The Riemannian orbibundle theorem

The goal of this post is to give something of an exposition of a paper by Farb and Weinberger which describes all closed, aspherical Riemannian manifolds M whose universal covers $\tilde{M}$ have lots of symmetry. More precisely, this means that $\rm{Isom}(\tilde{M})$ is not discrete. This is, I hope, a natural and interesting enough question for anyone interested in geometric topology. For instance, almost by definition a locally symmetric spaces has this property, and a huge amount of work has uncovered a vast wealth of interesting results in the theory of locally symmetric spaces, so one might wonder what other spaces are similar enough that analogous results could be proved. (This will rely on quite a lot of background material. My apologies to the reader for not being able to simplify it more).

First, we can imitate the construction of locally symmetric spaces in a bit more generality.

Definition: A manifold $M$ is locally homogeneous if its universal cover is isometric to $G/K$, where $G$ is a Lie group, $K$ is a maximal compact subgroup, and $G/K$ is endowed with a left $G$-invariant, $K$ bi-invariant metric.

These are the manifolds $M$ whose universal cover admits a transitive Lie group action whose isotropy subgroups are maximal compact, and are the basic examples of closed, aspherical, Riemannian manifolds whose universal covers have nondiscrete isometry groups. Of course one might also take a product of such a manifold with an arbitrary manifold. To find non-homogeneous examples which are not products, one can do the following construction.

Example: Let \(F\to M\to B\) be any Riemannian fibre bundle with the induced path metric on \(F\) locally homogeneous. Let \(f:B\to{\bf R}^{+}\) be any smooth function. Now at each point of \(M\) lying over \(b\), rescale the metric in the tangent space \(TM_{b}=TF_{b}\oplus TB_{b}\) by rescaling \(TF_{b}\) by \(f(b)\). Almost any \(f\) gives a metric on \(M\) with \(\dim(\operatorname{Isom}(\widetilde{M}))>0\) but with \(\widetilde{M}\) not homogeneous, indeed with each \(\operatorname{Isom}(\widetilde{M})\)-orbit a fibre. This construction can be further extended by scaling fibres using any smooth map from \(B\) to the moduli space of locally homogeneous metrics on \(F\); this moduli space is large for example when \(F\) is an \(n\)-dimensional torus.

Hence we see that there are many closed, aspherical, Riemannian manifolds whose universal covers admit a nontransitive action of a positive-dimensional Lie group. We now claim that we have found basically everything.

Before stating the general result, we need some terminology. A Riemannian orbifold \(B\) is a smooth orbifold where the local charts are modelled on quotients \(V/G\), where \(G\) is a finite group and \(V\) is a linear \(G\)-representation endowed with some \(G\)-invariant Riemannian metric. The orbifold \(B\) is good if it is the quotient of \(V\) by a properly discontinuous group action.

A Riemannian orbibundle is a smooth map \(M\longrightarrow B\) from a Riemannian manifold to a Riemannian orbifold locally modelled on the quotient map \(p:V\times_{G}F\longrightarrow V/G\), where \(F\) is a fixed smooth manifold with smooth \(G\)-action, and where \(V\times F\) has a \(G\)-invariant Riemannian metric such that projection to \(V\) is an orthogonal projection on each tangent space. Note that in this definition, the induced metric on the fibres of a Riemannian orbibundle may vary, and so a Riemannian orbibundle is not a fibre bundle structure in the Riemannian category. Note that similar objects already occur in the classification of geometric 3-manifolds in the guise of Seifert-fibred 3-manifolds.

Theorem 1: Let \(M\) be a closed, aspherical Riemannian manifold. Then either \(\operatorname{Isom}(\widetilde{M})\) is discrete, or there exists a good Riemannian orbifold  \(B\) such that \(M\) is isometric to an orbibundle \[F\longrightarrow M\longrightarrow B\] where each fibre \(F\), endowed with the induced metric, is isometric to a closed, aspherical, locally homogeneous Riemannian \(n\)-manifold, \(n>0\).

