V. V. Gorbatsevich, “On low dimensional bases of natural bundles for compact homogeneous spaces”, Izv. Math., 88:6 (2024), 1119

The article studies the bases of natural bundles, one of the elements of the fiber structure of compact homogeneous spaces; the consideration is carried out up to a finite covering. Note that arbitrary compact homogeneous spaces of dimensions $\leqslant 7$ were studied extensively and in detail by the author of the present paper. But here we will be interested not in the homogeneous spaces themselves and their dimensions, but only in the bases of natural bundles for them of small dimension ($\leqslant 5$). Homogeneous spaces with such bases can themselves have an arbitrarily large dimension.

We will denote Lie groups by capital Latin letters, and their Lie algebras, by the corresponding lowercase Latin letters. We will usually assume that Lie groups transitive on manifolds are connected (and sometimes simply connected), and their transitive actions are locally effective. By $\pi_k(\ast)$ we will denote homotopy groups. By a lattice in a Lie group, we will understand a discrete subgroup whose quotient space is compact (in other words, we will consider only uniform lattices). A smooth locally trivial fibration of a manifold $M$ over a base $B$ and with fiber $F$ will be denoted by $F\to M \to B$. A description of the properties of Lie groups and algebras we need can be found in [1]. Details about the structure and properties of homogeneous spaces can be found in [2], and the necessary information on lattices in Lie groups, in [3].

The bases $M_a$ of natural bundles of compact homogeneous spaces $M$ are aspheric compact manifolds (asphericity means that their homotopy groups $\pi_k$ are trivial for all $k \geqslant 2)$. Such manifolds $M_a$ are defined, up to homotopic equivalence (and often up to diffeomorphism, see [4]) by their fundamental group $\pi_1(M_a)$. Note that the fundamental group (considered up to an isomorphism) of a compact manifold is finitely presented (that is, the numbers of generators and relations are finite), and, therefore, the set of all such groups is countable, up to isomorphism. Therefore, the set of manifolds $M_a$ is countable (at least when considering them up to homotopy equivalence). In this article, special attention is paid to countability of some special classes of manifolds $M_a$.

We first present the definitions we need and formulate some results previously published by the author and used below.

Let $M$ be a compact manifold (all manifolds here are assumed to be finite dimensional and smooth). Suppose that there is a Lie group $G$ (always assumed here to be finite dimensional) that acts transitively on $M$. In this case, the manifold $M$ is called a homogeneous (with respect to the specified action of the Lie group $G$) space, and it can be represented as the quotient space $M=G/H$ of the Lie group $G$ with respect to some closed Lie subgroup $H \subset G $ (a stationary subgroup of some point from $M$).

Sometimes, when considering homogeneous spaces, one abstracts from specification of the corresponding transitive Lie group (some different Lie groups can act transitively and even effectively on the same manifold). In this case, we will talk about homogeneous manifolds — these being the manifolds on which there is a transitive action of some finite dimensional Lie group (not specified which one).

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 this Lie group to be simply connected and locally effective on $M$ (however, it will not always be effective). Let $K$ be some maximal compact subgroup of $G$. It is simply connected and semisimple (due to assumed simply connectedness of the Lie group $G$) and uniquely defined 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 all of them, as it turns out, will be of the same dimension. But they may have different orbital types (that is, the stationary subgroups corresponding to different points of the manifold $M$ will not always be conjugate in $K$).

An action of a compact group $K$ on a manifold $M$ is called equiorbital if all its orbits are of the same orbital type, that is, if, for all 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 is a compact homogeneous space $M' = G/H'$ (of the same Lie group $G$) covering it finitely (here $H'$ is some subgroup of finite index in $H$) and equiorbital with respect to the natural action of the maximal compact subgroup $K$. In this case, the orbital space $K \setminus M' $ of this action is a smooth manifold, and the natural map $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 fibrated, but the appropriate manifold $M'$ that finitely covers it). We denote the base of this bundle by $M_a$, and its fiber, by $M_c$.

In the general case, when constructing a natural bundle, the transition to $M'$ is necessary if we want to obtain the structure of a smooth manifold on $K \setminus M'$. Moreover, there is no canonical choice of a finite covering. However, the original manifold $M$ also has a kind of fibration — this will be a Seifert fibration 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.

So, the natural bundle $M_c \to M' \to M_a$ is a smooth locally trivial bundle for a suitable manifold $M'$ finitely covering the original compact homogeneous manifold $M$. This bundle over the manifold $M'$ is uniquely determined up to fiberwise homotopy equivalence.

The base of the natural bundle $M_a$ has the form $K\setminus M' $; this is an aspherical smooth manifold. Moreover, the universal covering manifold $M_a$ is diffeomorphic to a Euclidean space (note that not every aspherical manifold has this property, although the universal covering manifold for them will always be contractible). In this case, up to a diffeomorphism, the base of the natural bundle can be written (possibly again by passing to some finite covering) in the form of 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 a maximal compact Lie subgroup in $F$, and $\Gamma$ is a lattice (that is, a discrete subgroup with compact quotient space) in $F$ free from torsion. In this case, we can assume by passing, if necessary, to a subgroup of finite index in $\Gamma$ (which is equivalent to passing from $M_a$ to some manifold covering it finitely) that the intersection $\Gamma \cap C$ is contained in the center $Z(F)$ of the Lie group $F$. In addition, we can then assume that this intersection is trivial by factorizing the Lie group $F$ by it.

We will need the concept of commensurability of bases of natural bundles. In fact, this concept applies to arbitrary manifolds and even to topological spaces. Two topological spaces $X$ and $Y$ are said to be commensurable if there is a topological space $Z$ finitely covering both $X$ and $Y$. It is clear that commensurability is an equivalence relation. Any topological space that finitely covers some topological space is commensurable with it.

The concept of commensurability of groups is introduced similarly. We will call two groups commensurable if they contain subgroups of finite indices that are isomorphic to each other. It is clear that topological spaces are commensurable only if so are their fundamental groups. Note that sometimes commensurability in the study of lattices (and, in general, discrete subgroups in Lie groups) is understood as finiteness of the intersection indices of such groups in each of them, which is different from the definition used in the present paper.

Representation of the base $M_a$ of a natural bundle in the form $\Gamma \setminus F/C$ (where $C$ is a maximal compact subgroup in a connected Lie group $F$, and $\Gamma$ is a lattice in $F$, free from torsion) is called the standard representation of the base. The base of a natural bundle can have several different (and significantly different) standard representations.

