Manifold and Orbifold Constructions

Abstract

In this chapter, we collect some information on various constructions of manifolds, orbifolds, and their covers. Notably, we discuss the notions of fiber product (in the sense of Thurston) and compositum of manifolds over a common developable orbifold, the difference with the set-theoretic fiber product, the connected components of the fiber product, compositum of Galois covers, normal closure of a cover, and the relation between commensurability, arithmeticity, and the existence of certain covers in relation to Mostow rigidity.

In this chapter, we collect some information on various constructions of manifolds, orbifolds, and their covers. Notably, we discuss the notions of fiber product (in the sense of Thurston) and compositum of manifolds over a common developable orbifold, the difference with the set-theoretic fiber product, the connected components of the fiber product, compositum of Galois covers, normal closure of a cover, and the relation between commensurability, arithmeticity, and the existence of certain covers in relation to Mostow rigidity.

1 Riemannian Coverings and Their Equivalence

We fix the following notation for the remaining of the text: we assume that M ₁ and M ₂ are connected smooth oriented closed (i.e., compact with empty boundary) Riemannian manifolds that are finite Riemannian coverings of a developable Riemannian orbifold M ₀, i.e., there is a diagram

of the form (1.1). Recall that the connected space M ₀ being a developable orbifold means that it is the quotient of a connected and simply connected Riemannian manifold $\widetilde M_0$ by a cocompact discrete group of isometries Γ₀, that is the (orbifold) fundamental group of M ₀. For the theory of (good/developable) orbifolds, see, e.g., [95, Ch. 13] [25] [33, §1]. Given this setup, we can write $M_i = \Gamma _i \backslash \widetilde M_0$ for finite index subgroups Γ_i of Γ₀ acting without fixed points on $\widetilde M_0$.

Remark 2.1.1

In dimension 2, there is a relation with the theory of branched coverings of surfaces [78, §3] (compare [95, Thm. 13.3.6]). Consider an orbifold cover M ₁ → M ₀ of degree n, where M ₀ is a closed developable 2-orbifold with r elliptic (cone) points {x _i}: the local orbifold structure is that of rotation over an angle 2π∕e _i. Assume that M ₁ is a closed 2-manifold. To such an orbifold cover corresponds a branched degree n cover of the corresponding underlying closed surfaces Σ₁ → Σ₀ branched over the r given points such that x _i has exactly n∕e _i points above it with ramification index e _i (in particular, e _i divides n). A diagram (1.1) corresponds to a diagram of branched coverings of the corresponding surfaces

where, if p _j has degree d _j (j = 1, 2), we have the added data of r branch points {x _i} on Σ₀ and r integers e _i dividing gcd(d ₁, d ₂) such that there are exactly d _j∕e _i points above x _i in Σ_j for j = 1, 2 and i = 1, …, r.

Lemma 2.1.2

Suppose we have a diagram of manifolds/orbifolds

of the form (1.2) and Γ ₀, Γ ₁, Γ ₂ are as above. Suppose Γ is such that $M = \Gamma \backslash \widetilde M_0$ . Then the following are equivalent:

• (i) H ₁ and H ₂ are conjugate in G;

• (ii) Γ ₁ and Γ ₂ are conjugate in Γ ₀;

• (iii) M ₁ and M ₂ are equivalent Riemannian covers of M ₀ , i.e., there is an isometry φ such that following diagram commutes

  

  (2.1)

Proof

With H _i = Γ_i∕ Γ and Γ normal in Γ₀, we have that there exists a g ∈ G satisfying gH ₁ g ⁻¹ = H ₂ if and only if for any lift γ ₀ ∈ Γ₀ of g, $\gamma _0 \Gamma _1 \gamma _0^{-1} = \Gamma _2$. This shows the equivalence of (i) and (ii). To show the equivalence of (ii) and (iii), we argue as follows. If gH ₁ g ⁻¹ = H ₂ for g ∈ G, a (finite) group of isometries of M, then φ: M ₁ → M ₂ given by φ(H ₁ x) = H ₂ gx (for $x \in \widetilde M_0$) satisfies the requirements of (iii). Conversely, if M ₁ and M ₂ are equivalent Riemannian covers of M ₀, then the corresponding subgroups Γ₁ and Γ₂ of the orbifold fundamental group Γ₀ are conjugate. □