While this theorem is very nice in itself as a classification, the corollaries are just as spectacular, if not more so. But first, the proof. We will require some other lovely results from the literature. 

Preliminaries

The first result we help ourselves to is

Conner conjecture: Let $G$ be a compact Lie group, and let $Y$ have the homotopy type of a finite dimensional $G$-CW complex with finitely many orbit types. If $\tilde{H}^∗(Y; A) = 0$, then $\tilde{H}^∗(Y/G; A) = 0$

This was first proved by Oliver using classical Smith theory. A more modern viewpoint, based on the generalisation of Smith theory which goes by the name of Bredon (co-)homology, is exposed here.

Slice theorem: Given a manifold $M$ on which a Lie group $G$ acts by diffeomorphisms, for any $x \in M$, the map $G/G_{x}\to M,\,[g]\mapsto g\cdot x$ extends to an invariant neighborhood of $G/G_{x}$ (viewed as a zero section) in $G\times_{G_{x}}T_{x}M/T_{x}(G\cdot x)$ so that it defines an equivariant diffeomorphism from the neighborhood to its image, which contains the orbit of $x$.

The corollary we need is that the quotient $M/G$ admits a manifold structure when $G$ is compact and the action is free. This is like a tubular neighbourhood theorem for the image of the section, and when $G$ is compact one can prove the result in a similar way after first doing the common and useful trick of averaging the metric over the entire group (only well-defined in general when $G$ is compact) so that $G$ acts by isometries. The proof extends further to the case of proper actions of noncompact groups; one simply produces an invariant Riemannian metric by translating a compactly supported pseudometric, and this gives the required structure via exponentiation.

I remark here also that, although we won't need it, there are many different variations of the slice theorem depending on the setting one is interested in. For instance, there is an algebraic geometry version of this which goes by the name of Luna's slice theorem. Glossing over the technicalities, this states that given an affine variety $V$ on which a reductive algebraic group acts, for every $x \in V$ with closed orbit there is a slice at $x$, i.e. a subvariety $W$ such that $V$ looks locally like $G \times_{G_x} W$. This result is important in classifying representations of algebraic groups. See this survey and the references therein for more.

Myers-Steenrod Theorem: Let \(M\) be a Riemannian manifold. Then \(\operatorname{Isom}(M)\) is a Lie group, and acts properly on \(M\). If \(M\) is compact then \(\operatorname{Isom}(M)\) is compact.

Note that the Lie group \(\operatorname{Isom}(M)\) may have infinitely many components. For example, let \(M\) be the universal cover of a bumpy metric on the torus. An exposition of the proof can be found in these notes.

Although we won't need it, I can't pass up the opportunity to mention the second result which was proved by Myers and Steenrod in the same paper. This states that every distance-preserving surjective map (that is, an isometry of metric spaces) between two connected Riemannian manifolds is a smooth isometry of Riemannian manifolds. A proof of this can be found here. Palais gives a short proof of a generalisation of this: he proves that the smooth manifold can be reconstructed from the metric in a suitable sense. The point here seems to be that given a point $x \in M$ one can look at the geodesics passing through $x$ to recover the inner product on the tangent space, which determines the metric structure via the usual polarisation process, as well as the linear structure.

Throughout this post we will use the following notation:

•  \(M\) = a closed, aspherical Riemannian manifold
• \(\Gamma = \pi_{1}(M)\)
• \(I = \operatorname{Isom}(\widetilde{M})\) = the group of isometries of \(\widetilde{M}\)
• \(I_{0}\) = the connected component of \(I\) containing the identity 
• \(\Gamma_{0} = \Gamma \cap I_{0}\)

Here \(\widetilde{M}\) is endowed with the unique Riemannian metric for which the covering map \(\widetilde{M}\to M\) is a Riemannian covering. Hence \(\Gamma\) acts on \(\widetilde{M}\) isometrically by deck transformations, giving a natural inclusion \(\Gamma\to I\), where \(I = \operatorname{Isom}(\widetilde{M})\) is the isometry group of \(\widetilde{M}\).

