First cohomology of finite holomorphic bundles on a polydisc

Rescale the polydisc to $\mathrm{\mid }{z}_j\mathrm{\mid }<1$, and suppose its dimension $n$ is positive; dimension zero is immediate. Put $t={\sum }_j-\log (1-\mathrm{\mid }{z}_j{\mathrm{\mid }}^2)$ and $g={\sum }_j(1-\mathrm{\mid }{z}_j{\mathrm{\mid }}^2{)}^{-2}\mathrm{\mid }d{z}_j{\mathrm{\mid }}^2$. Direct differentiation gives $i\mathrm{\partial }\bar{\mathrm{\partial }}t$ equal to the Kähler form of g, up to the fixed convention factor, which we absorb into g. The g-norm of dt is bounded. The metric is complete: a path leaving every compact set has unbounded t, and its length bounds the variation of t divided by that gradient bound. The functions $\eta (t\mathrm{/}A)$, where $\eta =1$ on $[0,1]$ and $\eta =0$ on $[2,\mathrm{\infty })$, are compactly supported cutoffs with gradient at most $C\mathrm{/}A$.

Tensor $E$ with $K_P^{-1}$, calling the resulting bundle B. Identify an E-valued $(0,q)$ form $\alpha$ with $(-1{)}^{nq}(d{z}_1\wedge \cdots \wedge d{z}_n\wedge \alpha )\otimes (d{z}_1\wedge \cdots \wedge d{z}_n{)}^{-1}$. This frame-independent identification intertwines dbar exactly and is a B-valued $(n,q)$ form; give B the metric making this identification isometric for g. Choose a smooth convex increasing function $\chi$ of t so that the Nakano curvature of $B$ with metric multiplied by $e^{-\chi (t)}$ dominates the identity on $(n,1)$ coefficients. Here this inequality is constructed, not assumed: on each compact shell $k\le t\le k+1$ the negative part of the original curvature matrix has a finite bound $C_k$; choose ${\chi }^{\mathrm{\prime }}\ge 1+{\max }_{j\le k+1}{C}_j$ there and ${\chi }^{\mathrm{\prime }\mathrm{\prime }}\ge 0$. The added curvature is ${\chi }^{\mathrm{\prime }}i\mathrm{\partial }\bar{\mathrm{\partial }}t+{\chi }^{\mathrm{\prime }\mathrm{\prime }}i\mathrm{\partial }t\wedge \bar{\mathrm{\partial }}t$. Enlarging these increasing bounds if necessary also makes any specified smooth E-valued form f square integrable: bound its unweighted integral on each compact shell by $a_k<\mathrm{\infty }$ and arrange $\chi (k)\ge \log (1+a_k)+2k$. Smooth convex interpolation of these increasing bounds gives one such function on $[0,\mathrm{\infty })$.

Write $A=D_{\chi }^{\mathrm{\prime }}$, $B_0=\bar{\mathrm{\partial }}$, ${\mathrm{\Theta }}_{\chi }=A{B}_0+B_{0}A$, and let $\mathrm{\Lambda }$ be adjoint to wedge multiplication by the Kähler form. In holomorphic normal Kähler coordinates and a holomorphic frame Chern-normal for the weighted metric at the point, the wedge/contraction relations $\{{ϵ}_j,{\iota }_k\}={\delta }_{jk}$ give $[\mathrm{\Lambda },B_0]=-i{A}^{\ast }$ and $[\mathrm{\Lambda },A]=i{B}_0^{\ast }$. First derivatives of g and the connection coefficients vanish there; these are equalities of first-order operators, hence hold globally. Expanding ordinary products gives ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}-{\mathrm{\Delta }}^{\mathrm{\prime }}=-i\{B_0,[\mathrm{\Lambda },A]\}-i\{A,[\mathrm{\Lambda },B_0]\}=[i{\mathrm{\Theta }}_{\chi },\mathrm{\Lambda }]\mathrm{.}$ For compactly supported smooth $(n,1)$ forms, integration by parts therefore gives $\mathrm{\parallel }\bar{\mathrm{\partial }}u{\mathrm{\parallel }}^2+\mathrm{\parallel }{\bar{\mathrm{\partial }}}^{\ast }u{\mathrm{\parallel }}^2=\mathrm{\parallel }{D}_{\chi }^{\mathrm{\prime }}u{\mathrm{\parallel }}^2+\mathrm{\parallel }{D}_{\chi }^{\mathrm{\prime }\ast }u{\mathrm{\parallel }}^2+⟨[i{\mathrm{\Theta }}_{\chi },\mathrm{\Lambda }]u,u⟩\ge \mathrm{\parallel }u{\mathrm{\parallel }}^2\mathrm{.}$ Indeed $D_{\chi }^{\mathrm{\prime }}u=0$ by holomorphic degree. In a holomorphic frame $A_j={\mathrm{\partial }}_j+h^{-1}{\mathrm{\partial }}_{j}h-{\chi }_j$, so the curvature coefficient in $i{\mathrm{\Theta }}_{\chi }=i\sum R_{j\bar{k}}d{z}_j\wedge d{\bar{z}}_k$ is $R_{j\bar{k}}=-{\mathrm{\partial }}_{\bar{k}}(h^{-1}{\mathrm{\partial }}_{j}h)+{\chi }_{j\bar{k}}I$. The commutator quadratic form on $(n,1)$ coefficients is ${\sum }_{j,k}⟨R_{j\bar{k}}{u}_j,u_k⟩$ in the signed top-form convention. Its lower bound is step 2.1; in particular the added weight Hessian has positive sign. No boundary term occurs. This explicitly derives the identity, rather than taking a named estimate as a premise.

