V. V. Gorbatsevich, “On lattices in Lie groups of general type and some applications”, Izv. Math., 89:4 (2025), 681

We consider lattices in Lie groups. The results obtained will be applied on lattices to the study of fundamental groups of compact homogeneous spaces and their connections with the topological structure of such spaces.

Usually, lattices are considered either in semisimple Lie groups or in solvable ones (or even only in nilpotent ones). Here, we are interested in lattices in Lie groups of general type — these being Lie groups $G$ for which both the radical $R$ and the semisimple part $S$ included in the Levi decomposition $G=S \cdot R$ of the Lie group $G$ are non-trivial.

The concepts and properties of Lie groups and algebras used in this article can be found in [1], for lattices in Lie groups see [2], and for homogeneous spaces, [3]. Lie groups will be denoted by capital Latin letters, and their Lie algebras, by the corresponding lowercase Latin letters. By $P_0$ we will denote the connected component of the identity of the Lie group $P$ (usually, such Lie groups will be Lie subgroups in connected Lie groups). For a group $\Pi$, by $Z(\Pi)$ we denote its center, and by $N_\Pi(D)$ we denote the normalizer of the subgroup $D$ in $\Pi$. By $A \ltimes_\phi B$ we denote the semidirect product of groups $A$ and $B$ defined by the homomorphism $\phi \colon A\to \mathrm{Aut}(B)$ into the automorphism group $\mathrm{Aut}(B)$ of $B$. This semidirect product is a product of subgroups isomorphic to $A$ and $B$, where $B$ is a normal subgroup and the intersection $A\cap B$ is trivial. If some group $\Pi$ is represented as a product $\Pi = A\cdot B$ of a subgroup $A$ and a normal subgroup $B$, where the intersection $A \cap B$ is finite, then we say that the group $\Pi$ is decomposed into an almost semidirect product of groups $A$ and $B$. Next, two groups $\Pi$, $\Pi'$ are called weakly commensurable if they contain subgroups $\Pi_1$, $\Pi_1'$ of finite indices and if $\Pi_1$, $\Pi_1'$ contain finite normal subgroups $\Phi$, $\Phi'$, respectively, such that the group $\Pi_1/\Phi$ is isomorphic to $\Pi_1' /\Phi'$ (for more details, see [1]). By $M' \to M \to M''$ we denote a smooth locally trivial fibering of a smooth manifold $M$ over base $M''$ with fiber $M'$.

We first give some definitions and formulate the main results of the present paper.

Let $G$ be a Lie group (which we will usually assume to be connected). A discrete subgroup $\Gamma \subset G$ is called a lattice if the volume of the quotient space $G/\Gamma$ (with respect to a $G$-invariant measure) is finite. A subgroup $H \subset G$ is called a uniform subgroup if it is closed in $G$ and the quotient space $G/H$ is compact. A discrete uniform subgroup is called a uniform lattice. In this paper, we will mainly consider uniform lattices, although some results are true for general lattices as well.

A Lie group $G$ will be called almost simply connected if its fundamental group is finite. Another way to express this is that a Lie group is covered with finite sheets by a simply connected Lie group (its universal covering).

Let $\Gamma$ be a lattice in a connected Lie group $G$, and let $F$ be a closed subgroup (and then automatically a Lie subgroup) in $G$. The subgroup $\Gamma \cap F$ will be a discrete subgroup of $F$, but will not be a lattice there, in general. If $\Gamma$ is a uniform lattice in $G$, then the subgroup $\Gamma \cap F$ will be a lattice in $F$ if and only if the subgroup $\Gamma \cap F$ is uniform in $F$. The condition of uniformity of the intersection of $\Gamma \cap F$ in $F$ is equivalent to the condition of closedness of the set $\Gamma \cdot F$ (the set of products $\gamma \cdot f$, where $\gamma \in \Gamma$, $f \in F$) and, if the Lie subgroup $F$ is normal in $G$ (and then $\Gamma\cdot F$ is a Lie subgroup), then also to the condition of closedness of all orbits of the natural action of the Lie group $F$ on the manifold $G/\Gamma$.

Many particular conditions are known (on $\Gamma, G, F$) under which the subgroup $\Gamma \cap F$ is a lattice in $F$ (see, for example, [2], [4]–[11]).

This paper considers several results on the intersection of lattices with various Lie subgroups in Lie groups of general type. Based on these results, some properties of the fundamental groups of compact homogeneous spaces (for which the stationary subgroup is no longer necessarily discrete) and on the structure of such spaces are proved.

In § 1, we give some facts about intersections of lattices in Lie groups with radicals (a solvable radical and a nilradical) in these Lie groups.

Theorem 1. Let $G$ be a connected Lie group, $G=S\cdot R$ be its Levi decomposition ($R$ is the radical of $G$, and $S$ is a Levi subgroup, that is, a semisimple part), and suppose that $S$ has no compact factors. Further, let $\Gamma$ be a lattice in $G$. Assume that one of the following three conditions holds:

(i) the subgroup $S \cap R$ is finite;

(ii) the Lie group $G$ is almost simply connected (that is, its fundamental group is finite);

(iii) the Lie group $G$ is almost linear (that is, has a finite-dimensional linear representation with finite kernel).

Then $\Gamma$ has a subgroup of finite index smoothly deformable (inside $G$) into a lattice $D \subset G$ such that $D= (D\cap S) \cdot (D \cap R )$ is a semidirect product in case (i) and an almost semidirect product of $D\cap S$ and a normal subgroup $D\cap R$ (that is, in the latter case the intersection of $D \cap S$ with $D \cap R$ is finite).

Theorem 1, which will be proved in § 3, generalizes to some extent Mostow’s results from [7] and [9], which we formulate as Theorem 6 below.

We will also prove a generalization of one result from [8] for soluble Lie groups to the case of arbitrary Lie groups.

Next, we will consider the Levi–Mostow decomposition for lattices in Lie groups (in § 4) and for fundamental groups of compact homogeneous spaces (in § 5).

This theorem (with some generalization for assertion (ii)) will be proved in § 7.

We will also give some examples related not only to lattices, but also to compact homogeneous spaces of general type.

§ 2. On the intersection of lattices with radicals

The first result on the intersection of a lattice in a Lie group with some Lie subgroup in this Lie group was obtained by Mal’tsev [4], who he actually proved that in a nilpotent Lie group the intersection of a lattice with the commutator subgroup of this Lie group is a lattice. Later, for the case of solvable Lie groups, a similar result on the intersection of a lattice with a nilradical (in fact, not only for lattices, but for arbitrary uniform subgroups) was proved by Mostow [5]. However, these results were obtained only for particular (nilpotent and solvable) classes of Lie groups. Some subsequent results of this kind were obtained by Auslender [8]. Here, we are interested in the results for the case of Lie groups of general form (that is, with a non-trivial semisimple part).

Let $G$ be some connected Lie group, $G=S\cdot R$ be its Levi decomposition ($R$ be the solvable radical, and $S$ be the semisimple part, the Levi factor in $G$). We will also need the nilradical $N$ in $G$. For an arbitrary lattice $\Gamma$, the natural question is whether the intersections of this lattice with the radicals $R$, $N$ are lattices in them. The answer to this question has a rather long and almost detective story; for details, see [11].

A result of this kind was published in [10], but then a counterexample was given [12]. Then followed partial and incomplete versions of the statement in [10] about such intersections. In a sense, the end of this topic was the paper [11], where the following result was proved.

§ 3. On the intersection of lattices with a Levi subgroup

We first formulate some known results on intersections of lattices with a Levi subgroup.

Let $G=S\cdot R$ be the Levi decomposition of a connected Lie group $G$ ($R$ is the radical of $G$, $S$ is a Levi subgroup). Consider the almost direct decomposition $S= S_{\mathrm n}\cdot S_{\mathrm c}$, where $S_{\mathrm c}$ is the maximal connected compact normal subgroup of $S$, and $S_{\mathrm n}$ is the connected normal subgroup of $S$ complementary to $S_{\mathrm c}$ ($S_{\mathrm n}$ has no non-trivial connected compact normal subgroups), and the subgroup $S_{\mathrm n} \cap S_{\mathrm c}$ is finite.

