Hodge decomposition for Dolbeault forms on a compact Riemann surface

If $f$ has degree $q$ and lies in ${\mathcal{H}}^{0,q}(E{)}^{\perp }$, then ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}Gf=f$ and $Gf\perp \mathcal{H}$. Its other-degree component is therefore harmonic and orthogonal to the harmonic space, so it vanishes; thus $Gf$ has degree $q$. For $q=0$ the Green identity gives $f={\mathrm{\Delta }}_0^{\mathrm{\prime }\mathrm{\prime }}Gf={\bar{D}}^{\ast }\bar{D}Gf$, so $f\in \mathrm{ran}{\bar{D}}^{\ast }$. Conversely, if $f={\bar{D}}^{\ast }v$ and $u\in \ker \bar{D}={\mathcal{H}}^{0,0}(E)$, the adjoint identity gives $⟨u,f⟩=⟨\bar{D}u,v⟩=0$. Thus $\mathrm{ran}{\bar{D}}^{\ast }={\mathcal{H}}^{0,0}(E{)}^{\perp }$. For $q=1$, the Green identity gives $f={\mathrm{\Delta }}_1^{\mathrm{\prime }\mathrm{\prime }}Gf=\bar{D}{\bar{D}}^{\ast }Gf\in \mathrm{ran}\bar{D}$; and $\ker {\bar{D}}^{\ast }=(\mathrm{ran}\bar{D}{)}^{\perp }$ implies $\mathrm{ran}\bar{D}\subseteq {\mathcal{H}}^{0,1}(E{)}^{\perp }$. Hence $\mathrm{ran}\bar{D}={\mathcal{H}}^{0,1}(E{)}^{\perp }$. Both ranges are closed and give the stated degreewise $L^2$ splittings.

Write $f=h+v$ with $h\in \mathcal{H}$ and $v\in {\mathcal{H}}^{\perp }$. The Green identity gives ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}\tilde{G}f=v$, so $P_{\mathcal{H}}f=h$; thus the displayed formula is the orthogonal projection. By [F6], it preserves smooth forms and the harmonic summands are smooth. If $w\in {\mathrm{\Omega }}^{0,0}(E)$ is orthogonal to ${\mathcal{H}}^{0,0}(E)$, step 1.1 gives $w={\bar{D}}^{\ast }u$ for some $u\in \mathrm{dom}{\bar{D}}^{\ast }$. Since $w$ is smooth, it lies in $\mathrm{dom}\bar{D}$; hence $u\in \mathrm{dom}{\mathrm{\Delta }}_1^{\mathrm{\prime }\mathrm{\prime }}$ and ${\mathrm{\Delta }}_1^{\mathrm{\prime }\mathrm{\prime }}u=\bar{D}w$ is smooth. Also $u\in H^1$ by [F9]; the smooth-coefficient formula in [F8] is uniformly elliptic on compactly contained chart patches, and testing its distributional equation by integration by parts gives the local weak-solution identity. The smooth-data regularity in [F6] gives a smooth representative on each such patch. These representatives agree on overlaps because they represent the same section almost everywhere and are continuous, so they glue to a smooth section $\tilde{u}$ with $w={\bar{\mathrm{\partial }}}_E^{\ast }\tilde{u}$ by [F7]. Conversely every smooth ${\bar{\mathrm{\partial }}}_E^{\ast }$-image is orthogonal to $\ker \bar{D}$ by the adjoint identity. For degree one, step 1.1 gives every smooth form orthogonal to ${\mathcal{H}}^{0,1}(E)$ in $\mathrm{ran}\bar{D}$, and [F6] upgrades it to a smooth ${\bar{\mathrm{\partial }}}_E$-image; the adjoint identity gives the reverse orthogonality. This proves both smooth decompositions.

Fix $q$ and $k$, and choose a finite $L^2$-orthonormal basis $e_1,\dots ,e_m$ of ${\mathcal{H}}^{0,q}(E)$; if this space is zero, take the empty basis. Each $e_j$ is smooth by [F6], so $P_{q}f={\sum }_{j=1}^m⟨f,e_j{⟩}_{L^2}{e}_j$ and [F10]–[F11] give $\mathrm{\parallel }{P}_{q}f{\mathrm{\parallel }}_{H^k}\le {\sum }_j\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }{e}_j{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }{e}_j{\mathrm{\parallel }}_{H^k}\le C_{k,q}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^k}$. Thus $P_q$ extends to a bounded projection on $H_q^k$, whose range is ${\mathcal{H}}^{0,q}(E)$ and whose kernel is closed. The smooth complement is exactly the smooth range from step 2.1, so it is closed in the relative $H^k$ topology. If $f\in \ker P_q\subset H_q^k$, approximate it in $H^k$ by smooth $f_n$ and set $g_n=f_n-P_q{f}_n$; boundedness gives $g_n\to f$, and each $g_n$ is in that smooth complement. Therefore its $H^k$ closure is precisely $\ker P_q$, proving the topological splitting.

On a Riemann surface every smooth $(0,1)$-form is $\bar{\mathrm{\partial }}$-closed, so its degree-one Dolbeault class is taken modulo ${\bar{\mathrm{\partial }}}_E{\mathrm{\Omega }}^{0,0}(E)$. The degree-one smooth decomposition from step 2.1 gives a harmonic representative for each class. If two harmonic forms represent the same class, their difference lies both in ${\mathcal{H}}^{0,1}(E)$ and in its orthogonal complement, so its squared norm is zero and the representatives agree. By [F1], [F2], and [F7], ${\mathcal{H}}^{0,0}(E)=\ker \bar{D}\cap {\mathrm{\Omega }}^{0,0}(E)=H^0(X,E)$; the degree-zero splitting in step 2.1 is therefore the asserted unique holomorphic-section decomposition.