Kähler manifolds and groups, part 2

In this post I hope to start talking in a bit more depth about the global geometry of compact Kähler manifolds and their covers. Basic references for much of this post are the book Fundamental groups of compact Kähler manifolds by Amoros-Burger-Corlette-Kotschick-Toledo, and the paper Kähler hyperbolicity and L2 Hodge theory by Gromov. It turns out that there is a basic distinction in the world of compact Kähler manifolds between those that admit a holomorphic surjection with connected fibers to a compact Riemann surface of genus at least 2, and those that don’t. The existence or non-existence of such a fibration turns out to depend only on the fundamental group of the manifold, and in fact only on the algebraic structure of the cup product on $H^1$; thus one talks about fibered or nonfibered Kähler groups.

If X is a connected CW complex, by successively attaching cells of dimension 3 and higher to X we may obtain a CW complex Y for which the inclusion of X into Y induces an isomorphism on fundamental groups, while the universal cover of Y is contractible (i.e. Y is a $K(\pi,1)$ with $\pi$ the fundamental group of X). The (co)-homology of Y is (by definition) the group (co)-homology of the fundamental group of X. Since Y is obtained from X by attaching cells of dimension at least 3, the map induced by inclusion $H^*(Y) \to H^*(X)$ is an isomorphism in dimension 0 and 1, and an injection in dimension 2 (dually, the map $H_2(X) \to H_2(Y)$ is a surjection, whose kernel is the image of $\pi_2(X)$ under the Hurewicz map; so the cokernel of $H^2(Y) \to H^2(X)$ measures the pairing of the 2-dimensional cohomology of X with essential 2-spheres).

A surjective map f from a space X to a space S with connected fibers is surjective on fundamental groups. This basically follows from the long exact sequence in homotopy groups for a fibration; more prosaically, first note that 1-manifolds in S can be lifted locally to 1-manifolds in X, then distinct lifts of endpoints of small segments can be connected in their fibers in X. A surjection $f_*$ on fundamental groups induces an injection on $H^1$ in the other direction, and by naturality of cup product, if $V$ is a subspace of $H^1(S)$ on which the cup product vanishes identically — i.e. if it is isotropic — then $f^*V$ is also isotropic. If S is a closed oriented surface of genus g then cup product makes $H^1(S)$ into a symplectic vector space of (real) dimension 2g, and any Lagrangian subspace V is isotropic of dimension g. Thus: a surjective map with connected fibers from a space X to a closed Riemann surface S of genus at least 2 gives rise to an isotropic subspace of $H^1(X)$ of dimension at least 2.

So in a nutshell: the purpose of this blog post is to explain how the existence of isotropic subspaces in 1-dimensional cohomology of Kähler manifolds imposes very strong geometric constraints. This is true for “ordinary” cohomology on compact manifolds, and also for more exotic (i.e. $L_2$) cohomology on noncompact covers.

1. Fibered Kähler groups

For a compact Kähler manifold Hodge theory gives

$H^1(M) = H^{1,0}\oplus H^{0,1} = \Omega^1_h \oplus \overline{\Omega}^1_h$

(recall that the notation $\Omega^p_h$ means the holomorphic p-forms). In other words, every (complex) 1-dimensional cohomology class has a unique representative 1-form which is a linear combination of holomorphic and anti-holomorphic 1-forms. Since the wedge product of holomorphic 1-forms is holomorphic (the first miracle mentioned in the previous post!), for holomorphic 1-forms $\alpha,\beta$ we have

$[\alpha]\cup[\beta] = 0 \in H^2(M)$ if and only if $\alpha \wedge \beta = 0$ as forms.

This has the following classical application:

Theorem (Castelnuovo-de Franchis): Let M be a compact Kähler manifold, and let V be a subspace of the space of holomorphic 1-forms on M which is isotropic with respect to the pairing (on cohomology; but equivalently, on forms). Suppose that the dimension of V is at least 2. Then there exists a surjective holomorphic map f with connected fibers from M to a compact Riemann surface C of genus g such that V is pulled back by f from C.