Theorem 3 is equivalent to the discreteness and uniformity of the subgroup $\Gamma/\Gamma \cap (S_{\mathrm c} \cdot R)$ in the semisimple Lie group $S_{\mathrm n}' = S_{\mathrm n}/S_{\mathrm n} \cap S_{\mathrm c}$, locally isomorphic to $S_{\mathrm n}$. If $S_{\mathrm c}$ is trivial (that is, $S$ has no non-trivial connected compact normal subgroups), then $\Gamma \cap R$ is a uniform lattice in $R$. In general, $\Gamma \cap R$ is not a lattice in $R$ (even if the Lie group $G$ has no compact connected normal subgroups; see [12] and the examples below).

We begin the presentation of our new results with the proof of Theorem 1 (see above), which generalizes the following results of G. Mostow.

We prove our Theorem 1, which generalizes both results from the indicated articles of D. Mostow.

Proof of Theorem 1. First, note that condition (ii) or (iii) implies condition (i). For an almost linear Lie group, the center of the semisimple part is finite, and therefore, the intersection $S \cap R$ (which is contained, in particular, in the center of the Lie group $S$) is finite. For a simply connected Lie group, the intersection $S\cap R$ is trivial, and therefore, for a finitely covered Lie group, the analogous intersection is finite. So, it remains for us to prove that condition (i) implies the assertion of Theorem 1. We proceed with this.

Consider the universal covering $\pi \colon \widetilde G= \widetilde S \cdot \widetilde R \to G = S\cdot R$ of the Lie group $G$. Let $Z=\ker \pi |_{\widetilde S} \cdot \ker\pi|_{\widetilde R} $; this is some subgroup of the abelian central subgroup $\ker \pi \subset \widetilde Z(G)$. By the hypothesis, the intersection $S\cap R$ is finite, and therefore, the subgroup $Z$ has finite index in $\ker \pi$.

Let $\widetilde \Gamma = \pi^{-1}(\Gamma)$. Applying Theorem 6 to the pair $\widetilde G$, $\widetilde \Gamma$ in case (ii), we obtain in $\widetilde \Gamma$ a finite index subgroup $\widetilde \Gamma_1$ smoothly deformable inside $\widetilde G$ into a lattice $\widetilde D$ such that $\widetilde D=(\widetilde D\cap \widetilde S)\cdot (\widetilde D \cap\widetilde R)$, and the elements of $\widetilde \Gamma_1 \cap \widetilde R$ are fixed under this deformation. Moreover, all elements of $\widetilde \Gamma_1\cap \widetilde S$ will also be fixed by the construction of the deformation from [9] (in this paper, the deformation motion of the lattice elements is along unclosed curves). In particular, all elements of $\widetilde \Gamma_1 \cap \ker \pi_{\widetilde S}$ will be fixed under this deformation. Therefore, under the above deformation, all elements of the subgroup $\widetilde \Gamma_1 \cap Z$ are also fixed.

Let $G'= \widetilde G/Z$, then the deformation of the lattice $\widetilde \Gamma_1$ inside $\widetilde G$ induces, due to immobility of the subgroup $\widetilde \Gamma_1 \cap Z$, a deformation of the lattice $\Gamma'_1 = \widetilde \Gamma_1/ \widetilde \Gamma_1 \cap Z$ inside the Lie group $G'$. The index of the subgroup $Z$ in $\ker \pi$ is finite, therefore $Z'=\ker \pi/Z$ is a finite central subgroup of $G'$. Let us show that all elements of the subgroup $\Gamma'_1 \cap Z'$ will be fixed under the above deformation.

For any $z \in \Gamma'_1 \cap Z'$, there exists some neighborhood $U_z \subset G'$ in which there are no elements $g$ of the Lie group $G'$ having the same order as $z$, since otherwise in $G'$ there would exist elements (obtained by multiplying the elements $g$ by $z^{-1}$) of order bounded above, arbitrarily close to the identity element, which is obviously impossible. Therefore, under the above deformation, the elements of $\Gamma'_1\cap Z'$ must be immobile. It follows from this that, under the deformation of the lattice $\Gamma'_1$ inside $G'$, all elements of $\widetilde \Gamma_1 \cap \ker \pi$ will be immobile. But then this deformation induces a smooth deformation of the subgroup $\Gamma_1=\widetilde \Gamma_1/\widetilde \Gamma_1 \cap \ker \pi$ (which has finite index in $\Gamma$) inside the Lie group $G=\widetilde G/\ker \pi$ into a lattice $D=\pi(\widetilde D)$ such that $D=(D\cap S) \cdot (D \cap R)$. And since $(D\cap S) \cap (D \cap R) \subset S \cap R$, and the group $S \cap R$ is finite by assumption, the above decomposition for $D$ is an almost semidirect product $D=(D\cap S) \cdot (D \cap R)$. This proves Theorem 1.

In Theorem 1, two extreme in a sense cases were considered, viz., when the Lie group is almost simply connected and when it is almost linear. But there are many Lie groups that do not satisfy either of these two restrictions. Here is the simplest example of this kind.

Consider the Lie group $\widetilde S = \mathrm{SL}(2,\mathbb{R}) \times \widetilde {\mathrm{SL}(2,\mathbb{R})}$ (here $\widetilde {\mathrm{SL}(2,\mathbb{R})}$ is the universal covering for the Lie group $\mathrm{SL}(2,\mathbb{R})$). This Lie group is non-linear (since it is known that the Lie group $\widetilde {\mathrm{SL}(2,\mathbb{R})}$ will be non-linear), but its fundamental group is infinite (it is isomorphic to $\mathbb{Z}$). Therefore, it does not satisfy any of the conditions (i), (ii), (iii). Augmenting this Lie group $S$ with some simply connected radical, we also obtain non-semisimple Lie groups “intermediate” between those considered in Theorem 1. It is easy to construct examples in which one of conditions (ii), (iii) is not met.

However, there is one special case when the assertion of Theorem 1 is satisfied for all locally isomorphic Lie groups $G$. We will formulate it as a corollary of Theorem 1.

Proof. In fact, we will prove that if a Lie group $G$ satisfies the condition of this corollary, then it is either almost linear or almost simply connected. Now this corollary follows from Theorem 1.

So, let us consider in more detail the structure of the Lie group $G=S\cdot R$. Its semisimple part $S$ is simple, and therefore, as is known, its fundamental group is either finite or isomorphic to the direct sum of the group $\mathbb{Z}$ and some finite group (and in this case $S$ is almost simply connected).

By the hypothesis, the radical of the Lie group $G$ is simply connected, and therefore, the fundamental group $\pi_1(G)$ is isomorphic to the fundamental group of the Levi subgroup $S$. Therefore, $\pi_1(G)$ is either finite or contains, as a subgroup of finite index, a subgroup isomorphic to $\mathbb{Z}$. In the first case, the Lie group $G$ is almost simply connected.

The second case requires a more detailed consideration. Any simply connected solvable Lie group is known to be linear (that is, it has a faithful finite-dimensional linear representation). In our case, this is obviously true — as such a representation, we can choose, by virtue of the triviality of the center, the adjoint representation.

Consider the universal covering $\pi \colon \widetilde G\to G$. The group $\pi_1(G)$ is isomorphic to the kernel of this representation. But since the center $Z(R)$ is trivial, $\ker \pi \subset Z(\widetilde S)$. In our case, this implies that the center of the Lie group $S$ is finite. Therefore, such a Lie group $S$ is almost linear (that is, it has a finite-dimensional linear representation with finite kernel). It follows from [13] that then the entire Lie group $G$ is almost linear. This completes the proof of Corollary 1.

Note that there are also quite a few semisimple Lie groups $S$ for which the argument from Corollary 1 is applicable if such Lie groups are taken as Levi subgroups. For example, the class of such semisimple Lie groups includes all complex semisimple Lie groups $S$ (whose centers are always finite), as well as direct sums of a non-compact simple Lie group and semisimple groups with complex structure. There are many other examples. Furthermore, as is well known, among simply connected solvable Lie groups there are many one whose center is trivial. Note that such Lie groups always have a faithful linear representation (for example, like the adjoint representation). Let us give some examples of simply connected solvable Lie groups with trivial center.

