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Locally a U(1) connection whose gauge curvature has no $(2,0)$ or $(0,2)$ part can be written $A=\partial u+\bar{\partial }v$ by the Dolbeault-Poincaré lemma. Unitarity permits $v=-\bar{u}$. Writing $u=p+iq$, a gauge transformation removes $idq$ and leaves $A=\partial p-\bar{\partial }p$. Setting $f=-2p$ gives the displayed unitary form, whose gauge curvature is $\partial \bar{\partial }f$. Its remaining anti-self-duality of gauge curvature condition is the scalar Laplace equation. Alternatively a complex gauge transformation by $e^{-u}$ gives $A=\bar{\partial }(v-u)$; this second gauge generally does not preserve unitarity.

// Destination: complex-geometry.bigb

On a polydisc, $H_{BC}^{p,q}=0$ for $p,q>0$. A Poincaré lemma primitive for a closed pure-type form can be modified by exact terms until only bidegrees $(p-1,q)$ and $(p,q-1)$ remain. The Dolbeault-Poincaré lemma and conjugate Dolbeault-Poincaré lemma then make the original form $\partial \bar{\partial }$-exact.

On a disc $D\subset \mathbb{C}$, the Cauchy-Green operatoris a right inverse to $\partial /\partial \bar{z}$ in the interior: ${\partial }_{\bar{z}}Tg=g$. Applying it to coefficients gives the local homotopy operator used in the Dolbeault-Poincaré lemma.

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 $p$.

The Dolbeault-Poincaré lemma makes this a locally exact complex of sheaves, with degree-zero kernel the holomorphic differential forms. The sheaves of smooth forms are fine sheaves by a partition of unity. The acyclic resolution theorem therefore computes the cohomology of ${\Omega }^p$ by the global Dolbeault operator complex.

For the sheaf ${\Omega }^p$ of holomorphic $p$-forms on a complex manifold, the Dolbeault-Poincaré lemma makesa resolution by fine sheaves. Taking global sections therefore gives the natural isomorphism

For a closed real $(1,1)$ form, the Poincaré lemma gives a real one-form $\eta$ with $d\eta =\omega$. Its $(0,1)$ part is $\bar{\partial }$-closed, so the Dolbeault-Poincaré lemma gives ${\eta }^{0,1}=\bar{\partial }\psi$. Reality gives ${\eta }^{1,0}=\partial \bar{\psi }$, hence $\omega =i\partial \bar{\partial }(2\mathrm{Im}\psi )$. Positivity is a separate requirement for this form to define a metric.

Let $F_{+}=(F+\ast F)/2$ and $F_{-}=(F-\ast F)/2$, using the oriented Euclidean metric. For a self-dual two-form $\omega$,Apply this componentwise in the Lie algebra. Since the three ${\omega }_i$ span the self-dual subspace, their vanishing wedges force $F_{+}=0$. Thus $\ast F=-F$. The gauge-theory Bianchi identity is $D_{A}F=0$, so $D_A\ast F=-D_{A}F=0$, the Euclidean Yang-Mills equations. This proves the implication locally; finite action would additionally be needed to call the solution an instanton.

Use the complex orientation for $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The real and imaginary parts of $dw\wedge dz$, and $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. Accordingly the three gauge curvature equations areFor a spectral parameter $\lambda \in {\mathbb{CP}}^1$, consider the Lax pairTheir commutator isIt vanishes for every $\lambda$ exactly when all three coefficients vanish. The derivative directions ${\partial }_w-\lambda {\partial }_{\bar{z}}$ and ${\partial }_z+\lambda {\partial }_{\bar{w}}$ are mutually orthogonal null vectors for the complexified metric. Flatness on the corresponding two-planes makes the linear parallel-transport equations locally compatible. Conversely compatibility for an invertible fundamental matrix $\Psi$ forces the commutator to vanish; a single zero solution would not establish it. This is the Euclidean complex-coordinate anti-self-dual Lax pair.

