Subvarieties

A. Subvarieties of parallelizable manifolds

1. Survey

Let X be a connected complex parallelizable manifold, i.e. a quotient of a (simply-connected) complex Lie group G by a discrete cocompact subgroup Γ. In this chapter we are interested in the compact complex subspaces Z of X. A large part of the results in this chapter is joint work with A.T. Huckleberry, published in [14].

In the case of hypersurfaces, tori play a central role: the Albanese mapping α : X → Alb(X) is an equivariant surjective fibration α : X = G/Γ → G/J = Alb(X) and, for every complex hypersurface H in X, there exists a unique hypersurface H̃ in Alb(X) with H = α^{−1}(H̃) ([GR]). Thus, in a certain sense, higher-codimensional subvarieties are of more interest.

It is normally possible to find subgroups H in G with H · Γ closed, i.e. Z can arise as an orbit of a subgroup. After some preparations we indicate in §3 how such orbits arise. Namely the radical, the nilradical, the center and Cartan-subgroups always have closed orbits. For many parallelizable manifolds the orbits of the commutator group and more generally all subgroups of the normal and the central series are closed as well. On the other hand, there are no continuous families. Given a compact complex parallelizable manifold X and a point x ∈ X there exists at most countably many closed orbits of connected Lie subgroups through x.

If Z is a compact complex space with Kodaira-dimension κ(Z), the canonical ring induces a surjective meromorphic map φ_{K} : Z ⇀ V , where dim V = κ(Z) and the “generic” fibers F satisfy κ(F) = 0 (cf. [U]). One of our main goals here is, to group theoretically realize this “pluricanonical map” in the case of Z ↪ X = G/Γ. This is carried out in §8. For example, if either G is semisimple and X compact or G is solvable, then there is a closed complex subgroup H < G which, up to covering spaces, can be identified with Aut_{O}(Z)^{0} , where H · Γ is closed, and the canonical map is given by the quotient Z → Z/H. After a natural modification of Γ this fibration can be extended to a fibration of the ambient homogeneous space. It should be remarked that the results in §6 and §8 are known in the case of tori (cf. [U]).

Applying our results on the canonical map, in §9 we give upper bounds on the dimension of Z. For example, if G is simple, then codim Z ≥ \sqrt{dim G}.

Compact complex parallelizable manifolds

For nilmanifolds we prove that there are no transversal submanifolds in the following sense: Assume that X = G/Γ is a compact complex nilmanifold, Ψ : X → Alb(X) = G/G′Γ the projection onto the Albanese. Let Z be an irreducible closed complex subspace of X. If Ψ(Z) = Ψ(X), then Z = X. This is used to show that for certain nilmanifolds every irreducible closed complex subspace must be an orbit. In particular this holds, if Alb(X) is isogenous to a product of simple tori of algebraic dimension zero.

We study the Barlet cycle space, which is the complex-analytic analogue for the Chow scheme in algebraic geometry. It turns out that for G not nilpotent the cycle space has always non-compact irreducible components. (In contrast, for a compact Kähler manifold the cycle space has only compact irreducible components.)

In §14 we show that the natural version of the Bloch conjecture is true in this setting: Let f : ℂ → X be a holomorphic map. Then the complex-analytic Zariski closure Z = \overline{f(ℂ)} is the orbit of a subgroup H of G. In the case of tori, this is a basic result of Green and Griffiths ([GG]) which we in fact use. We also provide a hyperbolicity criterion: A closed complex subspace Z in a complex parallelizable manifold X is hyperbolic (in the sense of Kobayashi [15]) if and only if Z contains no compact complex tori.

Some of our results have been proved for subvarieties of X = G/Γ, where Γ is discrete but not necessarily cocompact. It would be desirable to have a complete theory in this case.

For example, if X = SL_{2}(ℂ)/Γ, we know of no irreducible subvarieties Z ⫋ X which are not orbits. We are only able to show that there are no topological obstructions to the existence of such subvarieties.

2. Non-compact parallelizable manifolds

Our investigations focus mainly on compact complex parallelizable manifolds. However, it turns out that some results on compact complex subspaces Z of compact complex parallelizable manifolds X actually use only the compactness of Z and not the compactness of X. Therefore we include a short discussion of non-compact complex parallelizable manifolds.

Let X be a complex parallelizable manifold, i.e. a (not necessarily compact) complex manifold whose tangent bundle is holomorphically trivial. If X admits only constant holomorphic functions this implies that the Lie algebra of holomorphic vector fields on X is finite–dimensional, in fact of the same dimension as X. We will see that in order to discuss compact complex subspaces of parallelizable manifolds it is enough to consider parallelizable manifolds X with O(X) = ℂ.

Proposition 1. Let X be a complex manifold, Z a submanifold with O(Z) = ℂ. Assume that X is locally homogeneous, i.e. the tangent bundle of X is spanned by global sections. Then X has a locally homogeneous complex submanifold Y with Z ⊂ Y and O(Y) = ℂ.

Corollary. Let X be a complex parallelizable manifold, Z a compact complex subspace. Then X contains a complex parallelizable submanifold X_{1} such that O(X_{1}) = ℂ and Z ⊂ X_{1} .

Proof. By the theorem there is a locally homogeneous submanifold X_{1} with Z ⊂ X_{1} and O(X_{1}) = ℂ. Now locally homogeneous means that T X_{1} is spanned by global sections. On the other hand the dual T^{∗}X_{1} is likewise spanned, because X_{1} is a submanifold of a manifold X with trivial cotangent bundle T^{∗}X. With O(X_{1}) = ℂ it follows that T X_{1} is trivial, i.e. X_{1} is parallelizable.

Proof of the proposition. Let X_{0} = X and define recursively X_{k+1} as the connected component of

X′_{k+1} := {x ∈ X_{k} : f(x) = f|_{Z} ∀f ∈ O(X_{k})}.

which contains Z. Then this sequence will terminate, and we set Y = X_{k} for k >> 0. Clearly O(Y) = ℂ. We have to show that Y is locally homogeneous. Assume that X_{k} is locally homogeneous. The Zariski tangent sheaf of X_{k+1} is defined by df = 0 for all f ∈ O(X_{k}). Let X be a holomorphic vector field on X_{k} which is tangent to X_{k+1} at some point p ∈ X_{k+1} . It follows that df(X)(p) = 0 and hence df(X)|_{X_{k+1}} ≡ 0 for all f ∈ O(X_{k}). This implies that X is tangential to X_{k+1} everywhere, i.e. X restricts to a holomorphic vector field on X_{k+1} . Thus for every point p ∈ X_{k+1} and every tangent vector v ∈ T_{p}X_{k+1} there exists a holomorphic vector field X on X_{k+1} such that X_{p} = v. It follows that X_{k+1} is locally homogeneous, too. Hence by induction on k it follows that Y is locally homogeneous.

It should be emphasized that in fact the tangent bundle of Y is spanned by the restrictions of global vector fields on X. In particular, if X is homogeneous, then so is Y .

3. Existence of closed orbits

Often submanifolds of a compact complex parallelizable manifold X ≃ G/Γ arise as closed orbits of Lie subgroups. Many canonically defined subgroups have closed orbits. In particular the radical, the nilradical, the center and all the groups of the ascending central series have closed orbits. In most cases

23 A.3 the orbits of the commutator group are closed as well. Besides characteristic subgroups, using centralizers there is also a way to construct non-normal Lie subgroups with closed orbits.

Many of the closedness results below are special cases of well-known results for real Lie groups. Usually, the statements for complex Lie groups are simpler, because for real Lie groups one often needs additional assumptions which are empty for complex Lie groups (e.g. exclusion of compact semisimple factors).

A main new result in this section is the theorem on the orbits of commutator groups.

The radical and the nilradical

Theorem. Let G be a connected complex Lie group and Γ be a discrete cocompact subgroup. Let R denote the radical, i.e. the maximal connected normal solvable Lie subgroup of G and N the nilradical, i.e. the maximal connected normal nilpotent Lie subgroup of G.

Then the R- and N -orbits on G/Γ are closed.

For a proof see [23] or [36]. The proof given in [26, 8.28] is wrong, see [36]. One may notice that in the complex category the orbits of the center are

always closed (see proposition 3 below). Since R/N is the connected component of the center of G/N, closedness of R-orbits is thereby an immediate corollary to the closedness of N -orbits.

Commutator groups

For a group G we always denote the commutator group by G′. This is the subgroup generated by all the commutators ghg^{−1}h^{−1} with g, h ∈ G. For a simply-connected complex Lie group G this is a connected closed complex Lie subgroup, namely the connected Lie subgroup corresponding to the commutator algebra of the Lie algebra of G. For most parallelizable manifolds the orbits of the commutator group are closed.

Theorem 1. Let G be a simply-connected complex Lie group, R the radical, Γ a discrete cocompact subgroup of G. Assume that no factor of the semisimple Lie group G/R is isomorphic to SL_{2}(ℂ).

Then the orbits of the commutator group G′ in X = G/Γ are closed. Furthermore (Γ ∩ G′)/Γ′ is a finite group isomorphic to the torsion part of the first homology group H_{1}(X, ℤ).

Before proving the theorem we provide an auxiliary tool.

Lemma 1. Given a complex Lie group H we say H has property (∗) if for every discrete cocompact subgroup Γ

(i) the H′ -orbits in H/Γ are closed and (ii) the quotient group (H′ ∩ Γ)/Γ′ is finite.

Assume that G is a complex Lie group with normal Lie subgroup A and both G/A′ and A have property (∗). Furthermore assume that, for every discrete cocompact subgroup Γ, the A-orbits in G/Γ are closed. Then G also has property (∗).

Proof. Since A has property (∗), it is clear that the A′ -orbits in G/Γ are closed. Hence Γ/(A′ ∩ Γ) is a discrete cocompact subgroup of G/A′. Now property (∗) for G/A′ implies that A′G′Γ = G′Γ is closed in G. Thus the G′ -orbits in G/Γ are closed.

Property (∗) for G/A′ and A imply that both (G′ ∩ Γ)/Γ′ · (A′ ∩ Γ) and (A′ ∩ Γ)/(A ∩ Γ)′ are finite. Hence (G′ ∩ Γ)/Γ′ · (A ∩ Γ)′ = (G′ ∩ Γ)/Γ′ is likewise finite.

Proof of the theorem. First note that without loss of generality we may assume that G is simply-connected. For nilpotent Lie groups the statement of the theorem is a result of Malcev [19]. Let us now consider the case where G is solvable. Here closedness of G′ -orbits has been proven by Barth and Otte [2]. Hence it suffices to prove finiteness of (G′ ∩ Γ)/Γ′. As usual let N denote the nilradical. Since the N - orbits are always closed, we may apply the preceding lemma and restrict our attention to the case where N is commutative. Since G′ ⊂ N, this implies in particular that G′ is commutative. Since G is simply-connected, G′ ≃ (ℂ^{d}, +). Moreover Γ_{1} = Γ∩G′ is a lattice in G′ by the result of Barth-Otte. Now for each γ ∈ Γ we obtain a ℂ-linear map ζ_{γ} : ℂ^{d} → ℂ^{d} given by ζ_{γ} : g ↦ γgγ^{−1}g^{−1} . Actually ζ_{γ}(g) = ρ(γ)(g) − g if ρ : G → Aut(G′) ≃ GL_{d}(ℂ) denotes the natural representation given by conjugation.

