FINITE-VOLUME HYPERBOLIC 4-MANIFOLDS

THAT SHARE A FUNDAMENTAL POLYHEDRON

Dubravko Ivanšić

Abstract. It is known that the volume function for hyperbolic manifolds of dimension ≥ 3 is finite-to-one. We show that the number of nonhomeomorphic hyperbolic 4-manifolds with the same volume can be made arbitrarily large. This is done by constructing a sequence of finite-sided finite-volume polyhedra with side-pairings that yield manifolds. In fact, we show that arbitrarily many nonhomeomorphic hyperbolic 4-manifolds may share a fundamental polyhedron. As a by-product of our examples, we also show in a constructive way that the set of volumes of hyperbolic 4-manifolds contains the set of even integral multiples of 4π^{2}/3. This is ”half” the set of possible values for volumes, which is the integral multiples of 4π^{2}/3 due to the Gauss-Bonnet formula Vol(M) = 4π^{2}/3 · χ(M).

0. Introduction and statement of results

The original aim of research that produced this paper was to construct noncompact hyperbolic 4-manifolds by means of side-pairings of polyhedra. Previous examples of hyperbolic manifolds with dimension higher than three were restricted to constructions via arithmetic groups (see, for example, [A2], [M]), or via ”interbreeding” arithmetic groups to get non-arithmetic ones ([G-P]) and there was only one (compact) example using side-pairings, that of Davis in [D]. We were able to produce a number of examples of side-pairings of hyperbolic 4-polyhedra and get, in addition, new information about volumes of hyperbolic 4-manifolds. Further research led to consideration of embedability of these manifolds as complements of surfaces in compact 4-manifolds — we deal with this in [I].

It is known (see [W]) that for every constant c > 0 there are only finitely many complete non-homeomorphic hyperbolic n-manifolds with volume < c, where n ≥ 4. For n = 3 the set of volumes is a well-ordered (infinite) set, but still only finitely many manifolds may have the same volume. We concern ourselves with whether there is a bound on the number of manifolds that have the same volume.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 51M10, 51M25.

This paper represents a part of the author’s PhD thesis. The author would like to thank his advisor, Dr. Boris Apanasov, for many helpful conversations and suggestions concerning these results. Research at MSRI is supported in part by NSF grant DMS-9022140.

In dimension 3, this has been answered by Wielenberg (see [Wi1]) for the noncompact case, and by Apanasov and Gutsul ([A-G]) for the compact one. In both papers, for N’s that can be made arbitrarily large, polyhedra are constructed in ℍ^{3} and different side-pairings are given on them whose quotient spaces are N nonhomeomorphic hyperbolic manifolds.

In this paper we prove the analogous result for the non-compact case in dimension 4, namely,

Theorem A. Given any number N, there exist more than N non-homeomorphic, non-compact, complete hyperbolic 4-manifolds of finite volume that share the same fundamental polyhedron in ℍ^{4}. In particular, they have the same volume.

The proof is by constructing polyhedra in ℍ^{4} with different side-pairings and utilizing Poincare’s polyhedron theorem to see that identifying paired sides yields complete hyperbolic 4-manifolds. The manifolds are then distinguished by how many ends they have. It is known (Theorem 5.39 in [A2] or [A1]) that a complete, hyperbolic, geometrically finite n-manifold has finitely many ends. If the manifold has finite volume, then all of the ends are standard cusp ends, that is, they are of the form E × [0, ∞), where E is a closed flat manifold. Furthermore, each end of the manifold corresponds to a cycle (equivalence class under identification by side-pairing) of ideal vertices of the polyhedron. We count classes of ideal vertices for each of the side-pairings that we construct and show that we arrive at different numbers for different side-pairings. Therefore, the resulting manifolds are nonhomeomorphic, because they have different numbers of ends.

We also give a geometric interpretation of the manifolds we construct. It turns out that each of the manifolds may be obtained by taking two basic manifolds, cutting them along a two-sided totally geodesic embedded 3-manifold and stringing several of these together by gluing them along the isometric cuts. In the process of justifying this interpretation we prove a convenient sufficient condition for when a plane intersecting a fundamental polyhedron for a group G is precisely invariant with respect to some subgroup J ⊂ G (Theorem 4.4).

After the constructions in this work have been completed the author became aware of two other preprints where non-compact hyperbolic 4-manifolds were obtained by side-pairings of polyhedra in ways different from the one here. Those constructions were then used to prove interesting results. One of the preprints is [N], where B. Nimershiem constructs classes of examples that are used to show that the set of all flat three manifolds that appear as cusps of hyperbolic four-manifolds is dense in the set of all flat three manifolds.

The other, by J. Ratcliffe and S. Tschantz ([R-T]), classifies all non-compact hyperbolic 4-manifolds of minimal volume. In addition to that, it is proved that the set of volumes is the positive integral multiples of 4π^{2}/3. A by-product of our construction is

Theorem B. The set of all volumes of hyperbolic 4-manifolds contains the even multiples of 4π^{2}/3.

FINITE-VOLUME HYPERBOLIC 4-MANIFOLDS … 3

This result is only ”half as good” as the quoted one, but an advantage is that we provide an explicit side-pairing to produce manifolds with the desired volumes, whereas Ratcliffe and Tschantz’s proof was not constructive and gave only their existence.

1. The polyhedron P and its side-pairings Φ_{1} and Φ_{2}

We use the upper half-space model of hyperbolic space to define a convex fourdimensional hyperbolic polyhedron P as an intersection of some hyperbolic halfspaces.

Recall that the Poincaré upper half-space model of hyperbolic n-space is ℍ^{n} = \{(x_{1}, …, x_{n−1}, t) ∈ ℝ^{n} | t > 0\} with the metric given by ds^{2} = \frac{dx_{1}^{2}+···+dx_{n−1}^{2}+dt^{2}}{t^{2}} . The boundary at infinity of a set S is the set of all points in ∂ℍ^{n} = ℝ^{n−1} ∪ {∞} that are in the (Euclidean) closure of S. In the upper-half-space model hyperbolic hyperplanes are either Euclidean half-spheres or Euclidean half-planes orthogonal to ∂ℍ^{n} and they are uniquely determined by their own boundaries at infinity, which are Euclidean (n − 2)-spheres or (n − 2)-planes in ℝ^{n−1} ∪ {∞}. We will say that the hyperplane is based, respectively, on a sphere or a plane. (In our case n = 4, so the hyperplanes will be based on 2-spheres and 2-planes in ℝ^{3}.) The angle between hyperplanes is the same as the angle between their boundaries at infinity.

Every hyperplane in ℍ^{n} determines two closed half-spaces: each contains the hyperplane and their interiors are disjoint. The (n − 2)-sphere or -plane on which the hyperplane is based divides ℝ^{n−1} into two closed sets, each of which is the boundary of one of the half-spaces that the hyperplane determines.

