HOLOMORPHIC SELF-MAPS OF PARALLELIZABLE MANIFOLDS

J. WINKELMANN ^{∗}

Mathematisches Institut NA 4/69 Ruhr-Universität Bochum D-44780 Bochum Germany

jw@cplx.ruhr-uni-bochum.de

Abstract. We investigate holomorphic self-maps of complex manifolds of the form G/Γ where G is a complex Lie group and Γ a lattice. We show that they are induced by automorphisms of G and that a surjective holomorphic self-map can be nonbijective only in the directions of the nilradical of G.

Survey

In this article we investigate automorphisms and more generally self-maps of complex-parallelizable manifolds of the form X = G/Γ, where G denotes a simply-connected complex Lie group and Γ a discrete subgroup. Mostly we assume that Γ is a lattice, i.e., G/Γ is of finite volume (with respect to the Haar measure of G.)

In particular we prove that, for a lattice Γ in a semisimple complex Lie group G, every surjective holomorphic self-map f of the complex manifold X = G/Γ with f(eΓ) = eΓ is already an automorphism and of the form

f : gΓ ↦ ϕ(g)Γ

where ϕ denotes an automorphism of G (as complex Lie group) with ϕ(Γ) = Γ. As a consequence we obtain that, for semisimple G, the automorphism group of such a complex manifold G/Γ has only finitely many connected components and its connected component is isomorphic to G/Z ∩ Γ where Z denotes the (finite) center of G.

If G is not semisimple, there may exist surjective holomorphic self-maps of G/Γ that are not bijective. However, they can be nonbijective only in the direction of the nilradical. To give a precise statement, let us assume that G is a simply-connected complex Lie group, N its nilradical, Γ a lattice in G, and f : G/Γ → G/Γ a surjective holomorphic map with f(eΓ) = eΓ.

∗The author wishes to thank the Université de Nancy for their invitation during which this manuscript was completed.

Received January 15, 1997. Accepted April 15, 1997.

Then there exists an automorphism ϕ of G with ϕ(Γ) ⊂ Γ such that f(gΓ) = ϕ(g)Γ and there is a commutative diagram of holomorphic mappings

G/Γ ⟶^{f} G/Γ
↓^{π} ↓^{π}
G/NΓ ⟶_{F}^{∼} G/NΓ

with F (gNΓ) = ϕ(g)NΓ and π(gΓ) = gNΓ such that F is biholo morphic.

For the case where Γ is cocompact, Theorem 1 and 2 are already contained in [12].

Notations and basic facts

In this paper complex manifolds are always assumed to be connected, Hausdorff, finite-dimensional and paracompact. We call a complex manifold parallelizable if it is biholomorphic to a quotient G/Γ of a complex Lie group G by a discrete subgroup Γ. There is no loss in generality in assuming G to be simply-connected (as we will usually do). For a compact complex manifold it is well-known that they are parallelizable in our sense if and only if the tangent bundle is holomorphically trivial ([11]).

A closed subgroup Γ in a Lie group G is called cocompact if the quotient G/Γ is compact. A discrete subgroup Γ is called a lattice if the quotient G/Γ admits a G-invariant probability measure. In this case we also say that G/Γ has finite volume. A discrete cocompact subgroup is necessarily a lattice, but the converse holds only for subgroups in solvable groups. (As general reference for discrete subgroups in Lie groups we mention [10],[7],[12].)

The theory of lattices in complex Lie groups is in some aspects easier than the theory of lattices in real Lie groups, mainly because there are no compact complex semisimple Lie groups. In particular, if G is a simplyconnected complex Lie group and Γ a lattice, then Γ is “Zariski-dense” in G in the following sense: For every homomorphism ρ of complex Lie groups from G to a complex linear-algebraic group H the Zariski-closure of ρ(G) and ρ(Γ) coincide [5]. Furthermore every Γ-invariant holomorphic function on G is constant. If R denotes the radical of G and N the nilradical, then both N/(N ∩ Γ) and R/(R ∩ Γ) are compact.

Maps to parallelizable manifolds are equivariant

We start with a basic lemma on holomorphic mappings between parallelizable manifolds.

Lemma 1. Let X be a complex manifold on which a simply-connected complex Lie group G acts. Let Y = H/Λ be a complex-parallelizable manifold, f : X → Y a holomorphic map and assume that every holomorphic function on X is constant. Then f is equivariant with respect to a homomorphism of complex Lie groups ϕ : G → H.