Fix the weight chosen in step 2.1. Let $H_q=L_{(0,q)}^2(P,E;g,h{e}^{-\chi (t)})$, with the bundle metric understood through the isometric identification in step 2.1, and put $H_q=\{0\}$ for $q>n$. For $q=0,1$ define the maximal distributional operator by $\mathrm{Dom}{D}_q=\{u\in H_q:\bar{\mathrm{\partial }}u\text{ in distributions is represented by an element of }{H}_{q+1}\},D_{q}u=\bar{\mathrm{\partial }}u\mathrm{.}$ If $q=n$ this is the zero operator on all of $H_n$. The space $C_c^{\mathrm{\infty }}(P,E\otimes {\mathrm{\Lambda }}^{0,q})$ is contained in this domain and is dense in $H_q$: first truncate an arbitrary $L^2$ form by the compact cutoffs of step 1.1, then use finitely many bundle charts, a smooth partition and coefficient convolution on that compact support. Smooth positive metrics and weights are bounded above and below there, so ordinary local $L^2$ approximation also converges in the weighted norm. These same local bounds show closedness: if $u_j\to u$ in $H_q$ and $D_q{u}_j\to v$ in $H_{q+1}$, both convergences hold in distributions on every compact chart; differentiation is continuous on distributions, so $\bar{\mathrm{\partial }}u=v$ and $(u,v)$ belongs to the graph. Set $T=D_0$ and $S=D_1$. Define $\mathrm{Dom}{T}^{\ast }$ to consist of $a\in H_1$ for which there is $b\in H_0$ satisfying $⟨Tv,a⟩=⟨v,b⟩$ for every $v\in \mathrm{Dom}T$, and put $T^{\ast }a=b$; $b$ is unique because $\mathrm{Dom}T$ is dense. For every compactly supported smooth $a$, the smooth weighted formal adjoint $T^{†}a$ has compact support and belongs to $H_0$. The distributional definition of $T$ gives $⟨Tv,a⟩=⟨v,T^{†}a⟩$ for all $v\in \mathrm{Dom}T$. Hence $C_c^{\mathrm{\infty }}(P,E\otimes {\mathrm{\Lambda }}^{0,1})\subseteq \mathrm{Dom}{T}^{\ast }$, proving that this adjoint domain is dense in $H_1$. Passing to limits in this pairing identity also proves that $T^{\ast }$ is closed. For $a\in \mathrm{Dom}{T}^{\ast }$, testing against compactly supported smooth $v$ identifies $T^{\ast }a$ with $T^{†}a$ distributionally; no converse equality of maximal formal-adjoint and Hilbert-adjoint domains is assumed. In the degree-one estimate write $\bar{\mathrm{\partial }}=S$ and ${\bar{\mathrm{\partial }}}^{\ast }=T^{\ast }$. The same estimate holds on $\mathrm{Dom}{T}^{\ast }\cap \mathrm{Dom}S$. Here is the simultaneous graph-domain justification. For a cutoff $\zeta$ and $a\in \mathrm{Dom}{T}^{\ast }$, the identity $T(\zeta v)=\zeta Tv+(\bar{\mathrm{\partial }}\zeta )\wedge v$ for $v\in \mathrm{Dom}T$ shows directly from the adjoint pairing that $\zeta a\in \mathrm{Dom}{T}^{\ast }$ with the formal-adjoint Leibniz rule. Multiplication by the cutoffs of step 1.1 therefore tends to the identity in both graph norms: the commutator with either first-order operator is contraction or wedge multiplication by its gradient, bounded by $C\mathrm{/}A$, and dominated convergence handles the other terms. On the resulting compact support, take finitely many coordinate and bundle frames, a subordinate partition, and convolve their coefficients. For a smooth first-order coefficient a the commutator of $a\mathrm{\partial }$ with convolution tends to zero in L2: its integral is a difference quotient of a times the approximate identity, and Taylor's formula proves the assertion for smooth inputs, while its uniformly bounded operator norm and density prove it for L2 inputs. Zero-order terms converge directly. This supplies compactly supported smooth approximants in the simultaneous graph norm. Apply step 3.1 and pass to the limit.