For k = ℝ, ℂ and a subset E ⊂ ℂ^{d}, let ⟨E⟩_{k} denote the k -subvectorspace spanned by E. Now ⟨Γ_{1}⟩_{ℝ} = ⟨Γ_{1}⟩_{ℂ} = G′, because Γ_{1} = Γ ∩ G′ is a lattice in G′. Since all the maps ζ_{γ} are ℂ-linear, this implies that ⟨H⟩_{ℝ} = ⟨H⟩_{ℂ} for H = ⨁_{γ∈Γ} ζ_{γ}(Γ_{1}). Now we fix an embedding G ↪ GL_{m}(ℂ) (which is possible if we require G to be simply-connected). Then [Γ, Γ′] has to be Zariski-dense in the unipotent group [G, G′]. It follows that ⟨H⟩_{ℂ} = [G, G′]. Thus H = [Γ, Γ ∩ G′] is a discrete cocompact subgroup in [G, G′]. Dividing through [G, G′] the statement now follows from Malcev’s theorem, since G/[G, G′] is clearly nilpotent. This completes the proof for the solvable case.

In the general case, using the result for solvable groups and the preceding lemma we may assume that the radical R is commutative. Now [Γ, R ∩ Γ] is Zariski-dense in [G, R] = R ∩ G′ and arguments similar to those above imply that ⟨[Γ, R ∩ Γ]⟩_{ℝ} is already a complex subvectorspace of R ∩ G′ ≃ (ℂ^{n}, +). It follows that [Γ, R ∩ Γ] is a discrete cocompact subgroup in R ∩ G′. In particular, the R ∩ G′ -orbits are closed

25 A.3

and [Γ, R ∩ Γ] ⊂ Γ′ ∩ R ∩ G′ is of finite index in R ∩ G′ ∩ Γ. Hence without loss of generality we may assume that R ∩ G′ = {e}. Then R is central and G ≃ S × R. We consider the projection π : G → G/R ≃ S. Now πΓ is discrete and cocompact in the semisimple complex Lie group S which has no factor isomorphic to SL_{2}(ℂ). Then (πΓ)′ is of finite index in πΓ (see Margulis [20]). Thus (πΓ)′ is cocompact in G/R. With (πΓ)′ = π(Γ′) and Γ′ ⊂ S this implies that Γ′ is cocompact in S. Hence the S -orbits in G/Γ are closed and Γ′ is a subgroup of finite index in S ∩ Γ.

Note that π_{1}(X) ≃ Γ, because G is simply-connected. Thus H_{1}(X, ℤ) ≃ Γ/Γ′ by the Hurewicz isomorphism. Furthermore G being simply-connected implies that G/G′ is simply-connected, hence isomorphic to some (ℂ^{d}, +) and therefore torsionfree. It follows that G′ ∩ Γ/Γ′ is the torsion part of Γ/Γ′.

Example.

(i) The exclusion of SL_{2}(ℂ)-factors is necessary. There exists a discrete cocompact subgroup Λ ⊂ SL_{2}(ℂ) with a surjective group homomorphism τ : Λ → ℤ. Let Γ ≃ Λ × ℤ^{2} be embedded in G = SL_{2}(ℂ) × ℂ by (λ, m, n) ↦ (λ, m+in+\sqrt{2}τ(λ)). Then the orbits of G′ = SL_{2}(ℂ) in G/Γ are not closed.

(ii) Let

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

m ∈ ℕ and Γ ⊂ G the set of those elements g = (x, y, z) ∈ G for which x, y, mz ∈ ℤ + iℤ. Then Γ is discrete, cocompact and Γ′ is a subgroup of index m^{2} in G′ ∩ Γ.

(iii) For real nilpotent Lie groups the similar statement is true; this is a result of Malcev. However, for arbitrary real solvable Lie groups no such result holds. For instance, let V = ℝ^{3} and Γ ⊂ V a lattice such that Γ ∩ W = {0} where W = {(x, y, 0) : x, y ∈ ℝ}. Let S^{1} ≃ ℝ/2πℤ act on V by (x, y, z) ↦ (cos θx + sin θy, cos θy − sin θx, z). Then {e} × Γ is discrete, cocompact in S^{1} × (V, +) and the G′ -orbits are not closed.

(iv) Cocompactness is decisive, even for complex nilpotent Lie groups. See [24] for an example of a nilpotent complex Lie group G and a discrete subgroup Γ such that every holomorphic function on G/Γ is constant and the G′ -orbits in G/Γ are not closed.

We will now generalize this result to subgroups in the derived and central series.

Corollary. Let G be a simply-connected complex Lie group, R the radical, Γ a discrete cocompact subgroup of G. Assume that no factor of the semisimple Lie group G/R is isomorphic to SL_{2}(ℂ). Let G^{(k)} denote the derived series and G^{k} the descending central series, i.e. G^{(1)} = G′ = G^{1} , G^{(k+1)} = [G^{(k)}, G^{(k)}] and G^{k+1} = [G, G^{k}]. Furthermore let G^{(∞)} = ∩_{k}G^{(k)} and G^{∞} = ∩_{k}G^{k}.

Then for all k ∈ ℕ ∪ {∞} both G^{(k)} and G^{k} have closed orbits in G/Γ.

Proof. First note that G/R ≃ G′/R_{G′} where R_{G′} denotes the radical of G′. Therefore we may apply induction in order to deduce that the G^{(k)} -orbits are closed for k ∈ ℕ. Observe that G is finite-dimensional and the G^{(k)} form a descending sequence of connected Lie subgroups. It follows that there exists a number m ∈ ℕ such that G^{(∞)} = G^{(m)} . Now H = G/G^{(∞)} is a simply-connected solvable complex Lie group with discrete cocompact subgroup Γ/(Γ ∩ G^{(∞)}). Clearly G^{n} ⊃ G^{(∞)} for all n ∈ ℕ ∪ {∞}. Hence closedness of the G^{n} -orbits follows by induction from the proposition below.

Proposition (Barth-Otte [2]). Let G be a simply-connected solvable complex Lie group, Γ a discrete cocompact subgroup, H a connected complex Lie subgroup with HΓ closed.

Then [H, G]Γ is closed.

Some Lie groups do not admit irreducible lattices

We have already seen that some complex Lie groups do not contain any discrete cocompact subgroups. Now the results on closed orbits imply that certain complex Lie groups do not contain interesting discrete cocompact subgroups. As usual, a discrete subgroup Γ in a simply-connected Lie group G is called reducible if G decomposes as a direct product of positive-dimensional Lie subgroups G = A × B such that (A ∩ Γ) × (B ∩ Γ) has finite index in Γ. It is called irreducible, if it is not reducible.

Proposition 2. Let S be a simply-connected semisimple complex Lie group, R ≠ {e} a simply-connected solvable complex Lie group, G = S × R. Assume that no factor of S is isomorphic to SL_{2}(ℂ).

Then every discrete cocompact subgroup in G is reducible.

Proof. Both the radical R and all the subgroups G^{k} have closed orbits in G/Γ for any discrete cocompact subgroup Γ ⊂ G. Now S = G^{k} for k large enough. Hence (S ∩ Γ) × (R ∩ Γ) is cocompact in G. It follows that (S ∩ Γ) × (R ∩ Γ) is of finite index in Γ.

Corollary. Let G be a simply-connected complex Lie group, R the radical. Assume {e} ≠ R ≠ G. Assume in addition that S = G/R is neither isomorphic to a SL_{n}(ℂ) nor to the simple group associated to the Dynkin diagram E_{8} . Assume furthermore that there exists a discrete cocompact subgroup Γ ⊂ G.

Then G ≃ S × R and Γ is reducible.

Proof. By the aforementioned result of Otte-Potters (cf. [25]) the existence of a discrete cocompact subgroup implies G = S ×R. Reducibility of Γ follows from the preceding proposition.

Centralizers.

Using centralizers is a very good way of constructing subgroups which have closed orbits. The following is a typical starting point:

Lemma 2. Let G be a (real) Lie group and Γ a discrete subgroup. Let Δ be a subset of Γ and C = C_{G}(Δ) its centralizer. Then C · Γ is closed.

This is stated in [26],[30] under the additional assumptions that G/Γ has a finite G-invariant measure and Δ is finite. However, the first of these two additional assumptions is simply superfluous, while the second can be circumvented.

Proof. Case Δ finite. Let c_{n} ∈ C, γ_{n} ∈ Γ with lim_{n}(c_{n}γ_{n}) = z. We have to show z ∈ CΓ. For any d ∈ Δ we have

z^{−1}dz = lim γ_{n}^{−1}c_{n}^{−1}dc_{n}γ_{n} = lim γ_{n}^{−1}dγ_{n}.

Since d ∈ Γ and Γ discrete, this implies that there is a number N(d) (depending on d ∈ Δ) such that z^{−1}dz = γ_{n}^{−1}dγ for all n > N(d). Let N = max_{d∈Δ} N(d). Then γ_{N} z^{−1}dzγ_{N}^{−1} = d for all d ∈ Δ. Hence zγ_{N}^{−1} ∈ C and thereby z ∈ CΓ.

General case. Let (Δ_{j})_{j∈J} denote the family of all finite subsets of Δ. All the C_{G}(Δ_{j})Γ are closed, hence H = ⋂_{j} (C_{G}(Δ_{j})Γ) is closed. Now it is easy to show that there exists a finite subset Δ_{0} ⊂ Δ such that dim(H) = dim(C_{G}(Δ_{0})Γ) = dim(C_{G}(Δ_{0})). It follows that C_{G}(Δ_{j})^{0} ⊃ C_{G}(Δ_{0})^{0} for all j ∈ J. Thereby C_{G}(Δ) ⊃ C_{G}(Δ_{0})^{0} = H^{0} . This implies H^{0} ⊂ (⋂_{j} C_{G}(Δ_{j})) Γ ⊂ (⋂_{j} C_{G}(Δ_{j}Γ) = H. Since H/H^{0} is discrete, it follows that C_{G}(Δ)Γ = (⋂ C_{G}(Δ_{j})) Γ is closed in G.

Note: For the case Δ finite it is enough to assume that G is a topological group with a topology fulfilling the first axiom of countability. However, in the proof of the general case we used the assumption that G is a Lie group.

If we already have a subgroup with a closed orbit, then it is often possible to build other groups with the same property.

Proposition 3. Let G be a connected complex Lie group, Γ a discrete cocompact subgroup and H a connected complex subgroup of G with closed orbit through the neutral point eΓ. Then both C = C_{G}(H) and C · H have closed orbits through eΓ.

Proof. Here there is no loss of generality to assume that G is simply-connected. In particular G may be realized as a complex linear group.

Now, by proposition I.5, C_{G}(H) = C_{G}(H ∩ Γ). Then it follows from lemma 2 above that CΓ is closed, i.e. compact.

Since the elements of C commute with those of H, we have the natural continuous surjective map

C/C ∩ Γ × H/H ∩ Γ → CH/CH ∩ Γ .

Since the orbits C/C ∩ Γ and H/H ∩ Γ are compact, it follows that the CH -orbit is likewise compact.