2 Compositum

We set $M_{00}:= \Gamma _{00} \backslash \widetilde M_0$ for Γ₀₀ := Γ₁ ∩ Γ₂. This is a finite connected common Riemannian cover of M ₁ and M ₂ (finiteness follows since [ Γ₀ : Γ₁ ∩ Γ₂] ≤ [ Γ₀ : Γ₁] ⋅ [ Γ₀ : Γ₂]), which we call the compositum $M_1 \bullet _{M_0} M_2$ of p ₁: M ₁ → M ₀ and p ₂: M ₂ → M ₀. Note that M ₀₀ is a manifold since Γ_i acts properly discontinuously without fixed points on $\widetilde M_0$. It is important to notice that the construction of M ₀₀ and the covering maps M ₀₀ → M _i, i = 1, 2, depend on the actual maps p ₁, p ₂, not just the spaces M ₀, M ₁, M ₂, but it is customary to leave out the maps from the notation if they are clear. If necessary, we will write (or $M_1 \bullet _{M_0,p_2} M_2$ if p ₁ is clear).

3 Fiber Product

Given a diagram (1.1), we can also form the (orbifold) fiber product $M_1 \times _{M_0} M_2$ (or, more precisely, , where we use the same convention as before concerning including or leaving out the maps from the notation). We follow the construction of Thurston [95, 13.2.4] as explained in [25, 4.6.1] (but where we are using left actions instead of right actions):

$$\displaystyle \begin{aligned} M_1 \times_{M_0} M_2 = \Gamma_0 \backslash \left( \widetilde M_0 \times \Gamma_1 \backslash \Gamma_0 \times \Gamma_2 \backslash \Gamma_0 \right), \end{aligned}$$

where the left action of γ ₀ ∈ Γ₀ is given by

$$\displaystyle \begin{aligned} \gamma_0 (x,\Gamma_1 \gamma_0^{(1)}, \Gamma_2 \gamma_0^{(2)}) = (\gamma_0(x),\Gamma_1 \gamma_0^{(1)} \gamma_0^{-1}, \Gamma_2 \gamma_0^{(2)}\gamma_0^{-1} ) \end{aligned}$$

for $x \in \widetilde M_0, \gamma _0^{(i)} \in \Gamma _0 \ (i=1,2)$. This has the universality property required for fiber products. Using the bijection

$$\displaystyle \begin{aligned} \left( \Gamma_1 \backslash \Gamma_0 \times \Gamma_2 \backslash \Gamma_0 \right) / \Gamma_0 \rightarrow \Gamma_1 \backslash \Gamma_0 / \Gamma_2 \colon [(\alpha,\beta)] \rightarrow [\alpha \beta^{-1}], \end{aligned}$$

we find an isometry

$$\displaystyle \begin{aligned} M_1 \times_{M_0} M_2 = \bigsqcup_{ \gamma \in \Gamma_1 \backslash \Gamma_0/ \Gamma_2} (\Gamma_1 \cap \gamma \Gamma_2 \gamma^{-1}) \backslash \widetilde M_0. \end{aligned}$$

In our situation, where M ₁ and M ₂ are manifolds, it follows that $M_1 \times _{M_0} M_2$ is also a (possibly disconnected) manifold, since elements of Γ₁ ∩ γ Γ₂ γ ⁻¹ act without fixed points on $\widetilde M_0$. The fiber product is a Riemannian manifold for the metric inherited from the universal covering (manifold) $\widetilde M_0$.

The fiber product contains the compositum as a connected component. However, it is not necessarily connected (similar to the tensor product of two number fields not always being isomorphic to their compositum); in fact, we see from the above that $M_1 \times _{M_0} M_2$ has | Γ₁∖ Γ₀∕ Γ₂| connected components. The components need not all be isometric to the compositum, but if M _i → M ₀ are Galois covers, then they are (since then, γ Γ_i γ ⁻¹ = Γ_i for all γ ∈ Γ₀, i = 1, 2).

