Corollaries of the Riemannian orbibundle theorem

In the last post we gave an account of the following:

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, \(n>0\).
 
However, with stronger hypothesis we can get more. We define a warped Riemannian product to be a smooth manifold \( X = Y \times Z \) where \( Z \) is a (locally) homogeneous space, \( f : Y \longrightarrow \mathcal{H}(Z) \) is a smooth function with target the space \(\mathcal{H}(Z)\) of all (locally) homogeneous metrics on \( Z \), and the metric on \( X \) is given by\[g_X(y,z) = g_Y \oplus f(y)g_Z\]

Recall also 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

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

Proposition 2: Suppose that \( I_0 \) is semisimple with finite center. Then \( M \) has a finite cover which is a Riemannian warped product \( N \times B \), where \( N \) is nonempty, locally symmetric with nonpositive curvature, and has no local torus factors. In particular, \(\pi_1(B) < \pi_1(M)\), and any nontrivial, normal abelian subgroup of \(\pi_1(M)\) lies in \(\pi_1(B)\).

In this setting by calling a connected Lie group \( G \) ``semisimple'' we mean only that the Lie algebra of \( G \) is semisimple. Thus the center \( Z(G) \) may be infinite. Such examples do exist (for example the universal cover of \( U(n, 1) \)), and must be taken into account. We also point out that \( G \) may in general have compact factors. For the connected component \( I_0 \) of the isometry group of the universal cover of a closed, aspherical Riemannian manifold, however, as stated last time\( I_0 \) has no nontrivial compact factor. Even so, the semisimple part \((I_0)^{\text{ss}}\) may have nontrivial compact factors coming from \( Z(I_0) \). The hypothesis that \( I_0 \) is semisimple with finite center is more common than one might guess. 

Proposition 3: Suppose \(\Gamma\) contains no infinite, normal abelian subgroup. Then \(I_0\) is semisimple with finite center.

We take these as blackboxes. We now reap the bountiful rewards on offer thanks to Theorem 1 and the above propositions.

Characterisations of locally symmetric manifolds}

We begin with a characterisation of locally symmetric manifolds among all closed Riemannian manifolds. The theme is that such manifolds are characterized by some simple properties of their fundamental group, together with the property that their universal covers have nontrivial symmetry (i.e. have nondiscrete isometry group). We say that a smooth manifold \( M \) is smoothly irreducible if \( M \) is not smoothly covered by a nontrivial finite product of smooth manifolds.

Theorem 4: Let \( M \) be any closed Riemannian \( n \)-manifold, \( n > 1 \). Then the following are equivalent:

1.  \( M \) is aspherical, smoothly irreducible, \(\pi_1(M)\) has no nontrivial, normal abelian subgroup, and \(\operatorname{Isom}(\widetilde{M})\) is not discrete.
2. \( M \) is isometric to an irreducible, locally-symmetric Riemannian manifold of nonpositive sectional curvature.

Proof: The fact that (2) implies (1) follows immediately from well-known properties of closed, locally symmetric Riemannian manifolds. Such \(M\) are aspherical by the Cartan-Hadamard theorem. Any normal abelian subgroup is trivial since the symmetric space \(\widetilde{M}\) has no Euclidean factors. The other two properties follow from the definitions.

To prove that (1) implies (2), we first quote Proposition III.3 followed by Proposition III.1. This gives that \(M\) has a finite-sheeted Riemannian cover \(M^{\prime}\) of \(M\) which is a smooth (indeed Riemannian warped) product \(M^{\prime}=N\times B\), where \(N\) is isometric to a nonempty, irreducible, locally symmetric, nonpositively curved manifold. But \(M^{\prime}\) is smoothly irreducible by hypothesis, so that \(B\) must be a single point. It follows that \(M^{\prime}=N\) is locally symmetric. Since the metric on \(M^{\prime}\) was lifted from \(M\), we have that \(M\) is locally symmetric. $\blacksquare$

Recall that the Mostow Rigidity Theorem states that a closed, aspherical manifold of dimension at least three admits at most one irreducible, nonpositively curved, locally symmetric metric up to homotheties of its local direct factors. For such locally symmetric manifolds \( M \), Theorem 4 has the following immediate consequence:

Up to homotheties of its local direct factors, the locally symmetric metric on \( M \) is the unique Riemannian metric with \(\operatorname{Isom}(\widetilde{M})\) not discrete.

Uniqueness within the set of nonpositively curved Riemannian metrics on \( M \) follows from Eberlein's work.

Combined with basic facts about word-hyperbolic groups, Theorem 4 provides the following characterization of closed, negatively curved, locally symmetric manifolds.

Corollary 5: Let \( M \) be any closed Riemannian \( n \)-manifold, \( n > 1 \). Then the following are equivalent:

1.  \( M \) is aspherical, \( \pi_1(M) \) is word-hyperbolic, and \( \operatorname{Isom}(\widetilde{M}) \) is not discrete.
2. \( M \) is isometric to a negatively curved, locally symmetric Riemannian manifold.

 Proof:  Again, (2) implies (1) follows immediately from the basic properties of closed, rank one locally symmetric manifolds.