Corollary. Let G be a complex Lie group, Γ a discrete cocompact subgroup. Let Z = C_{1} denote the center of G and define C_{k} recursively such that C_{k}/C_{k+1} is the center of G/C_{k+1} .

Then all the orbits of all the subgroups C_{k} are closed.

Proof. Take H = G for the proof for C_{1} and then use induction.

By a result of Barth and Otte (see [2]), if G is nilpotent, the compactness assumption on G/Γ may be weakened.

Proposition. Let G be a complex nilpotent Lie group, Γ a discrete subgroup such that every holomorphic function on G/Γ is constant. Let C_{k} be defined as in the corollary above. Then all the C_{k} -orbits in G/Γ are closed.

Abelian subgroups.

Our goal here is to show that there are abelian subgroups with closed orbits and that every abelian subgroup with a closed orbit is contained in a maximal abelian subgroup with a closed orbit. For this we recall the following fact.

Lemma 3. Let G be a connected non-commutative complex Lie group, and Z_{G} the connected component at the identity of its center. For g ∈ G let C(g) be the centralizer of g. Then

(+)  dim C(g) ≥ dim Z_{g} + 1 .

Proof. Consider the map μ : G × G → G × G, (g, h) ↦ (g, ghg^{−1}h^{−1}). Let C be the irreducible component of the preimage of G × {e} which contains G × {e}, and μ̃ : C → G × {e} ≃ G the induced map. Note that μ̃^{−1}(g) is a union of components of C(g), among which is always C(g)^{0} . Now there is an open set of elements g ∈ G which are in the image of exp and are not in Z_{G}. For such elements (+) holds. The result then follows from the semi-continuity of the fiber-dimension of μ̃.

Corollary. Let G be a connected complex Lie group and Γ a discrete cocompact subgroup. Then there exists a positive-dimensional connected complex abelian subgroup H with H · Γ closed, i.e. the H -orbit through the neutral point eΓ in G/Γ is closed.

Proof. We may assume that G is not abelian. Then there exists an element γ ∈ Γ which is not in the center of G (see I.§6). The above lemma shows that C := C_{G}(γ) is positive-dimensional. Since CΓ is closed, we may replace G/Γ by C/C ∩ Γ and apply induction on the dimension of the group.

We may now prove the desired result on abelian groups.

Proposition 4. Let G be a connected complex Lie group and Γ a cocompact discrete subgroup. Suppose that A is a connected complex abelian subgroup of G with AΓ closed (possibly A = {e}). Then there exists a maximal connected complex abelian subgroup B in G with A ⊂ B and BΓ closed. In particular dim(B) > 0.

Proof. It is enough to show that if A is as above and is maximal with the property AΓ closed, then Z is maximal abelian.

If A is not maximal, then C = C_{G}(A)^{0} ⫌ A. Since C has a closed orbit (proposition 3), unless A is itself central, i.e. C = G, we are finished by applying induction to the dimension of G. Of course we may assume that C (= G) is not abelian. Thus there exists γ ∈ Γ with A ⊂ Z_{G}^{0} ⊂ C_{G}(γ)^{0} ⫋ G. Now, lemma 3 shows that dim C(γ) ≥ dim Z_{G} + 1. Thus C_{G}(γ)^{0} ⫌ A and we may replace G by C_{G}(γ)^{0} and apply induction.

As an immediate consequence we see that, if X = G/Γ is not a simple torus, then X has non-trivial subvarieties, i.e. orbits of certain abelian subgroups. This statement has a rather strong converse.

Proposition 5. Let X be a compact complex parallelizable manifold. Then the following conditions are equivalent.

(i) X has no non-trivial analytic subspaces.

(ii) X is a simple torus without non-constant meromorphic functions.

30 A.3

Proof. By the above observations it is enough to prove (ii)⇒(i). This is well-known, but for the sake of completeness we give a proof.

Let X be a simple torus. Note that for any irreducible subvariety Z in X we have the addition map

A_{k} : \underbrace{Z × ⋯ × Z}_{k times} → X , (z_{1}, …, z_{k}) ↦ z_{1} + ⋯ + z_{k} .

The image is a compact analytic subset Z_{k}. If Z_{k} = Z_{k+1} for some k, then Z_{k} = Z_{k+i} for i ≥ 0 and Z_{k} is subtorus. Since X is simple, this can only happen if Z_{k} = X.

Recall that a divisor D in a simple torus is even ample. Thus, if we assume that X has a non-trivial analytic subspace, then it is enough to produce a hypersurface. By the above argument it is enough to produce a 1-dimensional subspace Z, because, if n := dim X, then Z_{n−1} is a hypersurface.

Now let Z be a non-trivial irreducible analytic subspace of minimal positive dimension. Note that Z is smooth. Regarding X as a quotient ℂ^{n}/Γ, for every p ∈ Z, the tangent space T_{p}Z can be regarded as a k -dimensional subspace of ℂ^{n}. If T_{p}Z ≠ T_{q}Z for some p, q ∈ Z, then there is an invariant k -form ω on ℂ^{n} so that its restriction to Z vanishes at p and not at q. In this case the minimality of Z implies that dim Z = k = 1 as desired. If T_{p}Z = T_{q}Z for all p, q ∈ Z, then Z is a torus. This is contrary to X being simple.

Cartan Subalgebras

A Lie subalgebra h of a Lie algebra g is called a Cartan subalgebra, if (i) h is nilpotent and (ii) h equals its normalizer N_{g}(h) = {v ∈ g : ad(v)(h) ⊂ h}.

This implies that h is maximal among the nilpotent Lie subalgebras of g. If g is the Lie algebra of a semisimple complex Lie group G, then h is a Cartan subalgebra if and only if the corresponding connected Lie subgroup H ⊂ G is a maximal torus. (”torus” in the usual sense for the theory of linear-algebraic groups, i.e. a torus is a subgroup isomorphic to (ℂ^{∗})^{k}.)

Proposition 6. Let G be a connected complex Lie group, Γ a discrete cocompact subgroup, h a Cartan subalgebra of Lie(G), H the corresponding connected Lie subgroup of G.

Then H has a closed orbit in G/Γ, i.e. there exists an element g ∈ G such that HgΓg^{−1} is closed in G.

Proof. Let Z denote the center of G. Then HZ is a Cartan subgroup of G in the sense of Mostow [22]. Observe that any Cartan subalgebra of a Lie algebra must

31 A.3

contain the center. Hence (HZ)^{0} = H^{0} = H. Now Theorem 5.2. of [22] states that there exists a closed (real-)analytic solvable subgroup F of G with F/F ∩ Γ closed and Γ ∩ F nilpotent such that F contains a conjugate of (HZ)^{0} = H. For simplicity we asssume H ⊂ F . The actual construction of F shows that for a complex Lie group G this subgroup F will be a complex Lie subgroup. Aided by the density theorem it follows that F^{0} is nilpotent. But H is maximal among connected nilpotent Lie subgroups of G, hence F^{0} = H.

A Cartan subalgebra in a Lie algebra is never an ideal unless it equals the whole Lie algebra. Therefore this results yields non-normal connected Lie subgroups with closed orbits for all X = G/Γ in the case where G is not nilpotent. This fact will be used in our examination of the cycle space.

Corollary (Mostow). Let S be a connected semisimple complex Lie group, Γ a discrete cocompact subgroup, T a ”maximal torus”, i.e. a maximal commutative reductive Lie subgroup.

Then T has a closed orbit in S/Γ.

4. On the number of closed orbits

The main goal of this section is to show that there is no non-trivial continuous family of compact orbits in G/Γ, where G is a complex Lie group and Γ is a discrete subgroup. In the case of real Lie groups such families can arise.

Example. Let E = SO_{2} ⋉ℝ^{2} be the group of orientation preserving Euclidean motions. Recall that SO_{2} ≅ S^{1} so that the universal cover G is the semi-direct product ℝ ⋉ ℝ^{2} . Let Γ be the center of G, i.e. a copy of ℤ in the first factor which realizes the universal covering of S^{1} . Let H be the first factor itself. Note that the H -orbit in G/Γ = E is just the first factor in E. Conjugating H yields the continuous family of semi-simple factors of E. Of course all such conjugates contain Γ.

The above phenomenon does not occur in the case of complex Lie groups.

Proposition 7. Let G be a complex Lie group and Γ a discrete subgroup. Then there exist only countably many connected complex Lie subgroups H in G with H/H ∩ Γ compact.

Proof. First note that it is enough to handle the case where G is simply-connected. Further note that, since the fundamental group of a connected compact manifold is finitely generated, if the orbit H/H ∩ Γ is compact, then H ∩ Γ is finitely generated. Hence it is enough to prove the following statement:

Let G be a simply-connected complex Lie group and Γ a finitely-generated discrete subgroup. Then there exist at most countably many connected complex subgroups H in G with Γ ⊂ H and H/Γ compact.

Since π_{1}(G) = 1, we may replace it by its Zariski-hull in some linear realization, i.e. we may assume that G is a linear algebraic group. Now H/Γ is compact. Hence, by the density result (see proposition I.5), A := \overline{Γ} = \overline{H}. Thus \overline{H}′ = H′ = (A)′ [4]. Thus we must only count the groups H with H′ = A′ and H/Γ compact; in particular H > B := \overline{A′Γ}, where here we take the complex analytic closure. Thus it is enough to count the complex groups H̃ in the abelian group Ã = A/B^{0} which contain the discrete group Γ̃ := B/B^{0} such that H̃/Γ̃ is compact. Thus it only remains to prove the proposition in the abelian case.

Now assume that G is abelian and simply-connected. Then any connected Lie subgroup H with H/Γ compact is just the linear span of H ∩ Γ and is therefore determined by H ∩ Γ which is a finitely generated subgroup of Γ. Of course Γ has only countably many such subgroups.

As we will verify later, every parallelizable submanifold is an orbit. Hence we obtain the following consequence.

Corollary. Let X be a compact complex parallelizable manifold.

Then up to translation by holomorphic automorphisms ϕ ∈ Aut_{O}(X) there exist only countably many complex parallelizable submanifolds.

In the semisimple case we can say more.

Corollary. Up to isomorphism there exist only countably many tori which admit an embedding into a compact quotient S/Γ of a semisimple complex Lie group by a discrete cocompact subgroup.

Proof. This is a consequence of the proposition, because we have already demonstrated that there exist only countably many non-equivalent quotients of complex semisimple Lie groups by discrete cocompact subgroups.

Thus it is a rather special property for a torus to be embeddable into a parallelizable manifold with semisimple automorphism group.

Proposition 8. Let T = ℂ^{g}/Λ be a torus, which is embedded (as submanifold) in a compact quotient of a semisimple complex Lie group S by a discrete subgroup Γ.

Then there exists a number k ∈ ℕ and a complex Lie group homomorphism ϕ : ℂ^{g} → (ℂ^{∗})^{k} with discrete kernel such that ϕ(Λ) ⊂ \binom{}{ℚ̄^{∗}}^{k}.