There are two projections

$$\displaystyle \begin{aligned} M_1 \times_{M_0} M_2 \twoheadrightarrow \Gamma_0 \backslash \left( \widetilde M_0 \times \Gamma_i \backslash \Gamma_0 \right) \cong M_i, \end{aligned}$$

where the latter isometry is given by

$$\displaystyle \begin{aligned} (\{\gamma_0(x),\Gamma_i \gamma_0^{(i)} \gamma_0^{-1}\}_{\gamma_0 \in \Gamma_0}) \mapsto \Gamma_i \gamma_0^{(i)} x, \end{aligned}$$

which is bijective, since Γ_i has no fixed points on $\widetilde M_0$.

Remark 2.3.1

There is a surjective map from the orbifold fiber product to the set-theoretic fiber product

$$\displaystyle \begin{aligned} \$ M_1 \times_{M_0} M_2 \rightarrow \{(x_1,x_2) \in M_1 \times M_2 \colon p_1(x_1) = p_2(x_2) \}, \\ \$ (\{\gamma_0(x),\Gamma_1 \gamma_0^{(1)} \gamma_0^{-1},\Gamma_2 \gamma_0^{(2)} \gamma_0^{-1}\}_{\gamma_0 \in \Gamma_0}) \mapsto (\Gamma_1 \gamma_0^{(1)} x, \Gamma_2 \gamma_0^{(2)} x). \end{aligned} $$

(2.2)

If the action of Γ₀ on $\widetilde M_0$ has fixed points, then this map is not necessarily injective, so the set-theoretic description of fiber product cannot be used in the orbifold setting, even if M ₁, M ₂ are manifolds. However, if M ₀ is itself a manifold (so Γ₀ acts without fixed points), then the map in (2.2) is an isometry of Riemannian manifolds (in particular, bijective), where the right hand side is a manifold since the projection maps are Riemannian submersions.

We can complete the diagram (1.1) into diagrams of Riemannian coverings

Lemma 2.3.2

If M _i → M ₀ are Galois covers, then $M_1 \times _{M_0} M_2$ is connected and hence isometric to the compositum $M_1 \bullet _{M_0} M_2$ if and only if Γ ₀ = 〈 Γ ₁, Γ ₂〉, the subgroup of Γ ₀ generated by Γ ₁ and Γ ₂ . This holds if the degrees of M _i → M ₀ are coprime.

Proof

The number of components of the compositum is the cardinality of the double coset space Γ₁∖ Γ₀∕ Γ₂, and this is 1 if and only if Γ₀ = Γ₁ Γ₂. Since Γ_i are normal subgroups of Γ₀, the product Γ₁ Γ₂ is a subgroup; in fact, Γ₁ Γ₂ = 〈 Γ₁, Γ₂〉. Hence the first statement holds. To see the second, note that we have a sequence of finite index group inclusions

$$\displaystyle \begin{aligned} \Gamma_i \hookrightarrow \langle \Gamma_1,\Gamma_2 \rangle \hookrightarrow \Gamma_0, \end{aligned}$$

so we find that [ Γ₀ : 〈 Γ₁, Γ₂〉] divides gcd([ Γ₀, Γ₁], [ Γ₀, Γ₂]). □

Lemma 2.3.3

If M _i → M ₀ are G _i -Galois of coprime degree, then

• (i) $M_1 \bullet _{M_0} M_2 \rightarrow M_0$ is (G ₁ × G ₂)-Galois;

• (ii) $M_1 \bullet _{M_0} M_2 \rightarrow M_1$ is G ₂ -Galois and $M_1 \bullet _{M_0} M_2 \rightarrow M_2$ is G ₁ -Galois.

