Deficiencies of lattices in connected Lie groups Jonathan A. Hillman School of Mathematics and Statistics, The University of Sydney, Sydney, NSW 2006, AUSTRALIA e-mail: jonh@maths.usyd.edu.au

ABSTRACT

We complete the determination of the groups of deficiency > 0 which occur as lattices in connected Lie groups. The torsion free groups among them are 3manifold groups. We show that any other torsion free 3-manifold group which is such a lattice is the group of an aspherical closed geometric 3-manifold. AMS Subject Classification (1991): Primary 20F05. Secondary 22E40. Key words and phrases: deficiency. lattice. Lie group.

Lott showed in [Lo99] that if Γ is a lattice in a connected Lie group G and the deficiency of Γ is positive (def(Γ) > 0) then either

(i) Γ has a finite normal subgroup F such that Γ/F is a lattice in P SL(2, ℝ); or

(ii) def(Γ) = 1 and Γ is isomorphic to a torsion-free nonuniform lattice in

R × P SL(2, ℝ) or P SL(2, ℂ); or (iii) Γ ≅ Z, Z^{2} or Z ×_{−1} Z (the fundamental group of the Klein bottle).

This was an improvement upon earlier work of Lubotzky, who assumed G simple and either of rank ≥ 2 or G = Sp(n, 1) or F_{4}, in which cases def(Γ) ≤ 0, or G = SO(n, 1) (for n ≥ 3) or SU(n, 1) (for n ≥ 2), in which cases def(Γ) ≤ 1 [Lu83]. We shall show that in case (i) the subgroup F must be trivial, and exclude the Klein bottle group. Excepting the lattices in P SL(2, ℝ) with finite abelianization, all the remaining possibilities have positive deficiency.

Theorem 1. Let Γ be a finitely presentable group with a nontrivial finite normal subgroup F such that Γ/F is a lattice in P SL(2, ℝ). Then def(Γ) ≤ 0. Proof. A group has a presentation of deficiency > 0 if and only if it is the fundamental group of a finite 2-complex with Euler characteristic ≤ 0. The latter property is clearly inherited by subgroups of finite index.

Let P be a cyclic subgroup of F of prime order p. The centralizer C_{Γ}(P) has finite index in Γ. Let E be a subgroup of C_{Γ}(P) containing P and such that E/P is torsion free. Then E/P is either a free group or is the fundamental group of an aspherical closed surface. Let D be a subgroup of E containing P and of index divisible by p, and let H = D/P. Then the class in H^{2}(E/P; P) ≅ H^{2}(E/P; 𝔽_{p}) corresponding to the central extension 1 → P → E → E/P → 1 has image 0 in H^{2}(H; 𝔽_{p}), and so D ≅ H × P.

Now the deficiency of D is at most β_{1}(D; 𝔽_{p}) − β_{2}(D; 𝔽_{p}) = −β_{2}(H; 𝔽_{p}), and so is nonpositive. Therefore def(Γ) ≤ 0, by the observation in the first paragraph of this proof. □

The estimate is best possible, in general. If H ≅ F (r) (the free group of rank r ≥ 2) or is a P D_{2}^{+}-group (the fundamental group of an aspherical closed orientable surface) and C is a nontrivial finite cyclic group then H × C is a lattice in P SL(2, ℝ)×SO(2), and has deficiency 0 or −1, respectively. (When C = Z/2Z there are such lattices in SL(2, ℝ)).

Theorem 2. Let G be a connected Lie group with a virtually abelian lattice Γ. Then G/Γ is compact. Proof. Let R = Rad(G) and p : G → G/R the natural epimorphism. Since Γ is amenable and G/Γ has finite volume G is amenable, by Proposition 4.1.11 of [Zi]. (This follows easily from the definition of amenability). Hence G/R is compact, by Corollary 4.1.9 of [Zi]. The closure of p(Γ) in G/R is a compact Lie subgroup, and so has finitely many components. On replacing Γ by a subgroup of finite index, if necessary, we may assume that p(Γ) is abelian, and hence that \overline{p(Γ)} is abelian. Let H = p^{−1}(\overline{p(Γ)}) and Γ_{o} = H_{o} ∩ Γ. Then Γ_{o} is a lattice in the connected solvable Lie group H_{o}, and therefore is cocompact, by Theorem 3.1 of [Ra]. It follows easily that Γ is cocompact. □

A crystallographic group is a discrete cocompact subgroup of Isom(ℝ^{n}) = R^{n} ⋊ O(n). It is orientable if it is a subgroup of R^{n} ⋊ SO(n).

Corollary. Let Γ be a crystallographic group. Then Γ is a lattice in a connected Lie group if and only if it is orientable. Proof. Suppose that Γ is a lattice in the connected Lie group G. Let K be a maximal compact subgroup of G. Then Γ acts discretely and cocompactly by left multiplication on the symmetric space X = G/K, which is diffeomorphic to R^{d} for some d. If Γ is torsion free the action is free, and so d = v.c.d.Γ. Since G is connected every element of G must preserve the orientation of X. The converse is clear. □

In particular, the Klein bottle group is not isomorphic to such a lattice. Lott’s list can be revised in the light of these results as follows. Either

(i) Γ is a lattice in P SL(2, ℝ); or (ii) Γ ≅ Z × F (r) for some r ≥ 0; or (iii) Γ is isomorphic to a torsion-free nonuniform lattice in P SL(2, ℂ).