By a polyhedron in ℍ^{n} we will mean a connected subset of ℍ^{n} with non-empty interior whose boundary is a locally finite collection of hyperplanes. (The polyhedra in our construction are going to be intersections of finitely many half-spaces, so they will also be convex.) A codimension-one side S of P is a subset of ∂P such that S = P ∩ X and S = cl_{X}(int_{X}S), where X is a hyperplane that bounds one of the defining half-spaces of P. Then S is an (n − 1)-dimensional convex polyhedron in X. Proceeding inductively we may define a codimension-i side of P to be a codimension-one side of a codimension-(i − 1) side of P. (For more details on polyhedra, consult [A2] or [E-P].)

Since every codimension-i side is a polyhedron in dimension n − i, we also call it an (n − i)-side. Codimension-one sides we will simply call sides, codimension-two sides we call edges, and we will use the term vertex for a 0-side of P. Vertices of P are also called finite vertices or real vertices as opposed to vertices at infinity or ideal vertices that are the isolated boundary points of P in ∂ℍ^{n}. To simplify notation, a hyperbolic hyperplane, the side of P lying on the hyperplane and the boundary at infinity of the hyperplane will be denoted by the same letter. (No confusion should arise here because our P ’s are convex.)

Consider the planes that bound the rectangular box R ⊂ ℝ^{3}, R = [−2, 2] × [−2, 2] × [−2\sqrt{2}, 2\sqrt{2}]. Add to them the 12 spheres of radius \sqrt{2} with centers (±1, ±1, j2\sqrt{2}) for j = −1, 0, 1 and the 18 spheres of the same radius with cen-Figure 1. Section of P for t = \sqrt{2}/2, z = \sqrt{2}/2 showing the real vertices

ters (j, k, ±\sqrt{2}) for j, k = −2, 0, 2. The upper part of figure 2, going from left to right, depicts intersections of these spheres with planes with constant z-coordinates −2\sqrt{2}, −\sqrt{2}, 0, \sqrt{2}, 2\sqrt{2}. Label the spheres by the letters A_{i}, A′_{i}, B_{i}, B′_{i}, C_{i}, C′_{i}, D_{i}, D′_{i} in either of the ways suggested by figure 2. Let X_{1}, X′_{1}, Y_{1}, Y′_{1}, Z_{1}, Z′_{1} be respectively the planes \{x = −2\}, \{x = 2\}, \{y = −2\}, \{y = 2\}, \{z = −2\sqrt{2}\}, \{z = 2\sqrt{2}\}.

Each of the planes that comprise the boundary of R and each of the above spheres determine a hyperplane in ℍ^{4} = {(x, y, z, t) ∈ ℝ^{4} | t > 0} that divides ℍ^{4} into two half-spaces. For the spheres we choose the half-spaces whose boundary at infinity is unbounded in ℝ^{3}, for the planes the half-spaces so that the intersection of their boundaries at infinity is the rectangular box R. The polyhedron P is defined as the intersection of those half-spaces. For later convenience, we set P_{−} = {(x, y, z, t) ∈ P | z ≤ 0}, P_{+} = {(x, y, z, t) ∈ P | z ≥ 0}.

The following observations about the spheres and planes that we just defined are easy to check.

(1) Any two spheres that intersect do so at an angle of π/2.

(2) Whenever the intersection is non-empty, spheres A_{i}, A′_{i} intersect planes X_{1}, X′_{1}, Y_{1}, Y′_{1} at angle π/4. Any other pair of spheres or planes with nonempty intersection intersects at angle π/2.

(3) R is completely covered by the closed balls bounded by the spheres. This means that P has finite volume and has only finitely many points in its boundary at infinity.

(4) P has 36 vertices at infinity, which correspond to points not covered by the open balls. Their position is illustrated in Figure 9.

It is not obvious right away that P also has finite vertices (that is, 0-sides that are in ℍ^{4}). This is because many sets of four hyperplanes bounding the polyhedron P meet at one point. For example, sides A_{1}, B_{1}, C_{1} and A′_{2} meet at the point (0, 1, −3\sqrt{2}/2, \sqrt{2}/2) ∈ ℍ^{4}, and sides A_{1}, C_{1}, D_{1} and Y′_{1} meet at the point (−1, 2, −3\sqrt{2}/2, \sqrt{2}/2). Figure 1 depicts the section of P where t = \sqrt{2}/2 and

Figure 2. Side-pairings Φ_{1} (top) and Φ_{2} (bottom)

z = j\sqrt{2}/2. Here j is any of −3, −1, 1, 3 as the section for every j is the same. We can see where four sides of P intersect in a vertex and what letters those sides are labeled by. Figure 1 shows the location of all the vertices in one section — there being 4 · 12 = 48 in all.

Now we are ready to define two ways to pair sides of P. A side-pairing Φ of P is a rule which associates to each side S of P a side S′ of P and an isometry s which sends S to S′. This rule is subject to the conditions that s(int P) ∩ int P = ∅, that the side associated to S′ must be S, that the isometry that takes S′ to S must be s^{−1}. For more details, see [E-P] or [R].

First, define the following isometries of ℝ^{3}:

q_{0} = reflection in plane {z = 0}
q_{1} = reflection in plane {x − y = 0}
q_{2} = reflection in plane {x + y = 0}
s_{1} = rotation by π about line {x + y = 0, z = 0}
s_{2} = rotation by π about line {x − y = 0, z = 0}
t_{0} = translation by 2\sqrt{2} in the z direction.

Use the same letters to denote the extensions of these maps to ℍ^{4}. (A Euclidean isometry f : ℝ^{3} → ℝ^{3} extends to a hyperbolic isometry given by (x, y, z, t) ↦ (f(x, y, z), t).)

Let i_{S} denote the reflection in the hyperplane S. By s we denote the hyperbolic isometry that pairs the sides S and S′ (it sends S to S′). We define Φ_{1} to be the side-pairing given in the upper-half-space model by

x_{1} = translation by 4 in the x direction
y_{1} = translation by 4 in the y direction
z_{1} = t_{0}^{2} = translation by 4\sqrt{2} in the z direction
b_{1} = q_{1} ∘ t_{0} ∘ i_{B_{1}}
a_{j} = q_{l} ∘ i_{A_{j}}, so that l ≡ j (mod 2)
c_{k} = q_{0} ∘ i_{C_{k}}
d_{k} = q_{1} ∘ t_{0} ∘ i_{D_{k}},

where j = 1, …, 6, k = 1, …, 4 and l = 1, 2. The upper half of Figure 2 shows which sides are paired.