For an Abelian U(1) connection, all commutators of the connection components vanish, and the gauge curvature equations therefore readLocally the first two equations say that $A^{1,0}$ is $\partial$-closed and $A^{0,1}$ is $\bar{\partial }$-closed. The Dolbeault-Poincaré lemma, and its complex conjugate, give scalar functions $u,v$ withSubstituting in the third equation givesIn a complex gauge transformation, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. ThenThis is the scalar Laplace equation, since for the stated metric $\Delta =2({\partial }_z{\partial }_{\bar{z}}+{\partial }_w{\partial }_{\bar{w}})$. The complex gauge transformation need not remain in $U(1)$.

A reduction using only a unitary gauge transformation is also available. Since $A$ is anti-Hermitian, $A_{\bar{w}}=-\overline{A_w}$ and $A_{\bar{z}}=-\overline{A_z}$, so the potential for the barred components can be chosen as $v=-\bar{u}$. Write $u=p+iq$ with real $p,q$. Then $A=\partial p-\bar{\partial }p+idq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this givesIts gauge curvature is $\partial \bar{\partial }f$. Thus the Abelian anti-self-dual connections from harmonic scalar potentials reduction is local and does not require silently enlarging the real gauge group.

The Dolbeault theorem identifieswhere ${\Omega }_X^p$ is the sheaf of holomorphic differential forms of degree $p$.

The local analytic ingredient is the Dolbeault-Poincaré lemma: a $\bar{\partial }$-closed smooth $(p,q)$-form with $q>0$ is locally $\bar{\partial }$-exact. Here is a local proof. On a polydisc, the one-variable Cauchy-Green operator in coordinate $z_j$ iswith a smooth cutoff supported in the coordinate disc and equal to one on a smaller disc. The fundamental-solution identity ${\partial }_{\bar{z}}(1/(\pi z))={\delta }_0$ gives ${\partial }_{{\bar{z}}_j}{T}_{j}g=g$ on that smaller disc. The operator is smooth in the parameters and commutes with derivatives in the other coordinates.

For a $(0,q)$-form, write $\alpha =d{\bar{z}}_m\wedge \beta +\gamma$, with neither $\beta$ nor $\gamma$ containing $d{\bar{z}}_m$. Subtract $\bar{\partial }(T_m\beta )$. The remainder contains no $d{\bar{z}}_m$, remains closed, and its coefficients are holomorphic in $z_m$. Repeat with $z_{m-1}$ and then the other coordinates, shrinking discs as needed. Each new integral preserves the holomorphic dependence already achieved. At the end the closed remainder has positive antiholomorphic degree but contains no antiholomorphic differential, so it is zero. Factoring out each holomorphic basis form $d{z}_I$ gives the same assertion for $(p,q)$-forms, with the fixed degree sign included in the primitive. At degree zero, the kernel of $\bar{\partial }$ consists precisely of holomorphic coefficients.

Thus the following is an exact Dolbeault resolution of holomorphic differential forms of sheaves:Each sheaf of smooth forms is a fine sheaf: multiplication by a smooth partition of unity supplies endomorphisms supported in its open cover. Fine sheaves on a paracompact manifold have zero positive sheaf cohomology; the Čech contraction sums a cochain against the partition of unity, giving $\delta h+h\delta =I$ in positive degrees. The general acyclic resolution theorem therefore computes ${\check{H}}^q(X,{\Omega }_X^p)$ as the cohomology of the global-section complex above. That complex is exactly the Dolbeault resolution of holomorphic differential forms, proving the claimed isomorphism. The sheaf-cohomology/Čech comparison and the acyclic-resolution principle are the stated general Čech properties used here.

