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I am new to the study of lie groups, nilmanifolds etc. but the following question can be described very basically I think.

Take a complex Heisenberg group $G$

$$ \left( \begin{array}{ccc} 1 & \mathbb{C} & \mathbb{C} \\ 0 & 1 & \mathbb{C} \\ 0 & 0 & 1 \end{array} \right)$$

and the action on $G$ by matrix multiplication of the discrete subgroup $\Gamma$ of matrices like above with entries in $\mathbb{Z}[i]$ (the gaussian integers). Its convenient to denote the matrix

$$ \left( \begin{array}{ccc} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{array} \right)$$

by $(x,y,z)$ and multiplication by $(a,b,c)\cdot(x,y,z) = (a+x,b+y+az, c+z)$.

Now my question is to understand the geometry of the quotient space $G/\Gamma$. The equivalence classes or orbits of this action are elements of the form $(z,\omega +(n+im)\gamma, \gamma)$, where $z,\omega, \gamma \in \mathbb{C}$ and $n,m\in \mathbb{N}$ and the real and imaginary parts of the three coordinates are between 0 and 1.

Question: Describe somehow the fundamental domain of this action (if that is even possible geometrically, anything otherwise is also helpful).

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The quotient is the Iwasawa manifold. It is a compact complex manifold of dimension 3. It is an important example of a non-Kahler manifold. Its cohomology has been completely calculated (at least by M. Schweitzer, see his unique paper on the Arxiv), and if you start googling you'll find references on more detailed studies of this manifold. I don't know the fundamental domain of the action of $G$, but it should look like some sort of a fibration over the unit disc in the $z$ coordinate (b/c of the $b + y + az$ factor). –  Gunnar Magnusson Jul 22 '11 at 5:42
    
Interestingly (depending on your interests) the tangent bundle of $X = G / \Gamma$ is trivial, which one sees by calculating explicit holomorphic 1-forms invariant under $\Gamma$. It is thus an example of a compact 3-fold with trivial canonical bundle which is neither a torus nor a Calabi-Yau manifold, so it gives a counterexample to the naive question "Does $K_X$ trivial and $X$ compact imply $X$ Kahler?", which is true in lower dimensions. –  Gunnar Magnusson Jul 22 '11 at 5:46
    
I you want, I have a small note explaining a little on this manifold (www-fourier.ujf-grenoble.fr/~gunnarm/files/…). It wasn't meant for publication so it doesn't have any references and is a little incomplete, but the ideas are mostly taken from Demailly's book. –  Gunnar Magnusson Jul 22 '11 at 5:55
    
Can you do the real case? –  Mariano Suárez-Alvarez Jul 22 '11 at 5:57
    
@gunnar Thanks. –  AnonymousCoward Jul 22 '11 at 6:01

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