The fiber of the natural bundle has the form $M_c = K/L$, where $L = K \cap H'$ is a stationary subgroup for some point $m_0 \in M'$ (it is a closed subgroup in the compact group Lee $K$). If we assume that the Lie group $G$ is simply connected (and we will usually do so), then the maximal compact subgroup $K$ will, as is well known, be simply connected, and, therefore, it will necessarily be semisimple. Hence $M_c$ is a homogeneous space of a compact semisimple Lie group. The fiber $M_c$ of the natural bundle is almost simply connected, that is, its fundamental group $\pi_1(M_c)$ is finite. An important special case is when this fiber is simply connected. A necessary and sufficient condition for the fiber $M_c$ to be simply connected (or, equivalently, for the subgroup $L = K \cap H'$ to be connected) is the condition that the fundamental group $\pi_1(M)$ of the manifold $M$ is torsion free (under the assumption that the action of $K$ on $M$ is equiorbital).

The structure group of a natural bundle is the compact Lie group $W\!=N_K (L)/L$, where $N_K(L)$ is the normalizer of the Lie subgroup $L$ in $K$. This Lie group $W$ (which is disconnected in general) can be considered as the group of all automorphisms of the homogeneous spaces $K/L$ (that is, diffeomorphisms that commute with the transitive action of the Lie group $K$). The natural action of the group $W$ on the fiber of the natural bundle is free.

The first section of this article is devoted to some general properties of bases of natural bundles. The second section describes bases of natural bundles of dimensions $\leqslant 3$, and in §§ 3 and 4, we describe the bases of dimensions 4 and 5, respectively. In § 5, we provide some information about compact homogeneous manifolds with a given base of a natural bundle.

§ 1. Structure of bases of natural bundles

The base of the natural bundle $M_a$, as mentioned above, has the form $K \setminus M' $ and it is an aspherical smooth manifold. Moreover, up to a diffeomorphism, the base of the natural bundle can be written (possibly again passing to some finite covering) in the form $\Gamma \setminus F/C$, where $F$ is some connected, although not necessarily simply connected, Lie group, $C$ is a maximal compact Lie subgroup in $F$, and $\Gamma$ is a torsion free lattice in $F$.

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

Proof. Let $M_c \to_p M \to M_a$ be some natural bundle for the homogeneous manifold $M$ (here, $p$ denotes its projection). Consider a finite covering $f: N \to M_a$ and a bundle over $N$ induced by the covering $f$. We denote the space of this bundle by $M'$. It is obviously finitely covers the original manifold $M$. Let us show that $M'$ is a homogeneous manifold, the manifold $N$ is the basis of its natural bundle, and its fiber is the same as the fiber of the original natural bundle for $M$.

Let $G$ be a simply connected Lie group transitive on the manifold $M$. It corresponds to the finite dimensional Lie algebra $L$ of vector fields on $M$. Due to the local diffeomorphism of the covering, these vector fields are uniquely transferred to $M^ \prime$, forming there a finite dimensional Lie algebra $L'$ isomorphic to $L$. The manifold $M'$ is compact (since it finitely covers the compact manifold $M$), and, therefore, the Lie algebra $L'$ generates a Lie group acting on $M'$. We can assume that this Lie group is simply connected, and, therefore, it will be isomorphic to the original Lie group $G$. In this case, the resulting action of the Lie group $G$ on $M'$ will be transitive, since all its orbits are obviously open (due to the local diffeomorphicity of the covering map) and, therefore, due to the connectedness of the manifold $M^ \prime$, there exists only one orbit, which coincides with $M'$. Thus, we have proved that the manifold $M'$ we constructed is homogeneous, and the same Lie group $G$is transitive on it as is transitive on the original manifold $M$. Moreover, it is easy to understand (using the construction of a natural bundle described above) that the manifold $N$ is diffeomorphic to the base of some natural bundle for $M'$. This proves the lemma.

As for the set of all commensurable bases of natural bundles, it is difficult to describe it completely, since the construction of a natural bundle includes a poorly controlled transition to a finitesheeted covering. Therefore, below we will limit ourselves to considering only representatives of each commensurability class of bases of natural bundles. Sometimes among these bases you can choose the ones that are in some sense minimal, that is, such bases that are not non-trivial coverings of some other bases. As for the maximal ones, they do not exist, since the transition to a subgroup of finite index in $\pi_1(M_a)$ will also give, by Lemma 1, some base of the natural bundle. Finding subgroups of finite and arbitrarily large indices in the case of homogeneous manifolds under consideration is always possible if the original homogeneous manifold has an infinite fundamental group.

For semisimple bases (when the solvable component degenerates to a point), we can assume that the Lie group $F$ in the standard representation is semisimple. But, as is known, in semisimple Lie groups there exist maximal lattices (and there may be several nonconjugate ones), and, therefore, there are minimal bases of natural bundles.

For bases $M_a$ with nilpotent fundamental group, it is easy to show that there is no minimal natural bundle among the bases. This follows from the connection between lattices in nilpotent Lie groups and the $\bf Q$-structure on the corresponding Lie algebras (established by A. I. Maltsev). As for the case of a solvable fundamental group, the situation is more complicated.

§ 2. Bases of natural bundles of dimensions $1$, $2$, and $3$

Here, we will describe the bases of natural bundles of dimensions 1, 2, and 3. Unlike higher dimensions, a very detailed description of such bases will be given.

For $\dim M_a = 1$, the situation is very simple. There is only one (up to diffeomorphism) compact one-dimensional manifold — the circle $S^1$. It itself is obviously a homogeneous manifold (and even a Lie group). Therefore, it is the only example of a one-dimensional base of a natural bundle. Moreover, it is a solvmanifold (and for it the semisimple component degenerates to a point).

The case $\dim M_a=2$ is somewhat more interesting. However, the result we need — a description of all bases of natural bundles — is already known (if this base is orientable).

Any compact two-dimensional manifold is uniquely determined (up to a diffeomorphism) by its orientability and genus $g$. Only two-dimensional compact surfaces of genus $g>0$ are aspherical.

If $g=1$, then the surface is diffeomorphic to the torus $T^2$ or to the Klein bottle $K^2$. The fact that the surface $T^2$ is homogeneous is obvious (it itself is a Lie group), but for $K^2$ this fact is less obvious. Moreover, in his classification of compact homogeneous surfaces, É. Cartan missed the Klein bottle. Its homogeneity was discovered only by Mostow in [5]. Let us briefly describe the representation of $K^2$ as a homogeneous space of the three-dimensional Lie group $E(2)_0$ of orientation preserving movements of the two-dimensional Euclidean plane. This Lie group is formed by the following matrices of order 3:

The manifold $K^2$ can be written as $\mathrm{E}(2)_0/H$, where the one-dimensional stationary subgroup $H$ consists of matrices of the form

Let us move on to consider other surfaces that claim to be bases of natural bundles. For simplicity, we will limit ourselves to considering only the orientable ones among them. Such a surface is uniquely determined, up to diffeomorphism, by its genus, which in our case should be $>1$. Such manifolds, unlike those discussed above, will never be homogeneous. But, despite this, they can be bases of natural bundles (and sometimes even of trivial bundles). This was first shown in [6] and then refined in [7].