To get another side-pairing, Φ_{2}, we alter Φ_{1} in the way the sides labeled by B’s, C’s and D’s are paired. Refer to the lower half of Figure 2 to see which sides are paired. We define the new pairings b_{1}, c_{k} and d_{k} by

b_{1} = q_{0} ∘ i_{B_{1}}
c_{k} = q_{1} ∘ t_{0} ∘ i_{C_{k}}
d_{k} = s_{l} ∘ q_{l} ∘ i_{D_{k}}, so that l ≡ k (mod 2),

where k = 1, …, 4 and l = 1, 2.

2. Two closed finite-volume hyperbolic 4-manifolds

In this section we prove

Theorem 2.1.

(i) The side-pairings Φ_{1} and Φ_{2} generate discrete torsion-free subgroups G_{1} and G_{2} of Isom ℍ^{4} whose fundamental polyhedron is P. Therefore, the quotient of ℍ^{4} by the action of either of the groups is a complete hyperbolic 4-manifold.

(ii) ℍ^{4}/G_{1} has seven ends while ℍ^{4}/G_{2} has eight. In particular, the two manifolds are not homeomorphic.

Proof. To prove assertion (i), we use Poincaré’s polyhedron theorem. For details, the reader can consult [E-P] and [R] which were our main references, while we will shortly state the version that we are going to use. Other versions of the polyhedron theorem may be found in [A2] and [Ma].

First of all, the maps defined above really do map a side of P isometrically onto a side of P. To verify this for Φ_{1}, notice that each of its side-pairings

Figure 3. Representing edges in P by diagrams

a_{1}, …, a_{6}, c_{1}, …, c_{4}, is of the form f ∘ i_{S} where S is a side and f, which preserves P, is an extension to ℍ^{4} of a Euclidean transformation on ℝ^{3}. The i_{S} keeps S fixed so f ∘ i_{S}(S) = f(S), and f, being an isometry of P, sends its sides to some other sides. We proceed similarly for the the side-pairings b_{1}, d_{1}, …, d_{4}: each of them is of the form f ∘ i_{S}, but f is now an isometry that takes P_{−} to P_{+}, so it sends sides of P_{−} to sides of P_{+}. But the sides B_{1}, D_{1}, …, D_{4} are sides of both P_{−} and P, and, likewise the sides B′_{1}, D′_{1}, …, D′_{4} are sides of both P_{+} and P. For the side-pairing x_{1} (y_{1} and z_{1} are done similarly), compose it with a reflection in the side X′_{1} to get an isometry of P whose image of X_{1} are the same as by x_{1}. Therefore x_{1} carries X_{1} to another side of P, namely X′_{1}. The claim is proved in the same way for the side-pairing Φ_{2}.

Remark 2.2. Recall that a horosphere in ℍ^{n} is either a Euclidean sphere tangent to ∂ℍ^{n} or a Euclidean hyperplane {t = c} parallel to ∂ℍ^{n}. The former are said to be centered at the point of tangency with ∂ℍ^{n}, the latter at ∞. Consider a set T of disjoint horospheres, each centered at a vertex at infinity of P. For the vertex ∞ choose, say, the horizontal plane {t = 3}. For the other vertices choose horospheres of the same radius that is small enough so that the horospheres intersect only those sides of P which contain the center of the horosphere. If the center of a horosphere is on the boundary of a hyperplane S, then i_{S} preserves the horosphere. This combined with an argument like in the preceding paragraph can be used to show that the side pairings Φ_{1} and Φ_{2} satisfy the

Consistent horosphere condition: there exists a set T of disjoint horospheres centered at ideal vertices of P so that if g is a side-pairing of a side that contains the center of a horosphere H ∈ T in its boundary, then g(H) is again a horosphere from T.

Another condition for Poincaré’s polyhedron theorem is the ”edge cycle condition”, called Cyclic in [E-P]. In general, a side-pairing on P induces an equivalence relation on P that is generated by the relation x ∼ s(x), where x ∈ ∂P ∩ S, S is a side of P and s its side-pairing. The equivalence class [x] of x under this equivalence relation, is called the cycle of x. The cycle of an i-side is defined analogously, so that it contains all the i-sides of P that are identified by a string of side-pairings.

Figure 4. Edge chase for Φ_{1}

Every edge (codimension-2 side) of P is the intersection of two uniquely determined sides of P. The dihedral angle at an edge is the angle in the interior of P that the two sides subtend. A cycle of edges can be obtained in the following way. Start with an edge E_{1}, which is the intersection of sides S_{1} and R_{1}, and let g_{1} be the isometry pairing R_{1} and some side S_{2} of P. We get that g_{1}(E_{1}) = E_{2}, where E_{2} is an edge determined by S_{2} and some other side R_{2}. Now let g_{2} be the isometry pairing R_{2} and some side S_{3}. Continuing in the same way we get a sequence of edges, sides and isometries {σ_{i} = (E_{i}, S_{i}, R_{i}, g_{i})}_{i=1,2...}. This procedure is commonly called ”edge-chasing”. We require that the above sequence have a period q (called first cycle length in [E-P]), that is σ_{q+1} = σ_{1} for some q. The cycle of edges will then consist of exactly E_{1}, …, E_{q}. Due to finite-sidedness of our P, this condition will automatically be satisfied.

It is clear that g_{q} ∘ ⋯ ∘ g_{1}(E_{1}) = E_{1}, but it may happen that the restriction of g_{q} ∘ ⋯ ∘ g_{1} on E_{1} is not the identity. The second part of the edge cycle condition is that there must be a k so that (g_{q} ∘ ⋯ ∘ g_{1})^{k} |_{E_{1}}= id. The number kq is called the second cycle length in [E-P].

Finally, to fulfill the edge cycle condition we must show that if θ_{i}, i = 1, …, q is the dihedral angle of edge E_{i}, then there is a non-zero integer m so that k(θ_{1} + ⋯ + θ_{q}) = 2π/m. We may now formulate Poincaré’s polyhedron theorem for the case of a hyperbolic polyhedron as follows.

Theorem 2.3. (Poincaré’s polyhedron theorem) Let Φ be a side-pairing on a polyhedron P ⊂ ℍ^{n} that satisfies both the edge cycle condition and the consistent horosphere condition. Then the side-parings of Φ generate a discrete group G ⊂ Isom ℍ^{n} whose fundamental polyhedron is P. □

Now we check the edge cycle condition for edges of P and the two side-pairings Φ_{1} and Φ_{2}. With notation as above, we will always have k = 1 and m = 1. Hence, the second cycle length will always be the same as the first cycle length and they will be 4 and 8 respectively for edges with dihedral angles π/2 and π/4. Therefore, the sum of dihedral angles will be exactly 2π.