Proof

We always have an exact sequence

$$\displaystyle \begin{aligned} 1 \rightarrow \Gamma_1 \cap \Gamma_2 \rightarrow \Gamma_0 \xrightarrow{\varphi} \Gamma_0/\Gamma_1 \times \Gamma_0/\Gamma_2. \end{aligned}$$

Now since [ Γ₀ : Γ₁ ∩ Γ₂] is divisible by the coprime integers [ Γ₀ : Γ₁] and [ Γ₀ : Γ₂], the map φ is surjective. This proves the first statement. The second statement follows from

$$\displaystyle \begin{aligned} \Gamma_1/(\Gamma_1 \cap \Gamma_2) \cong (\Gamma_0/(\Gamma_1 \cap \Gamma_2)) / (\Gamma_0/\Gamma_1) \cong \Gamma_0/\Gamma_2, \end{aligned}$$

and similarly with the indices 1 and 2 interchanged. □

4 Normal Closure

If $M_{00} \twoheadrightarrow M_0$ is a general finite covering of connected spaces, then there exists a connected space M and a sequence of coverings $M \twoheadrightarrow M_{00} \twoheadrightarrow M_0$ such that $M \twoheadrightarrow M_0$ and $M \twoheadrightarrow M_{00}$ are finite and Galois (i.e., the corresponding subgroup of the fundamental group is normal), cf. [101, Thm. 1]. We call the minimal such M the normal closure of $M_{00} \twoheadrightarrow M_0$. In terms of fundamental groups, if M ₀₀ corresponds to the subgroup Γ₀₀ of Γ₀, then M corresponds to the subgroup Γ of Γ₀ given as the intersection of all Γ₀-conjugates of Γ₀₀, the so-called (normal) core of Γ₀₀ in Γ₀. Alternatively, the normal core Γ is the kernel of the action of Γ₀ by permutation on the cosets of Γ₀₀ in Γ₀. In particular, since the index of Γ₀₀ in Γ₀ is finite, so is the index of the normal core, and M → M ₀ is indeed finite. In fact, by the above alternative description, the degree of M → M ₀, the index of the normal core in Γ₀, is bounded by the order of the group of permutations of the set Γ₀₀∖ Γ₀. The number of elements of this set is the degree of the map $M_{00} \twoheadrightarrow M_0$, and hence we find a bound

$$\displaystyle \begin{aligned}{}[\Gamma_0:\Gamma] \leq \deg(M_{00} \twoheadrightarrow M_0)! = [\Gamma_0:\Gamma_{00}]!. \end{aligned} $$

(2.3)

Proposition 2.4.1

Given manifolds M ₁ and M ₂ fitting into a diagram ( 1.1 ), there exists a diagram of the form (1.2) .

Proof

We start with a diagram (1.1) and add the normal closure of the compositum, to find a diagram of the form (1.2), where M corresponds to the normal core of Γ₁ ∩ Γ₂:

$$\displaystyle \begin{aligned} M:=\Gamma \backslash \widetilde M_0 \mbox{ for } \Gamma:= \bigcap\limits_{\gamma \in \Gamma_0 }\gamma (\Gamma_1 \cap \Gamma_2) \gamma^{-1}. \end{aligned}$$

5 Commensurability

Two manifolds M ₁ and M ₂ are called commensurable if they admit a common finite covering. Proposition 2.4.1 implies that if M ₁ and M ₂ have a common finite (developable orbifold) quotient as in diagram (1.1), then they are commensurable. We briefly look at the converse statement.

Lemma 2.5.1

Assume that M ₁ and M ₂ are commensurable with common finite covering M, let $\widetilde M$ denote the universal cover of M with isometry group $\mathscr {I}$ , and let Γ, Γ ₁, Γ ₂ denote the subgroups of $\mathscr I$ corresponding to M, M ₁, M ₂ respectively. Then a diagram of the form (1.1) exists if and only if Γ ₁ and Γ ₂ are of finite index in the subgroup of Γ generated by Γ ₁ and Γ ₂.