Proof. We have to show that the G-fundamental vector fields on X can be pushed forward to H-fundamental vector fields on Y. There are Hfundamental vector fields v_{1}, …, v_{n} on Y (with n = dim_{ℂ}(Y)) which give a global trivialization of T Y. Now let w be a fundamental G-vector field on X. Using the trivialization of T Y we obtain holomorphic functions g_{i} on X such that f_{∗}(w_{x}) = ∑_{i} g_{i}(x)v_{i} for every x ∈ X. Since every holomorphic function on X is constant, it follows that the pushed-forward tangent vectors f_{∗}(w_{x}) constitute a vector field ∑_{i} g_{i}v_{i} with g_{i} ∈ ℂ. Thus we obtain a homomorphism of Lie algebras f_{∗} : Lie G → Lie H. Since G was assumed to be simply-connected, there exists an associated homomorphism of Lie groups ϕ : G → H. □

We are now able to deduce a structure theorem for holomorphic mappings between complex parallelizable manifolds.

Theorem 1. Let f : X → Y be a holomorphic map between complex parallelizable manifolds X ≃ G/Γ, Y ≃ H/Λ. Assume that every holomorphic function on X is constant. Then there exists a homomorphism of complex Lie groups F : G → H with F (Γ) ⊂ Λ and an element x ∈ H such that f(gΓ) = x · F (g)Λ. Two such pairs (F, x), (F′, x′) correspond to the same map if and only if there exists an element λ ∈ Λ such that x′ = xλ and F′(g) = λ^{−1}F (g)λ.

Proof. Choose x ∈ G such that f(eΓ) = xΛ. By the above lemma there exists a homomorphism of complex Lie groups ϕ : G → H such that

f(gΓ) = f(g · eΓ) = ϕ(g) · xΛ.

Let F be defined by F (g) = x^{−1}ϕ(g)x. For γ ∈ Γ we obtain

xΛ = f(eΓ) = f(γΓ) = xF (Γ)Λ.

Hence F (Γ) ⊂ Λ.

Finally consider two maps f, f′ given by f(gΓ) = xF (g)Λ (resp. f′(gΓ) = x′F′(g)Λ). If f = f′, then f(eΓ) = f′(eΓ) implies x′ = xλ for some λ ∈ Λ. Now

ζ : g ↦ F (g)^{−1}x^{−1}x′f′(g)

is a continuous map defined on the connected group G. The equality f = f′ implies that ζ(g) ∈ Λ for all g ∈ G. Hence ζ is constant. Since ζ(e) = λ, it follows that F′ = λ^{−1}F λ. □

Corollary 1. Let G be a simply-connected complex Lie group, Γ a discrete subgroup such that X = G/Γ has no nonconstant holomorphic functions. Let f : X → X be a holomorphic self-map such that D(f) has maximal rank in one point.

Then f is surjective, D(f) has maximal rank everywhere and there exists an automorphism ϕ ∈ Aut(G) with ϕ(Γ) ⊂ Γ and an element x ∈ G such that f(gΓ) = xϕ(g)Γ.

Proof. Due to the theorem f is equivariant. This implies that f is surjective and D(f) has maximal rank everywhere as soon as D(f) has maximal rank at one point. The theorem furthermore implies that we can find a homomorphism of complex Lie groups ϕ : G → G with discrete fibers which has the desired properties. However, for a simply-connected Lie group such a ϕ is necessarily an automorphism. This follows from the usual duality between simply-connected Lie groups and Lie algebras because such a ϕ evidently induces an automorphism of the Lie algebra of G. □

Given a complex Lie group G and a discrete subgroup Γ, we denote the set of all holomorphic Lie group automorphisms ϕ of G with ϕ(Γ) = Γ by Aut(G, Γ).

Corollary 2. Let G be a simply-connected complex Lie group, Γ a discrete subgroup such that G/Γ has no nonconstant holomorphic functions. Let A_{0} = Aut(G, Γ) × G be the natural semidirect product given by the natural action of Aut(G, Γ) on G. Then Aut(G/Γ) = A_{0}/Γ where Γ is embedded in A_{0} by

γ ↦ (int_{γ}, γ)

(with int_{γ} denoting conjugation by γ).

The group of connected components of Aut(X)

We will now study the quotient group Aut(X)/Aut(X)^{0}, where Aut(X)^{0} denotes the connected component of e (with respect to the compact-open topology). We will describe this quotient group and show that it is always countable and that, if G is semisimple, it is finite.*

Our previous results imply that Aut(X)/Aut(X)^{0} is in fact isomorphic to Aut(G, Γ)/Int(Γ). Hence we have to consider the size of Aut(G, Γ).

