The Dolbeault Complex and Integral Solutions: Examples and Counterexamples

The coordinate Wirtinger derivatives give the displayed derivatives of $f$. [F2, F3, F5, given, algebra] From [F2]–[F3] and $z_k=x_k+i{y}_k$, one has ${\mathrm{\partial }}_{z_j}{z}_k={\delta }_{jk}$, ${\mathrm{\partial }}_{z_j}{\bar{z}}_k=0$, ${\mathrm{\partial }}_{{\bar{z}}_j}{z}_k=0$, and ${\mathrm{\partial }}_{{\bar{z}}_j}{\bar{z}}_k={\delta }_{jk}$. The scalar product rule [F5] therefore gives ${\mathrm{\partial }}_{z_1}f=2z_1{\bar{z}}_2$, ${\mathrm{\partial }}_{z_2}f=0$, ${\mathrm{\partial }}_{{\bar{z}}_1}f=2{\bar{z}}_1$, and ${\mathrm{\partial }}_{{\bar{z}}_2}f=z_1^2$. Wedge these coefficients with their corresponding one-forms to obtain the displayed $\mathrm{\partial }f$ and $\bar{\mathrm{\partial }}f$.

Applying the type coefficient formulas to $\eta =fd{z}_2$ yields the two displayed form derivatives. [F1, F4, step 1.1, given, algebra] The only type factor in $\eta$ is $d{z}_2$. Thus [F1]–[F4] give $\mathrm{\partial }\eta =(2z_1{\bar{z}}_{2}d{z}_1)\wedge d{z}_2$ and $\bar{\mathrm{\partial }}\eta =(2{\bar{z}}_{1}d{\bar{z}}_1+z_1^{2}d{\bar{z}}_2)\wedge d{z}_2$. Anticommutativity gives $d{\bar{z}}_2\wedge d{z}_2=-d{z}_2\wedge d{\bar{z}}_2$, so the sign in the last summand is as stated.

Their sum is the exterior derivative of this example. [F6, step 2.1, given, algebra] By [F6], $d\eta =\mathrm{\partial }\eta +\bar{\mathrm{\partial }}\eta$, so the two explicitly computed type components in step 2.1 add to $d\eta$ with the same wedge sign. ∎

The zero extension is $C_c^1(\mathbb{C})$, and its interior $\bar{\mathrm{\partial }}$ derivative is explicit. [F1, given, algebra] The function $h(t)=(\max \{t,0\}{)}^2$ is $C^1$ on $\mathbb{R}$, so $f(z)=h(1-\mathrm{\mid }z{\mathrm{\mid }}^2)$ is $C^1$. It is zero for $\mathrm{\mid }z\mathrm{\mid }\ge 1$, hence has support in the compact closed unit disc. On $D$, the Wirtinger formula [F1] gives ${\mathrm{\partial }}_{\bar{\zeta }}f=-2\zeta (1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)$. In particular $f=0$ on $\mathrm{\partial }D$ and $f(0)=1$.

The sphere measure in [F4] has total mass $2\pi$. [F3, F4, F5, F6, F7, F8, F9, algebra] For the closed unit disc $K\subset {\mathbb{R}}^2$, [F5] and [F6] give $\mathrm{cont}(K)=V_2(1)=\pi$, and [F7] makes $K$ Jordan measurable. By [F3], full AC supplies the countable-choice premise of [F8], so ${\lambda }_2(K)=\pi$. Also [F9] gives ${\lambda }_2(\{0\})=0$. The set in [F4] for $E=S^1$ is $K\setminus \{0\}$, so its measure is $\pi$ and [F4] yields $\sigma (S^1)=2\pi$.