The only lattices in P SL(2, ℝ) which do not have deficiency ≥ 1 are those with signature (0; e_{1}, … e_{k}; t) where t = 0 or 1. (See page 99 of [Ka]).

The torsion free groups of type (i) are the free groups of finite rank ≥ 2 and the P D_{2}^{+}-groups other than Z^{2}, and have deficiency > 1, while the groups of types (ii) and (iii) have deficiency 1, as observed in [Lo99]. All these groups are also fundamental groups of compact orientable 3-manifolds with nonempty boundary. (A torsion free nonuniform lattice in P SL(2, ℂ) is the fundamental group of a compact orientable ℍ^{3}-manifold whose boundary is a nonempty union of tori. Every such manifold is homotopy equivalent to a finite aspherical 2-complex with Euler characteristic 0).

The multiplicativity of the Euler characteristic in finite coverings appears to be of little use when the deficiency is ≤ 0, and the range of examples is much greater. There are already many examples in dimension 3. If M is a compact 3-manifold then π = π_{1}(M) has a presentation of deficiency 0, and has deficiency > 0 if and only if either ∂M has a component S with χ(S) ≤ 0 or M has S^{2} × S^{1} or S^{2} ˜×S^{1} as a summand. The cocompact lattices in connected 3-dimensional Lie groups were determined in [RV81]. As π_{2}(G) = 0 for any Lie group 3-dimensional coset spaces G/Γ have no S^{2} × S^{1} summands, and so such lattices have deficiency 0. More generally, we have the following result. We shall say that a compact manifold is geometric if its interior is homeomorphic to a manifold with a complete geometry of finite volume in the sense of Thurston [Th82].

Theorem 3. Let M be a compact 3-manifold with fundamental group π. Then M is aspherical and geometric if and only if it has no fake 3-cells and π is torsion free but not free and is a lattice in a Lie group with finitely many components. Proof. The conditions are clearly necessary. Suppose that they hold. On passing to a subgroup of finite index, if necessary, we may assume that M is orientable and π is a lattice in a connected Lie group G. Let G_{1} = Rad(G)K, where K is the maximal compact connected normal subgroup of a Levi subgroup of G, and let p : G → G_{2} = G/G_{1} be the canonical epimorphism. Then G_{1} ∩ π and p(π) are lattices in G_{1} and G_{2}, respectively [Au63].

Suppose first that G_{1} ∩π ≠ 1. Since G_{1} is an extension of a compact group by a solvable group G_{1}∩π is amenable, by Proposition 4.1.11 of [Zi]. Since the compact quotient G_{1}/Rad(G) ≅ K is semisimple G_{1} ∩ π is virtually solvable, by Tits’ theorem. Hence M is either Seifert fibred or is a 𝕊ol^{3}-manifold [Ga92]. Thus either M is a (closed) ℝ^{3}-, ℕil^{3}- or 𝕊ol^{3}-manifold, or its interior is a ℍ^{2} × 𝔼^{1}- or S̃L-manifold.

If π has no nontrivial solvable normal subgroup then we may assume that G = G_{2}, which is a semisimple Lie group whose Lie algebra has no compact factors. The group π therefore acts discretely with finite covolume on the symmetric space X of compact subgroups of G. Moreover π is irreducible. Therefore X must be typewise homogeneous, and has no Euclidean factor. (See §7 of Chapter IX of [Ma]).

If X has rank r(X) ≥ 2 then π is arithmetic, by Theorem IX.1.11 of [Ma]. Hence c.d.π = dim(X) − r_{ℚ}(G), where r_{ℚ}(G) ≤ r(X) is the ℚ-rank of G [BS73]. If moreover r_{ℚ}(G) > 0 then π is not a Poincaré duality group, and so c.d.π ≤ 2. However there are no symmetric spaces of rank ≥ 2 and dimension ≤ 4. (See §6 of Chapter X of [He]). Therefore we may assume that X has rank 1. If now X/π is compact then c.d.π = dim(X). If X/π is not compact then c.d.π = dim(X) − 1, and π contains parabolic subgroups of the same cohomological dimension. Since c.d.π ≤ 3 we may conclude that X = ℍ^{2} or ℍ^{3}, in either case. If X = ℍ^{2} then π is a P D_{2}^{+}-group (since it is not free) and the interior of M is a ℍ^{2} × 𝔼^{1}-manifold. If X = ℍ^{3} then M is homotopy equivalent to an ℍ^{3}-manifold, and therefore is homeomorphic to an ℍ^{3}-manifold, by [GMT96]. □

The fundamental group π = π_{1}(M) of any compact 3-manifold M is virtually torsion free. If π has nontrivial torsion then M is not aspherical. Therefore if moreover π is isomorphic to a lattice in a Lie group with finitely many components it must be virtually free, by the theorem. Hence it is a free product of finite 3-manifold groups and a free group. Finite cyclic groups are lattices in SO(2). The group π_{1}(RP^{3}♯RP^{3}) ≅ (Z/2Z)∗(Z/2Z) = Z ⋊(Z/2Z) is virtually Z and is a lattice in Isom(ℝ) = R ⋊ (Z/2Z). However, it is not a lattice in a connected Lie group, by the Corollary to Theorem 2. All other free products of cyclic groups are isomorphic to lattices in P SL(2, ℝ) with signature of the form (0; e_{1}, … e_{k}; t), where t > 0.

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