Take all Hilbert inner products to be linear in their first variable. Use the maximal operators $T:H_0\to H_1$ and $S:H_1\to H_2$ defined and proved closed and densely defined in step 4.1. For every $v\in \mathrm{Dom}T$, the distributional identity $\bar{\mathrm{\partial }}(Tv)={\bar{\mathrm{\partial }}}^{2}v=0$ gives $Tv\in \mathrm{Dom}S$ and $STv=0$; thus this composition is defined on all of $\mathrm{Dom}T$. In particular, $\ker S$ is closed: if $u_n\in \mathrm{Dom}S$, $u_n\to u$, and $S{u}_n=0$, closedness gives $u\in \mathrm{Dom}S$ and $Su=0$. We give the Hilbert projection argument used here and again below. For a Hilbert space $H$ and closed linear subspace $N\subseteq H$, fix $x\in H$ and put $d={\inf }_{n\in N}\mathrm{\parallel }x-n\mathrm{\parallel }$. By full AC choose $n_j\in N$ with $\mathrm{\parallel }x-n_j\mathrm{\parallel }\le d+1\mathrm{/}j$. The parallelogram identity and $(n_j+n_k)\mathrm{/}2\in N$ give $\mathrm{\parallel }{n}_j-n_k{\mathrm{\parallel }}^2\le 2(d+1\mathrm{/}j{)}^2+2(d+1\mathrm{/}k{)}^2-4d^2$, so $(n_j)$ is Cauchy. Completeness and closedness give a limit $n_x\in N$ attaining distance $d$. For each $h\in N$, minimality of $n_x$ applied to $n_x+\lambda h$, first for real and then for purely imaginary $\lambda$, gives $⟨x-n_x,h⟩=0$. This minimizer is unique, since the difference of two such minimizers lies in $N\cap N^{\perp }=\{0\}$. Thus every $x$ has a unique decomposition $x=n_x+(x-n_x)$ with $n_x\in N$ and $x-n_x\in N^{\perp }$. Uniqueness makes $P_{N}x=n_x$ linear, and Pythagoras gives $H=N\oplus N^{\perp }$ and $\mathrm{\parallel }{P}_N\mathrm{\parallel }\le 1$. Apply this construction to the closed subspace $N=\ker S$ in the degree-one Hilbert space and call the resulting projection $P$. If $u\in \mathrm{Dom}{T}^{\ast }$, then $Tv\in \ker S$ for every $v\in \mathrm{Dom}T$, so $⟨Tv,Pu⟩=⟨Tv,u⟩=⟨v,T^{\ast }u⟩$. Hence $Pu\in \mathrm{Dom}{T}^{\ast }$ and $T^{\ast }Pu=T^{\ast }u$. Also $Pu\in \mathrm{Dom}S$ and $SPu=0$. Step 4.1 therefore gives $\mathrm{\parallel }Pu\mathrm{\parallel }\le \mathrm{\parallel }{T}^{\ast }u\mathrm{\parallel }$.