Firstly, we make sure that all edges are in cycles of said length. For all edges that are intersections of sides based on planes (i.e. ”vertical” sides), this check is easy and boils down to checking the conditions of Poincaré’s polyhedron theorem for a rectangular Euclidean parallelepiped with parallel sides paired by Euclidean translations. The check for any of the edges of type A_{i} ∩ Z_{1} is also straightforward. For all the other edges, we use the diagrams in Figures 4 and 5 to simplify and visualize the task of verifying.

Notice that the intersection of two hyperbolic hyperplanes is a codimension-2 hyperbolic subspace whose boundary at infinity is the intersection of the boundaries at infinity of the hyperplanes. Therefore, every edge E of P lies on a codimension-2 subspace determined either by the intersection of a sphere and a plane or by the intersection of two spheres in ℝ^{3}. (The spheres and planes are the boundaries at infinity of the sides that determine E.) These intersections are circles and they are represented by segments in Figure 3. The letters next to each segment indicate which side-types have generated the edge represented by it. To get the left and right diagrams we take intersections of planes and spheres labeled by the letters in the diagrams and then project them to the plane {z = 0}. For the middle diagram, we first project to the plane {z = 0} the centers of those pairs of spheres whose labels are listed in it. Then we take the perpendicular bisector of the line joining those centers. To account for all the edges, we need several of these diagrams (Figure 4). Edge chasing for Φ_{1} and Φ_{2} is now performed on Figures 4 and 5 respectively. All edges in one cycle in each horizontal component of the pictures are labeled with the same letter. Some are not labeled because their cycles are similar to other labeled cycles. Also, Figure 5 omits some of the edges because their cycles are the same as for Φ_{1}.

For example, choose the edge A_{1} ∩A_{2}. The edge chase, yielding the cycle labeled o in upper part of Figure 4 is

Figure 5. Edge chase for Φ_{2}

A_{1} ∩ A_{2} ⟶^{a_{2}} A′_{2} ∩ A_{1} ⟶^{a_{1}} A′_{1} ∩ A′_{2} ⟶^{a_{2}^{−1}} A_{2} ∩ A′_{1} ⟶^{a_{1}^{−1}} A_{1} ∩ A_{2}

As another example, choose the edge A_{1} ∩D_{1}. The edge chase, yielding the cycle labeled k in lower part of Figure 4 is

A_{1} ∩ D_{1} ⟶^{d_{1}} D′_{1} ∩ A′_{3} ⟶^{a_{3}^{−1}} A_{3} ∩ D′_{3} ⟶^{d_{3}^{−1}} D_{3} ∩ A′_{1} ⟶^{a_{1}^{−1}} A_{1} ∩ D_{1}.

Next, we check that for transformations g_{1}, …, g_{q} obtained by edge-chasing we have g_{q} ∘ ⋯ ∘ g_{1}|_{E_{1}}=1. As before, g_{i} = f_{i} ∘ r_{i}, where f_{i} is the extension to ℍ^{4} of a Euclidean transformation on ℝ^{3} and r_{i} is either reflection in a hyperplane containing E_{i} or the identity. Let f = f_{q} ∘ ⋯ ∘ f_{1}. It is not difficult to see that f is always orientation preserving. Clearly g_{q} ∘ ⋯ ∘ g_{1}|_{E_{1}} = f|_{E_{1}}, and it will be enough to show that f = 1. We will need the following easy lemma.

Lemma 2.4. Let f be a nontrivial orientation-preserving Euclidean isometry of ℝ^{3} that preserves a circle. Then it is a rotation about a line called the axis of f. Moreover, we have:

(i) If we write f as Ux + u, where U is an orthogonal transformation and u ∈ ℝ^{3} then the axis of f is parallel to the axis of U.

Figure 6. Cycles of real vertices

(ii) The axis of f passes through the center of the circle and is either in the plane of the circle, or perpendicular to it. In the first case, the rotation is by angle π.

(iii) If f preserves a line l, then its axis either orthogonally intersects l or it is that line. □

Let f = f_{q} ∘ ⋯ ∘ f_{1} as above and suppose it is nontrivial. In what follows we interchangeably view f as a Euclidean isometry on ℝ^{3} or as a hyperbolic isometry of ℍ^{4}.

We know that f preserves the circle that is the base of the edge E = E_{1}. Also, f preserves the family of planes V = {(x, y, z) ∈ ℝ^{3} | x = 4k+2 or y = 4k+2, k ∈ ℤ} because it is a composite of maps from {x_{1}, y_{1}, z_{1}, q_{0}, q_{1}, q_{2}, s_{1}, s_{2}}. The rotational part U of f is a composite of maps from {q_{0}, q_{1}, q_{2}, s_{1}, s_{2}}, each of which preserves the axes l_{1} and l_{2} of s_{1} and s_{2}, so U preserves them too. Now, looking at possible positions of the circle we get one of the following cases:

Case 1. When E is one of the sides represented in the lower half of Figure 4 then, by (i) and (ii) of Lemma 2.4, the axis l of U must either have direction vector (±1, ±1, \sqrt{2}) or is in the plane perpendicular to that vector. Since U preserves l_{1} and l_{2}, by part (iii) of the lemma, both of l_{1} and l_{2} must be either perpendicular or identical to l. If l ⊥ l_{1} and l ⊥ l_{2}, then l is the z-axis, which contradicts the possible positions of l. If l is equal to either l_{1} or l_{2}, then one can see that the axis of f is going pass exactly through the segment that represents E in the middle diagram of Figure 3. However, it is clear that no rotation about these segments can preserve the family of planes V so we must have f = 1.

Case 2. When E is one of the sides represented in the upper half of Figure 4 the axis l of U lies, by parts (i) and (ii) of Lemma 2.4, in one of the the planes {x = 0} or {y = 0} or it is the x- or y-axis. Clearly l ≠ l_{1} and l ≠ l_{2}, so applying part (iii) of Lemma 2.4 again we get l ⊥ l_{1} and l ⊥ l_{2}, which means l is the z-axis. As long as E is not of the forms C_{i} ∩X_{1}, C_{i} ∩Y_{1}, D_{i} ∩X_{1} or D_{i} ∩Y_{1} the center of the circle on which E is based has odd x and y coordinates. However, a rotation about an axis parallel to z through such points cannot preserve the family of planes V and we again get f = 1.

Case 3. In the remaining cases, if E is of form D_{i} ∩ X_{1} or D_{i} ∩ Y_{1}, regard f as an isometry of ℍ^{4} and examine its action on vertices of P that are on E. Looking at Figure 1, it is clear that f, being by the above a rotation about an axis parallel to z, must send vertices that are on E to points whose either x or y coordinate falls out of [−2, 2], a contradiction with the fact that f preserves E. Finally, if E is of form C_{i} ∩ X_{1} or C_{i} ∩ Y_{1} we just compute f: it is always q_{0}^{2} = 1 for the side-pairing Φ_{1} and it is always q_{1}^{2} = 1 for the side-pairing Φ_{2}.