Now we turn to the case $\dim M_a=3$, which, unlike the cases $\dim M_a=1,2$, was not considered in detail by the author earlier.

First, we need to identify those three-dimensional compact aspherical manifolds that can, in principle, be bases of natural bundles. And since here the number of candidates is very large (it, in particular, is associated with a large number of different lattices), now we will consider the bases $M_a$ only up to commensurability. In this case, the basis of our consideration will be the statement formulated above about the possibility of representing the base of a natural bundle (considered up to commensurability) in the form $\Gamma \setminus F/C$.

So, in what follows, we consider only the bases of natural bundles representable in the standard form $\Gamma \setminus F/C$ (where $C$ is the maximal compact subgroup in the connected Lie group $F$, and $\Gamma$ torsion free lattice in $F$).

For $\dim M_a=3$, there should obviously be $\dim F- \dim C = 3$. Let $F = S \cdot R$ be the Levi decomposition of the Lie group $F$ (here, $R$ is a radical, and $S$ is a semisimple part of the Lie group $F$).

Let us first assume that $S = \{e\}$, that is, the Lie group $F$ is solvable. Then we can assume (considering the bases of natural bundles up to commensurability) that $M_a = R/\Gamma$, where $R$ is a solvable three-dimensional simply connected Lie group, and $\Gamma$ is a lattice in $R $ (see, for example, [2]).

Before considering this case in detail, we prove one general statement concerning arbitrary compact manifolds of the form $R/\Gamma$, where $R$ is an arbitrary simply connected solvable Lie group (not just three-dimensional), and $\Gamma$ is a lattice in it.

Proof. Compact solvmanifolds $R/\Gamma$ are known to be determined, up to a diffeomorphism, by their fundamental group (isomorphic to $\Gamma$). Therefore, we will study precisely these fundamental groups. Consideration of manifolds $M$ up to commensurability in this case is equivalent to that of lattices $\Gamma$ up to commensurability.

To begin with, let us note the well-known fact that lattices in solvable simply connected Lie groups are polycyclic groups, that is, they have a subnormal number of subgroups with cyclic factors. The number of lattices $\Gamma$ among them, considered up to isomorphism, is certainly at most countable (although it is easy to see that it is infinite, and, therefore, countable). But we consider lattices up to commensurability, and in principle it could turn out that, among the countable number of lattices $\Gamma$, that are pairwise incommensurable, there is only a finite number. We will show that this is not so. Moreover, we will first consider lattices in three-dimensional simply connected solvable Lie groups, and the general statement for an arbitrary dimension $\geqslant 3$ is easily deduced from this fact.

We will now consider groups that are represented as a semidirect product ${\mathbf{Z}}\times_{\phi} {\mathbf{Z}}^2$. The semidirect product here is uniquely determined by the homomorphism $\phi \colon {\mathbf{Z}}\to \mathrm{GL} (2, \mathbf{Z})$, which, in turn, is uniquely determined by the element $\phi(1)\in \mathrm{GL}((2,\mathbf{Z})$, that is, an integer invertible second order matrix over $\mathbf{Z}$, which we denote by $A$. We will be interested in two invariants of such matrices: the trace $t\operatorname{tr} A)$ and the determinant $\det(A)$. As for the determinant here, it is only $+1$ or $1$. Passing from $A$ to $A^2$ (which is equivalent to passing from $\Gamma$ to its subgroup of index 2, which does not change the commensurability class), we can restrict ourselves to the case $\det(A)=1$.

Let us now consider the role of the trace of the matrix $A$, which we denote as a parameter by $t$ ($t \in \mathbf{Z})$. It is clear, due to the restriction on the determinant that we introduced, that the characteristic polynomial of the matrix $A$ has the form $\lambda^2 - t\lambda+1$. For what follows, it is sufficient for us to consider only the cases $t \in \mathbf{N}$ and $t >2$ (the latter condition, as is easy to understand, ensures that the roots are irrational).

Consider the quadratic extension of the field $F$ of rational numbers generated by the adjunction of $\sqrt {t^2-4}$. For brevity, we put $s=t^2-4$. The elements of the field $F$ have the form $a+b\sqrt s$, where $a,b\in \mathbf{Q} $. The characteristic roots $\lambda_1, \lambda_2$ of the matrix $A$ belong to the field $F$, and they are mutually inverse (that is, $\lambda_1\cdot \lambda_2=1$). It is clear that in the representation $\lambda_1=a+b \sqrt s$ for $t>2$ the coefficients $a,b$ are non-zero (we do not need more from them here).

We will need one result related to quadratic fields. It is closely related to Dirichlet’s theorem on the units of a quadratic field.

Proof. It is clear that the expression $(a_1+b_1 \sqrt p)^m$ can be written as $a_{1,m}+b_{1,m} \sqrt p$, where $a_{1,m},b_{ 1,m}$ are some rational numbers. Let us show that $b_{1,m} \ne 0$.

Let us assume that this is not the case. Then $(a_1+b_1 \sqrt p)^m = a_{1,m} \in \mathbf{Q}$ is some rational number. Therefore, we have the relation $(a_1+b_1 \sqrt p) = a_{1,m}^{1/m}$. Thus, we have a linear relationship with rational coefficients between $1,\sqrt p$ and the root of degree $m$ of some integer $c$ (it is obtained based on the number $a_{1,m}^{1/m}$), and the coefficients of 1 and $c$ are non-zero. This kind of relationship has been studied in number theory. We will mention only one of the oldest works on this topic — an article by Besicovitch [8]. From the content of this article it follows that in our case there should be $m=2$, and, what is most important for us, $a_1=0$. But this contradicts our assumption. Thus, we have proved that $b_{1,m} \ne 0$. A similar expansion takes place, of course, for $(a_2+b_2 \sqrt q)^m$, but here we do not need to prove here that any coefficients are different from zero.

As a result, from the original equality, due to $m=n=2$, we obtain an equality of the form $a_{1_m}+b_{1,m} \sqrt p = a_{2,n}+b_{2,n} \sqrt q $. By [8], we have $p=q$, since the numbers $p,q$ are both natural and free from squares. This proves the lemma.

