Is every compact solvmanifold the quotient of a simply connected solvable Lie group by a discrete subgroup

differential-geometrylie-groupssmooth-manifoldssolvable-groups

Recall that a manifold is called a nilmanifold if it is a homogeneous space for a connected nilpotent Lie group. Mal'cev showed that every compact nilmanifold is diffeomorphic to the quotient of a simply connected nilpotent Lie group by a discrete subgroup acting cocompactly.

A manifold is called a solvmanifold if it is a homogeneous space for a connected solvable Lie group. Every nilpotent group is solvable, so every nilmanifold is a solvmanifold. Is there an anologue of Mal'cev's result for solvmanifolds? That is,

Is every compact solvmanifold diffeomorphic to the quotient of a simply connected solvable Lie group by a discrete subgroup acting cocompactly?

Best Answer

As it turns out, Malcev's theorem is false for solv-manifolds. An example is the Klein bottle $K^2$:

  1. The fact that $K^2$ is not diffeomorphic to the quotient of any solvable group by a discrete subgroup follows from the classification of 2-dimensional simply-connected (necessarily solvable) Lie groups: These are the abelian group ${\mathbb R}^2$ and the group $Aff_+({\mathbb R})$ of (orientation-preserving) affine automorphisms of the real line (this was discussed many times, for instance, here). The first group contains no subgroups isomorphic to $\pi_1(K^2)$ (since the latter is noncommutaive). The second group contains no noncyclic discrete subgroups.

  2. One can represent $K^2$ as the quotient of the solvable group $SE(2)$, the group of orientation-preserving isometries of the Euclidean plane $R^2$. Consider the closed subgroup $H< SE(2)$ consisting of motions which preserve the (vertical) lines of the form $x=n$, $n\in {\mathbb Z}$. (I am using the standard Cartesian coordinates.) The elements of $H$ can translate these lines horizontally (by an integer), vertically and also rotate by 180 degrees. It is clear that $SE(2)/H$ is a compact surface $S$ fibered over the circle (the fibration comes from the homomorphism $SE(2)\to SO(2)$). In order to prove that $S$ is the Klein bottle we just need to prove that it is nonorientable. Let $V< H$ denote the (normal) subgroup consisting of vertical translations. The quotient $SE(2)/V$ is homeomorphic to the cylinder $A$. The order two rotation $\tau\in H$ fixing the origin will reverse the orientation on $A$ (since it reverses the orientation on the horizontal lines in the plane). Thus, $S$ is nonorientable and, hence, is homeomorphic to the Klein bottle.

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