Thus, we have shown that f = 1 in all possible cases and the edge cycle condition has been verified.

Since the consistent horosphere condition is fulfilled by Remark 2.2, we may apply theorem 2.3 to get that the groups G_{1} and G_{2} generated by the side-pairings Φ_{1} and Φ_{2} are discrete, and that P is the fundamental polyhedron for both of them.

What we do not yet know is whether G_{1} and G_{2} are torsion-free, that is, whether ℍ^{4}/G_{i}, i = 1, 2 are hyperbolic manifolds and not just orbifolds.

Figure 7. Cycles of some 1-sides for Φ_{1}

Recall (see [R]) that the normalized solid angle at point x of a polyhedron P is defined as ω(x) = (Vol B(x, r) ∩ P)/ Vol B(x, r). Here B(x, r) is a hyperbolic ball about x or radius r, Vol is hyperbolic volume and r is taken small enough so that B(x, r) intersects only those sides of P on which x lies. Let [x] = {x_{1}, …, x_{n}} be the cycle of x for some side-pairing of P. We define the normalized solid angle sum of [x] as ω[x] = ∑_{y∈[x]} ω(y). We are going to use Theorem 11.1.1 from [R] which says that

Theorem 2.5. If ω[x] = 1 for every x ∈ P then the group G generated by the side-pairings of P is torsion-free. □

Knowing that P is a fundamental polyhedron for a discrete group G implies ω[x] ≤ 1 for every point of P. Really, for every x_{i} ∈ [x], choose an isometry g_{i} ∈ G taking x_{i} to x. (In general, there may be many ways to make the choices.) We now have a injective map from {x_{1}, …, x_{n}} to {translates of P under G containing x} given by x_{i} ↦ g_{i}(P). Since {g_{1}(P) ∩ B(x, r), …, g_{n}(P) ∩ B(x, r)} fill out maybe only a portion of B(x, r), we get ω[x] ≤ 1. (Note that the strict inequality will occur if and only if x is the fixed point of an element in G.)

Therefore, it is enough to see that ω[x] ≥ 1. For an x ∈ P that is in the interior of 3 or 4-sides of P, it is clear that ω[x] = 1. For an x in the interior of 2-sides, this is the edge-cycle condition. This leaves 0- and 1-sides to be checked.

Figure 8. Cycles of some 1-sides for Φ_{2}

Figure 6 shows the sections of P for t = \sqrt{2}/2 and z = −3\sqrt{2}/2,−\sqrt{2}/2,\sqrt{2}/2, 3\sqrt{2}/2. These sections contain all the vertices (0-sides) of P. All vertices in the same cycle (there are only two cycles) are labeled by the same letter — the cycles are the same for both Φ_{1} and Φ_{2}. Vertices in the cycle labeled a occur as the intersection of four hyperplanes, each pair of which meets at angle π/2. Normalize P so that a vertex x from cycle a is the origin in the ball model 𝔹^{4} of hyperbolic space. We see that the normalized solid angle at x is the same as the normalized solid angle at 0 ∈ 𝔹^{4} subtended by the four coordinate planes, and that is 1/16. Vertices in the cycle labeled b are always intersections of four hyperplanes where one pair of them intersects at angle π/4 and all other pairs intersect at angle π/2. Normalizing as before, we see that the normalized solid angle at x is the same as the one at 0 subtended by the hyperplanes {x_{2} = 0}, {x_{3} = 0}, {x_{4} = 0}, {x_{1} − x_{2} = 0}, and that is 1/32. Since cycles a and b contain 16 and 32 points respectively, we are done.

Now consider 1-sides. There are three cases depending on whether a 1-side F connects an ideal and a real vertex, two ideal vertices, or two real vertices.

For the first case, let F be a 1-side of P that is a geodesic half-line between one real and one ideal vertex of P. If for some x ∈ int F we have ω[x] < 1, then x is a fixed point of some nontrivial g ∈ G. The isometry g must preserve F — otherwise, we’d have g(F) ∩ F = {x} and this contradicts the fact that translates of P meet only along i-sides. However, this implies that the real vertex on F is fixed under g, a possibility we just proved cannot happen. So, we are left with checking 1-sides that have as endpoints either both real or both ideal vertices of P.

Every 1-side is the intersection of three different sides of P. Hence, to find all 1sides with both endpoints real or ideal, we have to find pairs of real or ideal vertices lying on the same three sides. Figure 7 schematically depicts those 1-sides of P. A boldface line segment joining the real or ideal vertices indicates the existence of a 1-side joining them. The three sides on which the 1-sides lie are easily deduced from their position in the picture. (For example, the 1-sides labeled c are intersections of sides labeled by D’s, X’s and Y ’s.) As before, the letters on the 1-sides indicate to which cycle of 1-sides they belong.

It takes a bit of checking to see that we have found all the needed 1-sides. For example, to see that no 1-side joins an ideal vertex in the plane {z = 0} to an ideal vertex in the plane \{z = \sqrt{2}\} we note that every vertex in the plane \{z = 0\} lies on only one of the B, C or D-sides, and some X, Y, Z or A-sides, while every vertex in the plane \{z = \sqrt{2}\} lies on only one of the A-sides and some number of B, C or D-sides. Therefore, any pair of vertices from those two hyperplanes cannot belong to the same three sides.

The 1-sides in the cycle labeled a are intersections of three sides meeting pairwise at angles π/2 (two sides labeled by A and one by Z) . Taking an x from a side in the cycle, and normalizing in 𝔹^{4} so that x = 0 and the three sides are the first three coordinate hyperplanes, we see that ω(x) = 1/8. Since there are 8 1-sides in the cycle, we get ω[x] ≥ 1.

Other cycles are checked in the same way. For an x on a 1-side in the cycles b, c, d, e we get normalized solid angles of respectively 1/16, 1/8, 1/8, 1/8 with 16, 8, 8, 8 1-sides in the cycle, so ω[x] ≥ 1. Thus, we have proved that side-pairings Φ_{1} and Φ_{2} give rise to hyperbolic 4-manifolds.

Assertion (ii) of Theorem 2.1 is verified by counting cycles of ideal vertices for Φ_{1} and Φ_{2}. The cycles are shown in Figure 9. □

3. Proof of Theorem A

The crucial observation that the proof uses is that the side-pairings Φ_{1} and Φ_{2} yield manifolds with respectively 7 and 8 ends. This is immediate from the fact Figure 9. Cycles of ideal vertices for Φ_{1} (top) and Φ_{2} (bottom). In both cases, the vertex ∞ is labeled by 1

that there are 7 and 8 classes of vertices at infinity for the side-pairings Φ_{1} and Φ_{2}, as shown in Figure 9.

