Flat torus dolbeault harmonic representatives

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.