The simplest example of this kind is a two-dimensional non-Abelian solvable Lie group. It is simply connected, its center is trivial, and it is the only non-Abelian two-dimensional Lie group. Let us now consider three-dimensional simply connected solvable Lie groups $R$. As is easy to see, all of them can be represented as semidirect products of the form $R=\mathbb{R} \ltimes _\phi \mathbb{R}^2$, where $\phi\colon \mathbb{R} \to \mathrm{GL}(2,\mathbb{R})$ is some homomorphism. Such a Lie group has a trivial center if and only if no non-identity element of the one-dimensional matrix group $\phi(\mathbb{R})$ has fixed points on $\mathbb{R}^2$ distinct from the origin. For example, this class includes the one-parameter diagonal subgroups that have $\exp(at)$ and $\exp(bt)$ on the diagonal for $a,b\in \mathbb{R}$; $a,b \ne 0$. Among the indicated solvable simply connected Lie groups $R$, there are many ones that have lattices (for such groups to exist one should require in addition that $a+b=0$ and the sum $e^a+e^{-a}$ be a natural number).

Let us now consider in more detail the intersection of an arbitrary uniform lattice $\Gamma \subset G$ with Levi subgroups of an arbitrary Lie group $G$ (according to A. I. Mal’tsev, all such Levi subgroups are conjugate in $G$). Generally speaking, the intersection $\Gamma \cap S$ will not always be a lattice in a given Lie subgroup $S$. We will illustrate this with two examples.

Example 1. Here, the Levi subgroup $S$ in the Lie group $G$ will have rank $0$, that is, it is compact.

Consider a Lie group $G=K\times \mathbb{R}$ (with one-dimensional radical), where $K$ is some compact semisimple non-trivial Lie group (which in this case is a Levi subgroup of $G$). In the Lie group $K$, we choose some one-dimensional closed subgroup $C$ (isomorphic to $\mathrm{SO}(2)$ or, which is more convenient for us, to the one-dimensional circle in $\mathbb{C}$) and consider the set $A=C\times \mathbb{R}$ — this is an Abelian two-dimensional Lie group diffeomorphic to a cylinder. It is clear that this Lie group is isomorphic to some closed subgroup of $G$. In $A$ we consider a one-dimensional subgroup $B$ of the form $(e^{it},t)$; $t\in \mathbb{R}$ is an “irrational winding” of the cylinder $A$. Now, considering $A$ as a closed subgroup of $G=K\times \mathbb{R}$, we obtain a one-dimensional subgroup of $G$ (which we also denote by $B$) isomorphic to $\mathbb{R}$ and whose intersection with the Levi subgroup of $K$ is trivial. Now consider the subgroup $\Gamma$ in $B$ for which $t\in \mathbb{Z}$. Clearly, $\Gamma$ is a discrete subgroup of $G$ isomorphic to $\mathbb{Z}$. Moreover, as is easy to verify, the quotient space $G/\Gamma$ is diffeomorphic to $K\times S^1$ (where $S^1$ is a circle) and, in particular, the manifold $G/\Gamma$ is compact. Therefore $\Gamma$ is a lattice in $G$, and its construction implies that its intersection with the Levi subgroup $K$ is trivial.

In the first example, we considered a Levi subgroup of rank $0$ (this group is compact); in the next example, we will deal with a Levi subgroup of real rank $1$. In one particular case, such an example was given in [2].

Example 2. Let $S$ be a Lie group locally isomorphic to $\mathrm{SO}(1,n)$, $\mathrm{SU}(1,n)$, $G=S \times \mathbb{R}^m$, and $\Gamma=D\times \mathbb{Z}^m$, where $D$ is a lattice in $S$ (our example is also suitable for the case of general and uniform lattices), and such that the Abelian group $D/[D,D]$ is infinite (such lattices exist, see, for example, [2]). The Levi subgroup of this Lie group $G$ is unique and coincides with $S$. Further, it is obvious that there exists a homomorphism $\phi\colon D\to\mathbb{R}^m$ such that the Abelian subgroup $\phi(D)+\mathbb{Z}^m$ (the sum of the subgroups $\phi(D)$ and $\mathbb{Z}^m$ in $\mathbb{R}^m$), which is also a subgroup of $\mathbb{R}^m$ due to the Abelianity of $\mathbb{R}^m$, is a non-closed subgroup of $\mathbb{R}^m$.

Now consider the following embedding of our group $\Gamma$ in $G$ as a subgroup:

$$ \begin{equation*} \Gamma=D \times \mathbb{Z}^m \ni (\gamma,z) \mapsto_\alpha (i(\gamma), \phi(\gamma)+z) \in S \times \mathbb{R}^m = G, \end{equation*} \notag $$

where $i\colon D \hookrightarrow S$ is a natural embedding of the subgroup $D$. It is easy to see that $\alpha(\Gamma)$ is a uniform lattice in our Lie group $G$. However, the subgroup $\alpha (\Gamma) \cap S$ in the Levi subgroup $S$ is not a lattice (its index in $i(D)$ is infinite).

Note that this example involves, in particular, Lie groups locally isomorphic to $\mathrm{SL}(2,\mathbb{R})$ and $\mathrm{SL}(2,\mathbb{C})$ (locally isomorphic to $\mathrm{SO}(1,3)$).

Before stating the next result, we note that a real Lie group has real rank ($\mathbb{R}$-rank) equal to $0$ if and only if it is compact. Further, a simple real Lie group has $\mathbb{R}$-rank equal to $1$ if and only if this Lie group is locally isomorphic to one of the following simple classical Lie groups $\mathrm{SU}(1,n)$ ($n\geqslant 1$), $\mathrm{SO}(1,n)$ ($n \geqslant 3$), $\mathrm{Sp}(1,n)$ ($n \geqslant 2$), or the exceptional Lie group $F_{4(-20)}$ of type $F_4$ of index $-20$ (see [14]). Note also that the Lie algebras $\mathrm{su}(1,1)$, $\mathrm{so}(1,2)$, and $\mathrm{sp}(1,\mathbb{R})$ are isomorphic to $\mathrm{sl}(2,\mathbb{R})$, and $\mathrm{sl}(2,\mathbb{C})$ is isomorphic to $\mathrm{so}(1,3)$. It is clear that a semisimple Lie group will have real rank equal to $1$ if and only if it is an almost direct product of a simple Lie group whose $\mathbb{R}$-rank is $1$ and a compact semisimple Lie group.

Let us formulate one general result concerning the intersection of lattices with a Levi subgroup.

It is important to note here that not all simple Lie groups of rank $1$ appear as “exceptional factors” in the theorem, but only $\mathrm{SO}(1,n)$ and $\mathrm{SU}(1,n)$. Further, the most essential thing in this theorem is the existence of a Levi factor $S$ for which the intersection with the lattice $\Gamma$ is a lattice in $S$. Therefore, the result proved in Theorem 1 gives us non-trivial information only in cases where the semisimple part has no factors locally isomorphic to $\mathrm{SO}(1,n)$ or $\mathrm{SU}(1,n)$.

In this case, when the semisimple part $S$ of the Lie group $G$ has no factors locally isomorphic to $\mathrm{SO}(1,n), \mathrm{SU}(1,n)$ (for which among the lattices in them there are those for which the center is infinite), Theorem 1 (i) applies to all Lie groups $G$ of this kind.

§ 4. Levy–Mostow decomposition

Let $\Gamma$ be a lattice in some connected Lie group $G=S\cdot R$. By Theorems 1 and 2, in (many), cases this lattice $\Gamma$ or its subgroup of finite index $\Gamma_1$ decomposes into a semidirect product $\Gamma_{\mathrm s} \ltimes \Gamma_{\mathrm r}$ of lattices $\Gamma_{\mathrm s}$ and $\Gamma_{\mathrm r}$ in the Lie groups $S$ (the Levi subgroup) and $R$ (the radical of the Lie group $G$), respectively. We will call such a decomposition the Levi–Mostow decomposition for the group $\Gamma$. In the general case, such a decomposition for the lattice $\Gamma$ may not exist; for example, it will not exist if the subgroup $S\cap R$ is infinite, see Example 3 below). The subgroup $S\cap R$ can be infinite only when so is the center of the semisimple Lie group $S$. This can take place for simple Lie groups of any rank. More specifically, this is possible if a simple Lie group is commensurable with a simply connected Lie group, which is the universal covering group for the Lie groups $\mathrm{SU}(p,q)$, $\mathrm{SO}(2,n)$, $\mathrm{Sp}(n,\mathbb{R})$, $\mathrm{SO}^\ast(2n)$ (the series of classical Lie groups) or the for two exceptional ones $E_6^{\mathrm{III}}$, $E_7^{\mathrm{VII}}$.

