The maximal Dolbeault operator and its Hilbert adjoint on a compact Riemann surface

For smooth $u\in C^{\mathrm{\infty }}(X,E)$ and a smooth test form $\phi \in C_c^{\mathrm{\infty }}(X,K\otimes E^{\ast })$, the product $u\wedge \phi$ contracts to a degree-one form. On a curve its exterior derivative is its $\bar{\mathrm{\partial }}$ part because the $(2,0)$ component vanishes. Stokes and the graded Leibniz rule give ${\int }_X{\bar{\mathrm{\partial }}}_{E}u\wedge \phi =-{\int }_{X}u\wedge {\bar{\mathrm{\partial }}}_{E^{\ast }}\phi$, so smooth $u$ has weak derivative ${\bar{\mathrm{\partial }}}_{E}u$. If two $L_1^2$ forms $v,v^{\mathrm{\prime }}$ satisfy the same weak identity, their difference pairs to zero with every test. By [F1], every test is ${\star }_{E}w$ for a smooth $w$; the star identity turns that pairing into $⟨v-v^{\mathrm{\prime }},w{⟩}_{L^2}$. Density of smooth $w$ in $L_1^2$ implies $v=v^{\mathrm{\prime }}$. Thus the weak derivative is unique and $\bar{D}$ extends ${\bar{\mathrm{\partial }}}_E$.

Let $B:H_0\to H_1$ be closed and densely defined between Hilbert spaces, with $B^{\ast }$ closed and densely defined, and set $T=B^{\ast }B$ on $M=\{x\in \mathrm{dom}B:Bx\in \mathrm{dom}{B}^{\ast }\}$. The graph inner product $⟨x,y{⟩}_B=⟨x,y{⟩}_{H_0}+⟨Bx,By{⟩}_{H_1}$ makes $\mathrm{dom}B$ Hilbert by [F4]. For any $f\in H_0$, Riesz applied to $x\mapsto ⟨x,f⟩$ gives $u\in \mathrm{dom}B$ with $⟨x,u⟩+⟨Bx,Bu⟩=⟨x,f⟩$ for every $x\in \mathrm{dom}B$. Hence $⟨Bx,Bu⟩=⟨x,f-u⟩$, so $Bu\in \mathrm{dom}{B}^{\ast }$ and $B^{\ast }Bu=f-u$; therefore $u\in M$ and $(T+I)u=f$, proving $\mathrm{ran}(T+I)=H_0$. If $y\perp M$, apply this construction with $f=y$ to obtain $u\in M$ and $(T+I)u=y$. Then $0=⟨u,y⟩=⟨u,(T+I)u⟩=\mathrm{\parallel }u{\mathrm{\parallel }}^2+\mathrm{\parallel }Bu{\mathrm{\parallel }}^2$, so $u=y=0$; hence $M^{\perp }=0$ and $M$ is dense by [F5]. For $x,y\in M$, $⟨Tx,y⟩=⟨Bx,By⟩=⟨x,Ty⟩$; also $⟨Tx,x⟩=\mathrm{\parallel }Bx{\mathrm{\parallel }}^2$, so $T$ is symmetric and nonnegative. If $x_n\to x$ and $T{x}_n\to y$, then $\mathrm{\parallel }B(x_n-x_m){\mathrm{\parallel }}^2=⟨x_n-x_m,T(x_n-x_m)⟩\le \mathrm{\parallel }{x}_n-x_m\mathrm{\parallel }\mathrm{\parallel }T{x}_n-T{x}_m\mathrm{\parallel }$, making $B{x}_n$ Cauchy. Closedness of $B$ gives $B{x}_n\to Bx$, and closedness of $B^{\ast }$ applied to $(B{x}_n,T{x}_n)$ gives $x\in M$ and $Tx=y$. Thus $T$ is closed.