Lemma 2. Let G be a simply-connected complex Lie group and Γ a discrete subgroup. Assume that every holomorphic function on G/Γ is constant. Then Aut(G, Γ) naturally embeds into Aut(Γ) and is countable.

Proof. Let ϕ ∈ Aut(G, Γ) and assume that ϕ acts trivially on Γ. Then

ζ(gΓ) = g · ϕ(g^{−1})

defines a holomorphic map from G/Γ to G with ζ(eΓ) = e. But G (being a simply-connected complex Lie group) is holomorphically separable. It follows that ζ ≡ e which implies ϕ = id. Hence Aut(G, Γ) embeds into Aut(Γ). On the other hand Aut(G, Γ) is a closed subgroup of the Lie group Aut(G). These facts together imply that Aut(G, Γ) is countable. □

*There are complex manifolds X for which Aut(X)/Aut(X)^{0} is not countable. For example, let S = {0} × ℤ ⊂ ℂ^{2}. Then Aut(X)/Aut(X)^{0} is uncountable for X = ℂ^{2} \ S, because the whole permutation group of S embeds into Aut(X)/Aut(X)^{0}.

Corollary 3. Let G be a simply-connected complex Lie group, Γ a discrete subgroup. Assume that every holomorphic function on G/Γ is constant. Then Aut(X)/Aut(X)^{0} is countable.

We will now restrict to the case where G is semisimple and prove that in this case Aut(X)/Aut(X)^{0} is even finite. Here it is crucial that almost all automorphisms of semisimple Lie groups are inner. In fact we prove finiteness of Aut(X)/Aut(X)^{0} for all Lie groups G with Out(G) = Aut(G)/Int(G) finite. But first let us make a general remark and then proceed to an example.

The remark is the following. For a simply-connected Lie group G the automorphism group Aut(G) is isomorphic to the automorphism group of its Lie algebra and thereby easily seen to be linear-algebraic. In particular, Aut(G) has only finitely many connected components. Thus in order to check finiteness of Out(G) it suffices to check that Aut(G) and Int(G) = G/Z have the same dimension.

Now let us produce an example of a nonsemisimple Lie group for which Out(G) is finite and which does admit a lattice.

Example. Let A ∈ SL(2, ℤ) with |trace(A)| ≠ 2. Then A generates an infinite cyclic discrete subgroup Λ in SL(2, ℂ). Let T denote the Zariskiclosure of Λ in SL(2, ℂ). Now T is a maximal torus of SL(2, ℂ) and the natural semidirect product G_{0} = T × (ℂ^{2}, +) admits a lattice, viz. Λ × (ℤ^{2}, +). One can easily check that Out(G) ≃ ℤ/2ℤ, where G denotes the universal covering of G_{0}.

Lemma 3. Let G be a complex Lie group, Γ a lattice and assume that Out(G) = Aut(G)/Int(G) is finite. Then Aut(G, Γ)/Int(Γ) is finite.

Proof. Let H denote the subgroup of Aut(G, Γ) containing inner automorphisms of G. Then Aut(G, Γ)/H ⊂ Out(G) and H ≃ N/(N ∩ Z), where Z denotes the center of G and N the normalizer of Γ ∈ G. Observe that N^{0} = Z^{0} and that N is a closed subgroup of G containing Γ. Hence N/Z^{0} is a lattice in G/Z^{0} which contains the lattice ΓZ^{0}/Z^{0}. It follows that N/ΓZ^{0} is finite. This completes the proof of the finiteness of Aut(G, Γ)/Int(Γ). □

Together, the preceding lemmata imply the following result.

Theorem 2. Let G be a complex Lie group for which Out(G) is finite (e.g. G semisimple), Γ a lattice and X = G/Γ. Then Aut(X)/Aut(X)^{0} is finite.

Self-maps

Here we will demonstrate that except in the directions of the nilradical, a surjective self-map of a parallelizable manifold is necessarily biholomorphic.

We start with the following observation.

Lemma 4. Let G be a Lie group, Γ a lattice and ϕ an automorphism of G with ϕ(Γ) ⊂ Γ. Then #Γ/ϕ(Γ) = | det Ad(ϕ)| where Ad(ϕ) denotes the associated automorphism of the Lie algebra Lie(G).