By Myers-Steenrod, \(I\) is a Lie group, possibly with infinitely many components. Let \(I_{0}\) denote the connected component of the identity of \(I\); note that \(I_{0}\) is normal in \(I\). If \(I\) is discrete, then we are done, so suppose that \(I\) is not discrete.  Myers-Steenrod again then gives that the dimension of \(I\) is positive, and so \(I_{0}\) is a connected, positive-dimensional Lie group.

We have the following exact sequences:

\[1\longrightarrow I_{0}\longrightarrow I\longrightarrow I/I_{0} \longrightarrow 1 \]

and

\[1\longrightarrow\Gamma_{0}\longrightarrow\Gamma\longrightarrow\Gamma/\Gamma_{0} \longrightarrow 1 \]

We are now in a position to give a proof of theorem 1. 

Proof of theorem 1

The proof proceeds in a series of steps. The first step is to construct what will end up as the locally homogeneous fibres of the orbibundle by showing that the quotient \(I_{0}/\Gamma_{0}\) is compact. Myers-Steenrod implies that the isometry group is a Lie group, which has some maximal compact subgroup $K$ that has identity component $K_0$. The fibre will eventually be \(K_0 \backslash I_{0}/\Gamma_{0}\). Then we show that $\widetilde{M} / \Gamma_0$ is a contractible manifold on which $\Gamma/\Gamma_0$ acts, and this gives the base orbifold. Now we somewhat elaborate, though the reader is free to skip over things which are unfamiliar, since ultimately I don't think the details here are too important for understanding the big picture.
 

Claim I: The quotient \(I_{0}/\Gamma_{0}\) is compact.

Proof: Let \(\operatorname{Fr}(\widetilde{M})\) denote the frame bundle over \(\widetilde{M}\). The isometry group \(I\) acts freely on \(\operatorname{Fr}(\widetilde{M})\). The \(I_{0}\) orbits in \(\operatorname{Fr}(\widetilde{M})\) give a smooth foliation of \(\operatorname{Fr}(\widetilde{M})\) whose leaves are diffeomorphic to \(I_{0}\). This foliation descends via the natural projection \(\operatorname{Fr}(\widetilde{M})\longrightarrow\operatorname{Fr}(M)\) to give a smooth foliation \(\mathcal{F}\) on \(\operatorname{Fr}(M)\), each of whose leaves is diffeomorphic to \(I_{0}/\Gamma_{0}\). Thus we must prove that each of these leaves is compact.

The quotient of \(\operatorname{Fr}(\widetilde{M})\) by the smallest subgroup of \(I\) containing both \(\Gamma\) and \(I_{0}\) is homeomorphic to the space of leaves of \(\mathcal{F}\). 

Subclaim: this quotient is a finite cover of \(\operatorname{Fr}(\widetilde{M})/I\).

Proof: The Milnor-Svarc principle says that if $G$ is a compactly generated topological group, generated by a compact subspace$ S \subset G$, and $G$ is endowed with the word metric, then $G$ is quasi-isometric to any proper, geodesic metric space $N$ on which it acts properly and cocompactly by isometries. Roughly speaking, $G$ and $N$ have the same coarse geometry when you squint. This is usually stated for $S$ finite but the proofs work almost verbatim for $S$ compact.

Applying this principle, the cocompactness of the actions of both \(\Gamma\) and of \(I\) on \(\widetilde{M}\) give that the inclusion \(\Gamma \longrightarrow I\) is a quasi-isometry. The quotient map \(I \longrightarrow I/I_0\) is clearly distance nonincreasing, and so the image \(\Gamma/\Gamma_0\) of \(\Gamma\) under this quotient map is \(C\)-dense in \(I/I_0\). As both groups are discrete, this clearly implies that the inclusion \(\Gamma/\Gamma_0 \longrightarrow I/I_0\) is of finite index. Thus the subclaim is proved. $\blacksquare$

Now note that \(\operatorname{Fr}(\widetilde{M})/I\) is clearly compact, and is a manifold since \(I\) is acting freely and properly. Hence the leaf-space of \(\mathcal{F}\) is also a compact manifold. Since each leaf of \(\mathcal{F}\) is the inverse image of a point under the map from \(\operatorname{Fr}(M)\) to the leaf space, we have that each leaf of \(\mathcal{F}\) is compact. $\blacksquare$

It will be useful to know that \(I_0\) cannot have compact factors.
 