Let us now consider a countable set of groups of the form ${\mathbf{Z}} \times_{\phi} {\mathbf{Z}}^2$ defined by homomorphisms $\phi$ such that the corresponding values of the parameters $t=\operatorname{tr} \phi(1)$ will be different and they will all be natural numbers free from squares and greater than 2. That is we are talking about the numbers $t=3,5,6,7,10,11 \dots$ . We denote the set of such natural numbers by $T$. For each such $t \in T$, we construct the corresponding group $\Gamma_{t}= {\mathbf{Z}}\cdot {\mathbf{Z}}^2$ as a semidirect product of $\mathbf{Z}$ and ${\mathbf{Z}}^ 2$ using the matrix $A$ corresponding to the parameter $t$. We obtain a countable number of groups, each of which is a lattice in some three-dimensional simply connected solvable Lie group $R_{t}$. Let us describe this Lie group.

We fix the parameter $t$. The corresponding eigenvalues of our matrices $\phi(1)$ are real and positive (since we have $t>2)$. But then the matrix $\phi(1)$, as is easy to understand, will be logarithmable, that is, it lies in the image of the exponential map $\operatorname{exp}\colon \mathrm{gl}(2,\mathbf{R}) \to \mathrm{GL}(2, \bf R) $. Let $\phi(1) = \operatorname{exp} (X)$, where $X \in \mathrm{gl}(2, \mathbf{R})$. Consider the one-parameter subgroup $\phi(\alpha) = \operatorname{exp} (\alpha X)$ (here $\alpha$ is a parameter) and with its help we form the semidirect product $R_{t} = \mathbf{R}\times_{ \phi} \mathbf{R}^2$. We obtain a three-dimensional solvable simply connected Lie group. Moreover, taking $\alpha \in \mathbf{Z}$, we obtain an embedding of the group $\Gamma_{t}$ into $R_{t}$ as a lattice. It generates the manifold $M_{t}$ that we need.

Let us now show that the resulting three-dimensional compact solvmanifolds $M_t$ are pairwise incommensurable. To do this, it is enough to prove that the corresponding lattices $\Gamma_{t}$ will be incommensurable.

Consider the matrices $A_t$ generating the groups $\Gamma_{t} $. By Lemma 1, any of their degrees will not be equal to each other if $t \in T$. But then the corresponding groups $\Gamma_{t}$ cannot be commensurable.

The solvmanifolds $M_t$ that we have constructed are themselves homogeneous manifolds, and, therefore, provide examples of bases of natural bundles. Moreover, as shown above, they are pairwise incommensurable and their number is countable.

Now let us recall that the proposition we prove speaks about the arbitrary dimension $\geqslant 3$ of the manifolds we are considering. But it is very easy to go from the three-dimensional case to the $n$-dimensional one. It is enough to take the direct product of two solvmanifolds (one of the manifolds $M_{t}$ and a torus $T^{n-3}$ of dimension $n-3$). We obtain, as is easily seen, a countable number of the solvmanifolds we need, which are pairwise incommensurable with each other. This completes the proof of Proposition 1.

Note that the restriction $n \geqslant 3$ on the dimension of the manifold in Proposition 1 is not accidental. As shown above, for dimensions 1 and 2, there is not only a finite, but even only equal to 1, number of pairwise incommensurable bases of natural bundles with a solvable fundamental group. Namely, these are the manifolds $T^1, T^2$, respectively (the one- and two-dimensional tori).

Now suppose that the Lie group $F$ is not solvable, that is, its Levi factor $S$ is non-trivial.

We need the following fact: the codimension of a maximal compact Lie subgroup $C$ in a non-compact semisimple Lie group $S$ is always $\geqslant 2$. This statement can be easily derived from the classification of simple non-compact Lie groups, although in principle it is not difficult to prove it directly. Moreover, the indicated codimension for a simple non-compact Lie group is equal to 2 only if this Lie group is commensurable with the Lie group $\mathrm{SL}(2, \mathbf{R})$. Or, in another way, such a Lie group is a finite covering over the simple Lie group $\mathrm{PSL}(2, \mathbf{R})$. Further, the indicated codimension is 3 for a simple Lie group $S$ only in two cases: when this simple Lie group is isomorphic to $\mathcal A= \widetilde {\mathrm{SL}(2,\mathbf{R})}$ (the universal covering for the three-dimensional Lie group $\mathrm{SL}(2, \mathbf{R})$; here $C=\{e\}$) or is it commensurable with $\mathrm{SL}(2, \mathbf{C})$ (the Lie group of dimension 6 in which the maximal compact subgroup is three-dimensional and is isomorphic to $\mathrm{SU}(2)$). Let us consider these possibilities in more detail.

Suppose that the Lie group $F$ is not semisimple. Then if $R$ is its radical, so the codimension of the intersection $C\cap R$ must be $\leqslant 1$. If this codimension is 0, then the compact Lie group has an Abelian compact normal subgroup, by which we can factorize the Lie group $F$. If this codimension is 1, then $R$ is a solvable Lie group having a compact subgroup of codimension 1. This subgroup, as is easy to understand, is normal in the radical and after factorization by it we find that the radical of the Lie group $F$ must be one-dimensional, and, therefore, the Lie group $F$ will be commensurable with $\mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}$.

If the Lie group $F$ is isomorphic to $\mathcal A$, then the corresponding base $M_a$ of the natural bundle is diffeomorphic to $\mathcal A/\Gamma$, and the description of all possible manifolds $M_a$ is reduced to consideration of lattices in the Lie group $\mathcal A$. Moreover, this consideration should be done up to commensurability of these lattices. In this case, the structure bundle for $M_a$ will have the form $S^1\to M_a\to F_g$, where the base $F_g$ of the structure bundle is an orientable compact surface of genus $g \geqslant 2$. In our case, it is enough to assume that $g=2$, that is, the bundle is over the pretzel. Such a bundle is given by its characteristic class $c\in H^2(F_2, \mathbf{Z}) = \mathbf{Z}$ , that is, some integer $c$. When considering the manifolds $M_a$ up to commensurability, we can assume that $c$ is 0 or 1. For $c=0$, we find that the space of such a bundle is commensurable with the manifold $S^1 \times F_2$ (here, it is a byproduct of our consideration, since it does not have the form $\mathcal A/\Gamma$). As the author showed a long time ago, the manifold $F_2$ can already be the basis of a natural bundle (see above). Therefore, so is $F_2\times S^1$. If the Lie group $F$ is isomorphic t $\mathcal A$, then, up to commensurability, there exists only one three-dimensional base of natural bundles. Let us denote it by $S^1(F_2)$ (this is the space of a non-trivial bundle on the circle over $F_2$).