Proof. The Haar measure on G can be realized by integrating a G-invariant volume form ω which can be considered as an element in ∧^{n} Lie(G)^{∗}. Thus

∫_{G/ϕ(Γ)} ω = ∫_{G/Γ} ϕ^{∗}ω

and ω = | det(Ad(ϕ))|ϕ^{∗}ω. □

Corollary 4. Let G be a semisimple Lie group, Γ a lattice and ϕ an automorphism of G with ϕ(Γ) ⊂ Γ. Then ϕ(Γ) = Γ.

Corollary 5. Let S be a semisimple complex Lie group, Γ a lattice and f a surjective holomorphic self-map of the complex manifold G/Γ. Then f is bijective, i.e., an automorphism.

However, for nonsemisimple Lie groups it is possible that ϕ(Γ) ⊊ Γ.

Example. For a commutative ring A define

G_{A} = \{( \begin{matrix}1 & x & z \\ & 1 & y \\ & & 1 \end{matrix} ) : x, y, z ∈ A\} ,

Then Γ = G_{ℤ+iℤ} is a lattice in the complex Lie group G = G_{ℂ} and

ϕ : ( \begin{matrix}1 & x & z \\ & 1 & y \\ & & 1 \end{matrix} ) ↦ ( \begin{matrix}1 & nx & n^{2}z \\ & 1 & ny \\ & & 1 \end{matrix} )

is an automorphism of G with ϕ(Γ) ⊊ Γ, if n ∈ ℕ with n ≥ 2.

More generally, we have the following construction.

Lemma 5. Let g be a nilpotent Lie algebra, defined over ℚ. Assume that g is graded, i.e., there are linear subspaces g_{k} of g (with k ∈ ℕ) such that g = ⨁_{k} g_{k} as a vector space and [g_{m}, g_{n}] ⊂ g_{m+n} for all m, n (again everything being defined over ℚ). Then the simply-connected nilpotent complex Lie group G corresponding to g ⊗_{ℚ} ℂ admits a lattice Γ and an automorphism ϕ such that ϕ(Γ) ⊊ Γ (provided g_{0} ⊊ g).

Proof. For any positive natural number N we can define an automorphism Φ_{N} of g if we set Φ_{N}(v) = N^{k}v for v ∈ g_{k} and extend linearly to the whole of g. The grading being defined over ℚ allows us to find a ℤ-module Λ inside g such that Λ ⊗ _{Z} ℚ ≃ g and Λ = ⨁_{k}(g_{k} ∩ Λ). Thus Φ_{N}(Λ) ⊂ Λ. By a result of Malcev [6] exp(Λ + iΛ) generates a lattice Γ in G. Evidently ϕ(Γ) ⊂ Γ if ϕ is an automorphism of G corresponding to Φ_{N} for some N.

On the other hand | det Φ_{N}| > 1 whenever N > 1 and g ≠ g_{0}. Therefore ϕ(Γ) ≠ Γ. □

Remark. There are nilpotent ℚ-Lie algebras which do not admit any nontrivial grading [4].

We prove that for arbitrary (not necessarily nilpotent) complex Lie groups the inclusion ϕ(Γ) ⊂ Γ can be nonbijective only in the directions of the nilradical.

We begin with a lemma on lattice-preserving automorphisms of nilpotent Lie groups.

Lemma 6. Let N be a simply-connected real nilpotent Lie group, Γ a lattice, and ϕ an automorphism of N with ϕ(Γ) ⊆ Γ. Let χ_{ϕ}(t) denote the characteristic polynomial of the induced automorphism of Lie(N). Then χ_{ϕ} ∈ ℤ[t].

Proof. First let us consider the case where N is abelian. In this case N ≃ ℝ^{n} and Γ ≃ ℤ^{n} in suitable coordinates. Hence ϕ ∈ M(n, ℤ) and evidently χ_{ϕ} ∈ ℤ[t].

Now the general proof can be deduced by induction. Induction is justified by Malcev’s result [6] that for every lattice Γ in N and every subgroup of the central series N^{k} the intersection N^{k} ∩ Γ is a lattice in N^{k}. □

Proposition 1. Let G be a simply-connected solvable real Lie group, Γ_{k} an sequence of lattices in G. Let I^{0} denote the connected component of the closure of the union ∪_{k}Γ_{k}. Then I^{0} is contained in the nilradical N of G.