Claim II: \(I_0\) has no nontrivial compact factor.

I ask the reader's understanding in omitting the proof of this claim. The point is that, with a suitable cohomology theory (locally finite homology) that sees the dimension of $\widetilde{M}$ and $M$, one can detect the following: a compact factor $K$ would have to be connected and positive dimensional, which means that $\widetilde{M}$ and $\widetilde{M}/K$ are quasi-isometric. One can bootstrap this to a continuous QI, which would then induce isomorphisms on the cohomology theory, violating the dimension condition.

The next step in the proof is to determine information which will help us construct the orbifold base space $B$ of the orbibundle.

Claim III: \(\widetilde{M}/I_0\) is contractible.

Proposition 2: Let \(G\) be a connected Lie group acting properly by diffeomorphisms on an aspherical manifold \(N\). Denote by \(K\) the maximal compact subgroup of \(G\). Then there exists an aspherical manifold \(Y\) such that \(\widetilde{M}\) is diffeomorphic to \(Y \times G/K\), the manifold \(Y\) has a \(K\)-action, and the original action is given by the product action. In particular, \(N/G\) is diffeomorphic to \(Y/K\).

Proof: Let \(EG\) be the classifying space for proper CW \(G\)-complexes, so that \(EG/G\) is the classifying space for proper \(G\)-bundles. Now \(G/K\) is an \(EG\) space. Hence there is a proper \(G\)-map \(\psi : \widetilde{M} \longrightarrow EG\). But \(EG\) has only one \(G\)-orbit, so \(\psi\) is surjective. Now let \(Y = \psi^{-1}([K])\), where \([K]\) denotes the identity coset of \(K\). Hence \(\widetilde{M}\) is diffeomorphic to \(G \times_K Y\), and we are done. $\blacksquare$

Combined with the Conner conjecture, this yields

Corollary 3: Let \(G\) be a connected Lie group acting properly by diffeomorphisms on a contractible manifold \(Y\). Then the underlying topological space of the orbit space \(Y/G\) is contractible.

We are now ready to construct, on the level of universal covers, the orbibundle, and in particular to prove that the base space \(B\) is a Riemannian orbifold. The crucial point is to understand stabilizers of the \(I_{0}\) action on \(\widetilde{M}\). For \(x\in \widetilde{M}\), denote the stabilizer of \(x\) under the \(I_{0}\) action by \(I_{x}:=\{g\in I_{0}:gx=x\}\). Let \(K_{0}\) denote the maximal compact subgroup of \(I_{0}\); it is unique up to conjugacy.

Claim IV: \(I_{x}=K_{0}\) for each \(x\in \widetilde{M}\). Hence the following hold:

1. \(\widetilde{M}/I_{0}\) is a manifold.
2. Each \(I_{0}\)-orbit in \(\widetilde{M}\) is isometric to the contractible, homogeneous manifold \(I_{0}/K_{0}\), endowed with some left-invariant Riemannian metric.
3. The natural quotient map gives a Riemannian fibration \[    I_{0}/K_{0}\longrightarrow \widetilde{M}\longrightarrow \widetilde{M}/I_{0}    \]

Proof: Clearly \(I_{x}\subseteq K_{0}\). Iwasawa proved that any maximal compact subgroup of a connected Lie group is connected. Hence it is enough to prove that \(\dim(I_{x})=\dim(K_{0})\).