Proof

If M ₀ exists and corresponds to a subgroup Γ₀ of $\mathscr I$, then $\widetilde M_0 = \widetilde M$ and 〈 Γ₁, Γ₂〉 is a finite index subgroup of Γ₀, and by assumption Γ₁ and Γ₂ have finite index in Γ₀. Conversely, if 〈 Γ₁, Γ₂〉 is of finite index in Γ, we can set M ₀ to be the corresponding orbifold $\langle \Gamma _1, \Gamma _2 \rangle \backslash \widetilde M$. □

Proposition 2.5.2

If $\widetilde M$ is a homogeneous space for a connected semisimple non-compact real Lie group with trivial center and no compact factors and M ₁ and M ₂ are quotients of $\widetilde M$ corresponding to commensurable irreducible uniform lattices Γ ₁ and Γ ₂ , then there exists a diagram of the form ( 1.1 ) if at least one of Γ ₁ and Γ ₂ is non-arithmetic. In this case both Γ ₁ and Γ ₂ are non-arithmetic.

Proof

Let $\mathscr I$ denote the isometry group of $\widetilde M$. Since Γ₁ and Γ₂ are commensurable, their commensurator in $\mathscr I$,

$$\displaystyle \begin{aligned} \mathscr C := \mathrm{Comm}_{\mathscr I}(\Gamma_i) = \{ g \in \mathscr I \colon [g\Gamma_i g^{-1}: \Gamma_i \cap g \Gamma_i g^{-1}] \cdot [\Gamma_i : \Gamma_i \cap g \Gamma_i g^{-1}] < \infty \} \end{aligned}$$

is the same (indeed, if Γ₁ and Γ₂ are commensurable and Γ₁ is commensurable to g Γ₁ g ⁻¹, then also g Γ₁ g ⁻¹ and g Γ₂ g ⁻¹ are commensurable and, therefore Γ₂ and g Γ₂ g ⁻¹ are commensurable since commensurability is an equivalence relation). Margulis’ theorem [66, Thm. (1), p. 2] (see also [102, Props. 6.2.4, 6.2.5 and Thm. 6.2.6]) states that either Γ_i is not arithmetic and of finite index in $\mathscr C$ or Γ_i is arithmetic and $\mathscr C$ is dense in $\mathscr I$ (to connect to the more general formulation in [66]: we consider a single semisimple group over the reals, and the “anisotropy condition” in loc. cit. is satisfied since we assume the group is not compact). This directly implies that either both lattices Γ₁ and Γ₂ are arithmetic or they are both not arithmetic.

Consider the sequence of inclusions

$$\displaystyle \begin{aligned} \Gamma_i \hookrightarrow \langle \Gamma_1, \Gamma_2 \rangle \hookrightarrow \mathscr C \end{aligned} $$

(2.4)

for i = 1, 2.

If neither of Γ_i is arithmetic, the composed inclusion is of finite index. Hence the same holds for the first inclusion, and we can set $M_0:= \langle \Gamma _1, \Gamma _2 \rangle \backslash \widetilde M$. □

Example 2.5.3

The proposition applies in particular to compact hyperbolic manifolds $M_i = \Gamma _i \backslash \mathbb {H}^n$, where $\mathbb {H}^n$ is hyperbolic n-space. $\lozenge$

Proposition 2.5.4

There exist commensurable compact hyperbolic Riemann manifolds M ₁ and M ₂ of dimensions 2 and 3, for which a diagram of the form ( 1.1 ) does not exist.

Proof

The uniform arithmetic isospectral, non-isometric lattices constructed by Vignéras satisfy this property [97]; for commensurability, see [96, Ch. IV]. Additional information can be found in [24, Prop. 3]. □

Remark 2.5.5

More generally, Alan Reid has shown that isospectral manifolds corresponding to arithmetic lattices are commensurable ([81], compare [61] for a quantitative statement). It is an open problem whether isospectral Riemann surfaces are always commensurable. Through work of Lubotzky, Samuels and Vishne, it is known that isospectrality does not imply commensurability [64] in general.

Open Problem Find further relations (other than Proposition 2.5.2) between commensurability and existence of a common finite quotient for special types of spaces, e.g., spaces whose universal covering is fixed and has infinite fundamental group. Example 2.5.3 concerns the case where the common universal covering is hyperbolic n-space.

Open Problem Find a criterion to decide precisely which pairs of arithmetic manifolds are common finite cover of some developable orbifold. See Remark 2.1.1 for a relation between this problem for surfaces and branching data.