Example 12.6.18.704C3B45EDAF On $G=\mathbb C^3$ take the multiplication
\[ z\cdot w=(z_1+w_1,z_2+w_2,z_3+w_3+z_1w_2), \]
the Gaussian lattice $\Gamma =\mathbb Z[\mathrm i]^3$, and $X=\Gamma \backslash G$. This is a compact non-Kähler manifold of dimension three, with invariant coframe
\[ \alpha _1=\mathrm dz_1,\qquad \alpha _2=\mathrm dz_2,\qquad \alpha _3=\mathrm dz_3-z_1\mathrm dz_2. \]
For its full nine-dimensional cone $W=\operatorname {Herm}(3)$ and the normalized invariant probability,
\[ H=\frac6{\pi ^3},\qquad \mathcal V_\mu (\theta _ A)=\frac6{\pi ^3}\det A,\qquad S=\{ A>0:\det A=\pi ^3/6\} . \]
Every positive background is nonclosed. The eight-dimensional unit-volume hypersurface is complete and proper, and its intrinsic distance is
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12.6.19
\begin{equation} d_ S(A,B)=\frac1{\sqrt3}\| \log (A^{-1/2}BA^{-1/2})\| _{\mathrm F}. \end{equation}
Proof.
The bilinear cross term makes the multiplication associative, with identity zero and inverse $(-z_1,-z_2,-z_3+z_1z_2)$. These operations preserve the Gaussian lattice. Its action by left multiplication is free and properly discontinuous: for a compact set $K\subset G$, elements $\gamma $ with $\gamma K\cap K\ne \varnothing $ belong to the compact set $KK^{-1}$, containing only finitely many lattice elements.
Choose Gaussian integers $\gamma _1,\gamma _2$ that move $z_1,z_2$ into the half-open unit square, and then $\gamma _3$ that moves $z_3+\gamma _1z_2$ into it. This gives a representative in the real six-dimensional unit cube. If two points of the half-open cube are equivalent, then $\gamma _1=\gamma _2=0$ and then $\gamma _3=0$. The cube is therefore a measurable fundamental domain, and $X$ is compact, being the image of its closure.
This coframe is left-invariant, descends to $X$, and has
\[ \mathrm d\alpha _3=-\alpha _1\wedge \alpha _2,\qquad \alpha _1\wedge \alpha _2\wedge \alpha _3 =\mathrm dz_1\wedge \mathrm dz_2\wedge \mathrm dz_3. \]
The density of $h_0^3$ is $6/\pi ^3$ times real Lebesgue volume on the fundamental cube, giving the stated $H$. The coefficient of $\alpha _1\wedge \alpha _2\wedge \bar\alpha _3$ in $\partial \theta _ A$ is $-\mathrm iA_{33}/(2\pi )$, so positivity makes every such background nonclosed.
To see that $X$ admits no Kähler metric at all, suppose that a positive closed $(1,1)$-form $\kappa $ existed, and average its coefficient functions in this coframe against $\mu $. Their mean is a positive matrix $A$. In the coefficients of $\partial \kappa $, derivative terms have mean zero by (12.6.8); the remaining means are exactly the coefficients of $\partial \theta _ A$. Closedness would force them to vanish, contradicting $A_{33}>0$. Thus $X$ is non-Kähler.
The assertions about the volume, the minimizers and completeness follow from proposition 12.6.5, theorem 12.6.12. As $W$ is the full matrix space, the equal-determinant exponential path stays in $S$, and lemma 12.6.9 gives (12.6.19). For $A_\star =aI$, $a=(\pi ^3/6)^{1/3}$, this becomes
\[ d_ S(A_\star ,A)^2=\frac13\sum _{j=1}^3 \left(\log \frac{\lambda _ j(A)}a\right)^2. \]
The three logarithms sum to zero.
For a concrete degeneration, take $A_\varepsilon =\operatorname {diag}(1,1,\varepsilon )$, $0<\varepsilon \leq 1$. As $\varepsilon \to 0$, the coefficients converge to the nonzero smooth semipositive form $\theta _{\operatorname {diag}(1,1,0)}$. The normalized matrices and their distances, on the other hand, are
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12.6.20
\begin{equation} \begin{gathered} \widehat A_\varepsilon =a\operatorname {diag}(\varepsilon ^{-1/3},\varepsilon ^{-1/3},\varepsilon ^{2/3}),\\ d_ S(A_\star ,\widehat A_\varepsilon )=\frac{\sqrt2}{3}|\log \varepsilon |. \end{gathered} \end{equation}
Thus, after normalization, the boundary form $\theta _{\operatorname {diag}(1,1,0)}$ lies at infinite distance.
$\square$
Pome
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