33 A.4

Proof. Any such torus is an orbit of a Lie subgroup A ⊂ S which is commutative and whose Zariski-closure in S is reductive. Therefore A is a Lie subgroup of a Cartan subgroup H ≃ (ℂ^{∗})^{k}. Now we have an embedding ϕ : A → (ℂ^{∗})^{k}. Moreover due to root theory for a given a ∈ A all the ϕ_{k}(a) are algebraic numbers as soon as all the eigenvalues of Ad(a) are algebraic numbers. Here Ad denotes the adjoint representation. Now every semisimple complex Lie group may be defined over ℚ̄. Furthermore, there exists a g ∈ G such that gΓg^{−1} ⊂ S(ℚ̄) ([26, Th. 7.67]). Hence for all γ ∈ Γ all the eigenvalues of Ad(γ) are algebraic. This completes the proof.

Corollary. Let T = ℂ/ ⟨1, τ⟩ be an elliptic curve which may be embedded in a compact quotient X = S/Γ of a semisimple complex Lie group by a discrete subgroup.

Then τ is not an algebraic number.

Proof. By the above result there exists α ∈ ℂ^{∗} such that both e^{α} and e^{ατ} are algebraic. On the other hand α and ατ are clearly linearly independent over ℚ. Bakers theorem now implies that α and ατ are linearly independent over ℚ̄. Thus τ ∉ ℚ̄.

More generally the proposition has the following consequence.

Corollary. Let T = ℂ^{g}/Λ be a simple torus, which is embedded (as submanifold) in a compact quotient of a semisimple complex Lie group S by a discrete subgroup Γ.

Then T does not admit complex multiplication (as defined in Chapter D).

Proof. This follows from proposition D.13 in combination with the corollary to theorem D.6.

5. On closed orbits in the reductive case

A complex linear algebraic group is reductive if and only if it is the complexification of a maximal compact subgroup. If G is reductive and Γ is a cocompact discrete subgroup, then every element of Γ is semi-simple [26]. This allows a more precise description of closed orbits.

Lemma 4. Let G be a reductive complex Lie group and Γ a cocompact discrete subgroup. Let H be a connected subgroup with a closed orbit in X = G/Γ. Then the algebraic Zariski-closure of H is reductive.

Proof. Let R denote the radical of H and \overline{R} algebraic Zariski-closure of R in G. It suffices to show that \overline{R} is reductive. Recall that the R-orbits are compact (see §3).

34 A.5

Note that \overline{R}′ is unipotent. Since Γ contains only semisimple elements, it follows that the intersection of R ∩ Γ with \overline{R}′ is trivial. Hence R ∩ Γ is abelian. This implies that R is abelian (see I.§6). Thus \overline{R} is abelian and therefore a product A × U with A ≃ (ℂ^{∗})^{k} and U × (ℂ)^{l}. Now R ∩ Γ ⊂ A. Using the compactness of R/R ∩ Γ this implies R ⊂ A. Hence \overline{R} ⊂ A, i.e. \overline{R} = A. Thus \overline{R} is reductive and hence \overline{H} is likewise reductive.

Note that the group H is perhaps not reductive.

Example. Let G = (ℂ^{∗})^{2} and H be a closed connected subgroup of G which is isomorphic to ℂ, e.g. the image of the map z ↦ (e^{z}, e^{iz}). Let Γ be a discrete subgroup of H which is isomorphic to ℤ × ℤ. Then Γ is cocompact in H and in G.

The following result is needed for our discussion of the extension of the “pluri-canonical map” to the ambient homogeneous space.

Lemma 5. Let G be a reductive complex Lie group and H a connected Lie subgroup whose algebraic Zariski-closure is reductive. Then the normalizer group N_{G}(H) is reductive.

Proof. Let \overline{H} denote the Zariski-closure of H and let Z be the center of \overline{H}. Observe that \overline{H} = Z · S with S semisimple and H = Z_{0} · S for some Zariski-dense subgroup Z_{0} of Z. Let N = N_{G}(\overline{H}) denote the normalizer. The reductivity of \overline{H} implies that N^{0} = \overline{H} · C_{G}(\overline{H}), where C_{G}(\overline{H}) denotes the centralizer. It follows that the center Z of H is contained in the center of N^{0} . This implies that H is normal in N^{0} . On the other hand N_{G}(H) ⊂ N, because H is Zariski-dense in \overline{H}. Thus N^{0} ⊂ N_{G}(H) ⊂ N. Therefore the reductivity of N implies that N_{G}(H) is reductive.

6. Subvarieties Z with κ(Z) = 0.

Let Z be a connected complex manifold, K = K_{Z} its canonical bundle, and R_{Z} = ⨁_{k=0}^{∞} Γ(Z, K^{k}) its pluricanonical ring. Unless R_{Z} = ℂ, in which case Γ(Z, K^{k}) = (0) for all k > 0, the Kodaira-dimension κ(Z) is defined to be the transcendence degree over ℂ of the quotient field of R_{Z}. If R_{Z} = ℂ, then κ(Z) := −∞. It follows that κ(Z) ≤ dim Z. If κ(Z) = dim Z, then Z is said to be of general type.

The Kodaira-dimension is a bimeromorphic invariant [33]. Therefore it can be defined for an irreducible compact complex space Z by κ(Z) := κ(Ẑ) where Ẑ → Z denotes any desingularization.

Using the pluricanonical ring, one constructs a pluri-canonical map (also called Iitaka-reduction). Its properties are summarized as follows (see [33] for details).

Theorem. Let Z be an irreducible reduced compact complex space. Then there exists a surjective meromorphic map ψ : Z ⇀ V onto a compact complex manifold and a subset U of Z such that

(i) U is the complement of at most countably many nowhere dense analytic subsets of Z.

(ii) For all u ∈ U the fiber F_{u} = {ψ^{−1}(ψ{u})} is irreducible and smooth. Furthermore κ(F_{u}) = 0 and dim F_{u} = dim Z − dim V .

(iii) dim V = κ(Z).

In general V need not be of general type.^{∗} Furthermore, often U ≠ Z. It is an unsolved problem whether or not U is open in Z. As we will see for compact subspaces of (not necessarily compact) parallelizable manifolds the situation is much better. For instance, U = Z in this case.

Recall that, given a compact complex subspace Z in a parallelizable manifold X, there exists a parallelizable submanifold X_{1} with Z ⊂ X_{1} ⊂ X such that O(X_{1}) = ℂ (see §2). Hence wlog O(X) = ℂ. Then g = Γ(X, T X) is a finite-dimensional Lie algebra and there is a canonical trivialization T X ≃ X × g. Using this trivialization for each z ∈ Z the Zariski-tangent space T_{z}Z may be regarded as a subvectorspace of g. This yields a meromorphic (holomorphic, if Z is smooth) map γ from Z to the Grassmannian manifold G(k, g), where k = dim_{ℂ}(Z), which we call the Gauss map for Z.

Lemma 6. The Gauss map γ defined as above is a meromorphic map.

Proof. The graph Λ of the Gauss map may be defined as follows: (z, p) ∈ Λ ⊂ Z × G(k, g) iff Xf_{z} = 0 for all f ∈ I_{Z,z} and X ∈ V_{p}, where I_{Z} is the ideal sheaf of Z and V_{p} the k -dimensional subvectorspace of T_{z}X ≃ g corresponding to p ∈ G(k, g). Thus it is clear that the graph is an analytic subset of Z × G(k, g).

The following observation is trivial but useful.

Lemma 7. Let V , W be two different k -dimensional subvectorspaces of a vector space g. Then there exists an alternating k -form ω such that ω|_{V} = 0 but ω|_{W} ≠ 0.

As a first step we give a description of a “generic pluricanonical fibre” in Z.

∗ Examples with V ≃ ℙ_{1}(ℂ) are provided by certain elliptic surfaces.

Theorem 2. Let X be a parallelizable complex manifold and Z an irreducible compact complex subspace. Then κ(Z) ≥ 0 and κ(Z) = 0 iff Z is a parallelizable submanifold.

Proof. Let Ẑ be a desingularization of the n-dimensional space Z. Pull-back of nforms along Ẑ → Z → X yields a map r : Λ^{n}g^{∗} → Γ(Ẑ, K). Therefore Γ(Ẑ, K) ≠ {0}, hence κ(Z) ≥ 0.

We may assume O(X) = ℂ. If there are points p, q ∈ Z_{reg} with T_{p}Z ≠ T_{q}Z (regarded as subvectorspaces of g) then there are n-forms ω, μ induced from Λ^{n}g^{∗} with ω(p) = 0 ≠ ω(q) and and μ(p) ≠ 0 = μ(q). Hence, κ(Z) = 0 implies that T_{p}Z = T_{q}Z for all p, q ∈ Z_{reg} . Since Z_{reg} is a not necessarily closed submanifold of X, it is clear that we obtain an integrability condition, i.e. h = T_{p}Z for p ∈ Z_{reg} is a Lie subalgebra of g. Locally the vector fields of this Lie subalgebra are integrable and stabilize Z_{reg}, hence Z. Moreover these local orbits are open. This implies that Z is locally homogeneous, therefore Z = Z_{reg}. Hence κ(Z) = 0 implies that Z is a parallelizable submanifold. Conversely parallelizability of Z obviously implies κ(Z) = 0.

If X is a quotient of a complex Lie group by a discrete subgroup Γ, then every parallelizable submanifold is an orbit of a Lie subgroup. Thus in this case Z ⊂ X is an orbit of a Lie subgroup H ⊂ G if and only κ(Z) = 0. As a consequence, we deduce that the analytic Zariski closure of an orbit is an orbit.

Corollary. Let X be a group-theoretically parallelizable complex manifold, i.e. X = G/Γ for a simply-connected complex Lie group G with discrete subgroup Γ, and H a subgroup of G. Let Z denote the analytic Zariski-closure of an orbit H(x) of H in X, i.e. Z is the smallest complex subspace of X containing H(x). Suppose that Z is compact. Then Z is the orbit of a complex Lie subgroup I of G.

Proof. Define I = Stab_{G}(Z). We have to show that I acts transitively on Z. Note that H acts transitively on the set of irreducible components of Z and that H ⊂ I. Thus we may replace Z by an irreducible component and H by the subgroup of H which stabilizes this component, i.e. we may assume that Z is irreducible. For every n ∈ ℕ the sections in Kn ; Ẑ induce a meromorphic map ψ_{n} : Z ⇀ ℙ_{N} . Since these maps are canonically defined, they are equivariant for all automorphisms of Z. Thus ψ_{n} is H -equivariant. Note that for n ≫ 0 the image is a variety of general type [33]. Varieties of general type have finite automorphism groups. Thus the H -orbits in the image must be finite. But there is a Zariski-dense H -orbit in Z, hence a Zariski-dense H -orbit in the image. It follows that the image must be a point. Therefore ψ_{n} must be constant, i.e. κ(Z) ≤ 0 which implies that Z is an orbit.

37 A.6

If Z is a subspace of Kodaira-dimension zero (i.e. an orbit), then every compact subspace close to Z is likewise an orbit.

Proposition 9. Let X = G/Γ be a group-theoretically parallelizable complex manifold and Z ⊂ X a closed orbit of a complex Lie subgroup H ⊂ G.

Then there exists an open neighbourhood U of Z in X and an open neighbourhood W of e in G such that for every compact complex subspace Y ⊂ U there exists an element g ∈ W with Y ⊂ gZ.

Proof. We start with proving the following assertion.

There exists a open neighbourhood W of e in G such that for every g ∈ W either gZ ∩ Z = ∅ or g ∈ H.