Proof. For g ∈ G let Ad(g) denote the adjoint automorphism of Lie(G) and denote the characteristic polynomial of Ad(g)|_{N} by χ_{g}. Now N ∩ G is a lattice in G ([9], Lemma 3.9). Thus for every k and every γ ∈ Γ_{k} the characteristic polynomial χ_{γ} of Ad(γ)|_{N} is contained in ℤ[x]. By continuity this implies that for every g ∈ I^{0} the characteristic polynomial of Ad(g)|_{N} is trivial, i.e., Ad(g)|_{N} is unipotent. Since N contains the commutator group of G, Ad(g)_{Lie N} is nilpotent if and only if g ∈ N. It follows that I^{0} is contained in the nilradical. □

Lemma 7. Let G denote a Lie group, Γ a lattice, ϕ an automorphism of G with ϕ(Γ) ⊂ Γ. Then either ϕ(Γ) = Γ or Γ̂ = ∪_{k≥0}ϕ^{−k}(Γ) is a nondiscrete subgroup of G.

Proof. Since Γ is a lattice, it is clear that either Γ̂ is a lattice or Γ̂ is not discrete. Let d = [ϕ(Γ) : Γ] = #(Γ/ϕ(Γ)). Then vol(G/ϕ^{−k}(Γ)) = d^{−k} · vol(G/Γ). Thus Γ̂ can be discrete only in the case where lim_{k} d^{−k} > 0, i.e., only if d = 1. □

Corollary 6. Let G be a simply-connected complex solvable Lie group, N its nilradical, Γ a lattice in G and ϕ an automorphism of G with ϕ(Γ) ⊂ Γ. Then ϕ(Γ)N = ΓN.

Proof. First note that ∪_{k}ϕ^{−k}(Γ) is a subgroup of G because it is the union of an ascending sequence of subgroups. Let I denote the closure of ∪_{k}ϕ^{−k}(Γ) in G and denote its connected component by I^{0}. Evidently I^{0} is normalized by Γ. Now Ad(G) and Ad(Γ) have the same Zariski-closure in Aut(Lie G). Thus Ad(G) must stabilize Lie I and consequently I^{0} is a normal Lie subgroup of G.

We will now consider the quotient Lie group G/I^{0}. Since Γ ⊂ I, the quotient G/I admits an invariant finite measure ([8]), hence I/I^{0} is a lattice in G/I^{0}. On the other hand I^{0}Γ is also of finite covolume and therefore Γ_{0} = Γ/(Γ ∩ I^{0}) is likewise a lattice. It follows that Γ_{0} is a lattice in G/I^{0} with ∪_{k}ϕ_{0}^{−k}(Γ_{0}) being discrete (where ϕ_{0} is the automorphism of G/I^{0} induced by ϕ.) Thanks to the preceding lemma we may now conclude ϕ_{0}(Γ_{0}) = Γ_{0}, i.e., ϕ(Γ)I^{0} = ΓI^{0}. This completes the proof since I^{0} ⊂ N. □

We are now in a position to prove the following theorem.

Theorem 3. Let X = G/Γ be a complex-parallelizable manifold with finite volume. Let f be a surjective holomorphic self-map. Let N denote the nilradical of G. Then there is a commutative diagram

G/Γ ⟶^{f} G/Γ
↓^{π} ↓^{π}
G/NΓ ⟶_{F}^{∼} G/NΓ,

where F is an automorphism of the complex manifold G/NΓ.

Proof. There is an automorphism ϕ of the complex Lie group G with ϕ(Γ) ⊂ Γ and an element a ∈ G such that f is given by

f : gΓ ↦ aϕ(g)Γ.

We have to show that ϕ(Γ)N = ΓN. Note that both the radical R and the nilradical N of G have closed orbits in G/Γ. Thus RΓ/R is a lattice in the semisimple group G/R, implying ϕ(RΓ) = RΓ. On the other hand R ∩ Γ is a lattice in R. Hence ϕ((R ∩ Γ)N) = (R ∩ Γ)N. Together these facts yield ϕ(NΓ) = NΓ as desired. □

Meromorphic maps

All results on holomorphic maps in this paper are also valid for meromorphic maps for the simple reason that every meromorphic map to a complex parallelizable manifold is already holomorphic.

Indeed, let X be a complex manifold, Y = G/Γ a complex parallelizable manifold and let f : X → Y be a meromorphic map. Then f is holomorphic outside an analytic set E of codimension two. Now, for any simply-connected open subset U ⊂ X, U \ E is still simply-connected. It follows that f|_{U\E} lifts to a holomorphic map to the universal covering G of X. Since G is Stein, this lift can be realized by holomorphic functions. By Hartog’s theorem it then follows that f was holomorphic on U in the first place.

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