Proof: Let $\alpha_1,\cdots, \alpha_g$ where $g \ge 2$ be a basis of V. Where two forms $\alpha_i, \alpha_j$ don’t vanish, the condition that $\alpha_i \wedge \alpha_j = 0$ says that they are proportional, and therefore the ratio $\alpha_i/\alpha_j$ is a holomorphic function. If we let U denote the open (and dense) subset of M where none of the $\alpha_i$ vanish, then the ratios $\alpha_i/\alpha_1$ define the coordinates of a holomorphic map to $\mathbb{P}^{g-1}$. Since $\alpha_1$ is closed, its kernel is tangent to a (complex) codimension 1 foliation $\mathcal{F}$ on U. Since the $\alpha_i$ are closed, the ratio $\alpha_i/\alpha_1$ is constant on the leaves of $\mathcal{F}$, so the image of U in $\mathbb{P}^{g-1}$ is 1-dimensional, and the map factors through a map to a compact Riemann surface D.

A priori a holomorphic map to a Riemann surface defined on an open set U does not extend to M; the simplest example to think of is the holomorphic function

$x/y: \mathbb{C}^2 \to \mathbb{P}^1$

where x and y are the two coordinate functions. This map is well defined away from the origin, where it is indeterminate. On the other hand, as we approach the origin radially along a (complex) line, the ratio $x/y$ is constant; so the map, defined on $\mathbb{C}^2 - 0$, extends over a copy of $\mathbb{P}^1$ obtained by blowing up the origin. In general therefore a map $f: U \to D$ extends to $f:M' \to D$ where M’ is obtained from M by blowing up along the indeterminacy of the map f, and the fibers of the blow-up map from M’ to M are all copies of $\mathbb{P}^1$.

Now, the map $f:M' \to D$ does not necessarily have connected fibers, but it is proper. So there is a (so-called) Stein factorization $M' \to C \to D$ for some intermediate compact Riemann surface C, where $M' \to C$ has connected fibers, and $C \to D$ is finite-to-one. As a set, the points of C are just the connected components of the point preimages of $M' \to D$. As a complex manifold, the charts on $C$ are modeled on the transverse holomorphic structure on the foliation $\mathcal{F}$. Notice that since (as remarked above) the 1-forms $\alpha_j$ are all locally constant on the leaves of $\mathcal{F}$, they descend to well-defined 1-forms on $C$ (which pull back to the $\alpha_j$ under the map). In particular, we deduce that $C$ has genus at least $g\ge 2$. But now we see that there was no indeterminacy at all, since the $\mathbb{P}^1$ fibers of the blow up $M' \to M$ admit no non-constant holomorphic map to a surface of positive genus, and therefore the map $M' \to C$ factors through $M \to C$ after all. qed

Now suppose M is a compact Kähler manifold, and let V be a subspace of $H^1(M)$ which is isotropic with respect to cup product, and of dimension at least 2. We can choose real harmonic 1-forms $\beta_i$ which are a basis for V, and take their holomorphic (1,0)-part $\alpha_i$.  Then $\alpha_i \wedge \alpha_j$ is holomorphic, and is equal to the (2,0)-part of $\beta_i \wedge \beta_j$. Since the holomorphic 2-forms inject into cohomology, it follows that $\alpha_i \wedge \alpha_j = 0$ as forms. It is straightforward to check that the $\alpha_i$ are linearly independent if the $\beta_i$ are, so we obtain an isotropic subspace of holomorphic 1-forms of the same dimension as V. Applying Castelnuovo-de Franchis, we see that M fibers over D as above (this observation is due to Catanese).

From this we easily deduce the following theorem of Siu-Beauville, proved originally by hard analytic methods (i.e. the theory of harmonic maps):

Corollary (Siu, Beauville): Let M be a compact Kähler manifold, and let $g \ge 2$. Then there is a holomorphic map with connected fibers from M to a compact Riemann surface C of genus at least g if and only if there is a surjective homomorphism $\pi_1(M) \to \pi_1(C )$.