Since Z is compact, there is a compact ”fundamental domain” F , i.e. a compact subset F ⊂ H with F Γ = HΓ. Now let W = {g : g(F) ⊂ G \ (HΓ \ H)}. Then

gZ ∩ Z ≠ ∅
⟺ gF Γ ∩ HΓ ≠ ∅
⟺ gF ∩ HΓ ≠ ∅
⟺ gF ∩ H ≠ ∅(for g ∈ W)
⟺ g ∈ H

Next let W_{1} be an open neighbourhood of e in G such that v · w ∈ W for all v, w ∈ W_{1} . Then W_{1} · Z is a disjoint union of translates of Z. A suitably choosen ”slice” T ⊂ W_{1} yields a biholomorphic map T ×Z →^{∼} V to some open neighbourhood of Z in X. Hence Z admits an open neighbourhood which is biholomorphically equivalent to a direct product Z × T with T holomorphically separable. It follows that every compact complex subspace in U = T · Z is contained in a translate of Z.

7. Moishezon subspaces, Kähler subspaces and subspaces of general type

Subspaces which are “at the opposite extreme” to κ(Z) = 0 are those of general type, i.e. subspaces with κ(Z) = dim Z. Of course such spaces are Moishezon, i.e. tr deg_{ℂ} M(Z) = dim Z. Now Moishezon spaces are in Fujiki’s Class C. This is by definition the class of compact complex spaces which are holomorphic images of Kähler manifolds. Equivalently, a compact complex space is in Class C if it is bimeromorphically equivalent to a Kähler manifold ([34]). Sometimes manifolds in class C are called weakly Kähler.

If Z is in class C, then it is possible to prove strong statements regarding its Albanese mapping. Of course Z may be singular, so the Albanese α : Z ⇀ Alb(Z) is defined by the Albanese of a desingularization Ẑ. Recall that a meromorphic mapping of a complex manifold into a space with a Stein universal cover, e.g. ℂ^{n}, is holomorphic. Thus the Albanese is defined independent of the desingularization.

Proposition 10. Let Z be a compact subspace of a complex parallelizable manifold X. Let ψ : Z → Alb(Z) be the Albanese. If Z is in Class C, then rank(dψ) = dim Z at all nonsingular points of Z, and in particular ψ : Z → ψ(Z) is an unramified covering if Z is smooth.

Proof. Let Z̃ be a desingularization. Since Z is in Class C, it follows that every holomorphic 1-form on Z̃ is closed and a pull-back from the Albanese torus. Therefore Ker(dψ) is the space of those tangent vectors which are annihilated by all holomorphic 1-forms on Z̃. Thus the first assertion is proven by considering holomorphic one-forms on X restricted to Z and lifted to Z̃. For the second assertion note that any proper locally biholomorphic map is an unramified covering.

Corollary. Let Z be a compact subspace in Class C which is in a parallelizable complex manifold X. Then dim Alb(Z) ≥ dim Z. Equality holds if and only if Z is a torus. In particular, if Z is of general type, then dim Alb(Z) > dim Z.

Proof. If Z is of general type, then it is of course in Class C – it is bimeromorphic to a projective algebraic manifold – and therefore, by the above, dim Alb(Z) ≥ dim Z. If dim Alb(Z) = dim Z, then dim Γ(Z, Ω^{1}) = dim Z. Hence κ(Z) = 0. Now theorem 2 implies that Z is smooth. Thus the proposition implies that Z is an unramified cover of the torus Alb(Z) and therefore itself a torus.

For smooth subspaces we have the following refinement of the above result.

Proposition 11. Let Z be a compact submanifold of a complex parallelizable manifold X. Then the following statements are equivalent:

(1) Every holomorphic 1-form on Z is closed;

(2) Z is in class C;

(3) Z is a Kähler manifold;

(4) The Albanese ψ : Z → ψ(Z) is an unramified covering.

Proof. (4) ⇒ (3) is proved by pulling back a Kähler form from Alb(Z). (3) ⇒ (2) follows immediately from the definitions, and (3) ⇒ (1) is well-known. (2) ⇒ (4) is contained in proposition 10 above.

Finally, to prove (1) ⇒ (3), let ω_{1} , …, ω_{n} be holomorphic 1-forms on X which form a basis of the space of sections of its cotangent bundle. Since their restrictions are closed, ω := ∑ ω_{i} ∧ ω̄_{i} defines a Kähler form on Z.

The statement that the 1-forms are closed can be translated into an extremely strong statement on the tangent space T_{p}Z, p ∈ Z_{reg} . For a complex parallelizable manifold X with O(X) = ℂ, there is a unique trivialization of the tangent bundle T X ≃ X ×Γ(X, T X). In this way T_{p}Z may be regarded as subvectorspace of the Lie algebra of holomorphic vector fields Γ(X, T X). We will now demonstrate that T_{p}Z forms an abelian subalgebra of Γ(X, T X) for compact subspaces Z of class C.

Proposition 12. Let Z be a compact subspace of class C of X, where X is a complex parallelizable manifold with O(X) = ℂ. Then, for all p ∈ Z_{reg} , the tangent space T_{p}Z corresponds to an abelian subalgebra of the Lie algebra of holomorphic vector fields Γ(X, T X).

Proof. Let v, w ∈ T_{p}Z and let ξ, η ∈ Γ(X, T X) be the corresponding vector fields (i.e. ξ_{o} = v and η_{p} = w). Let ı : Z → X be the natural injection. If ω is a holomorphic 1-form on X, then

0 = dı^{∗}(ω)_{p}(v, w) = dω_{p}(ı_{∗}(v), ı_{∗}(w)) = dω(ξ, η) = ξω(η) − ηω(ξ) − \frac{1}{2}ω([ξ, η]) .

Note that ω(η) and ω(xi) are global holomorphic functions on X, hence constant. Thus the above equations yield ω([ξ, η]) = 0. Since ω was choosen arbitrarily and T^{∗}X is trivial (hence spanned by global sections), it follows that [ξ, η] = 0.

8. The pluricanonical map

In this section we prove a structure theorem for the pluricanonical map of compact subvarieties of parallelizable manifolds.

Theorem 3. Let Z be an irreducible compact complex subvariety of X = G/Γ, where G is a simply-connected complex Lie group and Γ a discrete (not necessarily cocompact) subgroup. Then there exists a subgroup Γ_{0} ⊂ Γ and a complex Lie subgroup Γ_{0} ⊂ H ⊂ G such that:

(1) The embedding i : Z ↪ X can be lifted to an embedding i_{0} : Z ↪ G/Γ_{0} ;

(2) The projection π : G/Γ_{0} → G/H restricted to i_{0}(Z) yields a pluricanonical map ψ : Z → Y . All fibers of ψ are of Kodaira-dimension zero and moreover are isomorphic to H/Γ_{0} .

(3) With Γ ≃ π_{1}(X) the subgroup Γ_{0} ⊂ Γ is the image of the group homomorphism π_{1}(Z) → π_{1}(X) induced by the embedding map.

In particular all fibers of the pluricanonical map have Kodaira-dimension zero.

Proof. We may assume eΓ ∈ Z. Consider the canonical projection π : G → X and let Z_{0} be the connected component of π^{−1}(Z) which contains the identity e. Regard G to be acting on itself on the right. Let Γ_{0} := Stab_{Γ}(Z_{0}). Then Γ_{0} ⊂ Γ ≃ π_{1}(X) is simply the image of the group homomorphism π_{1}(Z) → π_{1}(X) induced by the embedding i : Z ↪ X. Thus the embedding i : Z ↪ X = G/Γ can be lifted to an embedding i_{0} : Z ↪ G/Γ_{0} . In this way we may regard Z as a subvariety of G/Γ_{0} .

Let ψ : Z ⇀ V be a pluricanonical map and U ⊂ Z the set of “generic” fibers (see §6). Let π_{0} : G → G/Γ_{0} be the natural map and, for g ∈ Z_{0} , let F_{π_{0}(g)} be the ψ -fiber through π_{0}(g) whenever it is defined. For g ∈ π_{0}^{−1}(U) it follows that κ(F_{π_{0}(g)}) = 0 and thus g^{−1}F_{π_{0}(g)} is an orbit of a connected Lie subgroup I(g) (theorem 2). Recall that there exists only countably many compact orbits of complex connected Lie subgroups through eΓ_{0} (proposition 7). Recall further that U is the complement in Z of a countable union of nowhere dense analytic subsets and that π_{0}^{−1}(U) is dense in the connected analytic subset Z_{0} ⊂ G. Together these facts imply that there exists a connected complex Lie subgroup I ⊂ G such that gIΓ_{0} is contained in a ψ -fiber for all g ∈ π_{0}^{−1}(Z) = Z_{0} . This implies in particular that gIΓ_{0} = IΓ_{0} for g ∈ Γ_{0} , i.e. Γ_{0} normalizes I. Moreover it follows that the pluricanonical map (up to some bimeromorphic modification of the base space) coincides with the fibration of G/Γ_{0} → G/IΓ_{0} restricted to Z.

For subvarieties of certain parallelizable manifolds we can say more.

Theorem 4. Assume in addition to the assumptions of the above theorem that either G is solvable or that G is reductive and X is compact.

Then

(1) The base space Y can be embedded into a (not necessarily compact) complex parallelizable manifold;

(2) The base space Y is of general type, i.e. κ(Y) = dim(Y);

(3) H^{0} is isomorphic to a covering group of Aut_{O}(Z)^{0} and ψ realizes the geometric quotient for the Aut_{O}(Z)^{0} -action on Z, i.e. the fibers of ψ are exactly the orbits of Aut_{O}(Z)^{0} ;

(4) For every non-singular point p ∈ Z \ Sing(Z) the tangent space T_{p}Z corresponds to a Lie subalgebra of Γ(X, T X).

Proof. We preserve the notation introduced with the preceding theorem. Then the pluricanonical map ψ : Z → Y can be realized as restriction of a fibration G/Γ_{0} →

41 A.8

G/H. Let N denote the normalizer of H^{0} in G. Then Γ_{0} ⊂ N, hence we obtain a map Z → G/N. We claim that this map is constant. Since N is the normalizer of a connected Lie subgroup, it follows that G/N can be embedded equivariantly in some Grassmann manifold. If G is solvable, this implies that G/N is holomorphically separable (see e.g. [12]). Now let us consider the second case, i.e. G is reductive and X is compact. Then it follows that N is reductive. This implies that G/N is Stein [21]. Thus in both cases G/N does not contain any positive-dimensional compact complex subvarieties, i.e. the map Z → G/N must be constant. We may therefore replace G by N and hence assume that H^{0} is normal in G.

Under this assumption G/H is a quotient manifold of G/H^{0} by its discrete subgroup H/H^{0} . In other words: G/H is parallelizable.

Now assume that Y ⊂ G/H is not of general type. Then we obtain a nontrivial pluricanonical map ψ′ : Y → W which is the restriction of some fibration G/H → G/I. This yields a surjective holomorphic map f : Z → W where all fibers are isomorphic to I/Γ_{0} . In particular, the fibers have Kodaira-dimension zero. However, for every connected holomorphic map f : Z → W of compact complex spaces the inequality κ(Z) ≤ κ(F) + dim(W) holds for a generic fiber F (see [33,Th.6.12]). Thus dim(Y) = κ(Z) ≤ dim(W). Hence Y must be of general type.