To prove that (1) implies (2), first note that no torsion-free word-hyperbolic group can virtually be a nontrivial product, since then it would contain a copy of \(\mathbf{Z}\times\mathbf{Z}\). It then follows from the previous theorem that \(M\) is locally symmetric. But every closed, locally symmetric manifold \(M\) either contains \(\mathbf{Z}\times\mathbf{Z}\) in its fundamental group, or \(M\) must be negatively curved; hence the latter must hold for \(M\). $\blacksquare$

Theorem 4 can also be combined with Margulis's Normal Subgroup Theorem to give a simple characterization in the higher rank case. We say that a group \(\Gamma\) is almost simple if every normal subgroup of \(\Gamma\) is finite or has finite index in \(\Gamma\).

Corollary 6: Let \( M \) be any closed Riemannian manifold. Then the following are equivalent:

1.  \( M \) is aspherical, \( \pi_1(M) \) is almost simple, and \( \operatorname{Isom}(\widetilde{M}) \) is not discrete.
2. \( M \) is isometric to a nonpositively curved, irreducible, locally symmetric Riemannian manifold of (real) rank at least 2.

Proof: This follows just as the proof of Corollary 5, but using the following fact: an irreducible, cocompact lattice in a noncompact semisimple Lie group \(G\) is almost simple if and only if \(\operatorname{rank}_{\mathbf{R}}(G)\geq 2\). The if direction is the statement of the Margulis Normal Subgroup Theorem. For the only if direction, first recall that cocompact lattices in rank one semisimple Lie groups are non-elementary word-hyperbolic. Such groups are never almost simple. It is known that all such groups are SQ-universal, i.e. any countable group embeds as a subgroup in some quotient. Alternatively, a theorem of Gromov-Olshanskii gives that all such groups have infinite torsion quotients. $\blacksquare$

The above results distinguish, by a few simple properties, the locally symmetric manifolds among all Riemannian manifolds. 

Manifolds with both closed and finite volume quotients

We can also apply our methods to answer the following fundamental question in Riemannian geometry: which contractible Riemannian manifolds \( X \) cover both a closed manifold and a (noncompact, complete) finite volume manifold?

This question has been answered for many (but not all) contractible homogeneous spaces \( X \). Recall that a contractible (Riemannian) homogeneous space \( X \) is the quotient of a connected Lie group \( H \) by a maximal compact subgroup, endowed with a left-invariant metric. Mostow proved that solvable \( H \) admit only cocompact lattices, while Borel proved that noncompact, semisimple \( H \) have both cocompact and noncocompact lattices. The case of arbitrary homogeneous spaces is more subtle, and as far as we can tell remains open.

Theorem 7: Let \( X \) be a contractible Riemannian manifold. Suppose that \( X \) Riemannian covers both a closed manifold and a noncompact, finite volume, complete manifold. Then \( X \) is isometric to a warped product \( Y \times X_0 \), where \( Y \) is a contractible manifold (possibly a point) and \( X_0 \) is a homogeneous space which admits both cocompact and noncocompact lattices. In particular, if \( X \) is not a Riemannian warped product then it is homogeneous.

Proof: Let \(\Gamma_1\) (resp. \(\Gamma_2\)) be a cocompact (resp. noncocompact) lattice in \(\text{Isom}(X)\). Since \(\text{Isom}(X)\) contains \(\Gamma_1\), it acts cocompactly on \(X\). Suppose \(\text{Isom}(X)\) were discrete, so that \(\text{Isom}(X)\) acts properly and cocompactly on \(X\). Then the (orbifold) quotient \(\text{Isom}(X)\setminus X\) would have finite volume. Since covolume is multiplicative in index, it would follow that \(\Gamma_2 < \text{Isom}(X)\) has finite index. But then \(\Gamma_2\) would act cocompactly since \(\text{Isom}(X)\) does, a contradiction. Hence \(\text{Isom}(X)\) is not discrete and we may apply Theorem 1.

We thus obtain a Riemannian orbibundle, which at the level of universal covers gives a Riemannian warped product structure \(Y \times X_0\), with \(Y\) the universal cover of \(B\), where the metric has the property that for each \(x \in X\), the metric on \(x \times X_0\) is an \(I_0\)-homogeneous metric, depending on \(x\).

Let \(\Lambda_i := \Gamma_i \cap I_0, i = 1, 2\). Claim I of the proof of Theorem 1 gives that \(\Lambda_1\) is cocompact. We must now prove that \(\Lambda_2\) is a noncocompact lattice.

To this end, first note that \(\pi := \Gamma_2 / \Lambda_2\) is a cocompact lattice in \(\text{Isom}(Y)\). First suppose \(B\) is not 1-dimensional. We can then perturb the metric on \(B\) to get a new universal (in the category of orbifolds) cover \(Y'\) with \(\text{Isom}(X) = \text{Isom}(Y' \times X_0)\) but with \(\text{Isom}(Y') = \pi_1^{\text{orb}}(B)\). Now \(\pi\) is a lattice in \(\text{Isom}(Y')\), so it has finite index in \(\pi_1(B)\).