Now let $\Gamma$ be an arbitrary abstract group. The abstract Levi–Mostow decomposition of $\Gamma$ is the decomposition $\Gamma_1=\Gamma_{\mathrm s} \cdot \Gamma_{\mathrm r}$ of $\Gamma_1$, weakly commensurable with $\Gamma$, into an almost semidirect product of lattices $\Gamma_{\mathrm s}$ and $\Gamma_{\mathrm r}$ in some semisimple and solvable Lie groups, respectively. It turns out that in the general case for lattices in Lie groups the abstract Levi–Mostow decomposition does not always exist (although in some sense this happens quite rarely).

Example 3 (an example of a lattice in a Lie group that does not have a Levi–Mostow decomposition). Consider the two Lie groups: $S=\mathrm{SL}(2,\mathbb{R})$ and $R=N(3,\mathbb{R})$ (the Lie group of real unipotent matrices of order three). As is easy to verify, in the group $\mathrm{Aut}(R)$ of automorphisms of the Lie group $N(3,\mathbb{R})$, the Levi subgroup is isomorphic to $\mathrm{SL}(2, \mathbb{R})$. By $i\colon \mathrm{SL}(2,\mathbb{R}) \hookrightarrow \mathrm{Aut}(R)$ we denote the corresponding embedding. By $\mu\colon \widetilde {\mathrm{SL}(2,\mathbb{R})}\to \mathrm{SL}(2,\mathbb{R})$ we denote the universal covering for the Lie group $\mathrm{SL}(2,\mathbb{R})$. The kernel $\ker \mu $ of the homomorphism $\mu $ is isomorphic to $\mathbb{Z}$. Let $G=\widetilde {\mathrm{SL}(2,\mathbb{R})} \ltimes_\phi N(3,\mathbb{R})$; this is the semidirect product defined by the homomorphism $\phi= i \circ \mu \colon \widetilde{\mathrm{SL}(2,\mathbb{R})} \to \mathrm{Aut}(N(3,\mathbb{R} )) $. Next, let $D=\mu^{-1}(\mathrm{SL}(2,\mathbb{Z})) $. The embedding $i$ can be chosen so that the subgroup $\mathrm{SL}(2,\mathbb{Z})\subset \mathrm{SL}(2,\mathbb{R})$ preserves the lattice $N(3,\mathbb{Z}) \subset N(3,\mathbb{R})$ (where $N(3,\mathbb{Z})$ is the subgroup of integers in $N(3,\mathbb{R})$). Finally, we set $\widetilde \Gamma = D\ltimes_{\psi} N(3,\mathbb{Z})$, where $\psi$ is the restriction of the homomorphism $\phi$ to $D$. As is easy to see, we obtain a (non-uniform, since $\mathrm{SL}(2,\mathbb{Z})$ lattice in $\mathrm{SL}(2,\mathbb{R}$)) in the Lie group $G$.

The group $\mathrm{SL}(2,\mathbb{R})$ acts trivially on the center $Z(N(3,\mathbb{Z}))$. Therefore, the center $Z(G)$ of the Lie group $G$ is the subgroup $Z(\widetilde {\mathrm{SL}(2,\mathbb{R})}) \times Z(N(3,\mathbb{R}))$ isomorphic to $\mathbb{Z} \times \mathbb{R}$. By $Z$ we denote a subgroup of $Z(G)$ that is isomorphic to $\mathbb{Z}$ and diagonally embedded in the subgroup $Z(\widetilde {\mathrm{SL}(2,\mathbb{R})} \times Z(N(3,\mathbb{R}))$. We set $\Gamma=\widetilde \Gamma/Z$. Since $Z\subset \widetilde \Gamma $, the group $\Gamma=\widetilde \Gamma/Z$ is obviously a lattice in the Lie group $G$. We show that this lattice $\Gamma$ does not even have an abstract Levy–Mostow decomposition.

We assume the opposite and let $\Gamma_1$ be some group weakly commensurable with $\Gamma$ and which has decompositions of the form $\Gamma_{\mathrm s} \cdot \Gamma_{\mathrm r}$. Clearly, in this case the group $\Gamma_{\mathrm s}$ must be weakly commensurable with $\mathrm{SL}(2,\mathbb{Z})$, and $\Gamma_{\mathrm r}$, with the lattice $N(3,\mathbb{Z})$.

Consider the virtual cohomological dimension $\operatorname{vcd} (\Gamma)$ of $\Gamma$ (on $\operatorname{vcd}$ of groups considered up to weak commensurability, see, for example, [2]). Since $\operatorname{vcd}(D)=1$, we have $\operatorname{vcd}(\Gamma) = \operatorname{vcd}(D\times_\phi N(3,\mathbb{Z}))= 4$. On the other hand, it is obvious that $\operatorname{vcd} (\Gamma) = \operatorname{vcd}( \widetilde \Gamma) -1$. Therefore, the group $\Gamma$ cannot be weakly commensurable with a group of the form $D\times_\phi N(3,\mathbb{Z})$. Therefore, our lattice $\Gamma$ cannot have a Levi–Mostow decomposition (even an abstract one).

In Example 3, we have constructed a non-uniform lattice. This is done for simplicity of exposition. It is also possible to construct uniform lattices for which a Levi–Mostow decomposition does not exist.

§ 5. On the fundamental group of compact homogeneous spaces

We now consider one application of the above results to the study of the fundamental group of arbitrary compact homogeneous spaces. First, we present some information about the structure of such spaces (for more details, see [3]).

Let $M = G/H$ be a compact homogeneous space of some connected Lie group $G$ (the Lie subgroup $H$ is always assumed to be closed). We can (and will) often assume that this Lie group is simply connected and locally effective on $M$ (however, it will not always be effective). If the Lie group itself, transitive on $M$, is inessential for us (and only the existence of a transitive action of some finite-dimensional Lie group is essential), then $M$ will be called a homogeneous manifold. Let $K$ be some maximal compact subgroup of $G$. It is simply connected and semisimple (due to the assumed simply connectedness of the Lie group $G$) and is uniquely determined up to conjugation in $G$. There is a natural action of the Lie group $K$ on the manifold $M$. In the general case, the orbits of this action are not isomorphic to each other, although here they all, as it turns out, will be of the same dimension. But they can have different orbit types (that is, the stationary subgroups corresponding to different points of the manifold $M$ will not always be conjugate in $K$).

The action of the compact group $K$ on the manifold $M$ is called equiorbital if all its orbits are of the same orbit type, that is, if for any points $m, m' \in M$ the corresponding stationary subgroups $K_m$, $K_{m'}$ are conjugate in $K$. It turns out that for an arbitrary compact homogeneous space $G/H$ there exists some compact homogeneous space $M' = G/H'$ (of the same Lie group $G$) that has a finite-sheeted covering (here $H'$ is some subgroup of finite index in $H$) and which is equiorbital with respect to the natural action of the maximal compact subgroup $K$. In this case, the orbit space $K \setminus M'$ of this action is a smooth manifold, and the natural mapping $M' \to K \setminus M' $ is a smooth, locally trivial bundle. This bundle is called the natural bundle for $M$ (although in fact it is often not the manifold $M$ itself that is bundled, but a suitable manifold $M'$ that has a finite-sheeted covering of it). We denote the base of this bundle by $M_{\mathrm a}$, and its fiber by $M_{\mathrm c}$.