Proof: A surjective map with connected fibers is surjective on fundamental groups. Conversely, a surjective map on fundamental groups pulls back $H^1(C )$ injectively, and pulls back a maximal isotropic subspace of $H^1(C )$ (which has dimension $g \ge 2$) to an isotropic subspace of $H^1(M)$. qed

Definition: A Kähler group is fibered if it surjects onto the fundamental group of a compact Riemann surface of genus at least 2; equivalently, if some (equivalently: every) compact Kähler manifold with that fundamental group holomorphically fibers over a compact Riemann surface of genus at least 2 with connected fibers.

Note that the condition of being fibered implies $b^1\ge 4$.

2. L2 cohomology

Perhaps the fundamental method in geometric group theory is to study a group via its cocompact isometric action on some (typically noncompact) space. If G is the fundamental group of a manifold M, then G acts as a deck group on the universal cover of M. The aim of geometric group theory is to perceive algebraic properties of the group G in the “global” geometry of this universal cover.

The most important tool for the study of differential forms on compact Riemannian manifolds is Hodge theory. To use this tool on noncompact manifolds one must impose additional (global) restrictions on the forms that one studies. Thus Hodge theory on noncompact manifolds is related directly not to ordinary cohomology, but to more refined, quantitative versions, of which one of the most important is $L_2$-cohomology.

If M is a smooth Riemannian manifold (not assumed to be compact), the pointwise inner product on forms gives rise to a global inner product which is well-defined on compactly supported forms. We say that a smooth form $\alpha$ is in $L_2$ if

$\|\alpha\|^2:=\int_M \alpha \wedge *\alpha < \infty$

Now, the $L_2$-forms do not usually form a chain complex, but we can pass to a subcomplex $L_2\Omega^*$ consisting of forms $\alpha$ for which both $\alpha$ and $d\alpha$ are $L_2$-forms. Since $d^2=0$ this is a complex, and we can define $L_2$ cohomology:

$L_2H^k(M):= (\text{ker}d:L_2\Omega^k \to L_2\Omega^{k+1})/dL_2\Omega^{k-1}$

In general, the image of d is not a closed subspace (in the $L_2$ topology), so we define the reduced $L_2$ cohomology to be:

$\overline{L_2H}^k(M):=(\text{ker}d:L_2\Omega^k \to L_2\Omega^{k+1})/\overline{dL_2\Omega^{k-1}}$

The advantage of working with reduced cohomology is that there is an $L_2$-analogue of the Hodge theorem. The operators $d$ and $\delta = \pm *d*$ still make sense on a noncompact Riemannian manifold, and so does $\Delta:=d\delta+\delta d$. We can define the harmonic forms to be those for which $\Delta\alpha=0$, and we denote by $\mathcal{H}^p_{(2)}$ the space of harmonic p-forms which are $L_2$.

Let’s impose some reasonable global conditions on our manifold M. We say that a (complete) Riemannian manifold has bounded geometry if it satisfies the following two conditions:

1. The curvature and its derivatives satisfy uniform 2-sided bounds: $|\nabla^k K| \le C_k < \infty$ for each k; and
2. The injectivity radius satisfies a uniform lower bound: $\text{inj} \ge \epsilon > 0$ everywhere.

Bounded geometry is the natural condition to impose to ensure that the manifold is “precompact” in Gromov-Hausdorff space; i.e. that for any sequence of points $p_i$ in $M$ the sequence of pointed metric spaces $(M,p_i)$ contain a subsequence which converge on compact subsets to a pointed Riemannian manifold $M',p'$. An equivalent way to think about it is that this is the condition which ensures that the Riemannian manifold $M$ can appear as a leaf in a compact lamination. The condition of bounded geometry is automatically satisfied for any cover (infinite or not) of a compact Riemannian manifold. Since this is essentially the only class of noncompact Riemannian manifolds we will consider, we hereafter assume that all our noncompact Riemannian manifolds have bounded geometry.

