Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 18 / 2

Use the convention that a Hermitian form is complex linear in its first argument. In the integral Hermitian forms on the square complex torus calculation, every Hermitian form on $\mathbb{C}$ has the form $H_t(z,w)=tz\overline{w}$, where $t=H_t(1,1)$ is real. For $\lambda =m+ni$ and $\mu =r+si$ in the period lattice,Consequently integrality holds precisely when $t\in \mathbb{Z}$: necessity follows from $E_t(1,i)=-t$, and sufficiency follows from the displayed formula. Addition of Hermitian forms corresponds to addition of $t$. Positivity is equivalent to $t>0$, since $H_t(z,z)=t|z{|}^2$. The answer isWith the conjugate-linear-first convention the formula is $t\overline{z}w$ and the sign of $E_t$ reverses. The positive integer parameter is unchanged.

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 ${\alpha }_1(m+ni)=(-1{)}^{mn}$.

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.

A holomorphic group endomorphism of the complex torus lifts to multiplication by a scalar $c\in \mathbb{C}$, as in Question 1. It descends exactly when $c\Lambda \subseteq \Lambda$. Applying this condition to $1$ gives $c\in \mathbb{Z}[i]$. Conversely any Gaussian integer $c$ preserves $\mathbb{Z}[i]$, since the latter is a ring. Addition of endomorphisms is addition of scalars, and composition is multiplication of scalars. Thus, as rings,The wording “endomorphism algebra” here denotes this integral endomorphism ring of an elliptic curve. If it instead meant the rational algebra ${\mathrm{End}}^0(X)=\mathrm{End}(X){\otimes }_{\mathbb{Z}}\mathbb{Q}$, that algebra would be $\mathbb{Q}(i)$.

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.