Flat complex three-torus O3/O7 parity data

The six vectors $e_j,i{e}_j$ form a real basis of ${\mathbb{C}}^3$, so $\mathrm{\Lambda }$ is a discrete full-rank lattice. Its translation action is free. For each compact $K\subset {\mathbb{C}}^3$, any translate $K+\lambda$ meeting $K$ has $\lambda \in K-K$, a bounded set containing only finitely many lattice points, so the action is properly discontinuous. The closed unit fundamental parallelepiped is compact and maps onto $X$. Small balls of radius below half the shortest nonzero lattice length give quotient charts with translation transition maps; these are holomorphic. Thus $X$ is a smooth compact complex threefold, the metric descends, $J$ is positive and closed, and $\mathrm{\Omega }$ descends nowhere zero.

Write $q=(x_1,y_1,x_2,y_2,x_3,y_3)$ for the unit-torus coordinates, let $\mathrm{\Delta }={\sum }_{j=1}^6{\mathrm{\partial }}_{q_j}^2$, and for $k\in {\mathbb{Z}}^6$ put $\widehat{f}(k)={\int }_{[0,1{]}^6}f(q)e^{-2\pi ik\cdot q}dq$. Periodic integration by parts gives $(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^N\widehat{f}(k)=\widehat{(1-\mathrm{\Delta }{)}^{N}f}(k)$ for every integer $N\ge 0$, hence $\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\le \mathrm{\parallel }(1-\mathrm{\Delta }{)}^{N}f{\mathrm{\parallel }}_{L^1}(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-N}$. Here $\mathrm{\mid }k{\mathrm{\mid }}^2={\sum }_{j=1}^6{k}_j^2$ and $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}={\max }_j\mathrm{\mid }{k}_j\mathrm{\mid }$. For $m\ge 1$, the shell $m\le \mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}<m+1$ has $(2m+1{)}^6-(2m-1{)}^6\le 728m^5$ points. Thus for each derivative order $r$, choosing $N$ with $2N-r>6$ makes the Fourier series ${\sum }_k\widehat{f}(k)e^{2\pi ik\cdot q}$ and its termwise derivatives of order at most $r$ absolutely and uniformly convergent. Uniform convergence of all these differentiated series, together with the fundamental theorem of calculus along coordinate segments, gives a smooth sum. To show it equals $f$, the kernels $F_N(t)=(N+1{)}^{-1}\mathrm{\mid }{\sum }_{j=0}^N{e}^{2\pi ijt}{\mathrm{\mid }}^2$ on $\mathbb{R}\mathrm{/}\mathbb{Z}$ are nonnegative and have integral one by expansion of the finite square. The geometric-sum formula gives $F_N(t)\le ((N+1){\sin }^2(\pi \delta ){)}^{-1}$ whenever the distance from $t$ to $\mathbb{Z}$ is at least $\delta \in (0,1\mathrm{/}2)$. Their sixfold products have mass outside the cube where all coordinate distances are less than $\delta$ at most the sum of the six one-dimensional tail masses, which tends to zero. Uniform continuity of $f$ therefore makes convolution with these product kernels converge uniformly to $f$. Expanding the finite squares shows each convolution is a trigonometric polynomial with coefficients given by finite Fejér weights times $\widehat{f}(k)$; consequently a continuous periodic function with all Fourier coefficients zero is zero. Termwise integration shows the uniformly convergent Fourier series has the same coefficients as $f$, so its sum is $f$. Applying the same argument to every derivative proves convergence in $C^{\mathrm{\infty }}$, componentwise for smooth $(p,q)$-forms.

Every smooth $(p,q)$-form on $X$ therefore has a Fourier expansion with constant $(p,q)$-form coefficients multiplying $e_k(x,y)=\exp (2\pi i{\sum }_j(m_j{x}_j+n_j{y}_j))$, for $k=(m,n)\in {\mathbb{Z}}^6$. On the $k$th mode, $\bar{\mathrm{\partial }}$ is wedge multiplication by ${\xi }_k^{0,1}=\pi {\sum }_j(i{m}_j-n_j)d{\bar{z}}_j$. If $k\mathrm{\ne }0$, write ${\lambda }_j=\pi (i{m}_j-n_j)$ and set $v_k={\sum }_j{\bar{\lambda }}_j({\sum }_{\mathrm{\ell }}\mathrm{\mid }{\lambda }_{\mathrm{\ell }}{\mathrm{\mid }}^2{)}^{-1}\mathrm{\partial }\mathrm{/}\mathrm{\partial }{\bar{z}}_j$; then ${\xi }_k^{0,1}(v_k)=1$ and $\mathrm{\parallel }{v}_k\mathrm{\parallel }=O(\mathrm{\parallel }k{\mathrm{\parallel }}^{-1})$. Contraction satisfies ${\iota }_{v_k}({\xi }_k^{0,1}\wedge \eta )+{\xi }_k^{0,1}\wedge {\iota }_{v_k}\eta =\eta$ for each constant coefficient form $\eta$; hence every $\bar{\mathrm{\partial }}$-closed nonzero Fourier mode is $\bar{\mathrm{\partial }}$-exact. The rapid coefficient decay in step 1.2 and the bound on $\mathrm{\parallel }{v}_k\mathrm{\parallel }$ show that the resulting primitive series and all its derivatives converge uniformly, so it is a smooth primitive.

Negation preserves the lattice and the flat metric, squares to the identity, and pulls back every $d{z}_j$ and $d{\bar{z}}_j$ to its negative. It therefore preserves $J$ and sends $\mathrm{\Omega }$ to $(-1{)}^3\mathrm{\Omega }=-\mathrm{\Omega }$.

A fixed point satisfies $2z_j\in \mathbb{Z}+i\mathbb{Z}$ for each $j$. In one factor $E$, the four solutions modulo the lattice are $0$, $\frac{1}{2}$, $\frac{i}{2}$, and $\frac{1+i}{2}$. The three coordinates are independent, so $\mathrm{\mid }\mathrm{Fix}(\sigma )\mathrm{\mid }=4^3=64$. Each is isolated because $D\sigma =-I$ has no $+1$ eigenvector; by [F1] this agrees with the complex-dimension-zero case of the fixed-locus conclusion.

The zero Fourier modes are constant forms and are $\bar{\mathrm{\partial }}$-closed. The average of a $\bar{\mathrm{\partial }}$ of a periodic form is zero, so no nonzero constant form is exact. Thus constant forms give all of $H^{p,q}(X)$. A basis consists of wedges of $p$ distinct $d{z}_j$ and $q$ distinct $d{\bar{z}}_j$, giving dimension $\left(\genfrac{}{}{0px}{}{3}{p}\right)\left(\genfrac{}{}{0px}{}{3}{q}\right)$. Negation multiplies each basis form by $(-1{)}^{p+q}$, so the stated parity dimensions follow; in particular the four listed dimensions are $9,0,0,9$.