The next possibility here is if the Lie group $F$ is locally isomorphic to $\mathrm{SL}(2,\mathbf{C})$. Here, we need to consider (up to commensurability) the lattices $\Gamma \subset \mathrm{SL}(2,\mathbf{C})$. In this case, the manifolds $M_a$ can be represented as factors of the three-dimensional hyperbolic space $H^3 = \mathrm{SL}(2, {\mathbf{C}})/ \mathrm{SU}(2)$, since the group $\mathrm{SL}_2(\mathbf{C})$ is locally isomorphic to the pseudo-orthogonal group $\mathrm{SO}(1,3)$. So, the study of manifolds $M_a$ in this case reduces to that of compact forms of three-dimensional hyperbolic space, considered up to commensurability. In this case, it is known that there exists a countable number of pairwise incommensurable lattices (considered up to conjugation) — see [9]. Therefore, the set of corresponding hyperbolic manifolds $M_a$, considered up to commensurability is also countable. In this case, as is known, it is possible to quite explicitly describe all arithmetic lattices $\Gamma \subset \mathrm{SO}(1,3)$.

It remains to consider the case where $F$ is commensurable with the Lie group $\mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}$. In this case, it turns out that the corresponding manifold $\Gamma \setminus F/C$ is commensurable with the manifold $F_2\times S^1$. This follows from the results of G. Mostow, which will be discussed below (see Proposition 2). We note, however, that for, the particular case now being considered, it is not difficult to give a direct, rather short proof.

Note that, in the case $\dim M_a=1$, the manifold $M_a$ coincides with $M_r$, which is the solvable component of the structure bundle. In the case, $\dim M_a=2$, the base $M_a$ is commensurable with either $M_r$ (the torus $T^2$) or with a semisimple component $M_s$ (diffeomorphic to $F_2$). For $\dim M_a=3$, either $M_a=M_r$ or $M_a= M_s$ (this will be for hyperbolic manifolds) or $M_r=S^1, M_s=F_2$,; in this case, $M_a$ is a base of general form (with non-trivial solvable and semisimple components).

In the study of the bases of natural bundles, it is useful to distinguish the indecomposable ones among them. We will call the base of a natural bundle (considered, as usual, up to commensurability) decomposable if it is commensurable with the direct product of two non-trivial (that is, positive dimension) bases of natural bundles.

Proof. To begin with, we note that a compact solvmanifold of dimension $\leqslant 3$ is decomposable only if it is commensurable with a torus. For example, for a three-dimensional compact solvmanifold, its decomposability would mean that it is commensurable with the direct product of a one-dimensional and two-dimensional manifold, which are the bases of natural bundles. But from Theorem 1 it is clear that then the two-dimensional manifold here is a torus (since $F_2$ cannot appear here due to the non-solvability of its fundamental group), and, therefore, we find that the original solvmanifold is commensurable with the torus. For smaller dimensions, a similar statement is obvious

The indecomposability of the surface $F_2$ is obvious.

By Theorem 1, we only have to consider the bases of natural bundles having the form $\Gamma \setminus \mathrm{SL}(2,{\mathbf{C}})/\mathrm{SU}(2)$. If such a manifold is decomposable, then it must be commensurable with the direct product of a two-dimensional manifold and a circle. Moreover, it is obvious that the indicated two-dimensional manifold must be commensurable with $F_g$. Hence the fundamental group $\pi_1(\Gamma \setminus \mathrm{SL}(2,{\mathbf{C}})/\mathrm{SU}(2))$ must have a subgroup of finite index whose center is isomorphic to $\mathbf{Z}$. However, for lattices in $\mathrm{SL}(2, \mathbf{C})$ the center is known to be finite. For example, this follows from the fact that if $\mathrm{SL}(2,\mathbf{C})$ is considered as an algebraic simple group, then the algebraic closure of the lattice in it must coincide with this group. But the center of the group $\mathrm{SL}(2,\mathbf{C})$ is finite. This proves the corollary.

Note that when we talk about the indecomposability of a manifold, we mean the absence (up to commensurability) of the decomposition of the manifold into a direct sum of manifolds of lower dimension, which themselves are the bases of some natural bundles. However, it is not difficult to show that the manifolds indicated in Corollary 1 are indecomposable in a stronger sense — they are, up to commensurability, indecomposable into the direct product of any manifolds.

§ 3. Bases of natural bundles of dimension $4$

Starting from dimension 4, the description of the bases of natural bundles (even considered up to commensurability) becomes noticeably more complicated.

Here, of course, there is also a countable number of pairwise incommensurable bases $M_a$ of natural bundles. We will indicate the general form of such manifolds, but it is very difficult to give a detailed description of all of them even up to commensurability.

Theorem 2. The base $M_a$ of the natural bundle for $\dim M_a=4$ is commensurable with one of the following manifolds.

a) Solvable ones: solvmanifolds of the form $R/\Gamma$, where $R$ is some solvable simply connected Lie group of dimension 4, and $\Gamma$ is a lattice in it. The set of such manifolds $M_a$, considered up to commensurability, is countable.

b) Semisimple ones:

- (i) $F_2 \times F_2$;
- (ii) of the form $\Gamma \setminus S/C$, where $S$ is a semisimple Lie group, $C$ is a maximal compact subgroup in $S$, and $S$ is commensurable with $\mathrm{SL}(2, \mathbf{R} ) \times \mathrm{SL} (2,\mathbf{R})$ (here, the lattice $\Gamma $ is irreducible) or with $\mathrm{SO}(1,4)$.

c) Of the general form: $\Gamma \setminus F/C$, where the Lie group $F$ is commensurable with $\mathcal A\times \mathbf{R}$ or with $\mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}^2$ or with $\mathrm{SL}(2,\mathbf{C}) \times \mathbf{R}$, and $\Gamma$ is a lattice in $F$.

Proof. The proof is carried out according to the same scheme as the consideration of the case $\dim M_a=3$ in the proof of the above Theorem 1.

We note only one special case where the Lie group $F$ could be commensurable with the semidirect product $\mathrm{SL}(2, \mathbf{R}) \times_{\tau} \mathbf{R}^2$ (here, $\tau$ is the tautological representation in $\mathbf{R}^2$). This Lie group has lattices, but not uniform ones, only those whose quotient space has finite measure. Let us prove this result.

Suppose that $\Gamma$ is some uniform lattice in $\mathrm{SL}(2, \mathbf{R}) \times_{\tau} \mathbf{R}^2$. Here, the semisimple part of the Lie group is isomorphic to $\mathrm{SL}(2,\mathbf{R})$, and the radical is Abelian and isomorphic to $\mathbf{R}^2$. Moreover, $\Gamma \cap \mathbf{R}^2$ is a lattice in $\mathbf{R}^2$ (it is isomorphic to ${\mathbf{Z}}^2$). The factor group $\Gamma /\Gamma \cap \mathbf{R}^2$ is a lattice in $\mathrm{SL}(2,\mathbf{R})$, and in its action on $\mathbf{R}^2$ preserving the lattice ${\mathbf{Z}}^2\subset \mathbf{R}^2$. But the subgroup in $\mathrm{GL}((2,\mathbf{R})$ that preserves the lattice in $\mathbf{R}^2$ is conjugate to the subgroup of integer matrices in $\mathrm{GL}((2,\mathbf{R})$, and this subgroup is not uniform lattice. This proves the theorem.

