Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 17 / 3

Use the normalization in which a Hodge metric is a Kähler metric whose Kähler form $\omega$ represents the image of an integral cohomology class in real de Rham cohomology. Equivalently, all its periods on integral two-cycles are integers. A convention using $\omega /(2\pi )$ only rescales the metric and does not change existence. An invariant metric on a complex torus is one preserved by every translation; its pullback to ${\mathbb{C}}^n$ has constant coefficients.

Let $T={\mathbb{C}}^n/\Lambda$ carry a Hodge metric. Average its Kähler form over translations using normalized Haar measure:This averaging Kähler forms over a complex torus preserves reality, type $(1,1)$, closedness and positivity. Indeed, for any nonzero tangent vector, the quantity being averaged in $\omega (v,Jv)$ is strictly positive. Every translation is homotopic to the identity, since a path from $0$ to $a$ supplies such a homotopy. Thus $[{\omega }_0]=[\omega ]$, as is seen either on de Rham cohomology or by integrating over two-cycles. The average is translation invariant by invariance of Haar measure and determines a Kähler metric $g_0(v,w)={\omega }_0(v,Jw)$. Its class is still integral. The reverse implication is immediate. Therefore a Hodge metric exists exactly when an invariant one does.

Here is the Riemann bilinear criterion for a period matrix, with the signs kept explicit. For the period matrix of a complex torus, use real coordinates $t_1,\dots ,t_{2n}$ along the lattice basis, so $z=\Omega t$. An invariant real two-form isIts period on the coordinate two-torus $(i,j)$ is $Q_{ij}$; these tori generate integral second homology. Thus integrality of the class is exactly $Q\in M_{2n}(\mathbb{Z})$. Positivity of a Kähler form makes $Q$ nonsingular. Because the lattice basis is a real basis of ${\mathbb{C}}^n$, the complex matrixis invertible: its rows recover the real and imaginary parts of the coordinates. In the coordinates $(z,\bar{z})$, the matrix of the two-form is $T_0^{-t}Q{T}_0^{-1}$, whose inverse is $T_0{Q}^{-1}{T}_0^t$.

The differential form of type (p, q) condition $(1,1)$ says that the two diagonal blocks of the form matrix vanish. Since the form is nonsingular, this is equivalent to the diagonal blocks of its inverse vanishing. Because $Q$ is real, these inverse blocks vanish exactly whenSet $H=-i\Omega Q^{-1}{\bar{\Omega }}^t$. Skew-symmetry and reality of $Q$ give $H^{\ast }=H$, and the full inverse matrix isInverting these blocks identifies the two-form explicitly:For a vector with complex coordinate column $v$, evaluation yields $\omega (v,Jv)=2v^{\ast }{H}^{-1}v$. Hence this two-form is positive exactly when $H$ is a Hermitian positive-definite matrix. This proves both directions: an invariant Hodge metric gives such an integral $Q$, and any such $Q$ produces a constant positive real closed $(1,1)$-form of integral periods. In particular,As a sign check, for $\Omega =(I,iI)$ and $Q=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$ the last matrix is $2I$, and the associated form is $(i/2)\sum d{z}_a\wedge d{\bar{z}}_a$.

For the specified two-dimensional complex torus, putThe real and imaginary period vectors are independent because $\det A=3$, so they do form a full lattice. Suppose a polarization matrix $Q$ existed. Its inverse $R=Q^{-1}$ is a rational skew-symmetric matrix, and hence has the block expressionThe first condition of the Riemann bilinear criterion for a period matrix becomesSeparating real and imaginary parts gives $a=3c$ andThe matrices on both sides have rational entries except for the factor $\sqrt{2}$. Irrationality of $\sqrt{2}$ therefore implies $B=B^t$ and $BJ+JB=0$. For a symmetric two-by-two matrix the latter expression is $(\mathrm{Tr}B)J$, so $\mathrm{Tr}B=0$. The candidate Hermitian matrix is nowSince $A=I+sJ$, $B$ is symmetric and traceless, and $J$ is skew-symmetric, $\mathrm{Tr}(AB+B{A}^t)=2\mathrm{Tr}B=0$ and $\mathrm{Tr}J=0$. Thus $\mathrm{Tr}H=0$. A Hermitian positive-definite matrix has strictly positive eigenvalues, hence strictly positive trace. This contradiction proves