Suppose we take an almost complex structure on $\mathbb{T}^{6}$ with $c\_{1} \neq 0$ (there should be infinitely many homotopy classes satisfying this requirement). Now pull it back to the universal cover $\mathbb{R}^6$, giving an almost complex structure $J$, it should be invariant by the deck group action of $\mathbb{Z}^{6}$.


My question is there a case where we can explicitly describe $J$ and the group action of $\mathbb{Z}^6$ in coordinates?


(I asked this question in stack exchange in August <https://math.stackexchange.com/questions/3789163/almost-complex-actions-on-mathbbr6> with no answer so I thought I would ask here.)



almost complex $\mathbb{Z}^{6}$-action
Here is a construction of a family of such examples that will work, but you will have to choose a particular map to get an explicit example.


Let's use coordinates $v\_1,v\_2,v\_3,v\_4,x,y$ (each periodic of period $2\pi$). Choose a smooth map $u=u(x,y):\mathbb{T}^2\to S^2$ that has nonzero degree. (This will ensure that it is not null-homotopic.) One possible example would be to make it a $2$-to-$1$ branched conformal double cover (i.e., it will have 4 branch points), the classical way to exhibit an elliptic curve as a branched double cover of $S^2=\mathbb{CP}^1$.


Write $u = (u\_1,u\_2,u\_3)$ and consider the following $2$-form on $\mathbb{T}^6$
$$
\begin{aligned}
\omega &= u\_1\,(\mathrm{d}v\_1\wedge\mathrm{d}v\_2+\mathrm{d}v\_3\wedge\mathrm{d}v\_4)
 +u\_2\,(\mathrm{d}v\_2\wedge\mathrm{d}v\_3+\mathrm{d}v\_1\wedge\mathrm{d}v\_4)\\
&\qquad +u\_3\,(\mathrm{d}v\_3\wedge\mathrm{d}v\_1+\mathrm{d}v\_2\wedge\mathrm{d}v\_4)
 + \mathrm{d}x\wedge\mathrm{d}y
\end{aligned}
$$
This $2$-form is compatible with the flat metric $g = {\mathrm{d}v\_1}^2 + {\mathrm{d}v\_2}^2 + {\mathrm{d}v\_3}^2 + {\mathrm{d}v\_4}^2 + {\mathrm{d}x}^2 + {\mathrm{d}y}^2$ and hence defines a $g$-orthogonal almost complex structure on $\mathbb{T}^6$ by the usual rule $\omega(w\_1,w\_2) = g(Jw\_1,w\_2)$ for all tangent vectors $w\_1,w\_2$.


I think this $J$ is probably not integrable for any nonconstant $u$ (and hence for $u:\mathbb{T}^2\to S^2$ of nonzero degree), but it will certainly be nonintegrable for `generic' $u$. The nonzero degree requirement on $u$ implies that $c\_1\not=0$ for this almost complex structure.