Proof. Let us start with indecomposability of the manifolds indicated in cases a) and b).

For compact solvmanifolds, their decomposability into a direct product of some manifolds of lower dimension is equivalent to their decomposability into a non-trivial product of the fundamental group of this manifold (see [10]). This proves, as is easy to understand, case a) of this corollary.

As for case b), decomposability here means representation as a direct product of a manifold of dimensions 1 and 3 or 2 and 2. The impossibility of such decompositions easily follows from the properties of lattices in semisimple Lie groups, and in the case of the manifold $\mathrm{SL}(2,\mathbf{R}) \times \mathrm{SL}(2,\mathbf{R})$ it is necessary to take into account that the lattices here are assumed to be irreducible.

We now proceed with decomposability of all other manifolds from Theorem 2. For solvmanifolds, we again use [10]. It remains to consider manifolds for which the Lie group $F$ is commensurable with $\mathcal A\times \mathbf{R}$ or with $ \mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}^2$.

We will deduce the commensurability of the corresponding manifolds from the following more general statement (which we will also use below when considering bases of dimension 5).

Proposition 2 (see [11] and [12]). Let $G$ be a connected Lie group and $G=S \cdot R$ be its Levi decomposition (where $S$ is its semisimple part and $R$ is its radical). Suppose that the semisimple Lie group $S$ has no compact factors. Next, let $\Gamma$ be some lattice in $G$.

Assume that one of the following conditions is met:

(i) the Lie group $G$ is simply connected;
(ii) the Lie group $G$ is almost linear (that is, it has a finite dimensional linear representation with finite kernel).

Then the group $\Gamma$ contains a subgroup $\Gamma_1$ of finite index, smoothly deformable (inside $G$) into a lattice $D \subset G $, for which there is a decomposition $D= (D\cap S) \cdot ( D\cap R)$ into the (almost) semidirect product of the lattice $D\cap S$ and of the normal subgroup $D\cap R$, which is a lattice in $R$. In addition, the manifolds $G/\Gamma_1$ and $G/D$ are diffeomorphic.

When we talk about an “almost semidirect” product, we mean the requirement that the intersection of the factors be discrete (or even finite).

This statement applies here to Lie groups $F$ commensurable with linear or simply connected Lie groups. In fact, it is also true in more general cases, but we will not dwell on this here and will consider only the special case we need here.

Now let the group $S$ be neither simply connected nor almost linear. In what follows, we will only need the case when the Lie group $S$ is isomorphic to $\mathcal A\times \mathrm{SL}(2,\mathbf{R})$. Its center is isomorphic to ${\mathbf{Z}} \times {\mathbf{Z}}_2$ and it does not have exact finite dimensional linear representations. Here, we will limit ourselves to considering only Lie groups of the form $G= \mathcal A\times \mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}^n$ (since it is precisely these that we will need below).

So, consider the Lie group $\mathcal A\times \mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}^n$ and the lattice $\Gamma$ in it. Since the Lie group $\mathcal A\times \mathrm{SL}(2,\mathbf{R})$ has no compact factors, the intersection $\Gamma \cap \mathbf{R}^n$ is a lattice in $\mathbf{R}^n $ (isomorphic to ${\mathbf{Z}}^n$). Let us now consider the commutator subgroup $[\Gamma,\Gamma]$. It is clear that it is contained in $S$. In this case, the center $Z(\Gamma)$ is almost entirely contained in $[\Gamma,\Gamma]$. But then from the Lie group $G$ we can, factoring by $Z(S)$, go to the linear Lie group $G^\ast = \mathrm{PSL}(2,\mathbf{R}) \times \mathrm{PSL}(2,\mathbf{R}) \times \mathbf{R}^n$ containing the lattice $\Gamma/\Gamma \cap Z(S)$. Moreover, the Lie group $G^\ast$ is linear, and, therefore, we can apply the result from [12] already mentioned above. Hence the manifold $\mathcal A\times \mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}^n / \Gamma$ will be commensurable with $S/D \times \mathbf{R}^n/{\mathbf{Z}}^n$.

Let us continue the proof of Corollary 2. From Proposition 2 it follows, in particular, that, for the Lie groups $\mathcal A\times \mathbf{R}$, $\mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}^2 $ and $\mathcal A\times \mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}$, the corresponding manifolds $G/\Gamma$ is decomposable. This completes the proof of Corollary 2.

§ 4. Bases of natural bundles of dimension $5$

Here, in contrast to the previous sections of the article, only a fairly general description of the bases of natural bundles will be given. This is due to the fact that for dimension 5 such a description becomes quite cumbersome, especially in the solvable case.

Theorem 3. The base $M_a$ of the natural bundle for $\dim M_a=5$ is commensurable with one of the following manifolds.

a) Solvable manifolds: solvmanifolds of the form $R/\Gamma$, where $R$ is a solvable and simply connected Lie group of dimension 5, and $\Gamma$ is a lattice in it. The set of such manifolds $M_a$, considered up to commensurability, is countable.

b) Semisimple manifolds: they are of the form $\Gamma \setminus S/K$, where $S$ is a semisimple Lie group, which is commensurable with one of the following Lie groups:

$$ \begin{equation*} \mathrm{SU}(1,2), \quad \mathrm{SL}(3,\mathbf{R}), \quad \mathrm{SO}(1,5), \quad \mathrm{SL}(2,{\mathbf{C}}) \times \mathrm{SL}(2,\mathbf{R}), \quad \mathcal A\times \mathrm{SL}(2,\mathbf{R}). \end{equation*} \notag $$

c) Of the general form: $\Gamma \setminus F/Q$ ($Q$ is a maximal subgroup in $F$, $\Gamma$ is a lattice in $F$), where the Lie group $G$ is commensurable with one of the following Lie groups:

$$ \begin{equation*} \begin{gathered} \, \mathcal A\times \mathbf{R}^2, \quad \mathrm{SL}(2, \mathbf{R}) \times R,\quad\mathrm{SL}(2, {\mathbf{C}}) \times \mathbf{R}^2, \quad \mathrm{SL}(2, \mathbf{R}) \times_{\mathrm{Ad}} \mathbf{R}^3, \\ \mathrm{SO}(1,4) \times \mathbf{R}, \quad\mathrm{SL}(2,\mathbf{R}) \times \mathrm{SL}(2,\mathbf{R}) \times \mathbf{R}, \end{gathered} \end{equation*} \notag $$

where $R$ is a simply connected three-dimensional Lie group having a lattice (see above for more details), and, for a Lie group $\mathrm{SL}(2, \mathbf{R}) \times _{\mathrm{Ad}} \mathbf{R}^3$, $\operatorname{Ad}$ denotes the adjoint representation (three-dimensional) of a Lie group on its Lie algebra.