Theorem (L2 Hodge theorem): Let M be a complete Riemannian manifold with bounded geometry. Then every cohomology class in $\overline{L_2H}^k$ has a unique representative $\alpha \in L_2\Omega^k$ minimizing $\|\alpha\|$. Such a form is harmonic; i.e. it is in $\mathcal{H}^k_{(2)}$. Moreover, there is an orthogonal decomposition

$L_2\Omega^k = \mathcal{H}^k_{(2)} \oplus \overline{dL_2\Omega^{k-1}} \oplus \overline{\delta L_2\Omega^{k+1}}$

One subtlety is that it is no longer true that $\delta$ is a formal adjoint to d, since integration by parts gives rise to a potentially nontrivial boundary term “at infinity”. But for an $L_2$ form $\alpha$, this boundary term vanishes, and one has

$\langle \Delta\alpha,\alpha\rangle = \langle d\alpha,d\alpha\rangle + \langle \delta\alpha,\delta\alpha\rangle$

(since a priori the forms $d\alpha$ and $\delta \alpha$ are not $L_2$, one first interprets this by using cutoff functions, and passing to a limit). In other words, a harmonic form which is also $L_2$ is closed and coclosed; conversely, any form which is closed and coclosed is harmonic (with no analytic conditions).

On a Kähler manifold the identity $\Delta = 2\Delta_{\overline{\partial}}$ still holds pointwise (since this is a consequence purely of the local properties of the metric), and so there is a further decomposition of $\mathcal{H}^k_{(2)}$ into components $\oplus_{p+q=k} \mathcal{H}^{p,q}_{(2)}$ which are individually harmonic. There is furthermore a Hodge decomposition

$L_2\Omega^{p,q} = \mathcal{H}^{p,q}_{(2)} \oplus \overline{\overline{\partial}L_2\Omega^{p,q-1}} \oplus \overline{\overline{\partial}^*L_2\Omega^{p,q+1}}$

and an $L_2$ form $\alpha$ satisfies $\Delta \alpha = 0$ if and only if $\overline{\partial}\alpha = 0$ and $\overline{\partial}^*\alpha=0$. Thus $\mathcal{H}^{p,0}_{(2)}$ consists precisely of holomorphic $L_2$ p-forms.

Example: A harmonic form which is not $L_2$ does not have to be in the kernel of d. For instance, a function is closed if and only if it is (locally) constant, but any nonconstant holomorphic function on a domain in $\mathbb{C}$ has harmonic real and imaginary parts. On the other hand, suppose that $\alpha \in \mathcal{H}^1_{(2),\text{ex}}$ is harmonic and $L_2$, and exact as a form, so that $\alpha = df$ for some smooth function f. Then we claim that f is actually harmonic (but not closed unless $\alpha = 0$). For, $\Delta$ and $d$ commute, so $\Delta f$ is a constant c, and by the Gaffney cutoff trick, it can be shown that c=0.

3. Kähler hyperbolicity

Gromov showed that under certain geometric conditions, the reduced $L_2$ cohomology of a Kähler manifold vanishes outside the middle dimension. To define this condition, one first introduces the notion of a bounded form; this is a form $\alpha$ for which $\|\alpha\|_\infty: = \sup_p |\alpha_p|$ is finite, where $|\alpha_p|$ denotes the (operator) norm of $\alpha$ at the point p.

Definition: A compact Kähler manifold M is Kähler hyperbolic if the pullback $\tilde{\omega}$ of the symplectic form $\omega$ to the universal cover $\tilde{M}$ satisfies $\tilde{\omega} = d\alpha$ for some bounded 1-form $\alpha$.

Suppose M is Kähler hyperbolic, and let $\beta$ be any harmonic $L_2$ form on $\tilde{M}$. Then $\beta$ is closed, and

$\tilde{\omega} \wedge \beta = d(\alpha\wedge\beta)$

Since $\alpha$ is bounded, the form $\alpha \wedge \beta$ is $L_2$. On the other hand, $\tilde{\omega}$ is bounded (because it is pulled back from a form on a compact manifold), so $\tilde{\omega} \wedge \beta$ is $L_2$. Now, (recalling the notation L for the operation of wedging with the Kähler form), the Kähler identity $[L,\Delta]=0$ is a purely local calculation, and therefore on any Kähler manifold (compact or not), wedge product with the Kähler form takes harmonic forms to harmonic forms. It follows that $\tilde{\omega}\wedge \beta$ is harmonic, $L_2$, and equal to the image of an $L_2$ form under d; thus it vanishes identically.