Choose a positive integer n, and let Q_{1}, …, Q_{n} be n copies of the polyhedron P. Call each of them a block. To each block, assign either of the side-pairings Φ_{1} or Φ_{2} and call it a block of type Φ_{1} or Φ_{2}. Now form a new polyhedron Q by attaching side Z_{1} of Q_{i} to side Z′_{1} of Q_{i−1}, i = 2, …, n, i.e. by stringing the blocks together in a linear fashion in the direction of the z-axis. Q has on it a side-pairing induced by the side-pairings on each block. Clearly, the sides that were attached have vanished, so they don’t fall under this rule: the remaining sides Z_{1} of Q_{1} and Z′_{1} of Q_{n} are paired by the translation t^{2n}. A moment’s reflection convinces us that the side-pairing on Q generates a torsion-free group G — the proof is basically an n-fold repetition of the proof of the same result for the side-pairings Φ_{1} and Φ_{2}. Except for i-sides, i = 0, 1, 2, that are contained in the sides Z′_{1} of Q_{1} and Z_{1} of Q_{n}, all the cycles are just inherited cycles from pairings on each block. The special cases are easily dealt with — they follow patterns established for Φ_{1} and Φ_{2}.

How many cycles of vertices at infinity does Q have? Let there be k blocks of type Φ_{1} and n − k blocks of type Φ_{2} among Q_{1}, …, Q_{n}. Starting with Q_{1}, …, Q_{l}, adding a block of type Φ_{1} onto Q_{l} will add three new cycles of vertices at infinity: there are seven cycles on Φ_{1} but four fall into cycles already existing on Q_{1}, …, Q_{l}. Similarly, adding a block of type Φ_{2} adds four new cycles of vertices at infinity. It is now easy to see that Q will have 4 + 3k + 4(n − k) = 4 + 4n − k cycles of vertices at infinity. We have complete freedom of choice for k, so by varying k from 0 to n we can get manifolds with anywhere from 4+3n to 4+4n ends. Therefore, we have obtained at least n + 1 nonhomeomorphic manifolds with the same fundamental polyhedron Q, which is what we set out to prove. (We likely get many more, since we completely ignored the various orderings of blocks of the two types that are possible when constructing Q.) □

4. A geometric interpretation of the construction

In this section we analyze the construction of the manifolds in the previous section from a gluing-of-manifolds perspective.

Let M = ℍ^{n}/G, where G is a discrete torsion-free subgroup of Isom ℍ^{n}. A totally geodesic hypersurface is a subset N ⊂ M so that for every x, y ∈ N every geodesic connecting x and y is also contained in N. We are interested in embedded totally geodesic hypersurfaces which are the ones for which p^{−1}(N) is a disjoint union of hyperplanes in ℍ^{n}, where p : ℍ^{n} → M is the standard projection. Let H be one of those hyperplanes, and J ⊂ G its stabilizer in G, that is the subgroup J = G_{H} = {j ∈ G | j(H) = H}. Then H is precisely invariant under J, i.e. g(H) = H when g ∈ J and g(H)∩H = ∅ when g ∈ G\J. In particular, we want to look at some hypersurfaces with Vol N < ∞, so they will correspond to subgroups G_{H} ⊂ G that act on a hyperplane H as a hyperbolic lattice.

Conversely, we may start with a subgroup J ⊂ G and a hyperplane H precisely invariant under J. Then H/J is an embedded totally geodesic hypersurface in M.

Let M_{i} = ℍ^{4}/G_{i}, where G_{i} is generated by the side-pairing Φ_{i}, i = 1, 2, and let H be the supporting hyperplane of side Z_{1} of the polyhedron P. Suppose we know that H is precisely invariant under the subgroup J ⊂ G_{1}, J =< a_{1}, a_{2}, x_{1}, x_{2} >. Then H/J is a two-sided totally geodesic hypersurface in M_{1} and we may cut M_{1} along this hypersurface to get a connected manifold M′_{1} that has two boundary components which are isometric 3-dimensional hyperbolic manifolds given by H/J. We may do the same with M_{2} to get M′_{2} which is also connected (the subgroup J in question doesn’t change).

Now let M = ℍ^{4}/G, where G is a subgroup generated by any of the side-pairings of the polyhedron Q defined in the previous section. We identify the block Q_{1} with P and its side-pairing with Φ_{1} or Φ_{2}. Let H_{i} be the hyperplane supporting the side Z_{1} of the block Q_{i}, i = 1, …, n. So, H_{i} = z_{1}^{i−1}(H) and it is kept invariant by the subgroup J_{i} ⊂ G, J_{i} = z_{1}^{i−1}Jz_{1}^{−(i−1)} (see Figure 10). Suppose we know that H_{i} is precisely invariant under the subgroup J_{i} ⊂ G . Then we may cut M along the hypersurfaces H_{i}/J_{i}. The hyperplanes H_{i} are exactly the ones that separate the polyhedron Q into blocks Q_{1}, …, Q_{n} (see Figure 10). Under identification of paired sides, we see that each block yields a submanifold of type M′_{1} or M′_{2}. Therefore, after cutting along H_{i}/J_{i}, i = 1, …, n, we will get n pieces, each isometric to either M′_{1} or M′_{2}. Thus, we will have shown

Figure 10. The planes H_{i} in the polyhedron Q

Proposition 4.1. Any of the manifolds constructed in section 3 are obtained by gluing n copies of either M′_{1} or M′_{2} so that each copy of M′_{1} or M′_{2} is glued to another copy along one of their two totally geodesic boundaries i.e. they are strung together in a circular fashion. Here M′_{1} and M′_{2} are manifolds that we obtain by cutting hyperbolic 4-manifolds M_{1} and M_{2} along a totally geodesic hypersurface. □

Remark 4.2. Let G_{0i}, i = 1, 2 be the subgroup generated by the same generators as G_{i}, but with z_{1} omitted. Then, by using Maskit’s combination theorems (see theorems 6.19 and 6.24 in [A2]) and fairly standard arguments (see, for example, [A3]) it is possible to show that

G = (G_{0i_{1}} ∗_{J_{2}} (z_{1}G_{0i_{2}}z_{1}^{−1}) ∗_{J_{3}} (z_{1}^{2}G_{0i_{3}}z_{1}^{−2}) ∗_{J_{4}} … ∗_{J_{n}} (z_{1}^{n−1}G_{0i_{n}}z_{1}^{−(n−1)})) ∗_{z_{1}^{n}} .