To prove (a) implies (c), shrink to a contractible coordinate neighborhood. The real Poincaré lemma gives a real one-form $\eta$ with $\omega =d\eta$. Write $\eta ={\eta }^{1,0}+{\eta }^{0,1}$. Because $\omega$ has type $(1,1)$, $\bar{\partial }{\eta }^{0,1}=0$ and $\partial {\eta }^{1,0}=0$. The Dolbeault-Poincaré lemma supplies a smooth function $\psi$ with ${\eta }^{0,1}=\bar{\partial }\psi$. Reality of $\eta$ gives ${\eta }^{1,0}=\partial \overline{\psi }$. ThusTaking the real function $f=2\mathrm{Im}\psi$ gives the local real potential for a closed (1,1)-form. Conversely, $d(i\partial \bar{\partial }f)=0$, by ${\partial }^2={\bar{\partial }}^2=0$ and anticommutation. Hence (a) and (c) are equivalent, completing all three conditions.

For the unheaded radial continuation, take the real potential $f$ as in (c). Rotation invariance makes $f$ constant on every circle of radius $r>0$, sois well defined and smooth by composition. This avoids treating a smooth function as if it had a convergent Taylor series. At nonzero $z$, put $t=\log |z{|}^2$. The identities ${\partial }_{z}t=1/z$ and ${\partial }_{\bar{z}}t=1/\bar{z}$ giveLet $q(z)=f_{z\bar{z}}(z)$, which is smooth on the whole plane. ThenAt the origin, rotation invariance gives the finite Taylor expansion $f(z)=f(0)+b|z{|}^2+O(|z{|}^4)$, with $b=q(0)$. Alternatively continuity of $q$ directly givesThe metric is positive away from the origin exactly when $u^{\prime \prime }(t)>0$ for all finite $t$, and it is positive at the origin exactly when this limit is positive. Smoothness at the origin is already supplied by the original smooth potential. Consequently the positivity criterion for a radial Kähler potential isClosedness is automatic from the potential, and positivity completes the Kähler condition. No completeness of this metric is asserted. The rotation-invariant Kähler potential on the complex plane is understood as real; if a complex radial potential is initially allowed with real $\omega$, its smooth radial imaginary part is harmonic and hence constant, so that constant can be removed.

We prove the Bott-Chern Poincaré lemma on a polydisc $B$. Let $p,q\ge 1$ and $d\alpha =0$, with $\alpha$ of type $(p,q)$. The allowed Dolbeault-Poincaré lemma first gives $\alpha =\bar{\partial }{b}_0$, where $b_0$ has type $(p,q-1)$. Since $\bar{\partial }\partial b_0=-\partial \alpha =0$, repeatedly applying the same lemma constructsZero spaces outside the allowed bidegrees cause no difficulty. The form $\partial b_{q-1}$ has type $(p+q,0)$, is holomorphic because its Dolbeault operator vanishes, and is $\partial$-closed. The holomorphic Poincaré lemma on a polydisc gives a holomorphic form $h$ of type $(p+q-1,0)$ with $\partial h=\partial b_{q-1}$. This holomorphic version follows from the radial homotopy formula: on positive-degree holomorphic forms the ordinary radial Poincaré lemma homotopy preserves holomorphic coefficients. If the displayed form is zero, take $h=0$.

Now $b_{q-1}-h$ is $\partial$-closed. The conjugate Dolbeault-Poincaré lemma, obtained by conjugating the allowed lemma, gives $b_{q-1}-h=\partial c_{q-1}$, since its holomorphic degree is positive. Moving backwards, if the last modified form is $\partial c_j$, thenApply the conjugate Dolbeault-Poincaré lemma again to obtain $b_{j-1}+\bar{\partial }{c}_j=\partial c_{j-1}$. The modification does not change its Dolbeault operator, so this process continues down to $b_0$. In the case $q=1$, subtract $h$ at this last step instead. Finally,Thus $\boxed{H_{BC}^{p,q}(B)=0(p,q\ge 1)}$.

This vanishing fails on general complex manifolds. On the complex projective line, let $\omega$ be its Fubini-Study form. It is a $d$-closed $(1,1)$-form with ${\int }_{{\mathbb{P}}^1}\omega >0$. If it were $\partial \bar{\partial }f$, it would be the exact differential form $d(\bar{\partial }f)$, whose integral is zero by Stokes theorem. Therefore $\boxed{H_{BC}^{1,1}({\mathbb{P}}^1)\ne 0}$.