Proof. The list of Lie groups $F$ appearing here is obtained by the same method that was used for lower dimensions.

Namely, we first select semisimple Lie groups that can give (in combination with some radical) the manifolds $M_a$ we need. Next, we add a possible form of maximal compact subgroups. Even for one simple Lie group, locally isomorphic Lie groups can have different dimensions of maximal compact subgroups. The simplest example is given by three-dimensional simple non-compact Lie groups. All of them are locally isomorphic to $\mathrm{SL}(2,\mathbf{R})$, but, for example, for $\mathrm{SL}(2,\mathbf{R})$ itself, the maximal compact subgroup is one-dimensional, and for $\mathcal A$ (the universal covering for $\mathrm{SL}( 2,\mathbf{R}$)), the maximal compact subgroup is trivial (and the Lie group $\mathcal A$ itself is diffeomorphic to $\mathbf{R}^3$).

As for the radical $R$, it must be $\dim R \leqslant 3$. We will show that the radical $R$ must be Abelian. And then it will be isomorphic to $\mathbf{R}$ or $\mathbf{R}^2$ (and in these two cases, the Levi decomposition will be direct) or $\mathbf{R}^3$. Note that in our cases the semisimple part of the Lie group $F$ does not have compact factors, and, therefore, $\Gamma \cap R$ will be a lattice in $R$. In particular, this implies that the radical $R$ is unimodular.

For $\dim R=1$, it is clear that the radical $R$ is isomorphic to $\mathbf{R}$, and the Levi decomposition is direct. For $\dim R=2$, there is only one simply connected unimodular Lie group (isomorphic to $\mathbf{R}^2$). But by proceeding as in the case $\dim M_a=4$ (Theorem 2), we find that the action of the Lie group $S$ on the radical must be trivial, and, therefore, the Levi decomposition will be direct.

Let us now consider the case $\dim R = 3$. From $\dim M_a=5$ it follows that $S$ will be commensurable with $\mathrm{SL}(2, \mathbf{R})$. Let us consider the action of the Lie algebra $L(S)$ on the Lie algebra $L(R)$ by conjugations. If this action is irreducible, then it is adjoint and the Lie group $R$ is necessarily Abelian. This case will be discussed below. If it is reducible, then it is trivial or is the direct sum of the trivial one-dimensional and the tautological representations. The last possibility, as in the proof of Theorem 2, is impossible. So, we obtain here the triviality of the Levi decomposition for the Lie group $F$. Let us now consider the case of a non-trivial action.

Let us assume that the three-dimensional radical is non-abelian. Then it is nilpotent (and then $\dim [R,R] =1$) or if it is nonnilpotent, we have $\dim[R,R]=2$. Let us consider both these cases.

Let the radical $R$ be nilpotent. Then $\dim [R,R] = 1$ and hence the Lie group $[R,R]$ coincides with the center $Z(R)$ and is isomorphic to $\mathbf{R}$. Since $\Gamma \cap R$ is a lattice in $R$, its intersection with $[R,R]$ will be a lattice (isomorphic to $\mathbf{Z})$. Consider the two-dimensional Abelian quotient group $R/Z(R)$ and in it the lattice $(\Gamma \cap R)/ (\Gamma \cap Z(R))$. The arguments given above in the proof of Theorem 2 in a similar case show that the Levi decomposition for the Lie group $R/Z(R)$ will be direct. But then it is easy to understand that the Levi decomposition for the Lie group $R$ must also be direct. But it is precisely such Lie groups that are listed in case c) in the theorem.

Let us now assume that the three-dimensional radical of a Lie group is not nilpotent. Then $[R,R]$ will be a nilradical, and, therefore, $\Gamma \cap [R,R]$ is a lattice in the two-dimensional Abelian Lie group $[R,R]$. An argument similar to that given when considering the previous case (when $R$ is nilpotent) again leads to the fact that the Levi decomposition for $F$ is direct.

For almost all Lie groups $G$listed in this theorem, the presence of lattices $\Gamma$ in them is beyond doubt. For semisimple Lie groups, we can take arithmetic (uniform) lattices, which, as is known, always exist. If the Lie group $G$ decomposes into the direct product of a semisimple Lie group and a simply connected Abelian group (for example, for $\mathcal A\times {\mathbf{R}^2}$), the required group can be obtained by the direct product of a lattice in a semisimple Lie group and an integer lattice in an Abelian simply connected Lie group. But the presence of a lattice in the Lie group $\mathrm{SL}(2, \mathbf{R}) \times _{\mathrm{Ad}} {\mathbf{R}^3}$ requires an explanation. Let us prove that lattices exist in this Lie group. It is important for us to obtain exactly a uniform lattice (but it is very easy to construct a lattice of finite covolume, starting from the lattice $\mathrm{SL}(2, \mathbf{Z})$ in $\mathrm{SL}(2, \mathbf{R})$). For this we need some information from number theory. For completeness of presentation, we present all the arguments in detail.

First, let us consider some integer quadratic form in three variables that does not represent 0 (for integer arguments). As a specific example, let us choose the form $x^2+2y^2+5z^2$.

Proof. Let us assume that the lemma is false and consider some integer equality $l^2+2m^2+5n^2=0$. Integers $l,m,n$ can be assumed not to have a common non-trivial factor (because otherwise we can divide by it).

From $l^2+2m^2+5n^2=0$ it follows that $l^2+2m^2\equiv 0 \mod 5$.

We first assume that $l$ is not 0 mod 5. Then $l$ is invertible mod 5, and hence, for $q=ml^{-1}$ there will be $2q^2 \equiv 1 \mod 5$. But this is impossible: for $q = 0, \pm 1,\pm 2$ (the full set of residues mod 5) we have, respectively, $2q^2 \equiv 0,2,3 \mod 5$. Therefore, for non-zero $l$ mod 5, we come to a contradiction with $2q^2 \equiv 1 \mod 5$. Now let $l \equiv 0 \mod 5$. Then, obviously, from $l^2+2m^2\equiv 0 \mod 5$ it follows that $m \equiv 0 \mod 5$. But then from $l^2+2m^2+5n^2=0$ it follows that $n \equiv 0 \mod 5$. As a result, we find that here all three integers $l,m,n$ are divisible by 5, which contradicts the assumption we made about the absence of a non-trivial common factor for these three numbers. The lemma is proved.

