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Homomorphisms between fundamental groups of K\"ahler manifolds

A group morphism is constructed, which can be realized as the induced morphism of fundamental groups from a holomorphic map between compact Kahler manifolds, but can not be realized by a holomorphic map between smooth projective varieties. And it is also proved that there exists no such example between abelian groups.


Introduction
The fundamental groups of compact Kähler manifolds and smooth complex projective varieties have been studied for more than twenty years. Arapura had a survey paper [A1] on this subject, and a more detailed introduction was given in the book [ABC]. A natural and unsolved question is whether the class of fundamental groups of compact Kähler manifolds and the class of fundamental groups of smooth complex projective varieties coincide. In [V], Voisin gave examples of compact Kähler manifolds of dimension four or higher, which do not have the homotopy type of any smooth complex projective variety. In this paper, we use Voison's method to answer a weaker question. If one considers not only the fundamental groups, but also the homomorphisms between them, we can give a negative answer.
First, we need some definitions. A group is called Kähler (or projective) if it is isomorphic to the fundamental group of a compact Kähler manifold (or smooth complex projective variety). A homomorphism between Kähler (or projective) groups is called Kähler (or projective) if it is the homomorphism of fundamental groups induced by a holomorphic map between compact Kähler manifolds (or smooth complex projective varieties). The main result of this paper is the following

Between abelian groups
In this section, we prove that for a homomorphism in the category of abelian groups, the condition of being Kähler is same as being projective.
Let me start with some basic properties about Kähler groups and Kähler homomorphisms.
Lemma 2.1. The abelianization of any Kähler group has even rank.
Proof. By Hodge theory, the first Betti number of a compact Kähler manifold must be even. And for any manifold X, is the abelianization of f , then Ker(f ), Coker(f ) and Im(f ) all have even rank.
Proof. This is because a holomorphic map between Kähler manifolds is compatible with the Hodge structure. In fact, suppose ϕ : X → Y is such a holomorphic map, then preserves the Hodge decomposition. And by Poincare duality, ϕ * : H 1 (Y, C) → H 1 (X, C) is dual to ϕ * : H 1 (X, C) → H 1 (Y, C). Therefore by the isomorphism in Lemma 2.1, the result follows.

Lemma 2.3. Every finite group is projective, hence Kähler.
This is observed by Serre, a proof can also be found in [S] page 227. There is a similar result for Kähler homomorphisms: Proof. Suppose A, B and C are the fundamental groups of compact Kähler manifolds X, Y and Z, and f is represented by a holomorphic map ϕ : X → Y , i.e., ϕ * : LetX,Ỹ andZ be the universal covers of X, Y and Z, then A, B and C act onX,Ỹ andZ respectively. Let It is a subgroup of A × C with finite index, so we can take the quotient ofX ×Z by this subgroup, which has a further quotient: The corresponding homomorphisms of fundamental groups are Notice that a finitely generated abelian group is Kähler if and only if it is projective, if and only if it has even rank. In fact, given any finitely generated abelian group of even rank, it is a direct sum of free part and torsion part. Therefore, according to Lemma 2.3, we can take the fiber product of an abelian variety and a smooth projective variety. And the fundamental group of the product is isomorphic to the given abelian group. We will prove that, for a homomorphism between abelian groups, the obstruction of being Kähler is the same.
Proposition 2.5. Let f : A → B be a homomorphism of abelian Kähler groups, then the following three conditions are equivalent: (i)f has even rank, i.e., one of Ker(f ), Coker(f ), Im(f ) has even rank, hence all of them.
is a homomorphism of abelian groups satisfying properties of (i).
We first prove the case that A and B are both free of rank 2r and 2s. Then f is represented by a matrix, which we can assume to be a Smith normal form by a proper choice of basis. Hence f factors into three steps, a projection to a direct summand, an injection with finite cokernel, an injection as a direct summand. Therefore, we will define a homomorphism φ : X → Y between some abelian varieties as a composition of three steps, a projection to a direct summand, a finite covering map, and an injection as a direct summand. And the induced map of the fundamental groups φ * : π(X) → π(Y ) will be isomorphic to f . We choose X to be the r-dimensional torus C r /(a 1 Z + a 2 iZ) + . . . + (a 2l−1 Z + a 2l iZ) + (Z + iZ) r−l The first step is projection to the torus of the first l components C l /(a 1 Z + a 2 iZ) + . . . + (a 2l−1 Z + a 2l iZ)) In the second step, we take the covering map onto torus C l /(Z + iZ) l , as a further quotient. Then we take the embedding to C s /(Z + iZ) s , which will be our Y . All the tori we chose are abelian varieties, since they are products of elliptic curves. The case that only B is free follows from the previous case and Lemma 2.3, since a homomorphism from a finite group to a free abelian group must be zero, and a finitely generated abelian group is a direct sum of it is free part and torsion part.
The general case follows from the previous case, and the previous lemma.