To this end we consider rational cohomological dimension \(\text{cd}_{\mathbf{Q}}\). By Claim III we have \(\widetilde{M}/I_{0}\) is contractible. Since \(\Gamma/\Gamma_{0}\) acts properly on \(\widetilde{M}/I_{0}\), we then have
\[\text{cd}_{\mathbf{Q}}(\Gamma/\Gamma_{0})\leq\dim(\widetilde{M}/I_{0})\]

Since \(K_{0}\) is maximal, we know \(I_{0}/K_{0}\) is contractible. By Claim I, we have that \(\Gamma_{0}\) is a uniform lattice in \(I_{0}\), and so
\[\text{cd}_{\mathbf{Q}}(\Gamma_{0})=\dim(I_{0}/K_{0})\]

Since \(\widetilde{M}\) is contractible and \(M=\widetilde{M}/\Gamma\) is a closed manifold, by general facts about cohomological dimension, we have
\[\dim(\widetilde{M})=\text{cd}_{\mathbf{Q}}(\Gamma)\leq\text{cd}_{\mathbf{Q}}(\Gamma_{0})+\text{cd}_{\mathbf{Q}}(\Gamma/\Gamma_{0})\]
which combined with the previous inequalities gives
\[\dim(\widetilde{M})\leq\dim(\widetilde{M}/I_{0})+\dim(I_{0}/K_{0})\]

But for each \(x\in \widetilde{M}\), we have
\[\dim(\widetilde{M})\geq\dim(\widetilde{M}/I_{0})+\dim(I_{0}/I_{x})\]
which combined with the previous inequality gives
\[\dim(I_{0}/I_{x})\leq\dim(I_{0}/K_{0})\]
and so \(\dim(I_{x})\geq\dim(K_{0})\), as desired. Thus \(I_{x}=K_{0}\).

It follows that each orbit \(I_{0}\cdot x\) is diffeomorphic to a common Euclidean space \(I_{0}/K_{0}\), so by the Slice Theorem it follows that \(\widetilde{M}/I_{0}\) is a manifold. $\blacksquare$

Finishing the proof: The action of \(\Gamma\) on \(\widetilde{M}\) induces actions of \(\Gamma_{0}\) on \(I_{0}/K_{0}\), and of \(\Gamma/\Gamma_{0}\) on \(\widetilde{M}/I_{0}\), compatible with the Riemannian fibration. By Myers-Steenrod, \(\Gamma/\Gamma_{0}\) acts properly discontinuously on \(\widetilde{M}/I_{0}\); we denote the quotient space of this action by \(B\). We thus have a Riemannian orbibundle:
\[F\longrightarrow M\longrightarrow B\]
where \(F\) denotes the closed, locally homogeneous Riemannian manifold \(\Gamma_{0}\backslash I_{0}/K_{0}\), endowed with the quotient metric of a left \(I_{0}\)-invariant metric on \(I_{0}/K_{0}\). $\blacksquare$

Theorem 1 is sharp

In the next post I will detail the wonderful corollaries of theorem 1. I would like to close by presenting the example in the Farb-Weinberger paper which shows that one cannot hope to prove a stronger theorem that says a finite cover fibres over a Riemannian manifold. We will rely on some group theoretic constructions.

Non-positively curved complexes with no finite quotients

At the turn of the millennium Burger and Mozes constructed examples of finitely presented simple groups which are amalgams of free groups over finite index subgroups that arise as fundamental groups of non-positively curved square complexes whose universal cover is a product of two trees. These have been extremely interesting examples that have spurred a huge amount of research and greatly broadened the horizons of group theorists in terms of what sort of groups there are out there. It is striking that two decades later there have been few significant improvements on this result. Most interesting to me is how they show that the group is simple: inspired by Margulis' normal subgroup theorem, they show that any normal subgroup of a sufficiently complicated lattice in the automorphism group of the product of two trees has to be either finite or finite index, and then do some magic to show that the resulting group isn't residually finite to get that the lattice is virtually simple. The normal subgroup theorem was subsequently generalised to the Bader-Shalom normal subgroup theorem (I wonder when I'll get round to writing that blog post...).