We thus have that each of \(\Gamma_i, i = 1, 2\) can be written as a group extension with kernel \(\Lambda_i\) and quotient a group with the same rational cohomological dimension \(\text{cd}_{\mathbf{Q}}\) as \(\pi_1(B)\). We consider rational cohomological dimension \(\text{cd}_{\mathbf{Q}}\) in order to deal with the fact that \(\pi_1(B)\) might not be virtually torsion-free.

Since each \(\Gamma_i\) acts properly on the contractible manifold \(X\), and since \(\Gamma_1\) acts cocompactly and \(\Gamma_2\) does not, we have that \(\text{cd}_{\mathbf{Q}}(\Gamma_2) < \text{cd}_{\mathbf{Q}}(\Gamma_1)\) (a classical result: think compact manifolds with boundary don't have a fundamental class, but closed ones do).

Now, if \(\Lambda_2\) were cocompact, then \(\Gamma_2\) would be an extension of fundamental groups of closed, aspherical manifolds, and so \(\text{cd}_{\mathbf{Q}}(\Gamma_2)\) would be the sum of the \(\text{cd}_{\mathbf{Q}}\) of the kernel and quotient; but this sum equals \(\text{cd}_{\mathbf{Q}}(\Gamma_1)\) by a theorem of Bieri, a contradiction. Hence \(\Lambda_2\) is not cocompact.

If \(B\) is one-dimensional, then no perturbation as above exists. To remedy this, we simply take the product of \(B\) with a closed, genus 2 surface, endowed with a Riemannian metric with trivial isometry group. We then run the rest of the argument verbatim. $\blacksquare$

Irreducible lattices in products

Let \(X = Y \times Z\) be a Riemannian product. Except in obvious cases, \(\text{Isom}(Y) \times \text{Isom}(Z) \hookrightarrow \text{Isom}(X)\) is a finite index inclusion. Recall that a lattice \(\Gamma\) in \(\text{Isom}(X)\) is irreducible if it is not virtually a product. Understanding which Lie groups admit irreducible lattices is a classical problem.

Theorem 8: Let \(X\) be a nontrivial Riemannian product, and suppose that \(\text{Isom}(X)\) admits an irreducible, cocompact lattice. Then \(X\) is isometric to a warped Riemannian product \(X = Y \times X_0\), where \(Y\) is a contractible manifold (possibly a point), \(X_0\) is a positive dimensional homogeneous space, and \(X_0\) admits an irreducible, cocompact lattice.

Note that the factor \(Y\) is necessary, as one can see by taking the product of a homogeneous space with the universal cover of any compact manifold. We begin the deduction of Theorem 7 from the other results in this paper by noting that its hypotheses imply that \(\text{Isom}(Z)\) is nondiscrete, so that our general result can be applied. As with Theorem 7, Theorem 8 is deduced from the other results in this paper by noting that its hypotheses imply that \(\text{Isom}(Z)\) is nondiscrete.

Proof: We first note that the hypothesis implies \(\text{Isom}(X)\) is not discrete. Since in addition \(X\) has a cocompact discrete subgroup, we may apply Theorem 1. Hence \(X\) is isometric to a warped Riemannian product \(X = Y \times X_0\). By the proof of Theorem 1, the group \(\text{Isom}(X_0)\) corresponds with connected component of the identity of \(\text{Isom}(X)\). In particular \(\text{Isom}(Y)\) must be discrete. The theorem follows easily. $\blacksquare$

The Singer and Hopf Conjectures

A well-known conjecture of Hopf-Chern-Thurston states that the Euler characteristic of any closed, aspherical manifold \(M^{2k}\) satisfies \((-1)^k \chi(M^{2k}) \geq 0\). A stronger conjecture of Singer posits that the \(L^2\)-cohomology of \(M^{2k}\) vanishes except in dimension \(k\). These conjectures are completely open except when \(k = 1\). 

Theorem 9: Let \(M^{2k}\) be any closed, aspherical, smooth manifold which is smoothly irreducible. If \(M\) admits some Riemannian metric so that the induced metric on the universal cover \(\widetilde{M}\) satisfies \([\text{Isom}(\widetilde{M}) : \pi_1(M)] = \infty\), then the Singer Conjecture (and hence the Hopf Conjecture) is true for \(M^{2k}\).
 

By general $L^2$ theory one can show the Singer Conjecture holds for products of surfaces and also for products of any 3-manifolds with \(S^1\). Thus Theorem 9 holds in dimension four without the assumption that \(M^4\) is smoothly irreducible.

Proof (relying on heavy results from the literature): If \(\pi_1(M^{2k})\) has no nontrivial normal abelian subgroups and is smoothly irreducible, then by Theorem 4 it admits a locally symmetric Riemannian metric. The Singer Conjecture is known for such manifolds. If \(\pi_1(M^{2k})\) does contain a nontrivial, normal abelian subgroup \(A\), then one may apply a theorem of Cheeger-Gromov which gives, even more generally for amenable \(A\), that Singer's Conjecture holds. $\blacksquare$