With the fixed chart/frame/partition data, each coefficient norm in the statement is the Euclidean $W^{k,2}$ norm of a compactly supported coefficient. On each nonempty compact support, the smooth positive metric and volume weights are bounded above and below, so these local norms are equivalent to the same coefficient norms measured against the Riemannian density; empty supports contribute zero. For a second choice, split each first partitioned term over the finitely many second charts meeting its compact support and insert the second partition. On each resulting compact overlap, coefficients transform by smooth frame and $d\bar{z}$ factors and by a smooth coordinate change $F$. For a scalar coefficient $f$, the first- and second-derivative formulas are $D_i(f\circ F)=(D_{a}f\circ F)D_i{F}_a$ and $D_{ij}(f\circ F)=(D_{ab}f\circ F)D_i{F}_a{D}_j{F}_b+(D_{a}f\circ F)D_{ij}{F}_a$; multiplying by the smooth frame and form transitions uses the Leibniz rule. All transition derivatives through order two are bounded on these compact supports by [F7], and [F6] gives bounded restriction and cutoff maps. Comparing the invariant density norms, summing finitely many terms gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k,1}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k,2}$ for $k=1,2$; reversing the choices gives the converse. Thus the norms are equivalent and their completions have the same smooth-form identification.

If $y\in \ker (T^{\ast }-i)$, choose $x\in M$ with $(T+I)x=y$ using the surjectivity in step 1.2. Since $T^{\ast }y=iy$, the first-variable-linear convention gives $\mathrm{\parallel }y{\mathrm{\parallel }}^2=⟨(T+I)x,y⟩=(1-i)⟨x,y⟩$, while $⟨x,y⟩=⟨x,(T+I)x⟩=\mathrm{\parallel }Bx{\mathrm{\parallel }}^2+\mathrm{\parallel }x{\mathrm{\parallel }}^2$ is real and nonnegative. Therefore $y=0$. The same argument with $-i$ gives $\ker (T^{\ast }+i)=0$. For $z=\pm i$, symmetry gives $\mathrm{\parallel }(T-z)x{\mathrm{\parallel }}^2=\mathrm{\parallel }Tx{\mathrm{\parallel }}^2+\mathrm{\parallel }x{\mathrm{\parallel }}^2$; if $(T-z)x_n$ converges, this estimate makes $x_n$ Cauchy and then $T{x}_n$ converges, so closedness of $T$ makes $\mathrm{ran}(T-z)$ closed. By [F3], the orthogonal complements of these two ranges are the opposite deficiency kernels, hence both ranges are dense and therefore equal $H_0$. The range criterion [F5] proves that $B^{\ast }B$ is self-adjoint.

Smooth forms are dense in $L_0^2$ by [F1], and step 1.1 shows they lie in $\mathrm{dom}\bar{D}$, so this domain is dense. If $u_n\to u$ in $L_0^2$ and $\bar{D}{u}_n\to v$ in $L_1^2$, then for every smooth test $\phi$, Cauchy–Schwarz makes the wedge pairings continuous and gives ${\int }_{X}v\wedge \phi ={\lim }_n{\int }_X\bar{D}{u}_n\wedge \phi =-{\lim }_n{\int }_X{u}_n\wedge {\bar{\mathrm{\partial }}}_{E^{\ast }}\phi =-{\int }_{X}u\wedge {\bar{\mathrm{\partial }}}_{E^{\ast }}\phi$. Thus $v$ is the weak derivative of $u$, so the graph of $\bar{D}$ is closed.

In the fixed chart data, each local $W^{k,2}$ norm contains the local $L^2$ norm, and the positive metric weights on compact supports give $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$; the derivative sums give $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le C^{\mathrm{\prime }}\mathrm{\parallel }u{\mathrm{\parallel }}_{H^2}$ for smooth $u$. These bounds extend the identity to continuous maps $H^2\to H^1\to L^2$. They are injective: if an $H^k$ Cauchy sequence of smooth forms tends to zero in $L^2$, each local coefficient tends to zero in $L^2$ while its derivatives through order $k$ have $L^2$ limits. Integration by parts against compactly supported smooth tests shows every such limit is a weak derivative of zero, hence vanishes by [F7]. The $H^k$ norm limit is therefore zero. Smooth forms are dense by the definition as completion.