We will take the above as a blackbox and use it to prove the following result due to Bridson:

Theorem 4: Every compact, connected, non-positively curved space \( X \) admits an isometric embedding into a compact, connected, non-positively curved space \(\overline{X}\) such that \(\overline{X}\) has no non-trivial finite-sheeted coverings. If \(X\) is a polyhedral complex of dimension \(n \geq 2\), then one can arrange for \(\overline{X}\) to be a complex of the same dimension.

We recall the following basic facts.

Proposition 5: Let \( X \) be a compact, connected, geodesic space of non-positive curvature. Fix \( x \in X \).

1.  Each homotopy class in \( \pi_1(X,x) \) contains a unique shortest loop based at \( x \). This based loop is the unique local geodesic in the given homotopy class.
2.  Each conjugacy class in \( \pi_1(X,x) \) is represented by a closed geodesic in \( X \) (i.e. a locally isometric embedding of a circle). In other words, every loop in \( X \) is freely homotopic to a closed geodesic (which need not pass through \( x \)). If two closed geodesics are freely homotopic then they have the same length.
3.  \( \pi_1(X,x) \) is torsion-free.
4. Metric graphs are non-positively curved. 
5. The induced path metric on the 1-point union of two non-positively curved spaces is again non-positively curved.
6. If \( X \) is a compact non-positively curved space, \( Z \) is a compact length space and \( i_1, i_2: Z \to X \) are locally isometric embeddings, then, when endowed with the induced path metric, the quotient of \( X \cup (Z \times [0,L]) \) by the equivalence relation generated by \( i_1(z) \sim (z,0) \) and \( i_2(z) \sim (z,L) \) is non-positively curved. Moreover, if \( L \) is greater than the diameter of \( X \), then \( X \) is isometrically embedded in the quotient.

A particular case of (6) that we shall need is where \( X \) is the disjoint union of spaces \( X_1 \) and \( X_2 \), and \( Z \) is a circle. In this case the quotient is obtained by joining \( X_1 \) to \( X_2 \) with a cylinder whose ends are attached along closed geodesics.

Proof: Choose a finite set of generators for \(\pi_1 X\), and let \(c_1, \ldots, c_N\) be closed geodesics in \(X\) representing the conjugacy classes of these elements. Burger-Mozes give us a (in fact many) compact non-positively curved 2-complex \(K_4\) whose fundamental group has no finite quotients; fix a closed geodesic \(c_0\) in \(K_4\). Take \(N\) copies of \(K_4\) and scale the metric on the \(i\)-th copy so that the length of \(c_0\) in the scaled metric is equal to the length \(l(c_i)\) of \(c_i\). Then glue the \(N\) copies of \(K_4\) to \(X\) using cylinders \(S_i \times [0, L]\) where \(S_i\) is a circle of length \(l(c_i)\); the ends of \(S_i \times [0, L]\) are attached by arc length parametrizations of \(c_0\) and \(c_i\) respectively. Call the resulting space \(\overline{X}\).

Part (6) of the previous proposition assures us that \(\overline{X}\) is non-positively curved, and if the length \(L\) of the gluing tubes is sufficiently large then the natural embedding \(X \hookrightarrow \overline{X}\) will be an isometry. $\blacksquare$

  For $n \geq 2$, let \(G_n = \mathbb{Z}/p\mathbb{Z} * \mathbb{Z}\). Then \(G_n\) acts properly discontinuously and cocompactly on its Bass-Serre tree \(T\). Using similar ideas, one can show that there is an amalgamated free product \(H_n = G_n*_F G_n\), with \(F\) a nonabelian free group, which has no finite quotients. Let \(Y_n\) be the universal cover of the standard Cayley 2-complex for \(H_n\). This will be contractible and \(H_n\) acts properly discontinuously and cocompactly on \(Y_n\). We will use $G_n$ in the next section

A very singular orbibundle

Theorem 6: For each prime $p$ there is a group \(\Gamma_p\) which acts cocompactly, properly discontinuously by diffeomorphisms on \(\mathbb{R}^n\) for some \(n\), but which is not virtually torsion free. Moreover, there exists \(\xi \in H^2(\Gamma_p, \mathbb{Z})\) which restricts to a nonzero class on some \(\mathbb{Z}/p\mathbb{Z} \in \Gamma_p\), and which also restricts to a nonzero class on every finite index subgroup.

