Some Remarks on Parallelizable Manifolds

by

K. Oeljeklaus & J. Winkelmann

1. A reduction Theorem

A complex manifold X is called parallelizable if its 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. If X is furthermore compact, then X can be realized as a quotient of a complex Lie group by a discrete subgroup. ([8]). In [5] compact submanifolds of parallelizable manifolds have been studied. Since most of the results in [5] require X to be a parallelizable manifold which is compact or at least has no non–constant holomorphic functions, it is desirable to reduce the case of an arbitrary parallelizable manifold to one with finite–dimensional Lie algebra. We will actually prove a more general result.

Theorem 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.

Proof. 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}). If X contained in the Zariski tangent sheaf in one point p, it follows that df(X)(p) = 0 and hence df(X)|_{X_{k+1}} ≡ 0. This implies that X is tangential everywhere. Since X_{k} is locally homogeneous, it follows that all the Zariski tangent spaces have the same dimension, in particular X_{k+1} is smooth and moreover locally homogeneous.

It should be noted that in fact the tangent bundle of Y is spanned by the restrictions of global vector fields on X. This implies that Y is homogeneous if X is.

The second author wishes to thank the Université de Lille for his invitation and their hospitality.

K. Oeljeklaus & J. Winkelmann

2. The normal bundle of a submanifold

By a result of Gieseker [3] a vector bundle N over a projective manifold V is ample if and only if for every curve C and any quotient line bundle L = N/Q the restriction of L to C is ample. Let A denote the vector space of global sections of N over V and A_{x} the subvectorspace of sections vanishing in x for each x ∈ V . Now Giesekers criterion may be reformulated as follows: N is ample if and only if ∑_{x∈C} A_{x} = A for every curve C. We will now apply this to our situation:

Definition. Let X be a parallelizable complex manifold, A the algebra of global holomorphic vector fields and Z a compact submanifold of X. Assume O(X) = C. Then every tangent space T_{x}Z for x ∈ Z may be considered as a subvectorspace of A. Now Z is called generating if A is the smallest Lie subalgebra of A containing all T_{x}Z and strongly generating if for any positive–dimensional submanifold V of Z the subvectorspace ⊕_{x∈V} T_{x}Z is already A.

Lemma 1. (i) ”generating” in our sense is equivalent to ”generateur” as defined by Hirschowitz [4]. (ii) ”Strongly generating” implies Z is generating, that Z is a projective manifold of general type and that the normal bundle is ample. (iii) Any submanifold of a strongly generating manifold is strongly generating. (iv) Any strongly generating submanifold is hyperbolic in the sense of Kobayashi.

Proof. (i) Left to the reader.

(ii) Note that by the results of [5] Z is either a projective manifold of general type or admits a submanifold M which is parallelizable. If Z is strongly generating such a M can not exist, hence Z is projective and we may apply the criterion of Gieseker to obtain that the normal bundle is ample.

(iii) Trivial. (iv) By [5] Z is hyperbolic iff no submanifold of Z is parallelizable. Now observe that if N is ample then H^{k}(Z, S^{μ}N ⊗ F) is vanishing for every coherent

sheaf F for μ ≥ μ(F) and k > 0. By Serre duality we obtain vanishing results for H^{k}(Z, S^{μ}N^{∗} ⊗ F̃) with k < dim(Z) and F̃ = K ⊗ F. For dim(Z) > 1 it follows that the embedding of strongly generating submanifolds is formally rigid, which gives rise to the hope for extension results. ([3]). Thus it is reasonable to expect that any strongly generating submanifold of dimension at least two admits an open neighbourhood with many meromorphic functions. This suggests that there are no strongly generating submanifolds of dimension two for S/Γ because of the following:

Proposition 1. Let S be semisimple complex Lie group and Γ a discrete subgroup, Z a generating compact submanifold of X = S/Γ. Then Z does not admit any meromorphically separable open neighbourhood.

Parallelizable Manifolds

Proof. Let Γ_{0} denote the image of π_{1}(Z) in Γ = π_{1}(X). Then Z may be lifted to a compact submanifold of S/Γ_{0} . Using the results of Hirschowitz [4] any meromorphic function in a neighbourhood U of Z may be extended to a meromorphic function on S/Γ_{0} . On the other hand Γ_{0} must be infinite, since S has no compact subspaces. Hence the algebraic Zariski–closure of Γ_{0} is positive–dimensional. But every meromorphic function on S/Γ_{0} is invariant under Γ̄_{0} , thus U can not be meromorphically separable.

3. A hyperbolicity criterion

Let Z be a compact subspace of a complex parallelizable manifold X = G/Γ. Z is hyperbolic if and only if it contains no parallelizable submanifold. This is proved in [5] using deep results of Green and Griffiths. We will now give an elementary proof. By the reparametrization results of Brody [1] 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 proposition below, because due to [5] the Zariski–closure of an orbit is always parallelizable provided it is compact.

Proposition 2. 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 the Liouville 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.

4. Nilmanifolds

Our goal is to prove the following:

Proposition 3. Let X = G/Γ a compact nilmanifold, i.e. a quotient of a complex nilpotent Lie group G by a discrete cocompact subgroup Γ. Let Y = X/G′ = G/G′Γ and π : X → Y the natural projection. Assume that Z is a subspace of X with π(Z) = Y .

Then Z = X.

Lemma 2. Let G be a nilpotent Lie group, Γ a discrete cocompact subgroup. Then Γ′ is of finite index in G′ ∩ Γ and cocompact in G′.

This is well-known, see [7]. It implies that we may choose a descending sequence of connected Lie subgroups A_{i} of G such that

- A_{1} = G′, [G, A_{k}] ⊂ A_{k+1} ,

- A_{k}/(A_{k} ∩ Γ) is compact,

- A_{k+1}Γ is normal in A_{k}Γ,

- A_{k} ∩ Γ′ is of finite index in A_{k} ∩ Γ,

- ∃N : A_{k} = {e} for k > N.

Lemma 3. Let X be a compact nilmanifold. Then there exists a sequence

G/Γ = X = X_{0} → X_{1} → … → X_{n} = G/G′Γ

of G-equivariant torus principal bundles. Furthermore H_{1}(X_{i}, ℚ) = H_{1}(X_{j}, ℚ) for all i, j.

Proof. Recall that H_{1}(W, ℚ) = π_{1}(W)/[π_{1}(W) : π_{1}(W)] ⊗_{ℤ} ℚ. Furthermore with X_{i} = G/A_{i}Γ we obtain π_{1}(X_{i}) = Γ/(Γ ∩ A_{i}), if we require G to be simply–connected. Now Γ′ ∩ A_{i} is of finite index in Γ ∩ A_{i} and torsion is killed by taking coefficients in ℚ.

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

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 4. 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.

In particular the torus principal bundles X_{i} → X_{i+1} constructed above are never almost trivial.

Lemma 5. Let E → B be 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.

Now we are in a position the prove the proposition.

Proof of the proposition. Let X = X_{0} → … → X_{n} = G/G′Γ be the above constructed tower of torus principal bundles. We may refine this tower such that all the tori occuring as fibers are simple, i.e. admit no subtori. 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 (not almost trivial) with a simple torus as fiber and Z ⊂ E with π(Z) = B. We want to show that Z = E. By the last lemma 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} denotes 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.

As a consequence we obtain

Corollary. Let X = G/Γ be a compact nilmanifold, Z a Moishezon subspace. Then π : Z → Y = G/G′Γ is not surjective.

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Lille,