Here ∗_{J_{i}} denotes the free product with amalgamation, while the last ∗_{z_{1}^{n}} is the HNN-

extension of the free amalgamated product in the parentheses by z_{1}^{n}, where z_{1}^{n} conjugates subgroups J_{1} and z_{1}^{n}J_{1}z_{1}^{−n}. The index i_{k} is 1 or 2 depending on whether the block Q_{k} is of type Φ_{1} or Φ_{2}.

Remark 4.3. The hyperbolic 3-manifold H/J has been described by Wielenberg, see Example 3 in [Wi2]. It is the complement of a certain four-component link in S^{3}. Thus, we are gluing 4-manifolds along boundaries that are link complements in S^{3}.

The only thing left to do is to verify that the hyperplanes H, H_{1}, …, H_{n} are precisely invariant, respectively, under the subgroups J, J_{1}, …, J_{n} ⊂ G. To this end we use the technical theorem stated below.

Theorem 4.4. Let P be a fundamental polyhedron for a discrete group G ⊂ Isom(ℍ^{n}) that is generated by some side-pairing of P. Let H be a hyperplane in ℍ^{n} so that int_{H}(H ∩P) ≠ ∅ and let J be a subgroup of G that keeps H invariant. Assume the following three conditions hold:

(1) H ∩ P is a fundamental polyhedron for the action of J in H.

(2) If H contains a side S of P and s is the side-pairing corresponding to S then s(H) ≠ H.

(3) Suppose H contains an edge E of P. Let {σ_{i} = (E_{i}, S_{i}, R_{i}, g_{i})}_{i=1,2...} be the sequence obtained by the edge-chase corresponding to E as in the edge cycle condition. (That is, E_{1} = E, S_{i} and R_{i} are the sides that determine E_{i}, E_{i} = S_{i} ∩ R_{i}, g_{i}(E_{i}) = E_{i+1} and g_{i} is the side-pairing that pairs R_{i} and S_{i+1}.) Let α be the angle between H and S_{1} and let θ_{i} be the dihedral angle of P at the edge E_{i}. If another edge E_{l+1} in the cycle of E is contained in H and β is the angle between H and S_{l}, then θ_{1} + ⋯ + θ_{l} + β − α = kπ must be satisfied for some integer k.

Then H is precisely invariant under J.

Proof. Let f ∈ G and suppose that K = f(H) ∩ H ≠ ∅. We want to show that f ∈ J. There are three cases depending on how K intersects the elements of the tiling {g(P), g ∈ G}. We will repeatedly use the fact that H ⊂ ∪_{j∈J} j(P) and f(H) ⊂ ∪_{j∈J} fj(P) which follows from the assumption that H ∩P is a fundamental polyhedron for J in H.

Case 1. There exists an x ∈ K so that x ∈ int g(P) for some g ∈ G. Since x ∈ H, there must be a translate of P under J that contains x. The only possible candidate is g, so we conclude g ∈ J. Likewise, since x ∈ f(H), there is a j ∈ J so that x ∈ fj(P). But this can only happen if fj = g, so f = gj^{−1} ∈ J.

Case 2. There exists an x ∈ K that is contained in the interior of a side R of some translate of P. Then R is common to exactly two translates of P. If R is not contained in H then H cuts into the interior of both of those translates. Again, the parts of H that are in the interiors of these translates must be covered by translates of P under J so the translates abutting R are of form j(P) and j′(P) for some j, j′ ∈ J. Furthermore, there must be a j′′ ∈ J so that x ∈ fj′′(P). Since x is in only two translates of P, this means that either fj′′ = j or fj′′ = j′. In both cases we get f ∈ J.

If, on the other hand, H does contain R, then at least one of the two translates of P abutting R is of form j(P), j ∈ J. Assuming f(H) ≠ H gives us that f(H) intersects the interior of j(P). The portion of f(H) in int j(P) must be in some fj′(P) for some j′ ∈ J so we get fj′ = j, which forces f ∈ J, contradicting f(H) ≠ H. Therefore f(H) = H. Now the two translates of P that abut R are of form j(P) and js^{−1}(P), where s is the side-pairing of the side S for which j(S) = R. One of those translates is also of form fj′(P) for some j′ ∈ J. If fj′ = j then f ∈ J. The other case, fj′ = js^{−1} implies s = j′^{−1}f^{−1}j, so s preserves H. This, however, contradicts assumption (2), because H contains S, since R ⊂ H, S = j^{−1}(R) and j^{−1}(H) = H.

Case 3. If neither case 1 nor 2 occurs, we get that K is contained in translates of edges of P, which are n − 2-dimensional. Since dim K ≥ n − 2 we get that K must be (n − 2)-dimensional, which implies f(H) ≠ H. Furthermore, there exists an x ∈ K and an edge E′ of some g(P) so that x is in the interior of E′. As before, one of the translates of P that contains x must be of the form j(P). Move everything by j^{−1} so that x is now on an edge E of P and E ⊂ j^{−1}f(H) ∩ H. The translates of P that abut E are P, g_{1}^{−1}(P), g_{1}^{−1}g_{2}^{−1}(P), …, so as before, there must be a j′ ∈ J and an integer l so that j^{−1}fj′ = g_{1}^{−1} ∘ ⋯ ∘ g_{l}^{−1}. But then j^{−1}f(H) = j^{−1}fj′(H) = g_{1}^{−1} ∘ ⋯ ∘ g_{l}^{−1}(H), so E ⊂ g_{1}^{−1} ∘ ⋯ ∘ g_{l}^{−1}(H) ∩ H. From E ⊂ g_{1}^{−1} ∘ ⋯ ∘ g_{l}^{−1}(H) we get that E_{l+1} = g_{l} ∘ ⋯ ∘ g_{1}(E) ⊂ H, so E_{l+1} is in the cycle of E and is contained in H. Let K^{⊥} be the 2-dimensional orthogonal complement of K through x. The intersections of translates of P that abut E with K^{⊥} are angles with rays emanating from a single vertex x. Intersections of H and g_{1}^{−1} ∘ ⋯ ∘ g_{l}^{−1}(H) with K^{⊥} are two lines and the angle between them is θ_{1} + ⋯ + θ_{l} + β − α. Condition 2) now says that this angle is kπ, so the lines are identical and so are the hyperplanes that they represent. From here it follows that f(H) = H, a contradiction with f(H) ≠ H. Therefore, case 3 never occurs and f ∈ J by cases 1 and 2. □

Remark 4.5. The proof of Theorem 4.4 did not use any hyperbolic space-specific properties, only the fact that P was a fundamental polyhedron. Therefore it also applies in the other two constant curvature settings, that is, for fundamental polyhedra of discrete isometry groups of the n-sphere and Euclidean n-space.

Remark 4.6. Notice that the group G in the theorem did not have to be torsionfree. However, if H contains an edge of P, condition (3) allows the number k that was defined in the edge cycle condition to only be 1 or 2.

