Hodge Theory on Compact Riemann Surfaces: Examples and Counterexamples

Translation by $m+in$ preserves the Euclidean metric, $d\bar{z}$, and $\sin (2\pi x)$, so these descend to the square torus. The lattice translation action is a covering-space action: disks of radius less than $1\mathrm{/}2$ have disjoint nontrivial lattice translates, so [F4] gives a covering quotient and holomorphic charts with translation transitions. The quotient map is open because the preimage of the image of an open set is the union of its translates. For inequivalent $z,w$, the distance from $z-w$ to the square lattice is positive: only finitely many lattice points lie in any bounded disk, and none equals $z-w$. Small disks about $z,w$ therefore project to disjoint neighborhoods, proving Hausdorffness. Images of a countable base of plane disks form a countable base of the quotient; it is nonempty and connected as a continuous image of $\mathbb{C}$. The image of the closed unit square covers the quotient and is compact, so these charts make it a compact Riemann surface. By [F1], ${\mathrm{\partial }}_{\bar{z}}\sin (2\pi x)=\pi \cos (2\pi x)$; this coefficient equals $\pi$ at $z=0$. Thus $\bar{\mathrm{\partial }}f$ is a nonzero smooth exact form.

If a smooth exact form $a={\bar{\mathrm{\partial }}}_{E}b$ on any compact Riemann surface with the stated metrics is harmonic, [F2] gives ${\bar{D}}^{\ast }a=0$ and $\bar{D}b=a$. The first-variable-linear adjoint identity yields $\mathrm{\parallel }a{\mathrm{\parallel }}^2=⟨\bar{D}b,a⟩=⟨b,{\bar{D}}^{\ast }a⟩=0$, so $a=0$. Applying this to step 1.1 proves that its exact form is not harmonic. The harmonic and exact summands therefore intersect only at zero, as also expressed by [F3]. Full AC is inherited through those operator and Hodge interfaces; the calculation selects no new family.

If $u\in \ker {\mathrm{\Delta }}_0^{\mathrm{\prime }\mathrm{\prime }}$, [F1] supplies a smooth representative with ${\bar{\mathrm{\partial }}}_{E}u=0$, so [F2] makes it holomorphic. Its modulus attains a maximum $M$ at a point $p$ by compactness and [F3]. Put $c=u(p)$ and $S=\{x:u(x)=c\}$. This set is closed by continuity and nonempty. At each $x\in S$, $\mathrm{\mid }u(x)\mathrm{\mid }=M$, so a connected coordinate disk about $x$ has an interior maximum of $\mathrm{\mid }u\mathrm{\mid }$; [F3] makes $u=c$ on that disk. Hence $S$ is open, and connectedness forces $S=X$. Thus every element of the operator kernel is an almost-everywhere constant.

Every constant is smooth, has zero Dolbeault derivative, belongs to the maximal domain, and has zero image in the adjoint domain; therefore it lies in the degree-zero Laplacian domain with ${\mathrm{\Delta }}_0^{\mathrm{\prime }\mathrm{\prime }}u=0$. In particular ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}1=0$ and $\ker {\mathrm{\Delta }}_0^{\mathrm{\prime }\mathrm{\prime }}=\mathbb{C}\cdot 1$, a one-dimensional space since $X$ is nonempty. By [F4], this is also $H^{0,0}(X,E)=H^0(X,\mathcal{O})$. The argument used no value of the Hermitian weight or of the compatible metric, so it proves the final assertion for every supplied smooth positive Hermitian metric. Full AC is inherited through [F1] and [F4].

Translation charts on $X$ have identity derivatives, so $d\bar{z}$ is a global nonvanishing form and every smooth $(0,1)$-form is $ud\bar{z}$ with a smooth lattice-periodic coefficient on the cover. For any positive smooth weight $\psi$, [F1] says it is harmonic precisely when ${\mathrm{\partial }}_z(\psi u)=0$. Then $\overline{\psi u}$ lifts to an entire function, bounded because it is periodic and bounded on the closed fundamental parallelogram. By [F2] it is constant. Conversely $u=c\mathrm{/}\psi$ is smooth and satisfies that adjoint equation, hence is harmonic. Thus the two spaces are $\mathbb{C}d\bar{z}$ and $\mathbb{C}{e}^{-\phi }d\bar{z}$; they differ because their equality would force the positive function $e^{-\phi }$ to be constant. Holomorphic functions on this torus likewise lift to bounded entire functions and are constant.

Set $b=e^{-\phi }d\bar{z}$. By [F4], $⟨d\bar{z},b{⟩}_{h_1}=2A$ and $⟨b,b{⟩}_{h_1}=2J$. Hence the orthogonal projection of $d\bar{z}$ onto $\mathbb{C}b$ is $(A\mathrm{/}J)b$ in the first-variable-linear convention. By [F3] the projection differs from $d\bar{z}$ by a smooth exact form, so the harmonic representative of $[d\bar{z}]$ for $h_1$ is precisely $(A\mathrm{/}J)e^{-\phi }d\bar{z}$. For $h_0$ it is $d\bar{z}$ itself. In particular the class is nonzero, since its harmonic representative is nonzero.