In general, when constructing the natural bundle, the transition to $M'$ is necessary if we want to obtain the structure of a smooth manifold on $K \setminus M'$. In this case, there is no canonical choice of a finite-sheeted covering here. However, for the original manifold $M$, there is also a bundle of sorts — this will be the Seifert bundle over the base $K \setminus M$, which is an orbifold (and therefore, generally speaking, a manifold with singularities). We will not consider this generalization of the natural bundle here.

Thus, the natural bundle $M_{\mathrm c} \to M' \to M_{\mathrm a}$ is a smooth locally trivial bundle for a suitable manifold $M'$, a finite-sheeted covering of the original compact homogeneous manifold $M$. This bundle over $M'$ is uniquely determined up to fiberwise homotopy equivalence.

The base of the natural bundle $M_{\mathrm a}$ has the form $K\setminus M'$ (the quotient manifold) and it is an aspherical smooth manifold. Moreover, the universal covering manifold of $M_{\mathrm a}$ is diffeomorphic to Euclidean space (note that not every aspherical manifold has this property, although the universal covering manifold for them will always be contractible). Here, up to a diffeomorphism, the base of the natural bundle can be written (possibly again by passing to some finite-sheeted covering) as the space of double cosets $\Gamma \setminus F/C$, where $F$ is some connected, although not necessarily simply connected, Lie group (in fact, it is closely related to the original transitive Lie group), $C$ is the maximal compact Lie subgroup of $F$, and $\Gamma$ is a torsion-free lattice (uniform, that is, discrete subgroup with compact quotient space) of $F$. Here we can assume, passing if necessary to a subgroup of finite index in $\Gamma$ (which is equivalent to passing from $M_{\mathrm a}$ to some manifold that covers it with finite sheets), that the intersection $\Gamma \cap C$ is contained in the center $Z(F)$ of the Lie group $F$. Moreover, we can then assume that this intersection is trivial by factorizing the Lie group $F$ by it.

A representation of the base $M_{\mathrm a}$ of the natural bundle in the form $\Gamma \setminus F/C$ (where $C$ is the maximal compact subgroup of the connected Lie group $F$ and $\Gamma$ is a torsion-free lattice in $F$) is called the standard representation of the base. The base of the natural bundle may have several different (and essentially different) standard representations.

For the base $M_{\mathrm a}$ of the natural bundle (or for some manifold $M'_{\mathrm a}$ corresponding to a suitable homogeneous space $M''$ covering $M'$ with finite sheets), one can, in turn, construct another useful bundle, viz., the structure bundle $M_{\mathrm r} \to M_{\mathrm a} \to M_{\mathrm s}$. This is a smooth locally trivial fiber bundle whose fiber has the form $M_{\mathrm r}=R/D$ (where $D$ is a lattice in some simply connected solvable Lie group $R$) and whose base $M_{\mathrm s}$ has the form $U\setminus S/\Pi$, where $S$ is some semisimple connected Lie group with finite center and without compact factors, $U$ is a maximal compact Lie subgroup of $S$, and $\Pi$ is a lattice in $S$. Moreover, $U \setminus S$ is a symmetric space of negative curvature, and $U \setminus S/\Pi$ is a locally symmetric space, which is a compact geometric form for the symmetric space $U\setminus S$. The structure fiber bundle can be viewed as a certain analogue of the Levi decomposition for Lie groups, constructed for compact homogeneous manifolds (more precisely, for the bases of their natural fiber bundles). The base $M_{\mathrm s}$ is called a semisimple component of the base $M_{\mathrm a}$ of the natural bundle, and the fiber $M_{\mathrm r}$ is called a solvable component. If both of these components are non-trivial (that is, have positive dimensions), then $M_{\mathrm a}$ will be called a base of mixed (or general) type.

For an arbitrary compact homogeneous space $M=G/H$ of a Lie group $G$, its fundamental group $\pi_1(M)$ is weakly commensurable with the group $\pi_{\mathrm a}=\pi_1(M_{\mathrm a})$ of the torsion-free base $M_{\mathrm a}$ of the natural bundle. Moreover, there is an exact sequence of groups $\{e\}\to \pi_{\mathrm r} \to \pi_{\mathrm a} \to \pi_{\mathrm s} \to \{e\}$, where $\pi_{\mathrm r}$ is the largest solvable normal subgroup in $\pi_{\mathrm a}$ (it can be proved that such a normal subgroup exists), and the group $\pi_{\mathrm s}$ is semisimple, that is, it has no non-trivial solvable normal subgroups (see, for example, [16]). For example, consider a compact homogeneous space of the form $M=G/\Gamma$, where $\Gamma$ is the uniform lattice in the Lie group $G$. Suppose that the center of the Levi subgroup $S$ in $G$ is trivial. Hence, if the Lie group $G$ has no compact semisimple compact normal subgroups, then $\pi_{\mathrm r}= \Gamma\cap R$ (where $R$ is the radical of the Lie group $G$), and the group $\pi_{\mathrm s}$ is isomorphic to $\Gamma/\Gamma\cap R$. In this case, the above exact sequence splits (and $\pi_{\mathrm a}$ is a semidirect product of the groups $\pi_{\mathrm s}$ and $\pi_{\mathrm r}$), since by Theorem 1 in this case, the intersection $S\cap R$ is trivial (since it must be contained in the center of the Lie group $G$). In the general case, the above exact sequence does not split, even if there is a Levi–Mostow decomposition for the lattice $\Gamma$. For example, let $\Gamma$ be a uniform lattice in the Lie group $G=\widetilde {\mathrm{SL}(2,\mathbb{R})}$ (the universal covering lattice for the Lie group $\mathrm{SL}(2,\mathbb{R})$). Then here we have $\pi_{\mathrm r} \simeq \mathbb{Z}$, $\pi_{\mathrm s} \simeq \Gamma/\Gamma\cap Z(G)$, where $\Gamma \cap Z(G) \simeq \mathbb{Z}$. It is easy to show that in the case under consideration the exact sequence $\{e\}\to \pi_{\mathrm r} \to \pi_{\mathrm a} \to \pi_{\mathrm s} \to \{e\}$ does not split, even if we pass from the lattice $\Gamma$ to some of its subgroups of finite index.

Proof. As was shown in [16], when considering a homogeneous space $G/H$ up to a finite-sheeted covering (under which the base of the natural bundle can be replaced by its finite-sheeted covering), one can assume, without changing the Levi factor of the Lie group $G$, that the action of the Lie group on $M$ is regular (this notion was introduced in [16], its details are not important for us here). Further, it was shown there that the group $\pi_{\mathrm a}$ can be embedded as a uniform lattice in some Lie group $F$ finitely covered by the Lie group $(N_G(H))_0/H_0 \times \mathbb{R}^m$ for some $m\in \mathbb{R}$.

Let $F=S\cdot R$ be a Levi decomposition (where $R$ is the radical of the Lie group $F$). From the conditions of the theorem and the above it follows that the center of the Levi factor of this Lie group is finite. But then the intersection $S\cap R$ is also finite (since it is obvious that $S\cap R\subset Z(S)$). Therefore, Theorem 1 applies in our situation. Applying this result, we see that some subgroup $\Pi$ of finite index in $\pi_{\mathrm a}$ decomposes into an almost semidirect product of the groups $\Pi \cap S$ and $\Pi \cap R$.

But since the manifold $M_{\mathrm a}$ (the base of the natural bundle) is aspherical, the group $\pi_1(M_{\mathrm a}) =\pi_{\mathrm a}$ has no torsion. The subgroup $(\Pi\cap S) \cap(\Pi \cap R)$ is contained in $S\cap R$ and therefore, it is finite. Since $\Pi$ is torsion-free, we obtain $(\Pi\cap S) \cap(\Pi \cap R) =\{ e\}$, and therefore, $\Pi$ is a semidirect product of the groups $\Pi \cap S$ and $\Pi\cap R$.