Let us now proceed with a direct construction of a lattice in $\mathrm{SL}(2, \mathbf{R}) \times _{\mathrm{Ad}} \mathbf{R}^3$. Consider the subgroup $\mathrm{SO}(Q)$ in $\mathrm{GL}((3,\mathbf{R})$ preserving the form $Q= x^2+2y^2+5z^2$. This subgroup is isomorphic to the Lie group $\mathrm{SO}(1,2)$, which in turn is isomorphic to $\mathrm{SL}(2, \mathbf{R})$. We need to find how this subgroup is embedded in $\mathrm{SL}(3, \mathbf{R})$. To do this, consider three dimensional representations of the group $\mathrm{SL}(2,\mathbf{R})$. There are only two such representations (non-trivial, up to equivalence) one reducible (the direct sum of the standard tautological two dimensional representation and the trivial representation) and the adjoint $\mathrm{Ad}$. In the construction we are considering, the first case is impossible for example, due to the fact that then the centralizer of the subgroup $\mathrm{SO}(Q)$ would be non-trivial, which, as is easy to understand, is not true in our case. Therefore, the embedding of our group $\mathrm{SO}(Q)$ is obtained using the adjoint representation $\mathrm{Ad}$.

By the above lemma the form $Q$ does not represent 0, and so the group $\mathrm{SO}(Q)_{\mathbf{Z}}$ of integer points will, as is known, be a uniform lattice in the group $\mathrm{SO}(Q)$ (isomorphic to $\mathrm{SL} (2, \mathbf{R})$). Taking the standard integer lattice ${\mathbf{Z}}^3 \subset \mathbf{R}^3$, consider the corresponding lattice $\mathrm{SO}(Q)_{\mathbf{Z}} \times_{\mathrm{Ad}} {\mathbf{Z}}^3$. This completes the construction of the required lattice in $\mathrm{SL}(2, \mathbf{R}) \times _{\mathrm{Ad}} \mathbf{R}^3$. The theorem is proved.

Among the manifolds listed in the theorem, we can distinguish those that are incommensurable with direct products. As can be verified, these will be manifolds of the form $\Gamma \setminus F/Q$ for which the Lie group $F$ is either solvable and incommensurable with the direct product of two Lie groups, or the Lie group $F$ is commensurable with $\mathrm{SU}(1,2)$, $\mathrm{SL}(3,\mathbf{R})$, $\mathrm{SO}(1,5)$ or with $\mathcal A\times \mathrm{SL}(2,\mathbf{R})$ (provided that the lattice $\Gamma$ in this case will be irreducible). The fact that the remaining manifolds are decomposable (up to commensurability) may be proved in the same way as was done in the previous section, with the addition of reasoning after the proof given there.

§ 5. Homogeneous spaces with a given base of the natural bundle

In this section, we will consider the following question: which compact homogeneous manifolds $M$ correspond to the bases $M_a$ of the dimensions considered above (as well as more general ones)?

Note that, for $\dim M_a=1$, the base $M_a$ is solvable (that is, $M_a$ coincides with its solvable component), and, for $\dim M_a=2$, the base $M_a$ is either solvable or semisimple. Compact homogeneous manifolds with a base of mixed type appear only when $\dim M_a = 3$: these are the manifolds $M_a$ that have a standard form $\mathcal A/\Gamma$ or are commensurable with $F_2 \times S^1$. Naturally, at higher dimensions the number of bases of each of the three different types increases noticeably.

Let us start with the case when the base of the natural bundle is solvable. Then it has a standard representation of the form $R/\Gamma$, where $R$ is a solvable simply connected Lie group, and $\Gamma$ is a lattice in it. In this case, we will not need to impose dimensional restrictions on $M_a$ here. In [10], for any compact manifold of the form $R/\Gamma$ (considered up to commensurability), all possible compact homogeneous manifolds with such a base of the natural bundle are described. In particular, any manifold of the form $K/L$ can appear here as a fiber of a natural bundle, where $K$ is some compact semisimple Lie group, and $L$ is its closed subgroup. Since the article [10] is currently difficult to access, we will briefly present here some details about homogeneous manifolds with a given base of the form $R/\Gamma$.

Let some compact solvmanifold $M_a=R/\Gamma$ be given. Additionally, we need to define several objects: the homogeneous space $M_c=K/L$ (the assumed fiber of the natural bundle, whose base is commensurable with $M_a$), where $K$ is some compact simply connected (and, therefore, semisimple) Lie group, and $L$ is a closed Lie subgroup of $K$. Let us further put $F =N_K(L)/L$ ($N_K(L)$ is the normalizer of the subgroup $L$ into $K$). Here, $F$ is some compact Lie group. Let us choose some torus $T$ (that is, a compact connected Abelian subgroup) in $F$ and denote its dimension by $n$. And finally, we choose and fix some cohomology class $c \in H^2(M_a, {\mathbf{Z}}^n)$.

There exists a compact homogeneous manifold whose base is finitely covered by the original solvmanifold, its fiber is the given homogeneous manifold $M_c$, the structure group of this bundle is reduced to the given torus $T$, and the given characteristic class is that of the corresponding principal $T$-bundle. So, there is a description of all (up to commensurability) homogeneous manifolds with a base of a natural bundle of the indicated type.

So, there is a description of all (up to commensurability) homogeneous manifolds with a base of a natural bundle of the indicated type.

Further, if the base of a natural bundle is semisimple, then the construction of all possible homogeneous spaces with such a base was given in [7]. It turns out that not every homogeneous manifold in the form $K/L$, even up to commensurability, can be realized as a fiber of a natural bundle with a given semisimple base. For example, in [7], Part II, it is shown that for $M_a=F_2$ (the pretzel) the fiber of the natural bundle cannot be manifolds diffeomorphic to the sphere $S^n$ for $n \ne 3$.

There still remain bases $M_a$ of general form, that is, those for which both components of the structure bundle (solvable and semisimple) are non-trivial. Some such bases of dimensions $\leqslant 5$, which are already described above, can be decomposed (up to commensurability) into the direct product of semisimple and solvable components, and, therefore, a fairly large class of compact homogeneous manifolds with such bases can be constructed by the direct product of two homogeneous manifolds, which bases of natural bundles have given semisimple and solvable components, respectively. However, not all homogeneous manifolds $M$ with bases of this type are obtained in this way. Apparently, in this case, a sufficiently complete description of all homogeneous manifolds $M$ is difficult to achieve.

Introduction

§ 1. Structure of bases of natural bundles

§ 2. Bases of natural bundles of dimensions $1$, $2$, and $3$

§ 3. Bases of natural bundles of dimension $4$

§ 4. Bases of natural bundles of dimension $5$

§ 5. Homogeneous spaces with a given base of the natural bundle

Bibliography