The smooth operator ${\bar{\mathrm{\partial }}}_E$ is determined by the holomorphic transitions, so the same vector space $\ker {\bar{\mathrm{\partial }}}_E$ in degree zero and the same quotient ${\mathrm{\Omega }}^{0,1}(E)\mathrm{/}{\bar{\mathrm{\partial }}}_E{\mathrm{\Omega }}^{0,0}(E)$ in degree one define the cohomology for both metrics. The harmonic isomorphisms of [F3] and step 1.1 therefore give $H^{0,0}(X,E)=\mathbb{C}$ and $\dim H^{0,1}(X,E)=1$ for either metric. For any compact Riemann surface and fixed holomorphic bundle the same kernel/quotient observation proves that $h^{0,0},h^{0,1}$ are invariant under changing either supplied metric; only the harmonic representative can change. Full AC is inherited through [F1] and [F3]; the explicit projection uses no additional choice.

A holomorphic function on the sphere is entire in the $z$ chart and bounded near infinity, since its $w$-chart expression is continuous at $0$. It is also bounded on each closed disk, so is bounded on all of $\mathbb{C}$. By [F2] it is constant, and all constants are global holomorphic sections of the trivial bundle. Thus [F3] gives $H^{0,0}(X,E)=\mathbb{C}\cdot 1$.

Write a global holomorphic differential as $f(z)dz$ on $\mathbb{C}$, with $f$ entire. In the other chart it is $a(w)dw$, where $a(w)=-w^{-2}f(1\mathrm{/}w)$ for $w\mathrm{\ne }0$ by [F1]. Holomorphy at $w=0$ bounds $a$ on a small closed disk. Consequently $\mathrm{\mid }f(z)\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^{-2}$ for sufficiently large $\mathrm{\mid }z\mathrm{\mid }$. The entire function $f$ is bounded on a closed disk and on its exterior, so [F2] makes it constant; the displayed decay forces that constant to vanish. Therefore $H^0(X,K)=0$.

The trivial dual bundle identifies $K\otimes E^{\ast }$ holomorphically with $K$. By [F3] and step 1.2, the degree-one harmonic space and hence $H^{0,1}(X,E)$ are zero. Every smooth $(0,1)$-form on a curve is closed because there are no $(0,2)$-forms; its zero cohomology class says exactly that it is ${\bar{\mathrm{\partial }}}_E$ of a global smooth function. This proves the stated exactness for any supplied compatible metric. Full AC is inherited through duality and the harmonic representative interfaces.

The real-linear isomorphism $T(s,t)=s{\omega }_1+t{\omega }_2$ has continuous inverse. Its inverse is bounded, so for some $c>0$, $\mathrm{\mid }T(m,n)\mathrm{\mid }\ge c\sqrt{m^2+n^2}$; every nonzero period therefore has length at least $c$. Disks of radius less than $c\mathrm{/}2$ have pairwise disjoint lattice translates and give the covering-action neighborhoods in [F1]. The quotient map is open since the preimage of the image of an open set is the union of its translates. The quotient is Hausdorff: for inequivalent $z,w$, only finitely many lattice points lie in any bounded disk by the bound just proved, so ${\inf }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }z-w-\lambda \mathrm{\mid }>0$; sufficiently small disks about $z,w$ project to disjoint neighborhoods. Images of a countable base of disks in $\mathbb{C}$ give a countable base of the quotient. It is nonempty and connected as a continuous image of $\mathbb{C}$. The projected closed parallelogram $T([0,1{]}^2)$ covers it by subtracting integer parts of the real coordinates, and is compact, making $X$ compact. Local inverses of the quotient map give charts with translation transitions. These are holomorphic and smooth, establishing the asserted compact Riemann surface and the descended flat compatible metric. Translation invariance also descends $dz,d\bar{z}$; no fundamental parallelogram is treated as a single global chart.

By [F2], a harmonic function is smooth and its lift is entire and lattice-periodic. It is bounded on the compact parallelogram and hence everywhere, so [F3] makes it constant. Conversely constants have zero Dolbeault derivative and are harmonic by [F2]. Every smooth degree-one form is $ud\bar{z}$ with a periodic smooth coefficient, since $d\bar{z}$ is a global frame. Its adjoint vanishes exactly when ${\mathrm{\partial }}_{z}u=0$, so $\bar{u}$ is an entire periodic function on the cover. The same boundedness and [F3] make $u$ constant; conversely constant coefficients have zero adjoint and are harmonic. Thus ${\mathcal{H}}^{0,0}(E)=\mathbb{C}1$ and ${\mathcal{H}}^{0,1}(E)=\mathbb{C}d\bar{z}$.

By [F4] each smooth Dolbeault class has a unique representative in the space computed in step 2.1, so the quotient is spanned by $[d\bar{z}]$. This class is nonzero: if $d\bar{z}=\bar{\mathrm{\partial }}f$ with smooth $f$, the adjoint identity gives $\mathrm{\parallel }d\bar{z}{\mathrm{\parallel }}^2=⟨f,{\bar{\mathrm{\partial }}}^{\ast }d\bar{z}⟩=0$, contradicting its everywhere nonzero pointwise norm and positive volume. Hence $H^{0,1}(X,E)=\mathbb{C}[d\bar{z}]$ has dimension one and $d\bar{z}$ is its harmonic representative. A holomorphic differential lifts to $a(z)dz$ with entire periodic coefficient $a$, so [F3] makes $a$ constant; conversely $dz$ descends and is holomorphic. Thus $H^0(X,K)=\mathbb{C}dz$, consistently with the conjugate-linear star formula in [F4]. Taking ${\omega }_1=1,{\omega }_2=i$ specializes every argument to the square torus. Full AC is inherited through [F2] and [F4]; the geometric construction and Liouville computation introduce no further choice.