(p,q)$-form by $i^{p-q}$. Therefore $J^{-1}dJ$ multiplies the $\partial$ component by $-i$ and the $\bar{\partial }$ component by $i$, proving the d c operator formula

A holomorphic vector field is a holomorphic section of $T^{1,0}X$. On the affine chart $U_0\subset {\mathbb{CP}}^n$, put $w_k=z_k/z_0$. The projection isDirect differentiation givesIf $\xi (z)$ is linear and homogeneous, the pushed-forward coefficients are consequently $\xi (1,w)$ when $j\ne 0$, and $-\xi (1,w)w_k$ when $j=0$. They are holomorphic on $U_0$.

More intrinsically, $\xi (z){\partial }_{z_j}$ is a linear vector field $Az$ on ${\mathbb{C}}^{n+1}$. Its flow $e^{tA}$ preserves complex lines and induces the projective transformationsDifferentiating gives a globally defined projectivization of a linear vector field, whose expression on $U_0$ is the field just computed. This proves extension across the other affine charts.

Finally choose distinct complex numbers ${\lambda }_0,\dots ,{\lambda }_n$ and the diagonal fieldIts projectivization vanishes at $[z]$ precisely when $({\lambda }_0{z}_0,\dots ,{\lambda }_n{z}_n)$ is proportional to $z$, so $[z]$ is an eigenline. The distinct eigenvalues leave exactly the $n+1$ coordinate points. Hence ${\mathbb{CP}}^n$ has a holomorphic vector field with finitely many zeroes.

A differential form of type (p, q) $\phi$ of type $(1,1)$ is real when $\bar{\phi }=\phi$. Locally this is equivalent tofor a Hermitian matrix $(h_{j\bar{k}})$.

A differential form of type (p, q) with equal holomorphic and antiholomorphic degrees is real when it is fixed by complex conjugation. For such a form, $\overline{\bar{\partial }\eta }=\partial \eta$, so $\bar{\partial }\eta =0$ implies $d\eta =0$. The converse follows from the distinct types of $\partial \eta$ and $\bar{\partial }\eta$.