The center $Z(S)$ of the Lie group $S$ is finite and the subgroup $\Pi \cap S$ is torsion-free, and hence $\Pi\cap Z(S) = \{e\}$. And since the maximal solvable normal subgroup of $\Pi \cap S$ is necessarily contained in $Z(S)$, it follows from the above that this normal subgroup must be trivial. This means that the group $\Pi \cap S$ is semisimple (that is, has no non-trivial solvable normal subgroups). Therefore, $\Pi \cap S$ is isomorphic to $\pi_{\mathrm s}$. The group $\Pi \cap R$ is obviously isomorphic to $\pi_{\mathrm r}$. Therefore, the group $\pi_{\mathrm a}$ decomposes into a semidirect product of the subgroup $\pi_{\mathrm s}$ and a normal subgroup $\pi_{\mathrm r}$. This proves the proposition.

In terms of the structural bundle, Proposition 1 means that this bundle has a section.

The conditions imposed in Proposition 1 on the Lie group $(N_G(H_0))/H_0$ (the absence of compact factors for its Levi subgroup and the finiteness of its center) are essential. To show this, consider compact homogeneous spaces $G/H$ with a discrete stationary subgroup $H$. In this case, the Lie group $(N_G(H))_0/H_0$ is simply the quotient space of the Lie group $G$ by the lattice (which in this case, is the subgroup $H$). But above we have already considered examples showing that in the general case of a group $G$, a lattice in it (isomorphic to the fundamental group of the corresponding homogeneous space, if we assume that the Lie group $G$ is simply connected) cannot always be represented as a semidirect product of subgroups of the desired form.

§ 6. Intersection of lattices with Lie subgroups of the form $S\cdot N$

Let $R$ be a solvable Lie group and $N$ its nilradical. If $\Gamma$ is a lattice in $R$, then, as proved in [5], it is always uniform and the intersection $ \Gamma \cap N$ is a lattice in $N$. This statement is sometimes called Mostow’s structure theorem. It serves as a basis for studying lattices in solvable Lie groups. In fact, a more general statement is proved in [5]: if $H$ is an arbitrary (not necessarily discrete) uniform subgroup in a solvable Lie group $R$, then the subgroup $H \cap N$ is uniform in $N$. Moreover, if $H_0$ does not contain non-discrete normal subgroups of the Lie group $R$, then we have the inclusion $H_0 \subset N$.

In this section, we will prove Theorem 2 from the introduction. This result is a generalization of the indicated result of G. Mostow to the case of lattices in arbitrary Lie groups. But then an example will be given showing that the analogue of G. Mostow’s result on arbitrary uniform subgroups in solvable Lie groups to the case of uniform subgroups in arbitrary Lie groups is already incorrect in the general case (a corresponding counterexample will be constructed).

Proof of Theorem 2. The fact that for the lattice $\Gamma$ the intersection $\Gamma \cap S\cdot N$ is a lattice in $S \cdot N$ is equivalent, as is known from the general theory of lattices in Lie groups, to the closedness of the subgroup $\Gamma \cdot S\cdot N$ in the Lie group $G$.

We can assume that the Lie group $G$ is simply connected. If this is not the case, we pass to its universal covering and consider there the corresponding lattice (the preimage of the lattice $\Gamma$ under the covering). The corresponding intersection of this lattice with a subgroup of the form $S_1\cdot N_1$ is a lattice if and only if the lattice is $\Gamma \cap S\cdot N$. Therefore, in what follows we restrict ourselves to the case when the Lie group $G$ is simply connected (then $S$, $R$, $N$ will be simply connected).

Let $C$ be the maximal compact semisimple connected normal subgroup of $G$ (such a $C$ in $G$ is unique). We put $G_1=G/C$, $\Gamma_1= \Gamma/\ C\cap \Gamma$; this is a lattice in $G_1$. The Levi decomposition for $G_1$ (similarly to $G$, it will also be assumed to be simply connected) has the form $G_1=S_1 \cdot R$ (the radical under the factorization under consideration). It is clear that the subgroup $\Gamma\cdot S\cdot R$ is closed in $G$ if and only if this holds for the corresponding objects in $G_1$. Therefore, we can restrict ourselves to the case when the Lie group has no non-trivial connected compact normal subgroups. But then $\Gamma \cap R$ and $\Gamma \cap N$ are (uniform) lattices in the Lie groups $N$ and $R$, respectively (see [17]).

Now consider the homomorphism $\phi \colon G\to \mathrm{Aut}(N)$ induced by the adjoint action of the Lie group $G$ on its nilradical $N$. The subgroup $\phi(\Gamma)$ preserves the lattice $\Gamma\cap N$ in $N$. We will show that the subgroup $\phi(\Gamma)$ is discrete (some more general results of this kind will be discussed later after the completion of this proof). To do this, we note that any automorphism of a lattice in a nilpotent simply connected Lie group (which is $N$) extends, and uniquely, to an automorphism of the entire enclosing nilpotent Lie group. In particular, an automorphism of the Lie group $N$ that is sufficiently close to the identity one acts on the lattice trivially leaves all its elements fixed. Therefore, the group of automorphisms of the Lie group $N$ that preserve the lattice $\Gamma\cap N$ is discrete. And therefore its subgroup $\phi(\Gamma)$ is also discrete in $\mathrm{Aut}(N)$.

The Lie group $\mathrm{Aut}(N)$ has the structure of an algebraic Lie group (it is isomorphic to the (algebraic) group of automorphisms of the corresponding Lie algebra). For the image $\phi(G)$ of the Lie group $G$, consider its algebraic closure $W$ in the Lie group $\mathrm{Aut}(N)$. It is clear that the semisimple part $P$ of the Lie group $W$ and the semisimple part of $\phi(G)$ are the same. The radical of the Lie group $W$ is obtained by splitting the radical of the Lie group $\phi(G)$. It follows that the preimage of $\phi^{-1} (P\cdot U)$ (where $U$ is the unipotent radical of $W$) coincides locally with $S \cdot N$ (that is, its connected component of the identity coincides with $S\cdot N$).

The subgroup $P\cdot U$ is algebraic and therefore closed in $\mathrm{Aut}(N)$. The discrete subgroup $\phi(\Gamma)$ normalizes $P\cdot U$ and therefore, the product $\phi(\Gamma) \cdot P \cdot U$ is a closed subgroup in $\mathrm{Aut}(N)$. But then its preimage under the mapping $\phi$ is also a closed subgroup in the Lie group $G$. Now by the above the subgroup $\Gamma \cdot S \cdot N$ is closed in $G$. This, as noted at the very beginning of this proof, is equivalent to saying that the intersection $\Gamma \cap S\cdot N$ is a lattice in $S\cdot N$, which completes our proof of Theorem 2.

In connection with the above property (viz., discreteness of their stabilizer in the group of automorphisms of the enclosing nilpotent Lie group) of lattices in nilpotent Lie groups, the natural question is whether the stabilizer of a lattice in an arbitrary Lie group will be discrete. Of course, in the very general case this is not true — it suffices to consider a compact simply connected semisimple Lie group $K$, and in it, a trivial subgroup (obviously, uniform). It is clear that the stabilizer of this subgroup in $\mathrm{Aut}(K)$ will not be discrete, viz., it coincides with the group of all automorphisms of this Lie group (and its connected component is isomorphic to $K$). But it turns out that even if we exclude from consideration the Lie groups whose semisimple part contains compact components, then the stabilizer of the lattice will not, generally speaking, be discrete. Although in some cases (for example, for nilpotent Lie groups and for semisimple groups that do not have compact components) the stabilizer is still discrete.

Consider an example of non-discreteness of the lattice stabilizer. Let $R$ be a three-dimensional solvable Lie group which is the universal covering group for the Lie group $E_2$ of orientation-preserving isometries of the two-dimensional Euclidean plane. This Lie group $R$ can be represented as a semidirect product $R=\mathbb{R} \ltimes_\alpha \mathbb{R}^2$ defined by the homomorphism $\mathbb{R} \to \mathrm{GL}(2,\mathbb{R})$ whose image is a subgroup of $\mathrm{SO}(2) \subset \mathrm{GL}(2,\mathbb{R})$. Consider in $R$ the lattice $\Gamma = \mathbb{Z} \cdot \mathbb{Z}^2$ defined by the restriction of the homomorphism $\phi$ to $\mathbb{Z}$ such that $\operatorname{Im}(\phi) \subset \mathrm{SL}(2, \mathbb{Z})$. It is easy to verify that the set of automorphisms of the Lie group $R$ whose restrictions to $\Gamma$ are trivial is isomorphic to $\mathbb{R}^2$ and is therefore not discrete.