For a holomorphic vector bundle $E$, the sheaf of vector-bundle-valued differential forms iswhere sections are smooth and restrictions are the usual ones. In a holomorphic local frame of $E$, the Dolbeault operator ${\bar{\partial }}_E$ acts coefficientwise and satisfies ${\bar{\partial }}_E^2=0$. Its Dolbeault cohomology with values in a holomorphic vector bundle is the cohomology of the global section complex ${\mathcal{A}}^{p,\bullet }(X,E)$.

The bundle-valued Dolbeault theorem identifies this with sheaf cohomology:Here ${\Omega }_X^p$ is the sheaf of holomorphic differential forms, and $\mathcal{O}(E)$ is the sheaf of holomorphic sections of a vector bundle. Equivalently, the complexis a fine sheaf resolution, by the local Dolbeault-Poincaré lemma and smooth partitions of unity. In particular $p=0$ computes $H^q(X,\mathcal{O}(E))$; for $p>0$ the holomorphic differential-form factor must be retained.

Complex conjugation interchanges the Dolbeault operator and conjugate Dolbeault operator. Thus $\partial \alpha =0$ implies $\bar{\partial }\bar{\alpha }=0$, and $\bar{\alpha }$ has type $(q,p)$. Its antiholomorphic degree is $p\ge 1$, so the allowed Dolbeault-Poincaré lemma on the polydisc gives a form $\gamma \in {\mathcal{A}}^{q,p-1}$ satisfying $\bar{\partial }\gamma =\bar{\alpha }$.

Conjugate this identity and put $\beta =\bar{\gamma }$. This proves the conjugate Dolbeault-Poincaré lemma:

A pure-type $d$-closed form is both $\partial$-closed and $\bar{\partial }$-closed, because these two derivatives have different types. Let $k=p+q$. The smooth Poincaré lemma on the polydisc gives $\eta \in {\mathcal{A}}_{\mathbb{C}}^{k-1}$ with $d\eta =\alpha$. We will modify $\eta$ by exact terms without changing $d\eta$.

Write $\eta ={\sum }_r{\eta }^{r,k-1-r}$. Starting with the smallest holomorphic degree, eliminate every component with $r<p-1$. Once the lower components are zero, the $(r,k-r)$ component of $d\eta$ says $\bar{\partial }{\eta }^{r,k-1-r}=0$, since $r<p$. Its antiholomorphic degree is positive, so the Dolbeault-Poincaré lemma supplies ${\nu }^{r,k-2-r}$ with $\bar{\partial }\nu ={\eta }^{r,k-1-r}$. Replace $\eta$ by $\eta -d\nu$: this removes the component at $r$ and changes only the next holomorphic degree.

Next, work downwards from the largest holomorphic degree and eliminate the components with $r>p$. Once the higher components are zero, the corresponding component of $d\eta$ gives $\partial {\eta }^{r,k-1-r}=0$. By part (b), write this as $\partial {\nu }^{r-1,k-1-r}$ and again subtract $d\nu$. This affects only the next lower holomorphic degree. The two finite procedures leaveThe components of $d\eta =\alpha$ outside type $(p,q)$ now give $\bar{\partial }u=0$ and $\partial v=0$, while its $(p,q)$ component gives $\alpha =\partial u+\bar{\partial }v$.

Since $q\ge 1$, the Dolbeault-Poincaré lemma gives $u=\bar{\partial }a$ with $a$ of type $(p-1,q-1)$. Since $p\ge 1$, part (b) gives $v=\partial b$ of the same type. The anticommutation identity therefore yieldsThis proves the Bott-Chern Poincaré lemma using precisely the permitted primitives:

Let ${\Omega }^p$ be the sheaf of holomorphic $p$-forms and ${\mathcal{A}}^{p,q}$ the sheaf of smooth $(p,q)$-forms. The Dolbeault-Poincaré lemma makesan exact resolution by fine sheaves, which are acyclic for global sections. Its global cochain complex computes Dolbeault cohomology, proving the Dolbeault theorem

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