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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.

For the intended complex unit circle, put $u=\alpha (1)$ and $v=\alpha (i)$, with $|u|=|v|=1$. The complete family of semicharacters of a complex lattice isTo obtain this, the identity at zero gives $\alpha (0)=1$; on each coordinate axis the alternating form vanishes, so $\alpha (m)=u^m$ and $\alpha (ni)=v^n$, also for negative integers. Combining the axes gives the factor $(-1{)}^{mn}$. Conversely, for $\lambda =m+ni$, $\mu =r+si$, the quotient of the proposed values is $(-1{)}^{ms+nr}$, which equals $e^{i\pi (nr-ms)}$. Hence every displayed formula is a semicharacter of a complex lattice and none are missing.

The original PDF actually puts $z\in \mathbb{Z}$ in its definition of $C_1$. Under that literal reading $C_1=\{1,-1\}$, so there are exactly four solutions, given by the same formula with $u,v\in \{1,-1\}$. Under either reading the specified choice is ${\alpha }_1(m+ni)=(-1{)}^{mn}$.

Pulling back the factors of automorphy in the Appell–Humbert theorem by $z\mapsto cz$, where $c=a+bi$, pulls back both the Hermitian form and the semicharacter of a complex lattice. Thus the explicit answer isWriting $\lambda =m+ni$ makes the second formula completely explicit:The second equality is an equality of signs, obtained by reducing the exponents modulo $2$. The Hermitian form follows from $H_1(cz,cw)=|c{|}^2{H}_1(z,w)$. Its integer parameter equals the norm of the Gaussian integer, agreeing with the degree of this nonzero isogeny of elliptic curves. The formulas also include $c=0$, when the pulled-back line bundle is trivial.