Examples. We give several applications of the theorem that include the claims of precise invariantness needed for Proposition 4.2. All except example 4.8 have as P the polyhedron defined in section 1.

Example 4.7. Let G = G_{1} or G_{2}, H=the hyperplane based on the plane {z = 0}, J =< x_{1}, y_{1}, a_{3}, a_{4} >. Clearly H is invariant under J. By applying (now in dimension 3) Poincaré’s polyhedron theorem to H ∩ P and restrictions of x_{1}, y_{1}, a_{3}, a_{4} to H we may easily see that H ∩ P is a fundamental polyhedron for J in H. (Here conditions (2) and (3) from the theorem do not apply.) Therefore, H/J is a totally geodesic hypersurface embedded in M_{1} or M_{2}.

Example 4.8. Let G = G_{1} or G_{2}, H = Z_{1} and J =< x_{1}, y_{1}, a_{1}, a_{2} >. As in Example 4.7 we check that H ∩ P is a fundamental polyhedron for J in H. Here we also need to verify condition (3) of Theorem 4.4. (Condition (2) clearly holds.) Taking, for example, E = Z_{1} ∩ A_{1} whose cycle is {Z_{1} ∩ A_{1}, A′_{1} ∩ Z_{1}, Z′_{1} ∩ A_{5}, A′_{5} ∩ Z′_{1}, } we see that α = 0, l = 1 and β = π/2, so condition (3) is satisfied. Using Theorem 4.4 gives us that H/J is a totally geodesic hypersurface embedded in M_{1} or M_{2}.

Example 4.9. It is now easy to see that the hyperplanes H_{1}, …, H_{n} (in above notation) are precisely invariant under the subgroups J_{1}, …, J_{n} ⊂ G. The proof for H_{1} and J_{1} corresponds to the one in Example 4.8, while the other cases correspond to Example 4.7. This completes the proof of Proposition 4.2.

Example 4.10. Let G = G_{1} or G_{2} and let H_{1}, H_{2} be the hyperplanes based respectively on the planes {x − y = 0} and {x + y = 0}. We may use Theorem 4.4 to verify that H_{1} is precisely invariant under < a_{2}, a_{4}, a_{6}, b_{1}, d_{2}, d_{4}, y_{1}x_{1} > and that H_{2} is precisely invariant under < a_{1}, a_{3}, a_{5}, b_{1}, d_{1}, d_{3}, y_{1}^{−1}x_{1} >. Again, condition (1) of Theorem 4.4 is verified by the Poincaré polyhedron theorem in dimension 3. Note that condition (3) of that same theorem applies.

5. Proof of Theorem B

We will show that the polyhedron P has volume 2 · 4π^{2}/3. Then every manifold obtained from any of the side-pairings of Q described in section 3 will have volume 2n · 4π^{2}/3. The Gauss-Bonnet formula (see [G], page 84, and [H]) applied to a non-compact hyperbolic 4-manifold M gives Vol M = χ(M) · 4π^{2}/3. Here χ(M) is the Euler characteristic of the compact part of M, i.e. the manifold with boundary obtained by retracting every end E × [0, ∞) of M to E × {0}. It will be enough to show that the manifold obtained from either of the side-pairings Φ_{1} or Φ_{2} has Euler characteristic 2.

Recall that for a finite CW-complex X, χ(X) may be computed either as an alternating sum of the numbers of i-cells in X, or as the alternating sum of the ranks of the i-th homology groups of X.

Let P be finite-sided n-dimensional hyperbolic polyhedron with a side-pairing defined on it. Now let X be the CW-complex obtained from P in the obvious way, with 0-cells the real and ideal vertices of P, 1-cells the 1-sides of P together with their points at infinity, and so on. Then X inherits identifications by side-pairings of P, which give rise to a quotient space Y, also a CW-complex (even if P yields a manifold by identification, Y will not be one). We then have

Lemma 5.1. If M is obtained from a side-pairing of P and Y is as above, then

χ(M) = χ(Y) − number of ends of M.

In other words, we may compute χ(M) directly from the polyhedron by taking the alternating sum of numbers of cycles of i-sides and ignoring the cycles of ideal vertices.

Proof. Using M to also denote the compact part of the hyperbolic manifold, we see that Y = M ∪V, where V is a disjoint union of cones over Euclidean manifolds that are the boundary of M, so that M ∩ V is a disjoint union of Euclidean manifolds. Consider the absolute Mayer-Vietoris sequence for M and V (see [Do], Proposition 8.15). Enumerate the terms so that the k-th homologies of M ∪ V, the sum of M and V, and M ∩ V correspond to indices 3k, 3k + 1 and 3k + 2 respectively (k ≥ 0). Let c_{j} and z_{j} denote respectively the rank of the j-th term and the rank of the kernel of the homomorphism joining the j-th and the (j − 1)-st terms of the sequence. By exactness of that sequence we have c_{j} = z_{j−1} + z_{j}. Use this equality to see that ∑(−1)^{k}c_{3k}−∑(−1)^{k}c_{3k+1}+∑(−1)^{k}c_{3k+2} = 0, which is exactly χ(M ∪V)−(χ(M)+χ(V))+χ(M ∩V) = 0. Since V is contractible, χ(V) =number of components of V =number of of boundary components of M. The fact that M ∩V is a disjoint union of Euclidean manifolds implies χ(M ∩ V) = 0 which yields the desired formula. □

Now to finish the proof of Theorem B, we just have to count cycles of i-sides for Φ_{1} or Φ_{2}. The polyhedron P from section 1 has one 4-side, 36 3-sides, 168 2-sides, 216 1-sides and 48 real 0-sides. Each 3-side is paired to exactly one other one, which yields 18 cycles of 3-sides. Among 2-sides, there are 24 with dihedral angle π/4, giving 3 cycles, and 144 with dihedral angle π/2, yielding 36 cycles. Among 1-sides, there are 80 with normalized solid angle 1/16 giving 5 cycles and 136 with normalized solid angle 1/8, yielding 17 cycles. From Figure 6 we know there are 2 cycles of 0-vertices. Thus, χ(M) = 2 − (17 + 5) + (36 + 3) − 18 + 1 = 2, which completes the proof. □

Remark 5.2. Note that the same reasoning as in the above paragraph may be used to see that, when the side-pairing of an n-polyhedron P yields a manifold M, χ(M) depends only on the alternating sum of normalized solid angles of P over all the i-sides of P. In particular, it doesn’t depend on the side-pairing of P. Indeed, the sum of normalized solid angles for each cycle of an i-side is exactly 1, which is contributed to the count of cycles of i-sides.

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University of Oklahoma, Mathematics Department, Norman, OK 73019-0315

   E-mail address: divansic@aardvark.ucs.ou.edu