Applying Cauchy–Pompeiu at $z=0$ gives the asserted value of the area term. [F2, F3, F10, step 1.1, step 1.2, given, algebra] The unit disc is a bounded $C^1$ domain, and step 1.1 proves the needed $C^1$ hypothesis for $f$. Full AC [F3] supplies the premise of [F2]. Since $f=0$ on $\mathrm{\partial }D$, its boundary term vanishes. For $\zeta \mathrm{\ne }0$, step 1.1 gives $({\mathrm{\partial }}_{\bar{\zeta }}f)\mathrm{/}\zeta =-2(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)$, which extends continuously to $-2$ at $0$. Put $q(x)=\max \{1-\mathrm{\mid }x{\mathrm{\mid }}^2,0\}$, a nonnegative Borel function on ${\mathbb{R}}^2$. By [F10] and the sphere mass from step 1.2, ${\int }_D(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)dA(\zeta )={\int }_{{\mathbb{R}}^2}qd{\lambda }_2=2\pi {\int }_0^1(1-r^2)rdr=\frac{\pi }{2}\mathrm{.}$ Thus the area term equals $-\frac{1}{\pi }{\int }_D\frac{{\mathrm{\partial }}_{\bar{\zeta }}f}{\zeta }dA=\frac{2}{\pi }{\int }_D(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2)dA=1=f(0)$, as claimed. ∎

The unit ball is bounded with smooth boundary, $0\in B$, and the constant function $1$ is holomorphic and smooth on $\bar{B}$; the full-AC hypothesis supplies the stated premise of [F2] by [F3]. Applying [F2] at $z=0$ to $f=1$ gives $1={\int }_{\mathrm{\partial }B}{\mathrm{\Omega }}_n(\zeta ,0)-{\int }_B\bar{\mathrm{\partial }}1\wedge {\mathrm{\Omega }}_n(\zeta ,0)={\int }_{\mathrm{\partial }B}{\mathrm{\Omega }}_n(\zeta ,0)$, since the holomorphic case in [F2] has zero interior term.

Fix $j$ and define $R_t(\zeta )=e^{it}\zeta$ on $\mathrm{\partial }B$; by the explicit kernel [F1], each coefficient gains $e^{-it}$ while its wedge part, with $n$ factors $d{\zeta }_k$ and $n-1$ factors $d{\bar{\zeta }}_k$, gains $e^{it}$, so $R_t^{\ast }{\mathrm{\Omega }}_n(\zeta ,0)={\mathrm{\Omega }}_n(\zeta ,0)$ and $R_t^{\ast }({\zeta }_j{\mathrm{\Omega }}_n(\zeta ,0))=e^{it}{\zeta }_j{\mathrm{\Omega }}_n(\zeta ,0)$.

The restriction of $R_t$ is an orientation-preserving diffeomorphism of the oriented sphere $\mathrm{\partial }B$: its ambient real determinant is $1$ and it carries outward radial normals to outward radial normals. The form ${\zeta }_j{\mathrm{\Omega }}_n(\zeta ,0)$ is smooth on the compact sphere, hence compactly supported there. With $I_j={\int }_{\mathrm{\partial }B}{\zeta }_j{\mathrm{\Omega }}_n(\zeta ,0)$, [F4] and step 1.2 give $I_j={\int }_{\mathrm{\partial }B}{R}_t^{\ast }({\zeta }_j{\mathrm{\Omega }}_n)=e^{it}{I}_j$. Taking $t=\pi$ yields $I_j=-I_j$, hence $I_j=0$. This calculation also covers $n=1$, since the kernel has one holomorphic differential and no antiholomorphic differentials in that case.

The integrals therefore equal $1(0)=1$ and $z_j(0)=0$ for the constant and coordinate functions, respectively; both are holomorphic on $B$, so these explicit values agree with the holomorphic cases of [F2]. [F2, step 1.1, step 2.1, given, algebra] $\mathrm{\square }$

By [F1], ${\mathrm{\partial }}_{{\bar{z}}_1}u=2{\bar{z}}_1$ and ${\mathrm{\partial }}_{{\bar{z}}_2}u=z_1$: the cross derivatives ${\mathrm{\partial }}_{{\bar{z}}_1}(z_1{\bar{z}}_2)$ and ${\mathrm{\partial }}_{{\bar{z}}_2}({\bar{z}}_1^2)$ are zero. The scalar case of [F2] therefore gives $\bar{\mathrm{\partial }}u=2{\bar{z}}_{1}d{\bar{z}}_1+z_{1}d{\bar{z}}_2=g$.