But if V is a real vector space of dimension 2n, and $\omega$ is a nondegenerate 2-form on V, then wedging with $\omega$ is injective on $\Lambda^* V$ below the middle dimension (this is the linear algebra fact which underpins the Hard Lefschetz Theorem for compact Kähler manifolds). Thus the operator L is injective on harmonic $L_2$-forms below the middle dimension. Dualizing, the operator $\Lambda$ is injective above the middle dimension, and we deduce the following:

Theorem (Gromov): If M is compact and Kähler hyperbolic, the reduced $L_2$ cohomology of the universal cover $\tilde{M}$ vanishes outside the middle dimension.

Example: If M is any compact manifold with $K<0$ then for any closed form $\beta$ on M the pullback of $\beta$ to the universal cover is d of a bounded form. This is proved by the Poincaré Lemma, since for a complete simply-connected manifold with $K \le -C < 0$, coning a submanifold along geodesics to a point gives a cone whose volume is bounded by the volume of the submanifold times a constant. So every Kähler manifold with a metric of strict negative curvature is Kähler hyperbolic. More generally, if M is merely nonpositively curved, and the flat planes are isotropic for the Kähler form, then the manifold is still Kähler hyperbolic. This applies (for example) to Kähler manifolds which are compact and locally symmetric of noncompact type. Generalizing in another direction, if M is Kähler with $\pi_2(M)=0$ and $\pi_1(M)$ word-hyperbolic, then M is Kähler hyperbolic.

4. Calibrations

The previous section shows that $\overline{L_2H}^1(\tilde{M})$ vanishes whenever M is Kähler hyperbolic of complex dimension at least 2, where $\tilde{M}$ denotes the universal cover of M. In fact, it turns out that one can completely understand the fundamental groups of Kähler manifolds for which $\overline{L_2H}^1(\tilde{M})$ is nonzero: it turns out that such groups are always virtually equal to the fundamental group of a closed Riemann surface of genus at least 2.

So let’s suppose M is a compact Kähler manifold, that $\tilde{M}$ is its universal cover, and let’s suppose that $\overline{L_2H}^1(\tilde{M})$ is nonzero. Since $\tilde{M}$ is simply-connected, every $L_2$ harmonic form (which is necessarily closed) is actually exact. Let $\alpha$ be a nonzero harmonic $L_2$ form, and let $\varphi$ denote its (1,0)-part, which is an $L_2$ holomorphic 1-form. Since $\varphi$ is also exact, we can write $\varphi = dg$ for some holomorphic function $g$ on $\tilde{M}$. By the coarea formula we compute

$\int_{\mathbb C} \text{vol}(g^{-1}(z))dg(z) = \|dg\|_2^2 < \infty$

or in other words, most of the level sets $g^{-1}(z)$ have finite volume. On the other hand, these level sets are complete holomorphic submanifolds, and holomorphic submanifolds of Kähler manifolds turn out to enjoy a very strong geometric property, which we now explain.

On a Kähler manifold, the symplectic form $\omega$ is a calibrating form. This means that it satisfies the following two properties:

1. it is closed; and
2. it satisfies a pointwise estimate $\omega^k(A) \le \text{vol}(A)$ for all real 2k-planes A, with equality if and only if A is a complex subspace.

It follows that if S is a holomorphic submanifold of complex dimension k, and S’ is a real 2k dimensional submanifold obtained from S by a compactly supported variation so that S and S’ are in the same (relative) homology class, there is an inequality

$\text{vol}(S) = \int_S \omega^k = \int_{S'} \omega^k \le \text{vol}(S')$

In other words, holomorphic submanifolds of Kähler manifolds are absolute volume minimizers in their homology classes (amongst compactly supported variations). From this one deduces the following:

Lemma: Let M be a Kähler manifold with bounded geometry. Then for each k there is a constant C so that if S is a complete holomorphic submanifold of complex dimension k, there is an estimate

$\text{diam}(S) \le C\cdot \text{vol}(S)$

Proof: It suffices to show that for some fixed $\epsilon$ (taken to be the injectivity radius, say), there is a constant $C$ so that the volume of $S\cap B_\epsilon(p )$ is at least $C$ for any point p in S. A Kähler manifold with bounded geometry is uniformly holomorphically bilipschitz to flat $\mathbb{C}^n$ in balls of size smaller than the injectivity radius, so we need only prove this estimate for holomorphic submanifolds of $\mathbb{C}^n$.