Since H^{0} is normal in G, it follows that hgΓ_{0} ∈ gH for all gΓ_{0} ∈ Z. Since gH is the fiber of the pluricanonical map through gΓ_{0} ∈ Z, this implies that the (left) H -action on G/Γ_{0} stabilizes Z. Hence H^{0} ↪ Aut_{O}(Z)^{0} and furthermore it follows that the fibers of the pluricanonical map are just the orbits of H^{0} .

We have to show that Aut_{O}(Z)^{0} is not larger than H^{0} . Recall that a proper connected holomorphic map is equivariant for any action of a connected complex Lie group [27]. Moreover there exists a group homomorphism ρ : Aut_{O}(Z)^{0} → G such that every automorphism ϕ ∈ Aut_{O}(Z)^{0} is given by left multiplication with ρ(ϕ) (see proposition B.1). Therefore H^{0} ≠ Aut_{O}(Z)^{0} implies that there exists a positive-dimensional connected complex Lie group acting non-trivially on Y . This is impossible, because Y is of general type [33,Cor.14.3].

Finally we prove statement (4). Observe that Lie(H) ⊂ T_{p}Z ⊂ Lie(G). Now T_{p}(Z)/ Lie(H) corresponds to the tangent T_{q}Y for q = ψ(p). Since Y is of general type, it follows from proposition 12 that T_{q}Y is an (abelian) Lie subalgebra of Lie(G)/ Lie(H). This implies that T_{p}(Z) is a Lie subalgebra of Lie(G).

9. Dimension bounds for subvarieties

Here we restrict to the case where X = G/Γ is compact, Γ is discrete and G is semi-simple.

Our goal here is to prove that the subvarieties of X have a high codimension. In order to handle the case of orbits, we begin with the following elementary rough estimate.

42 A.9 Lemma 8. Let G be a simple linear-algebraic group and H a proper connected Lie subgroup which is not parabolic. Then there exists a proper parabolic Lie subgroup P of G with dim H < dim P and codim_{G} H ≥ \sqrt{dim G}.

Proof. For any connected Lie subgroup H the commutator group H′ is algebraic and equals the commutator group H̄′ of the Zariski-closure H̄ [4]. Since G = G′, it follows that a proper connected Lie subgroup H can not be Zariski-dense in G. Thus we may replace H by its Zariski-closure and thereby assume that H is algebraic. Furthermore we may assume that H is maximal in the sense that it is not contained in any higher-dimensional non-parabolic subgroup of G except G itself.

Since H is algebraic, there exists a equivariant embedding of G/H into some projective space ℙ_{N} . Hence it is clear that there exists a number n ∈ ℕ and an equivariant morphism ϕ : G/H → ℙ_{n} with finite fibers such that for every k < n every equivariant morphism from G/H to ℙ_{k} has positive-dimensional fibers. We fix such a ϕ. Since H is not parabolic, G/H is not compact. Hence there exists a closed G-orbit G/Q in the boundary of ϕ(G/H). If G ≠ Q, the proof is complete. Assume G = Q, i.e. G has a fixed point p in ℙ_{n}. Complete reducibility of representations of G implies that there is a complementary stable hyperplane H ≃ ℙ_{n−1} in ℙ_{n} and an equivariant projection ℙ_{n} \ {p} → ℙ_{n−1} . The minimality of G/H → ℙ_{n} implies that the induced morphism G/H → ℙ_{n−1} has one-dimensional fibers. Thus we obtain a fibration G/H → G/I with H ⊂ I ⊂ G and dim I = dim H + 1. Now H being non-parabolic implies dim H ≤ dim G − 2, hence I ≠ G. Thus by our maximality assumption I is parabolic. The estimate codim_{G} H ≥ \sqrt{dim G} now follows from a case by case check of the dimensions of the maximal parabolic subgroups (use e.g. [32]).

Remark. Maximal connected Lie subgroups of reductive subgroups have been classified by Dynkin [6], [7]. Using this classification one could deduce sharp bounds on the dimension of Lie subgroups for every given semisimple Lie group. However, for our purposes here, this would not yield a result of qualitative difference.

Combining the above lemma with statement (4) of theorem 4 yields the following result.

Theorem 5. Let G be a simple complex Lie group, Γ a discrete cocompact subgroup, and Z a proper complex subspace of X = G/Γ. Then codim(Z) ≥ \sqrt{dim G}.

If Z is in Class C, then for every p ∈ Z_{reg} the tangent space T_{p}Z is an abelian subalgebra of g. As a consequence dim Z is bounded from above by the maximum dimension of an abelian subgroup.

43 A.9 Theorem 6. Let G be a semi-simple complex Lie group, Γ a discrete subgroup, X := G/Γ, and Z a compact subvariety of X which is in Class C. Then dim Z ≤ \frac{1}{3} dim X.

Proof. The upper bound for the dimension of abelian Lie subgroups of type A_{n} (SL_{n+1}(C)) is a classical result of Schur [29]. Malcev [18] obtained similar results for all semi-simple Lie groups. They are summarized in the table below, where a(G) denotes the maximal possible dimension for an abelian Lie subgroup.

Group A_{n}  B_{3} B_{4} B_{n} (n > 4)  C_{n}  D_{4}
dim G (n + 1)^{2} − 1 21 36 2n^{2} + n  2n^{2} + n 28
a(G) [(n + 1)^{2}/4] 5 7 \frac{1}{2}n(n − 1) + 1 \frac{1}{2}n(n + 1) 6
Group D_{n} (n > 4) E_{6} E_{7} E_{8} F_{4} G_{2}
dim G 2n^{2} − n  78 133 248 52 14
a(G) \frac{1}{2}n(n − 1) 16 27 36 9 3

Here [n] denotes the greatest integer smaller or equal to n.

From this it follows that a(G) ≤ \frac{1}{4}(dim G + rank(G)) and a(G) ≤ \frac{1}{3} dim G for any complex semi-simple Lie group G.

10. Transversal Submanifolds in Nilmanifolds

We will demonstrate that for nilmanifolds there are no submanifolds transversal to the Albanese fibration. This may be regarded as a geometric analog for the following well-known fact in group theory: Let G be a nilpotent group, H a subgroup such that H · G′ = G. Then H = G.

Theorem 7. Let X = G/Γ a compact nilmanifold, i.e. a quotient of a connected complex nilpotent Lie group G by a discrete cocompact subgroup Γ. Let Y = X/G′ = G/G′Γ = Alb(X) and π : X → Y the natural projection. Assume that Z is a complex subspace of X with π(Z) = Y .

Then Z = X.

We start with some preparation for the proof.

Definition. Let E → B be a torus principal bundle with structure group T. Then E is called almost trivial if there exists a finite subgroup Λ of the torus T such that E/Λ → B is trivial.

Lemma 9. Let B be a complex manifold, E → B a torus principal bundle and Z ⊂ E an irreducible complex subspace with dim(Z) = dim(B) such that the projection of Z to B is surjective.

Then E is almost trivial.

Proof. Generically π : Z → B must be locally injective. Let S denote the union of all positive–dimensional components of the fibers of the projection π : Z → B. Then S is a nowhere dense analytic subset and M = π(S) is of codimension at least two in B. We claim: If E \ π^{−1}(M) → B \ M is almost trivial, then E → B is almost trivial, too. This holds, because triviality of E/Λ → B is equivalent to the existence of a section. For small enough U ⊂ B the restriction of E to U is trivial, hence a section is simply a map σ : U → T . However the universal covering of T is ℂ^{n} hence any holomorphic map σ_{0} : U \ M → T extends through M. Thus we may neglect M and assume that all fibers of E → B are finite. Let N denote the number of points in a generic fiber. Let Λ be the (finite) subgroup of T containing all elements g with Ng = 0. Now for any N –tuple a_{1}, …, a_{n} in T the mean value \frac{1}{N} ∑_{i} a_{i} is well-defined up to translation with elements in Λ. Note that taking the mean value is an operation invariant under an affine change of coordinates. Hence the mean value of all points in a given fiber of Z → B is a well-defined point in E/Λ. Thus we obtain a section in the bundle E/Λ → B, i.e. E is almost trivial.

In the following h_{1}(W) always denotes dim_{ℚ}(H_{1}(W, ℚ)).

Lemma 10. Let E → B be an almost trivial torus principal bundle. Then h_{1}(E) > h_{1}(B).

Proof. The covering E → E/Λ induces a short exact sequence

1 → π_{1}(E) → π_{1}(E/Λ) → Λ → 1.

Since Λ is finite, it follows h_{1}(E) ≥ h_{1}(E/Λ). Furthermore h_{1}(E/Λ) = h_{1}(T/Λ) + h_{1}(B) with h_{1}(T/Λ) = h_{1}(T) > 0.

Now we are in a position the prove the theorem.

Proof of the theorem. It is well-known [19] that all the groups G^{k} of the (descending) central series have closed orbits. Hence we obtain a tower of torus principal bundles

X = X_{0} → … → X_{n} = G/G′Γ

given by X_{k} = G/G^{n+1−k}, where n is the smallest number for which G^{n+1} = {e}. We may refine this tower such that all the tori occuring as fibers are simple, i.e. admit no subtori. Now (Γ ∩ G′)/Γ′ is finite, hence h_{1}(X) = h_{1}(X_{n}). Observe that for every surjective proper holomorphic map between complex manifolds f : V → W the quotient π_{1}(W)/f_{∗}(π_{1}(V)) is finite. Therefore h_{1}(X_{k+1}) ≤ h_{1}(X_{k}) for all k. Since h_{1}(X_{0}) = h_{1}(X_{n}), it follows that all the numbers h_{1}(X_{k}) coincide. By the above lemma this implies in particular that none of the torus principal bundles X_{k} → X_{k+1} can be almost trivial.

We will inductively show that Z is mapped surjectively on the X_{i}. By assumption it is mapped surjectively on X_{n}. Now let E → B be a torus–principal bundle

45 A.10

(not almost trivial) with a simple torus as fiber and Z ⊂ E with π(Z) = B. We want to show that Z = E. By lemma 9 we know already that the fibers can not be finite. Now assume that the fibers of Z → B are generically d–dimensional. For x ∈ Z let Z_{x} denote the fiber of Z → B through x. Thanks to the principal action of the torus we may regard T_{x}(Z_{x}) as a subvectorspace of the Lie algebra of the torus. Thus we obtain a meromorphic Gauss map form Z to some Grassmann manifold. Assume that generically Z_{x} is neither finite nor the whole torus T . Since T admits no subtori, this implies that the Gauss map is not constant along the fiber Z_{x}. Using the pull-back of a meromorphic function on the Grassmann manifold, it follows that Z admits a meromorphic function which generically is not constant on the fibers of Z → B. But this implies that any level set Z^{1} of this function is mapped surjectively on B. Inductively it follows that there exists a Z^{k} such that Z^{k} → B is surjective and generically finite. However this contradicts E not being almost trivial. Hence Z = E.

For some nilmanifolds every subspace is an orbit.

Proposition 13. Let G be a simply-connected complex nilpotent Lie group, Γ a discrete cocompact subgroup, X = G/Γ, A = Alb(X) = G/G′Γ.