Proof: Fix the prime $p$, embed \(Y_p\) in Euclidean space, and equivariantly thicken the image. The action on this thickening gives a properly discontinuous, cocompact \(H_p\)-action on a contractible 6-dimensional manifold. Since the action is simplicial, we can perform the reflection group construction (this was previously mentioned on the blog; see https://link.springer.com/book/10.1007/978-3-031-91303-7) equivariantly to build a cocompact action of a group \(\Lambda_p\) on a contractible manifold, and with \(H_p\) being a retract of \(\Lambda_p\). This action can in fact be made smooth. Stallings proved that if $M, N$ are contractible piecewise linear manifolds of dimensions $x, y,$ where $x, y \geq 1$ and $x+y \geq 5$, then $M\times N$ is piecewise linearly homeomorphic to $\mathbb{R}^{x+y}$. Hence the group \(\Gamma_p := \Lambda_p \times \mathbb{Z}\) acts on \(\mathbb{R}^n\) properly discontinuously and cocompactly by diffeomorphisms.

Note that any finite index subgroup of \(\Gamma_p\) intersects \(H_p\) in a finite index subgroup of \(H\), which must therefore be all of \(H_p\) since \(H_p\) has no finite index subgroups. Hence the fixed \(\mathbb{Z}/p\mathbb{Z}\) subgroup of \(\Gamma_p\) must lie in each finite index subgroup of \(\Gamma_p\).

By construction there is a surjection \(\Gamma_p \longrightarrow G_p\). Let \(\xi \in H^2(\Gamma_p, \mathbb{Z})\) denote the pullback of the class generating \(H^2(G_p, \mathbb{Z})\). As the amalgamating subgroup \(F\) is free, the amalgamation dimension of \(H^2(H, \mathbb{Z})\) is at most one greater than \(H^2(G_p, \mathbb{Z})\). Since \(H_p\) is a retract of \(\Lambda_p\), we know that \(\xi \in H^2(\Lambda_p, \mathbb{Z})\) is also nonzero. It follows that \(\xi\) pulls back to a nonzero class in \(\Gamma_p\). $\blacksquare$

We now build a (7-dimensional) Riemmanian manifold $M$ with the property that $M$ is a Riemannian orbibundle, but no finite cover of $M$ is a Riemannian fibre bundle.  Let \(\Gamma_p, \xi\) be given as in the previous theorem. Let \(\tilde{\Gamma_p}\) denote the central extension of \(\Gamma_p\) given by the cocycle \(\xi \in H^2(\Gamma_p, \mathbb{Z})\). Since this cocycle vanishes in \(H^2(\Gamma_p, \mathbb{R})\), we have that \(\tilde{\Gamma_p}\) lies in \(\Gamma_p \times \mathbb{R}\). Now fix any \(\Gamma_p\)-invariant metric on \(\mathbb{R}^n\), and extend this to any \((\Gamma_p \times \mathbb{R})\)-invariant metric on \(\mathbb{R}^n \times \mathbb{R} = \mathbb{R}^{n+1}\); call the resulting Riemannian manifold \(Y\).

Now \(\Gamma_p \times \mathbb{R}\) acts properly discontinuously and cocompactly by isometries on \(Y \approx \mathbb{R}^{n+1}\). The quotient \(M\) clearly satisfies the claimed properties. Note too that \(\operatorname{Isom}(M) = \operatorname{Isom}(Y)\) contains \(\mathbb{R}\), and so is not discrete.