Now take any smooth closed E-valued $(0,1)$ form f and choose the weight of step 2.1 also for f. Since $f\in \ker S$, the functional $\mathrm{\ell }(T^{\ast }u)=⟨u,f⟩=⟨Pu,f⟩$ is well defined and bounded by $\mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }{T}^{\ast }u\mathrm{\parallel }$. Extend it continuously to $N=\overline{\mathrm{ran}{T}^{\ast }}$ in the degree-zero Hilbert space and denote this extension by $\mathrm{\ell }$. The projection proved in step 5.1 gives the orthogonal decomposition into $N$ and $N^{\perp }$, so $\tilde{\mathrm{\ell }}(x)=\mathrm{\ell }(P_{N}x)$ is a bounded linear functional on the whole degree-zero Hilbert space. We prove its representing vector directly. If $\tilde{\mathrm{\ell }}=0$, take $v=0$. Otherwise $K=\ker \tilde{\mathrm{\ell }}$ is a closed hyperplane; the projection construction of step 5.1 supplies its orthogonal projection $P_K$. Choose $x$ with $\tilde{\mathrm{\ell }}(x)\mathrm{\ne }0$ and set $e=x-P_{K}x$. Then $e\mathrm{\ne }0$, $e\perp K$, and $\tilde{\mathrm{\ell }}(e)=\tilde{\mathrm{\ell }}(x)$. For every y, $y-(\tilde{\mathrm{\ell }}(y)\mathrm{/}\tilde{\mathrm{\ell }}(e))e\in K$, so $v=\overline{\stackrel{~}{\mathrm{\ell }}(e)}e\mathrm{/}\mathrm{\parallel }e{\mathrm{\parallel }}^2$ satisfies $\tilde{\mathrm{\ell }}(y)=⟨y,v⟩$. Consequently $⟨T^{\ast }u,v⟩=⟨u,f⟩$, or equivalently $⟨v,T^{\ast }u⟩=⟨f,u⟩$, for every $u\in \mathrm{Dom}{T}^{\ast }$. Since step 4.1 proves that $\mathrm{Dom}{T}^{\ast }$ is dense in $H_1$, its adjoint $T^{\ast \ast }$ is defined; the displayed pairing identity says exactly that $v\in \mathrm{Dom}{T}^{\ast \ast }$ and $T^{\ast \ast }v=f$. For completeness, if T is densely defined and closed, its graph satisfies $\mathrm{graph}(T{)}^{\perp }=\{(-T^{\ast }y,y):y\in \mathrm{Dom}{T}^{\ast }\}$. A pair $(x,z)$ is orthogonal to every $(-T^{\ast }y,y)$ exactly when $⟨T^{\ast }y,x⟩=⟨y,z⟩$ for every $y\in \mathrm{Dom}{T}^{\ast }$, which is the defining condition $(x,z)\in \mathrm{graph}(T^{\ast \ast })$. Since the graph of T is closed, $\mathrm{graph}(T{)}^{\perp \perp }=\mathrm{graph}(T)$, so $T^{\ast \ast }=T$. Hence $Tv=f$. This constructs the required global distributional solution without invoking an unnamed Hilbert-space representation theorem.

This v is smooth. In each relatively compact local holomorphic frame its coefficients satisfy the scalar equations ${\mathrm{\partial }}_{{\bar{z}}_j}v=f_j$. Cut off and mollify. The Euclidean constant-coefficient identity $\mathrm{\parallel }\mathrm{\nabla }a{\mathrm{\parallel }}_2^2=4{\sum }_j\mathrm{\parallel }{\mathrm{\partial }}_{{\bar{z}}_j}a{\mathrm{\parallel }}_2^2$ bounds the first weak derivatives locally; cutoff errors are L2. Differentiating and repeating gives every Sobolev order because the right sides are smooth. Fourier Cauchy–Schwarz, or the repeated one-variable fundamental theorem of calculus on a cube, then gives smooth representatives. Hence every smooth closed degree-one form is globally exact. Local exactness in F2 and the smooth fine resolution in F3 give $H^1(P,\mathcal{O}(E))=0$. For a bundle short exact sequence, lift a last-term section smoothly; its dbar is a closed kernel-bundle form. Subtract the global solution just constructed. The corrected lift is holomorphic, proving the last assertion as well.