Non-projective example
In this section, we construct a Kähler homomorphism which is not projective.
In [V], Voisin proved the following: Lemma 3.1. Suppose an n(≥ 2)-dimensional complex torus T has an endomorphism φ, then φ * acts on H 1 (T, Z). Let p be the characteristic polynomial of φ * . If p is irreducible, has no real roots and the Galois group of its splitting field acts as the symmetric group on the roots, then the torus T is not an abelian variety.
She also gave the method to construct such a torus. A concrete example for n = 2 can be found in [H].
Suppose (T, φ) satisfies the assumption of the Lemma, and W is a compact Kähler manifold with π 1 (W ) isomorphic to the symmetric group S 3 . Denote the universal cover of W byW . Let S 3 act on (T × T × T ) ×W by permutation on the first part and by deck transformation on the second. Since the action is fixed point free, we get a Kähler manifold as the quotient (T × T × T ) ×W /S 3 . Fixing any point P onW , we define a sequence of maps: where we let the first map µ ′ : (a, b) → (a + b, φ(a) + b, b), and i is mapping to the fiber over P ∈W , and π is the quotient map.
Let µ be the composition π•i•µ ′ , then the induced map µ * of fundamental groups is Kähler. But it cannot be projective.
Proof. Suppose ν : X → Y is a holomorphic map between compact Kähler manifolds, which induces the same homomorphism of fundamental groups as µ. We will show that Y cannot be projective. Since we can takeỸ as the 6-fold covering of Y corresponding to the surjection Z 2n × Z 2n × Z 2n ⋊ S 3 → S 3 . S 3 will act onỸ . Notice the composition of µ * and the surjection Z 2n × Z 2n × Z 2n ⋊ S 3 → S 3 is trivial. Hence ν : X → Y factors through X ν ′ →Ỹ → Y . The choice of ν ′ is not unique, but we will specify which one we want later.
Consider the induced map of Albanese: Alb(ν ′ ) : Alb(X) → Alb(Ỹ ), where Alb(Ỹ ) has an S 3 action on it induced by the action onỸ . Let Σ be the fixed locus of Alb(Ỹ ) by this S 3 action. Then Σ is an n-dimensional subtorus. Also let Γ 1 , Γ 2 and Γ 3 be the fixed loci by the three order 2 elements s 1 = (23), s 2 = (31), s 3 = (12) of S 3 respectively. Therefore Γ i corresponds to the permutation s i ∈ S 3 fixing the i-th T . Therefore, s 3 gives an isomorphism between Γ 1 and Γ 2 . Let Γ 4 be the image of Alb(ν ′ ).
Since s 3 permutes Γ 1 and Γ 2 , and fixes Σ, it induces an isomorphism between Γ 1 and Γ 2 , which we also call s 3 by abusing of notation. Now, consider the composition: It is same as φ via identifying Γ 1 with T as real tori. Here Γ 4 plays the role of the graph.
By the previous lemma, Γ 1 is not an abelian variety. Consequently, none of Γ 1 , Alb(Ỹ ),Ỹ or Y can be a projective variety. Therefore µ * is not a projective homomorphism.