Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 126 / 1

Let $X=V/\Gamma$ be a complex torus. A Riemann form on a complex torus is a Hermitian form $H$ on $V$ whose imaginary parttakes integer values on $\Gamma \times \Gamma$. Equivalently, $E$ is an integral alternating form satisfyingfor every nonzero $v$. A polarisation of a complex torus is such a positive Riemann form, or equivalently its integral cohomology class. By the Appell–Humbert theorem, it is the first Chern class of an ample holomorphic line bundle. A polarisation is principal when the homomorphism $\Gamma \to {\Gamma }^{\ast }$ induced by $E$ is an isomorphism.

The rank assumption says that ${\Gamma }^{\prime }=\Gamma \cap V^{\prime }$ is a full lattice in the real vector space underlying $V^{\prime }$. Thus $X^{\prime }=V^{\prime }/{\Gamma }^{\prime }$ is compact, and the inclusion $V^{\prime }\hookrightarrow V$ descends to an injective holomorphic homomorphism $X^{\prime }\hookrightarrow X$. It is therefore a complex subtorus.

Conversely, let $j:Y\hookrightarrow X$ be a subtorus. Its differential at the identity identifies the universal cover of $Y$ with a complex subspace $V^{\prime }\subseteq V$. Lifting $j$ to universal covers shows that the period lattice of $Y$ is precisely $V^{\prime }\cap \Gamma$. Compactness of $Y$ makes this a full lattice of rank $2{\dim }_{\mathbb{C}}{V}^{\prime }$, so every subtorus has the stated form.

Now let $H$ be a polarisation on $X$. Its restriction to $V^{\prime }$ is still positive definite, and its imaginary part remains integral on ${\Gamma }^{\prime }$, so it polarises $X^{\prime }$. Define the Hermitian orthogonal complementBecause $V^{\prime }$ is spanned over $\mathbb{R}$ by lattice vectors and $E=\mathrm{Im}H$ is integral on $\Gamma$, the real equations $E(v,{\gamma }^{\prime })=0$ for ${\gamma }^{\prime }\in {\Gamma }^{\prime }$ make $V^{\prime \prime }$ rational with respect to $\Gamma$. Hence ${\Gamma }^{\prime \prime }=V^{\prime \prime }\cap \Gamma$ is a full lattice in $V^{\prime \prime }$, and $X^{\prime \prime }=V^{\prime \prime }/{\Gamma }^{\prime \prime }$ is a subtorus. Since $V=V^{\prime }\oplus V^{\prime \prime }$, the lattice ${\Gamma }^{\prime }+{\Gamma }^{\prime \prime }$ has finite index in $\Gamma$. Consequently the addition mapis an isogeny of complex tori: it is surjective and has finite kernel. Therefore $X=X^{\prime }+X^{\prime \prime }$ and $X^{\prime }\cap X^{\prime \prime }$ is finite.

Write a lattice vector asA one-dimensional subtorus would give two $\mathbb{Z}$-independent lattice vectors $u,u^{\prime }$ spanning one complex line, so their determinant would vanish. Separating the real and imaginary parts and using the $\mathbb{Q}$-linear independence of $1,\sqrt{2},\sqrt{3}$ givesIf either $b$ or $b^{\prime }$ is nonzero, these relations make $(a,b,c,d)$ and $(a^{\prime },b^{\prime },c^{\prime },d^{\prime })$ rationally proportional, contradicting their lattice independence. Thus $b=b^{\prime }=0$. If either $d$ or $d^{\prime }$ is nonzero, the last two relevant relations again make all four coordinates proportional. Hence $d=d^{\prime }=0$, and both vectors lie in $\mathbb{C}(1,0)$. Conversely, $(1,0)$ and $(i,0)$ lie in $\Gamma$, sois a one-dimensional subtorus. It is the unique one.

If $X$ admitted a polarisation, part (ii) would give a one-dimensional complementary subtorus $X^{\prime \prime }$ with $X=X^{\prime }+X^{\prime \prime }$. Uniqueness would force $X^{\prime \prime }=X^{\prime }$, making $X^{\prime }+X^{\prime \prime }=X^{\prime }\ne X$, a contradiction. Thus $X$ is a nonprojective complex torus and has no polarisation.