 Solution

The Appell–Humbert theorem identifies the positive Hermitian form $H_1$ with a holomorphic line bundle whose First Chern class is represented, in our linear-first convention, byIts integral over the square fundamental domain is $1$, so $\mathrm{deg}L=1$. In particular the oriented Chern pairing is $-E_1(1,i)=1$; using $E_1(1,i)$ without adjusting the Hermitian convention would give the wrong sign.

The complex torus has a nowhere-vanishing differential $dz$, so its canonical divisor is trivial and its genus is one. A nonzero section of a line bundle of negative degree would have an effective divisor of negative degree, which is impossible. Thus $H^0(X,L^{-1})=0$. The Riemann-Roch theorem and Serre duality giveTherefore $\boxed{h^0(X,L)=1}$. The choice of semicharacter of a complex lattice changes the degree-zero component of the line bundle, but not this calculation.