For $g=g_{1}d{\bar{z}}_1+g_{2}d{\bar{z}}_2$, where $g_1=2{\bar{z}}_1$ and $g_2=z_1$, [F1] gives ${\mathrm{\partial }}_{{\bar{z}}_1}{g}_1=2$ and ${\mathrm{\partial }}_{{\bar{z}}_2}{g}_1={\mathrm{\partial }}_{{\bar{z}}_1}{g}_2={\mathrm{\partial }}_{{\bar{z}}_2}{g}_2=0$. Hence [F2] gives $\bar{\mathrm{\partial }}g=2d{\bar{z}}_1\wedge d{\bar{z}}_1=0$ by alternation; in particular, both mixed cross derivatives vanish.

The proposed witness has a nonzero $\bar{\mathrm{\partial }}$ derivative at every point. [F1, given, algebra] Applying the coefficient formula [F1] to $g={\bar{z}}_{1}d{\bar{z}}_2$ differentiates its coefficient once in each barred coordinate. Only the $j=1$ derivative is nonzero, and it equals $1$, so $\bar{\mathrm{\partial }}g=d{\bar{z}}_1\wedge d{\bar{z}}_2\mathrm{.}$ Because the two coordinate covectors are distinct members of the local wedge basis, this $(0,2)$-form is nonzero at every point of ${\mathbb{C}}^2$; hence $g{\mathrm{\mid }}_U$ is not $\bar{\mathrm{\partial }}$-closed on any nonempty open $U$.

Nonclosedness contradicts the necessary condition for having a potential. [F2, step 1.1, given, algebra] If a smooth $u:U\to \mathbb{C}$ satisfied $\bar{\mathrm{\partial }}u=g{\mathrm{\mid }}_U$, applying $\bar{\mathrm{\partial }}$ and using [F2] would give $0={\bar{\mathrm{\partial }}}^{2}u=\bar{\mathrm{\partial }}g{\mathrm{\mid }}_U=d{\bar{z}}_1\wedge d{\bar{z}}_2$. The last form is nonzero at every point of the nonempty set $U$ by step 1.1, a contradiction. Thus no such $u$ exists. ∎

The singleton $\{0\}$ is compact, so [F3] gives $\chi =1$ near $0$ with support $K_{\chi }\subset B(0,1)$. By [F4], $K_{\chi }$ is closed; it is bounded because it lies in the unit ball, hence compact by [F5]. Since $f$ is holomorphic, [F14] makes it complex differentiable at each point; [F12] makes it smooth, and [F13] gives $\bar{\mathrm{\partial }}f=0$ (the dictionary [F5] identifies the library's coordinates $0,1$ with $z_1,z_2$ here). Since $\chi =1$ near $0$, $f_0$ is identically zero near the puncture and therefore smooth on $B(0,2)$. The bidegree formula [F6] makes $g=\bar{\mathrm{\partial }}{f}_0$ a smooth $(0,1)$ form. On $G$, the product rule [F19] gives $g=-f\bar{\mathrm{\partial }}\chi$. Outside $K_{\chi }$, $\chi$ vanishes on a neighborhood, so $g=0$ there; hence $\mathrm{supp}g\subseteq K_{\chi }\subset B(0,1)$. The support is closed by [F4] and bounded, so [F5] makes it compact. Thus $g$ is zero near $\mathrm{\partial }B(0,2)$, its zero extension is smooth, and [F7] gives $\bar{\mathrm{\partial }}g=0$ on all of ${\mathbb{C}}^2$.