Assume that there exists a sequence of subtori

{0} = A_{0} ⊂ A_{1} ⊂ … ⊂ A_{n} = A

such that A_{k+1}/A_{k} is a simple torus of algebraic dimension zero for all k. Then every closed irreducible complex subspace Z ⊂ X is a parallelizable submanifold, i.e. an orbit.

Proof. Let Z be a closed irreducible complex subspace of X. Let Y ⊂ X be a minimal connected nilmanifold containing Z. The results of Chapter D imply that Alb(Y) likewise fulfills the condition stated above for Alb(X). This implies that every closed irreducible complex subspace of Alb(Y) is a subtorus. Now minimality of Y implies that π : Z → Alb(Y) is surjective. Theorem 7 now yields Z = Y , i.e. Z is parallelizable.

11. The Cycle Space

For a complex space X the cycle space C in the sense of Barlet [1] parametrizes all pure-dimensional compact analytic cycles, i.e. formal linear combinations ∑_{i} n_{i}Z_{i} with n_{i} ∈ ℕ and each Z_{i} being a compact irreducible reduced complex subspace of the same dimension in X. It corresponds to the Chow scheme in algebraic geometry. We will now study cycle spaces for parallelizable manifolds.

46 A.11 Proposition 14. Let X = G/Γ be a group–theoretically parallelizable manifold. Let Z be a compact parallelizable submanifold and [Z] the corresponding point in the cycle space. Let S be the irreducible component of the cycle space C_{X} containing [Z]. Then G acts transitively on S, i.e. [C] ∈ S if and only if C = gZ for some g ∈ G.

Proof. Recall that Z = H/(H ∩ Γ) for some connected Lie subgroup H. Proposition 9 implies immediately that [Z] has an open neighbourhood S^{∗} in S which contains only translates of Z. Now let V ⊂ X × S be the associated universal cycle. On a Zariski–open subset U of V we may define a Gauss map in the following way: To (x, [C]) associate the tangent space T_{x}C regarded as a subvectorspace of the Lie algebra g. U may be choosen in such a way that it intersects all fibers of V → S. We obtain a holomorphic map from U to a Grassmann manifold which depends only on [C] but not on x provided [C] ∈ S^{∗}. From the identity principle it follows that γ never depends on x. It follows that any cycle in S has the form ∑_{i} n_{i}Z_{i} with Z_{i} parallelizable and of the same dimension as Z. Now observe that the G–orbit through [Z] in S is open and has dimension codim(Z). The G–orbit through a cycle ∑_{i} n_{i}Z_{i} has dimension greater or equal to the codimension of the Z_{i} in X. Since an irreducible space S can admit at most one open orbit of a complex Lie group G, it follows that the irreducible component S equals the G–orbit through [Z]. Moreover it follows that S is actually a connected component of the cycle space, because for every irreducible component S′ of C_{X} the intersection S ∩ S′ must be G-stable.

We will now start to investigate compactness properties of the cycle space.

Proposition 15. With the above notation we obtain S = G/I with I = H · (Γ ∩ N_{G}(H)). Furthermore S is compact iff the following two conditions are fulfilled: H is normal in G and G/Γ is compact.

Proof. Since H/(H ∩ Γ) is compact, it follows that S is compact iff G/(Γ ∩ N_{G}(H)) is compact. Thus S is compact iff Γ_{0} = Γ ∩ N_{G}(H) is of finite index in Γ and G/Γ compact. It remains to show that Γ_{0} being of finite index in Γ already implies that H is normal in G. Recall that N_{G}(H) may be realized as the kernel of a linear representation ρ of G. Now ρ(Γ) is Zariski–dense in ρ(G) (prop. I.5). This completes the proof.

Thus we obtain many new examples of compact complex manifolds for which the cycle space has non-compact irreducible components. (For large classes of compact complex manifolds it is known that all the irreducible components of the cycle space are compact, e.g. this is true for surfaces, Kähler manifolds and more generally all manifolds in class C.)

47 A.11 Corollary. Let G be a simply-connected complex Lie group, Γ a discrete cocompact subgroup. Assume that G is not nilpotent.

Then the cycle space C_{X} of X = G/Γ has a non-compact irreducible component.

Proof. There is a Lie subgroup H with closed orbit Z = H/(H ∩Γ) such that Lie(H) is a Cartan subalgebra of Lie(G) (see §3). By definition of a Cartan subalgebra Lie(H) is never an ideal in Lie(G) unless Lie(G) is nilpotent.

For tori every irreducible component of the cycle space is compact, because tori are Kähler. For some nilmanifolds X there are non-compact irreducible components of C_{X}, e.g. this is true for the standard Iwasawa manifold.

12. Fundamental groups of subspaces

A compact complex parallelizable manifold has a universal covering which is biholomorphic to a simply-connected complex Lie group and thereby biholomorphic to a Stein manifold. It follows that a compact complex parallelizable manifold can not contain simply-connected closed complex subspaces. Moreover it follows that for any closed complex subspace the embedding map is topologically non-trivial.

Lemma 11. Let X be a compact complex parallelizable manifold, Z a closed complex subspace. Then the embedding map i : Z → X induces a group homomorphism i_{∗} : π_{1}(Z) → π_{1}(X) with infinite image.

Proof. Write X = G/Γ with G simply-connected, Γ ≃ π_{1}(X). By standard covering theory the embedding i : Z → X can be lifted to an embedding j : Z → G/(i_{∗}π_{1}(Z)). This yields the assertion, because G/Λ is Stein for every finite subgroup Λ.

If i_{∗}π_{1}(Z) is commutative, then Z is contained in a torus.

Proposition 16. Let G be a simply-connected complex linear-algebraic group, Γ a discrete cocompact subgroup, Z ⊂ X = G/Γ a closed complex subspace.

Assume that i_{∗}π_{1}(Z) is a commutative subgroup of π_{1}(X) ≃ Γ, where i : Z → X denotes the embedding.

Then there exists a submanifold T which is biholomorphic to a compact complex torus such that Z ⊂ T ⊂ X.

Proof. We may assume eΓ ∈ Z. Let C denote the centralizer of i_{∗}π_{1}(Z) in G and A the center of C. Clearly i_{∗}π_{1}(Z) ⊂ A. Since the orbit of C through eΓ is closed, it follows that the A-orbit through eΓ is closed as well. Now i : Z → X = G/Γ can be lifted to an embedding j : Z → G/(i_{∗}π_{1}(Z)) which yields a holomorphic map Z → G/A. A result of Rosenlicht ([28,Th.3]) states that the quotient of a linearalgebraic group by a nilpotent algebraic group is always quasi-affine. Hence G/A is quasi-affine and therefore every holomorphic map from the compact complex space Z to G/A is constant. Thus Z is contained in the A-orbit A/A ∩ Γ ⊂ X. But A is abelian, hence A/A ∩ Γ is a compact complex torus.

For G = SL_{2}(ℂ) this can be improved.

Corollary. Let G = SL_{2}(ℂ), Γ a discrete cocompact subgroup and Z ⊂ X = G/Γ a closed complex subspace.

Then either i_{∗}π_{1}(Z) is Zariski-dense in G or Z is a one-dimensional compact complex torus.

Proof. We may assume that the Zariski-closure H of i_{∗}π_{1}(Z) is not all of G. It is easy to check that a Borel subgroup in SL_{2}(ℂ) does not contain any discrete Zariskidense subgroup. Hence H is one-dimensional and therefore contains a commutative subgroup of finite index. Similar arguments to those above show that Z is contained in an H -orbit. Hence it coincides with a connected component of an H -orbit.

13. More on subspaces of general type

The easiest way to construct subspaces of general type in a compact complex parallelizable manifolds is to take an orbit which is a compact complex torus and look for non-trivial subspaces of this torus. Under an (admittedly rather strong) additional topological assumption we have shown in the preceding section that this is the only way to construct subspaces of general type. Thus one might pose the question:

Does there exists a compact complex-parallelizable manifold X with a closed complex subspace of general type Z such that Z is not contained in any torus T ⊂ X? The author has been informed by Campana and Flenner that they recently

constructed such an example where X is a nilmanifold. Their result is the following.

Theorem (Campana-Flenner). Let E be an elliptic curve with complex multiplication, C ⊂ E × E a smooth curve of genus ≥ 2. Then there exists a three-dimensional compact complex nilmanifold X with Alb(X) ≃ E × E such that the embedding i : C ↪ E × E ≃ Alb(X) may be lifted to an embedding j : C ↪ X.

This statement implies in particular that j : C → X induces a surjection of the Albanese manifolds. Hence, any torus T ⊂ X with j(C) ⊂ T must map surjectively on Alb(X). But there are no such tori in X (see theorem 7). Hence there is no torus T with j(C) ⊂ T ⊂ X.

However, for G = SL_{2}(ℂ) the question is still open. If there exists a discrete cocompact subgroup Γ such that X = SL_{2}(ℂ)/Γ contains an irreducible compact complex subspace which is not a torus, then this subspace would of course be a (possibly singular) compact Riemann surface of genus g ≥ 2. We have already seen that the embedding of a compact Riemann surface C with g(C) ≥ 2 (if it exists) must be non-trivial on the homotopy level. This is not a useful obstruction, because there are many group homomorphisms from π_{1}(C) to Γ. (π_{1}(C) is almost free; to be more precise: it is generated by 2g generators a_{1}, …, a_{g}, b_{1}, …, b_{g} subject only to the relation [a_{1}, b_{1}][a_{2}, b_{2}] … [a_{g}, b_{g}] = e.) Furthermore every group homomorphism from π_{1}(C) to Γ is induced by a continuous mapping.

Proposition 17. Let C be a compact Riemann surface of genus g(C) ≥ 1, Γ a torsion-free discrete subgroup of SL_{2}(ℂ).

Then the set of homotopy classes of continuous maps from C to SL_{2}(ℂ)/Γ is in one-to-one correspondence to Hom(π_{1}(C), Γ).

For this we need two auxiliary results on Riemann surfaces.

Lemma 12. Let C be a Riemann surface, K a connected, simply-connected real Lie group. Then every K -principal bundle over C is (topologically) trivial.

Proof. Let E ⟶^{K} C be a K -principal bundle. Assume C ≄ ℙ_{1}(ℂ). Let π : C̃ → C denote the universal covering. Then C̃ is contractible, hence π^{∗}E is trivial. Thus π^{∗}E is given by a factor of automorphy α : Γ × C̃ → G. This is a map α fulfilling

α(γγ̃, x) = α(γ, γ̃x)α(γ̃, x).

