= Solution
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>:
$$
\omega_0=\int_T t_a^*\omega\,da.
$$
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 \b[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 $t_1,\ldots,t_{2n}$ along the lattice basis, so $z=\Omega t$. An invariant real two-form is
$$
\omega=\frac12\sum_{i,j}Q_{ij}\,dt_i\wedge dt_j,\qquad Q^t=-Q.
$$
Its 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 matrix
$$
T_0=\begin{pmatrix}\Omega\\\bar\Omega\end{pmatrix}
$$
is 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 $T_0^{-t}QT_0^{-1}$, whose inverse is $T_0Q^{-1}T_0^t$.
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 when
$$
\Omega Q^{-1}\Omega^t=0.
$$
Set $H=-i\Omega Q^{-1}\bar\Omega^t$. Skew-symmetry and reality of $Q$ give $H^*=H$, and the full inverse matrix is
$$
T_0Q^{-1}T_0^t=
\begin{pmatrix}0&iH\\-iH^t&0\end{pmatrix}.
$$
Inverting these blocks identifies the two-form explicitly:
$$
\omega=i\sum_{a,b}(H^{-1})_{ba}\,dz_a\wedge d\bar z_b.
$$
For a vector with complex coordinate column $v$, evaluation yields $\omega(v,Jv)=2v^*H^{-1}v$. 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,
$$
\boxed{T\text{ admits a Hodge metric}\iff
\exists Q\in M_{2n}(\mathbb Z),\
Q^t=-Q,\ \det Q\ne0,\
\Omega Q^{-1}\Omega^t=0,\
-i\Omega Q^{-1}\bar\Omega^t>0.}
$$
As a sign check, for $\Omega=(I,iI)$ and $Q=\begin{pmatrix}0&I\\-I&0\end{pmatrix}$ the last matrix is $2I$, and the associated form is $(i/2)\sum dz_a\wedge d\bar z_a$.
For the specified two-dimensional <complex torus>, put
$$
J=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad
s=\sqrt2,\qquad A=I+sJ.
$$
The 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 expression
$$
R=\begin{pmatrix}aJ&B\\-B^t&cJ\end{pmatrix},
\qquad a,c\in\mathbb Q,\quad B\in M_2(\mathbb Q).
$$
The first condition of the <Riemann bilinear criterion for a period matrix> becomes
$$
0=\Omega R\Omega^t=(a-3c)J+i(BA^t-AB^t).
$$
Separating real and imaginary parts gives $a=3c$ and
$$
B-B^t=s(BJ+JB^t).
$$
The matrices on both sides have rational entries except for the factor $\sqrt2$. Irrationality of $\sqrt2$ therefore implies $B=B^t$ and $BJ+JB=0$. For a symmetric two-by-two matrix the latter expression is $(\operatorname{Tr}B)J$, so $\operatorname{Tr}B=0$. The candidate <Hermitian matrix> is now
$$
H=-i\Omega R\bar\Omega^t
=AB+BA^t-6icJ.
$$
Since $A=I+sJ$, $B$ is symmetric and traceless, and $J$ is skew-symmetric, $\operatorname{Tr}(AB+BA^t)=2\operatorname{Tr}B=0$ and $\operatorname{Tr}J=0$. Thus $\operatorname{Tr}H=0$. A <Hermitian positive-definite matrix> has strictly positive <eigenvalues>, hence strictly positive <trace>. This contradiction proves
$$
\boxed{\mathbb C^2/\Lambda\text{ admits no Hodge metric}.}
$$
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