The set $G$ is open: if $z\in G$ and $r=\mathrm{\parallel }z\mathrm{\parallel }$, choose $\delta =\frac{1}{2}\min (r,2-r)>0$. For $w\in B(z,\delta )$, [F16] gives $0<\mathrm{\parallel }w\mathrm{\parallel }<2$, and [F17] makes this ball an open neighborhood contained in $G$. The ball notation and norm topology are those of [F8]. To connect two points of $G$, move each radially to the unit sphere; these paths remain at positive norm below $2$ and are continuous by [F18]. Join their endpoints by a path in the unit sphere using [F9] and [F20]. Thus $G$ is path-connected, hence connected by [F10].

Let $E=\{z\in {\mathbb{C}}^2:\mathrm{\parallel }z\mathrm{\parallel }>1\}$. It lies in ${\mathbb{C}}^2\setminus \mathrm{supp}g$ by step 1.1 and is unbounded. For each $z\in E$, a radial path joins $z$ to $2z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ while keeping the norm greater than $1$; [F18] ensures the path is continuous. The radius-$2$ sphere is path-connected by [F9] and [F20] after rescaling the unit sphere in ${\mathbb{R}}^4\cong {\mathbb{C}}^2$; the ball and sphere notation is that of [F8]. Thus $E$ is path-connected and connected by [F10]. It lies in a connected component by [F11]; that component is unbounded because it contains $E$, and therefore is the unique unbounded component named in [F1].

The full AC assumption [F2] permits applying [F1] to the smooth compactly supported closed $(0,1)$ form $g$ from step 1.1, giving $u\in C_c^{\mathrm{\infty }}({\mathbb{C}}^2)$ with $\bar{\mathrm{\partial }}u=g$ and $u=0$ on the unique unbounded component identified in step 2.1. Therefore $u=0$ on $E$. Since $\chi =0$ on $E$, the correction $h=u+\chi f$ vanishes on $A=\{1<\mathrm{\parallel }z\mathrm{\parallel }<2\}\subset G$. The annulus is nonempty because $(3\mathrm{/}2,0)\in A$. For any $z\in A$, set $ϵ=\frac{1}{2}\min (\mathrm{\parallel }z\mathrm{\parallel }-1,2-\mathrm{\parallel }z\mathrm{\parallel })>0$; [F16] gives $B(z,ϵ)\subset A$, and [F17] says this metric ball, with notation from [F8], is open. Thus $A$ is open.

On $G$, [F19] and step 1.1 give $\bar{\mathrm{\partial }}h=\bar{\mathrm{\partial }}u+\bar{\mathrm{\partial }}(\chi f)=g+f\bar{\mathrm{\partial }}\chi +\chi \bar{\mathrm{\partial }}f=0$. The function $h$ is smooth, so [F13]–[F14] make it holomorphic on $G$. By step 1.2, $G$ is connected and open; since $h=0$ on $A$, [F15] gives $h=0$ throughout $G$. Thus on $G$, $F=f_0-u=(1-\chi )f+\chi f=f$.

On $B(0,2)$, $F=f_0-u$ is smooth and $\bar{\mathrm{\partial }}F=\bar{\mathrm{\partial }}{f}_0-g=0$ by step 3.1. The Cauchy–Riemann criterion [F13], followed by the definition [F14], makes $F$ holomorphic on the whole ball. For $f\equiv 1$, step 1.1 gives $g=-\bar{\mathrm{\partial }}\chi$ and step 4.1 gives $u=-\chi$ on $G$, so the formula yields $F=1$ there. Every neighborhood of $0$ in the ball contains nonzero points of $G$, so continuity gives $F(0)=1$. If $f\equiv 0$, then $f_0=g=0$ and the unique solution in [F1] is $u=0$, since the zero function is a compactly supported solution; hence $F=0$. [F1, F13, F14, step 1.1, step 3.1, step 4.1, given, algebra] $\mathrm{\square }$