Note in connection with Theorem 2 that the subgroup $S \cdot N$ is a characteristic Lie subgroup of the Lie group $G$. In some cases, it is the only non-trivial normal subgroup of the Lie group $G$ with lattice $\Gamma$ whose intersection is a lattice. For example, let $G=S\ltimes _\phi \mathbb{R}^n$ be a Lie group with Levi factor $S$, which is some simple Lie group, and Abelian radical, which is isomorphic to $\mathbb{R}^n$. This Lie group is determined by a linear $n$-dimensional representation of $\phi$. As is known, uniform lattices can be constructed in Lie groups of this type. But if the representation $\phi$ is irreducible, then there are no proper normal Lie subgroups of $G$ whose intersection with $\Gamma$ is a lattice in this Lie subgroup. Here, $N=\mathbb{R}^n$, and there is nothing, save $S\cdot N$, suitable for our purposes here.

Having considered, in particular, the question of intersections of lattices with Lie subgroups of the form $S\cdot N$, the question naturally arises of whether it is possible to generalize this result to intersections of lattices with some Lie subgroups of the form $S\cdot U$, where $U$ is some characteristic Lie subgroup of $N$ (for example, a member of the lower or upper row). In particular, is the intersection of the original lattice with a Lie subgroup of the form $S \cdot Z(N)$ (where $Z(N)$ is the center of the nilradical of $N$) a lattice? But it turns out that the nilradical is unique among all the Lie subgroups it contains. Let us explain this. Let $U$ be some normal subgroup in $N$ that is invariant under the natural action of the semisimple part $S$. If $U \ne N$, then there exist, generally speaking, several different Lie subgroups of the form $S \cdot U$ for different Levi factors of $S$. The point is that all Levi factors are conjugate to each other by some elements from $N$, but it is not necessarily sufficient to restrict ourselves to elements from a smaller subgroup $U \subset N$. For example, for an Abelian radical (which will also be a nilradical), conjugations by its different elements will usually yield different semisimple parts.

All this indicates that Lie subgroups of the form $S\cdot N$ are in some sense unique, and therefore, Theorem 2, associated with them, is practically unimprovable in its generality. However, it turns out that a weaker assertion holds not only for Lie subgroups of the form $S\cdot N$, but for more general ones. But for this we will have to use the argument from [7] (which is quite difficult) to the full extent. Namely, it is shown there that when a lattice is deformed, its elements are multiplied by elements from the center $Z(N)$ of the nilradical (more precisely, even from the maximal simply connected subgroup lying in it). This leads us to the following assertion.

Let us now consider some consequences of Theorem 2 related to the topological structure of compact homogeneous spaces with a discrete stationary subgroup.

Similarly to this corollary, there are results which involve, instead of $Z(N)$, any other term of the lower central series of the nilradical $N$.

Corollaries 2 and 3 are sometimes capable of reducing the study of the topological structure of arbitrary compact homogeneous spaces with discrete stationary subgroup to the consideration of simpler special cases.

The following corollary of the theorem concerns the structure of the fundamental group of some compact homogeneous spaces $M$. The fundamental group $\pi_1(M)$ is weakly commensurable with the group $\pi_{\mathrm a}=\pi_1(M_{\mathrm a})$. By choosing a suitable base $M_{\mathrm a}$ of the natural bundle for a compact homogeneous space $M$, we can assume that $M_{\mathrm a}$ is diffeomorphic to a manifold of the form $K \backslash G/\Gamma $, where $\Gamma$ is the uniform lattice in some connected Lie group $G$, and $K$ is the maximal compact Lie subgroup of $G$. An application of Corollary 2 to the pair $(G, \Gamma)$ produces the following result.

We will now show by example that the generalization of Mostow’s structure theorem for arbitrary uniform subgroups (of positive dimension), as similar to that proved in Theorem 2 for the case of discrete uniform subgroups, is generally incorrect. In other words, if $H$ is a uniform subgroup of a Lie group $G$, then if it is not discrete, then its intersection $H\cap S\cdot N$ will not always be a uniform subgroup of $S\cdot N$, even if $G$ has no compact semisimple factors. If $G$ is solvable, then, according to [5], for arbitrary uniform Lie subgroups (and not only for discrete ones), their intersection with the nilradical (the Levi factor is trivial here) will be a uniform subgroup.

Let us now give an example. Consider the Lie group $G=\mathrm{GL}(2,\mathbb{R}) \ltimes _\tau \mathbb{R}^2$, where $\tau$ is the natural action of $\mathrm{GL}(2,\mathbb{R})$ on $\mathbb{R}^2$. The semisimple part of this Lie group is isomorphic to $\mathrm{SL}(2,\mathbb{R})$, and the nilradical is abelian and coincides with $\mathbb{R}^2$. The commutator subgroup of this Lie group is $\mathrm{SL}(2,\mathbb{R}) \cdot \mathbb{R}^2$.

In $\mathrm{GL}(2,\mathbb{R})$, consider the subgroup $W$ formed by matrices of the form

Consider the subgroup $H= W\ltimes _\tau \mathbb{R}^2$. It is clear that $H$ is a closed Lie subgroup of $G$, and this subgroup is uniform. Now consider a subgroup of the form $S\cdot N$; in this case, it has the form $\mathrm{SL}(2,\mathbb{R}) \ltimes_\tau \mathbb{R}^2$. The intersection $H\cap S\cdot N$ in this case is contained in the subgroup $T$ and even there it is not uniform. Therefore, it is not uniform in $G$.

§ 7. Intersection of lattices with commutator subgroups of Lie groups

Let $\Gamma$ be a uniform lattice in a connected Lie group $G$. Consider the commutator subgroup $[G,G]$ of the Lie group $G$ and consider the subgroup $\Gamma \cap [G,G]$ in $[G,G]$. This subgroup is discrete, but in general it is not a lattice in $[G,G]$, even if the Lie group $G$ is solvable (see, for example, [8]). However, in this section we prove Theorem 3 from the introduction, which indicates two cases, where the subgroup $\Gamma \cap [G,G]$ is still a lattice in $[G,G]$.

The assertion we need will be extracted from the paper [18], which is devoted to the consideration of complex solvable Lie groups and their subgroups, more general than the uniform ones.

Proof of Theorem 3. According to [18], if $G$ is a complex solvable Lie group and $H$ is a closed Lie subgroup of it, then, under some additional conditions on $H$, its intersection with $[G,G]$ is a uniform subgroup. But a careful analysis of the proofs in the paper [18] allows us to extract from them (with some additions) the statement we need. Let us show how to do this.

In Lemma 3.6 of [18] it is proved there that for a complex solvable Lie group (which can obviously be assumed to be simply connected) the intersection $\Gamma \cap [G,G]$ is a lattice in $[G,G]$. Recall that lattices in solvable Lie groups are always uniform.

The proof in this lemma involves Lemma 2.5 (concerning nilpotent Lie groups), Proposition 3.2 (the proof of which is based on Propositions 1.1, and Proposition 2.1, which is of a general nature), and Lemma 3.5 (which in turn refers to Lemmas 3.3 and 3.4). Corollary 2.3 and Lemma 3.4 are of a general nature, and Lemma 3.3 refers to Lemma 1.1.

The proof of Lemma 3.6 uses induction on the length of a solvable series of the Lie group $G$, and thus uses the solvability of the Lie group $G$.

Let us now consider in more detail Lemmas 3.3 and 3.4 of [18]. In Lemma 3.4, the solvability of the Lie group is not used in any way, while in Lemma 3.3 it is “implicitly” used in a reference to Lemma 1.1. And now the most interesting part begins. The proof of Lemma 1.1 itself employs the solvability of the Lie group $G$, but this lemma itself concerns subgroups more general than lattices. But after the proof of this lemma, the paper [18] contains an additional remark important to us. Namely, in the paragraph following the proof of Corollary 1.2, it is proved that the statement of Lemma 1.1 we need for what follows is true not only for solvable Lie groups $G$, but also for arbitrary ones, provided that the subgroup $F$ is discrete and uniform. This statement is the only one in [18] that considers arbitrary (rather than just solvable) Lie groups. Thus, both Lemmas 3.4 and 3.5 are applicable in the case we need here.

