6

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?

Michael Albanese
  • 104,029

1 Answers1

5

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.

Moishe Kohan
  • 111,854
  • Why is the last sentence of 1 true? Also, is it clear that $SE(2)$ is solvable? – Michael Albanese Mar 03 '20 at 23:35
  • It is a standard fact in hyperbolic geometry that if $f, g$ are elements of $PSL(2,R)$ which have exactly one fixed point in common and $f$ is hyperbolic then the group generated by $f, g$ is non-discrete. See for instance, Theorem 5.1.2 of Beardon's book "Geometry of discrete groups". (Also, use the fact that $Aff_+(R)$ is naturally embedded in $PSL(2,R)$.) Now, it is an exercise to see that if $f, g$ are in ${\mathbb R}$ and generate a discrete subgroup, then this subgroup is cyclic. – Moishe Kohan Mar 04 '20 at 00:56
  • Lastly, the fact that $G=SE(2)$ is solvable is clear: The commutator of any two elements is a translation and all translations commute. Hence, $[[G,G], [G,G]]=1$, i.e. $G$ is 2-step solvable. – Moishe Kohan Mar 04 '20 at 00:57