The associated K -principal bundle over C is defined by C̃ × K/ ∼ with (x, k) ∼ (γx, α(γ, x)k) for all x ∈ C̃, γ ∈ Γ, k ∈ K. This bundle is trivial if and only if there exists a map ϕ : C̃ → K such that

ϕ(x) = ϕ(γx) = α(γ, x)  (∗)

for all x ∈ C̃, γ ∈ Γ. (The trivialization is induced by (x, k) ↦ (Γx, ϕ(x)^{−1}k).) Thus we have to construct such a map ϕ. Recall that C may be equipped with the structure of a real two-dimensional CW -complex. This is a disjoint decomposition of C into cells fulfilling certain additional properties, cf. e.g. [17]. Since cells are simplyconnected, for each cell σ ⊂ C the pre-image π^{−1}(σ) ⊂ C̃ is a disjoint union of cells which are freely permuted by the Γ-action. Thus we have a natural induced structure of a CW -complex on C̃ such that for each cell σ belonging to this structure and every γ ∈ Γ \ {e} the transform γ(σ) is again a cell belonging to this CW -complex structure. Moreover σ ∩ γ(σ) = ∅. Let C̃_{k} denote the k -skeleton, i.e. the union of all cells of dimension ≤ k. The above considerations imply that C̃_{k} contains a closed CW -subcomplex C̃_{k}^{∗} with C̃_{k−1} ⊂ C̃_{k}^{∗} such that for each k -cell σ there exists a unique γ ∈ Γ for which γ(σ) ⊂ C̃_{k}^{∗}. In other words: The natural map Γ × (C̃_{k}^{∗} \ C̃_{k−1}) → C̃_{k} \ C̃_{k−1} is bijective. Now we claim that we can construct the desired map ϕ : C̃ → K by starting with an arbitrary map from C̃_{0}^{∗} → K and then extending this map along the inclusions

C̃_{0}^{∗} ⊂ C̃_{0} ⊂ C̃_{1}^{∗} ⊂ C̃_{1} ⊂ C̃_{2}^{∗} ⊂ C̃_{2} = C̃.

To extend from C̃_{k}^{∗} to C̃_{k} we simply use the desired relation (∗). For the inclusion C̃_{k−1} ⊂ C̃_{k}^{∗} it suffices to take any continuous extension. However, we have to show that there exists such a continuous extension. Recall that K is required to be connected and simply-connected. This implies that for any relative CW -complex (X, A) with dim_{ℝ}(X) ≤ 2 any continuous map f : A → K extends to a continuous map f̃: X → K.

The proof for the case C ≃ ℙ_{1}(ℂ) follows easily from the fact that ℙ_{1}(ℂ) is obtained from S^{1} by the suspension functor. We omit the details, since we do not need this case anyway. (Since ℙ_{1}(ℂ) is simply-connected, it is clearly impossible to embed ℙ_{1}(ℂ) into a quotient of a complex Lie group by a discrete subgroup.)

Lemma 13. Let C be a Riemann surface. Then every continuous map from C to SU_{2}(ℂ) is homotopic to a constant map.

Proof. Every continuous map between differentiable manifolds is homotopic to a differentiable map. Hence every continuous map from C to SU_{2}(ℂ) is homotopic to a non-surjective map. Now SU_{2}(ℂ) is homeomorphic to S^{3} , hence every non-surjective map is homotopic to a constant map.

Proof of the proposition. Let K ≃ SU_{2}(ℂ) denote a maximal compact subgroup of SL_{2}(ℂ). Since Γ is torsion-free and discrete, Γ ∩ K = {e}. Thus we have a K - principal bundle π : X = SL_{2}(ℂ)/Γ ⟶^{K} M = K\SL_{2}(ℂ)/Γ and M is an Eilenberg-MacLane space K(Γ, 1). Hence [C, M] = Hom(π_{1}(C), Γ). Now for every continuous map f : C → M we have the pull-backed K -principal bundle over C which is trivial. Therefore f lifts to a map from C to M. Finally we have to show that two maps f, g : C → X are homotopic if and only if π∘f and π∘g are homotopic. By homotopy lifting, π ∘ f ∼ π ∘ g implies f ∼ h for some map h : C → X with π ∘ h = π ∘ g. Thus h(x) = R_{ρ(x)} (g(x)) (x ∈ C) for some map ρ : C → K, where R denotes the principal K -right action on X. Now ρ is homotopic to a constant map. Hence f ∼ h ∼ g.

51 A.14 14. Holomorphic maps f : ℂ → X

The following result (“Bloch’s Conjecture”) was proved by Green and Griffiths [8]: Let f : ℂ → X be a holomorphic map of the complex numbers into a compact complex torus. Then the complex-analytic Zariski closure W = \overline{f(ℂ)} is a subtorus. (The complex-analytic Zariski-closure of a subset S in a complex space X is the smallest closed complex-analytic subset of X containing S.) We will generalize this to holomorphic mappings from ℂ to a compact parallelizable manifold.

Theorem 8. Let X be a compact parallelizable complex manifold and f : ℂ → X a holomorphic map. Then the complex-analytic Zariski closure W = \overline{f(ℂ)} is an orbit of a complex Lie subgroup of G := Aut_{O}(X)^{0} .

Proof. By the Theorem of Green and Griffiths α : W → Alb(W) is surjective. Hence it suffices to prove the following statement:

Let X = G/Γ be a compact complex parallelizable manifold, Z an irreducible compact complex subspace. Then either Z is parallelizable or Z → Alb(Z) is not surjective. Thus we will now assume that Z is a compact complex subspace of X = G/Γ

with Z → Alb(Z) surjective and deduce that this implies parallelizability of Z. First we note that the corollary to proposition 10 implies that Z is not of general type, i.e. κ(Z) > 0. Consider the map ψ : Z → Y as given in theorem 3. Then the image of f̃ = ψ ∘ f : ℂ → Y is dense in Y with respect to the analytic Zariski-topology. If we are in a position to apply theorem 4, then it follows that Y is of general type and Y → Alb(Y) is surjective. Hence Y is a point, i.e. Z is parallelizable. This completes the proof for the case where G is reductive or solvable.

It remains to handle the case where G is neither solvable nor reductive. Here we assume that G is minimal in the following sense: If G̃ is a complex Lie subgroup of G with Z contained in a closed G̃-orbit, then G̃ = G.

We begin by considering the radical fibration τ : G/Γ → G/RΓ = S/Λ, where R is the radical and S a maximal connected semi-simple Lie subgroup. Since S is reductive it follows that the image of Z under τ is an orbit. By minimality of G it follows that Z is mapped surjectively on S/Λ. Going back to the proof of theorem 4 we note that in order to deduce the statements of theorem 4 it is enough to show that G/N is holomorphically separable (with N = N_{G}(H^{0}) defined as in §8). Now, the manifold G/N is naturally embedded in a Graßmann manifold. Let \overline{G} denote the algebraic Zariski-closure of G in the automorphism group of this Graßmann manifold and \overline{N} the isotropy group of \overline{G} at the neutral point eN. Furthermore let \overline{R} denote the radical of \overline{G}. Then the \overline{R}-orbits in \overline{G}/\overline{N} are closed and \overline{R} \overline{N}/\overline{R} is a closed algebraic subgroup of \overline{G}/\overline{R} = S. Recall that Z may be embedded in G/Γ_{0} with

52 A.14

Γ_{0} ⊂ N. It follows that Λ_{0} = RΓ_{0}/R is a discrete cocompact subgroup of S with Λ_{0} ⊂ RN/R. By Proposition I.5 this implies that Λ_{0} is Zariski dense. Hence RN/R is also Zariski-dense in S. Thus \overline{R} \overline{N}/\overline{R} = S, i.e. \overline{G} = \overline{R} \overline{N}. Therefore \overline{R} acts transitively on \overline{G}/\overline{N}. It follows that G/N is a subset of an orbit of a solvable group in a Graßmann manifold. Since such orbits are holomorphically separable, it follows that G/N is holomorphically separable. This completes the proof.

The theorem can be generalized to deal with holomorphic maps from arbitrary complex Lie groups instead of ℂ.

Proposition 18. Let X be a compact parallelizable complex manifold, H a connected complex Lie group and f : H → X a holomorphic map. Then the complex-analytic Zariski closure W = \overline{f(H)} is an orbit of a complex Lie subgroup of G := Aut_{O}(X)^{0} .

Proof. For every element v in the Lie algebra Lie(H) let S_{v} denote the set {f(exp(vt) : t ∈ ℂ\}. By the theorem the complex-analytic Zariski closures \overline{S_{v}} are orbits of complex Lie subgroups of G containing f(e). Recall that there exist only countably many connected Lie subgroups of G with compact orbits through a given point (Prop. 7). Therefore there exists an element v_{0} ∈ Lie(H) and a subset Ω ⊂ Lie(H) such that Ω is dense in the analytic Zariski-topology and \overline{S_{v_{0}}} = \overline{S_{v}} for all v ∈ Ω. Density of Ω implies that S_{v} ⊂ Z for all v ∈ Lie(H), hence density of exp(Lie H) in H implies that f(H) ⊂ \overline{S_{v_{0}}} . Hence \overline{f(H)} = \overline{S_{v_{0}}} and the proof is complete.

As a consequence of the theorem we obtain a characterization of hyperbolic submanifolds.

Corollary. Let X = G/Γ be a compact complex parallelizable manifold and Z a complex subspace. Then Z is hyperbolic in the sense of Kobayashi if and only if it contains no compact complex torus.

Proof. By a result of Brody ([5]), Z is hyperbolic if and only if every holomorphic map from ℂ to Z is constant. Hence Z is hyperbolic if and only if it contains no parallelizable complex submanifold. Finally recall that every compact complex parallelizable manifold contains a compact complex torus (proposition 4).

For submanifolds of parallelizable manifolds we thus obtain an affirmative answer to the following general question.

Question: Given a compact complex manifold Z, are the following conditions equivalent?

(1) Z is hyperbolic in the sense of Kobayashi;

(2) There is no non-constant holomorphic map from compact complex torus to Z.

53 A.14

Serge Lang [16] conjectured that the answer is positive is true at least for projective manifolds. In general it is obvious that (1) implies (2). Furthermore (2) implies that every curve C ⊂ Z is of genus ≥ 2 and therefore hyperbolic (note that condition (2) excludes rational curves, because there is a surjective map from a torus to ℙ_{1}(ℂ)).

We would like to remark that this characterization of hyperbolic submanifolds can be proved in a more elementary way, in particular without using the result of Green and Griffiths.

By the reparametrization results of Brody [5] it suffices to prove the following:

Let h be a hermitian metric on Z and f : ℂ → Z be a non–constant holomorphic map with bounded derivative (with respect to h). Then Z contains a parallelizable submanifold.

But this is an immediate consequence of the lemma below, because (see corollary to theorem 2) the complex-analytic Zariski–closure of an orbit is parallelizable provided it is compact.

Lemma 14. Let G, H be complex Lie groups, f : G → H be a holomorphic map with f(e_{G}) = e_{H}. Assume that the derivative T f is bounded with respect to some G– resp. H –invariant hermitian metrics on G and H.

Then f is already a Lie group homomorphism.

Proof. Consider T f : T G → T H. It induces a map F : G × g → h which is linear in the second variable and such that F (·, v) : G → h is bounded. By Liouville’s theorem F (·, v) is constant for all v. Thus we obtain a linear map ϕ : g → h and a dual map ϕ^{∗} : h^{∗} → g^{∗}. The latter is induced by pulling back holomorphic 1-forms. Hence

α([ϕX, ϕY]) = dα(ϕX, ϕY) = ϕ^{∗}dα(X, Y) = dϕ^{∗}α(X, Y) = ϕ^{∗}([X, Y]) = α(ϕ[X, Y])

for all α ∈ h^{∗}. Thus [ϕX, ϕY] = ϕ[X, Y], i.e. ϕ is a Lie algebra homomorphism.

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