However, the proof of assertion (ii) in Lemma 3.6 about the uniformity of the intersection $\Gamma \cap [G,G]$ in $[G,G]$ is carried out by induction on $\dim G$. In this case, we use the fact that, for a solvable Lie group $G$, we always have $\dim [G,G] <\dim G$. But, for arbitrary Lie groups, such an inequality is generally false, that is, it may turn out that $[G,G]=G$. However, in this case, the radical $R$ of the Lie group $G$ must be nilpotent. The point is that $[G,R] \subset N$ is always true. Therefore, in addition to the proof in [18], we need to consider the case when the radical of the Lie group $G$ is nilpotent. Let us do this.

The Levi decomposition of the Lie group $G$ in this case has the form $G=S\cdot N$. In this case, the intersection $\Gamma \cap N$ will be a lattice in $N$, and in our case (when $S$ has no components of rank $1$, that is, it is locally isomorphic to $\mathrm{SL}(2,\mathbb{C})$), the intersection $\Gamma \cap S$ will also be a lattice in some Levi factor $S$.

Consider the adjoint action of the Lie group $G$ on $N$ defined by the homomorphism $\phi\colon G \to \mathrm{Aut}(N)$. The presence of a lattice in $N$ means that $N$ has the structure of an algebraic group defined over $\mathbb{Q}$. The action of the subgroup $\Gamma$ on $N$ preserves the lattice $\Gamma \cap N$. Therefore, the group $\phi(S)$ is defined over $\mathbb{Q}$. But then the subgroup $[S,N]$ is also defined over $\mathbb{Q}$ and, in particular, its intersection with $\Gamma\cap N$ will be uniform in it. But obviously $[G,G]= S\cdot [G,N]$, and therefore, the subgroup $\Gamma\cap [G,G]$ is also a lattice in $[G,G]$ (and uniform). This argument complements the proof of Lemma 3.6 in case (i) for non-solvable Lie groups, which completes the proof of Theorem 3 in case (i).

We proceed with the proof of assertion (ii) of our theorem. Here there is no restriction on the Levi subgroup, so we will apply a more complicated method of the proof with the help of the results from [9]. We can assume that the Lie group $G$ is simply connected. By Theorem 1, $\Gamma$ contains a finite index subgroup $\Gamma_1$ that is smoothly deformable inside $G$ into a uniform subgroup $D$ of finite index such that $D = (D\cap S) \cdot (D\cap R)$. Under this deformation, elements from $\Gamma_1$ are multiplied by some elements from the subgroup $Z(N)$ (the center of the nilradical $N$ of the Lie group $G$).

Let us show that in our case $Z(N)\subset [N,N]$. If this is not so, then it is easy to see that there is a decomposition $Z(N)=Z_1\times C$ (where $C$ is a one-parameter subgroup of $N$ generated by an element of $Z(N)\setminus [N,N]$, and $Z_1$ is some subgroup). But this contradicts our assumption in (ii). Therefore, $Z(N)\subset [N,N]$, and so $Q\subset [N,N] \subset [G,G]$. It follows that, under the above deformation of the lattice $\Gamma_1$, the images of all its elements under the natural epimorphism $G\to G/[G,G]$ are fixed. In particular, the subgroups $\Gamma_1/\Gamma_1\cap[G,G]$ and $D/D\cap[G,G]$ coincide. This means that the lattice deformation in $G$ that under consideration generates a deformation in $[G,G]$. Moreover, since it is the lattice that is deformed (that is, a discrete subgroup with a compact quotient space), it is easy to see that inside $[G,G]$ we also obtain a deformation (inverse to the one indicated above) of the lattice $D\cap [G,G]$ into a subgroup $\Gamma_1\cap [G,G]$, which is also a lattice. Therefore, $\Gamma \cap [G,G]$ is a lattice in $[G,G]$. This completes the proof of our Theorem 3.

We now give an example showing the importance of the conditions imposed on the Lie group $G$ in Theorem 3.

Let $H_{\mathrm n}(\mathbb{C})$ be the Heisenberg group, $\dim H_{\mathrm n}(\mathbb{C}) = 2N+1$. Its Lie algebra $h_{\mathrm n}(\mathbb{C})$ in a suitable basis $X_1, \dots, X_{\mathrm n}, Y_1,\dots, Y_{\mathrm n}, Z$ is defined by the relations $[X_i, Y_i]=Z$ (where $i=1,2,\dots,n$). In the Lie group of automorphisms $\mathrm{Aut}H_{\mathrm n}(\mathbb{C})$, the Levi subgroup is isomorphic to the symplectic group $\mathrm{Sp}_{\mathrm n}(\mathbb{C})$.

Let $\phi \colon \mathrm{SL}(2, \mathbb{C}) \to \mathrm{Sp}(n,\mathbb{C})$ be some non-trivial (and therefore, locally faithful here) representation of the Lie group $\mathrm{SL}(2,\mathbb{C})$. Consider the Lie group $G=(\mathrm{SL}(2,\mathbb{C}) \ltimes_\phi H_{\mathrm n}(\mathbb{C}))\times \mathbb{C}$ constructed with the help of $\phi$. As is known from the description of the representations of the group $\mathrm{SL}(2,\mathbb{C}$), one can find a representation $\phi$ and a uniform lattice $D$ in $\mathrm{SL}(2,\mathbb{C})$ such that $\phi(D) =(\mathrm{SL}(2,\mathbb{C}) \cap \mathrm{Sp}(n,{\mathbb{C})})|_ {\mathbb{Z}[i]}$, where $\mathbb{Z} [i]$ is the ring of Gaussian integers (see, for example, [19]). We choose the lattice $\Gamma$ such that the group $D/[D,D]$ is infinite (such groups are known to exist). Then the group $H^1(\Gamma, \mathbb{C})$ of one-dimensional cohomology with trivial coefficients will be non-trivial.

We set $\Gamma = D \times_\phi H_{\mathrm n}(\mathbb{C})|_{\mathbb{Z} [i]} \times \mathbb{Z}^2$, where $\mathbb{Z}^2$ is the standard integer lattice in $\mathbb{C}$. We construct an embedding of $\Gamma$ in $G$ as a uniform lattice.

Let $c \in H^1(D,\mathbb{C})$ be some non-trivial cohomology class. It can be viewed as a homomorphism $c\colon D \to \mathbb{C}$ whose image $\operatorname{Im}(c)$ is infinite. In this case, we can make it so that the subgroup $\operatorname{Im}(c)+\mathbb{Z}$ is not closed in $\mathbb{C}$.

For an element $\gamma=(d,h,z)\in D \times_\phi H_{\mathrm n}|_{\mathbb{Z}[i]} \times \mathbb{Z}^2$ we set $i({\gamma})=(d,h,z+c(d))\in \mathrm{SL}(2,\mathbb{C}) \times_\phi H_{\mathrm n}(\mathbb{C})\times \mathbb{C}$, where $d\in D,h\in H_{\mathrm n}|_{\mathbb{Z}[i]}, z\in \mathbb{Z}^2$. As is easy to understand, we obtain the embedding $i\colon \Gamma \hookrightarrow G$, and it is easily seen that $i(\Gamma)$ is a uniform lattice in $G$. It is obvious that $i(\Gamma)\cap S = i(\Gamma) \cap \mathrm{SL}(2,\mathbb{C})$ is not a lattice in $\mathrm{SL}(2,\mathbb{C})$. Further, from the constructions of the group $\Gamma,G$ and the embedding $i$, it is clear that $i(\Gamma) \cap [G,G]$ is not a lattice in $[G,G]$.

In this example, $S=\mathrm{SL}(2,\mathbb{C}), R=H_{\mathrm n}(\mathbb{C})\times \mathbb{C}$, that is, neither condition (i) nor condition (ii) in Theorem 3 are satisfied.

Theorem 3 has a corollary concerning the series of commutators $D^k(G)$, where $D^1(G)=[G,G]$, $D^{k+1}(G) =[D^k(G),D^k(G)]$.

§ 1. Introduction