Put $\mathcal{H}=L_0^2\oplus L_1^2$ and define the single-space operator $A(u,w)=(0,\bar{D}u)$ on $\mathrm{dom}A=\mathrm{dom}\bar{D}\oplus L_1^2$. By step 2.2, $A$ is densely defined and closed. For $(a,b)\in \mathcal{H}$, the adjoint identity $⟨A(u,w),(a,b){⟩}_{\mathcal{H}}=⟨(u,w),A^{\ast }(a,b){⟩}_{\mathcal{H}}$ for all $(u,w)\in \mathrm{dom}A$ holds exactly when $b\in \mathrm{dom}{\bar{D}}^{\ast }$ and $A^{\ast }(a,b)=({\bar{D}}^{\ast }b,0)$: vary $w$ first, then apply the adjoint identity on $L_0^2,L_1^2$. If $b_n\to b$ and ${\bar{D}}^{\ast }{b}_n\to c$, closedness of $A^{\ast }$ applied to $(0,b_n)\to (0,b)$ proves that ${\bar{D}}^{\ast }$ is closed. Since $A$ is closable, [F3] makes $\mathrm{dom}{A}^{\ast }=L_0^2\oplus \mathrm{dom}{\bar{D}}^{\ast }$ dense in $\mathcal{H}$, so $\mathrm{dom}{\bar{D}}^{\ast }$ is dense in $L_1^2$. Also $\ker A^{\ast }=\mathrm{ran}(A{)}^{\perp }$ by [F3]; with $\mathrm{ran}A=\{0\}\oplus \mathrm{ran}\bar{D}$ and $A^{\ast }(a,b)=({\bar{D}}^{\ast }b,0)$, intersecting with $\{0\}\oplus L_1^2$ yields $\ker {\bar{D}}^{\ast }=(\mathrm{ran}\bar{D}{)}^{\perp }$. Finally, $A^{\ast \ast }=A$ because $A$ is closed; the same block calculation gives $A^{\ast \ast }(u,w)=(0,({\bar{D}}^{\ast }{)}^{\ast }u)$, hence $({\bar{D}}^{\ast }{)}^{\ast }=\bar{D}$.

Apply steps 1.2 and 2.1 to $B=\bar{D}$ and to $B={\bar{D}}^{\ast }$. Step 3.1 gives the closed dense adjoints needed for both applications and $({\bar{D}}^{\ast }{)}^{\ast }=\bar{D}$, so both ${\bar{D}}^{\ast }\bar{D}$ on $L_0^2$ and $\bar{D}{\bar{D}}^{\ast }$ on $L_1^2$ are self-adjoint nonnegative operators. Their direct sum on the stated domain is densely defined and symmetric; its shifts by $\pm i$ are onto because the corresponding shifts of each summand are onto, so [F5] makes the direct sum self-adjoint. The adjoint identity gives $⟨{\bar{D}}^{\ast }\bar{D}{u}_0,u_0⟩=\mathrm{\parallel }\bar{D}{u}_0{\mathrm{\parallel }}^2$ and $⟨\bar{D}{\bar{D}}^{\ast }{u}_1,u_1⟩=\mathrm{\parallel }{\bar{D}}^{\ast }{u}_1{\mathrm{\parallel }}^2$. Adding these identities proves the energy formula; it vanishes exactly when $\bar{D}{u}_0=0$ and ${\bar{D}}^{\ast }{u}_1=0$, which proves the harmonic-kernel description.

Steps 1.1–4.1 establish the unique weak maximal operator, its densely defined closed Hilbert adjoint, the self-adjoint nonnegative block Laplacian and its energy kernel, and the choice-independent Sobolev completions with continuous inclusions. Full AC is used through the stated Sobolev restriction and cutoff localization interface [F6]; the other interfaces named in [F1]–[F5] and the partition construction use only Countable Choice.