But actually, the estimate follows just from the fact that S is a minimal surface. If S is a complete minimal surface of real dimension N in a Euclidean space, passing through the origin (say), then the Monotonicity Formula says that for any $R>r>0$ there is an inequality

$R^{-N}\text{vol}(S\cap B_R(0)) \ge r^{-N}\text{vol}(S\cap B_r(0))$

This can be proved directly by using the vanishing of the mean curvature, but there is a softer proof that $R^{-N}\text{vol}(S\cap B_R(0))\ge C > 0$ where C is the volume of the unit ball in Euclidean N dimensional space, which is enough for our purposes. To see this, observe that C is the limit of $f(R ):=R^{-N}\text{vol}(S\cap B_R(0))$ as R goes to zero. Suppose on some interval $[0,T]$ that $f(R ) < C$ somewhere, WLOG achieving its minimum at $T$. The value of $f( R)$ on $[0,T]$ gives a lower bound for the volume of $S\cap B_T(0)$, by the coarea formula. But the cone on $S \cap \partial B_T(0)$ evidently has less volume than this, in violation of the fact that S is calibrated. The estimate, and the proof follow. qed

It follows from this estimate that some of the fibers of $g:\tilde{M} \to \mathbb{C}$ are compact. The components of these fibers are the leaves of a foliation, and since the foliation is defined locally by a closed 1-form, the set of compact leaves is open; but these leaves are all locally homologous and thus have locally constant volume and therefore uniformly bounded diameter, so the set of compact leaves is closed, and therefore every leaf is compact. The space of leaves is 1 (complex) dimensional, and we thereby obtain a proper holomorphic map with connected fibers $h:\tilde{M} \to S$ to a Riemann surface S. Note that the group of holomorphic automorphisms of $\tilde{M}$ (which includes the deck group $\pi_1(M)$) must permute leaves of the foliation; for, since the leaves are compact, if their image were not contained in a leaf, the map to $S$ would be nonconstant, in contradiction of the fact that a holomorphic map from a compact holomorphic manifold to a noncompact one must be constant.

In summary, the deck group $\pi_1(M)$ acts on $\tilde{M}$ permuting the fibers of the map h, and thus descends to an action on S. Because the fibers have uniformly bounded diameter, and the action of the deck group on $\tilde{M}$ is cocompact and proper, the action on S is also cocompact and proper. Since the map h is surjective with connected fibers, S is simply-connected; since the reduced $L_2$-cohomology class $\varphi$ is pulled back from S, it follows that S is the unit disk, and therefore $\pi_1(M)$ contains a finite index subgroup which acts freely, and is isomorphic to the fundamental group of a closed Riemann surface of genus at least 2.

Now, it turns out that for a compact manifold M, the 1-dimensional $L_2$-cohomology of the universal cover depends only on the fundamental group G of M, and is equal to $\overline{\ell_2 H}^1(G)$, where the (reduced) $\ell_2$ cohomology groups may be defined directly from the bar complex. We have therefore proved the following theorem of Gromov:

Theorem (Gromov): Let G be a Kähler group with $\overline{\ell_2 H}^1(G)\ne 0$. Then G is commensurable with the fundamental group of a closed Riemann surface of genus at least 2.

5. Ends

To apply Gromov’s theorem (and its generalizations) it is important to have some interesting examples of groups with $\overline{\ell_2 H}^1(G)\ne 0$. Let X be a locally compact topological space. Then for every compact set K we have the set $\pi_0(X-K)$ of components of X-K, and an inclusion $K \to L$ induces $\pi_0(X-L) \to \pi_0(X-K)$. The space of ends of X (introduced by Freudenthal) is the inverse limit:

$\mathcal{E}(X):=\lim_{\leftarrow} \pi_0(X-K)$

taken with respect to the directed system of complements of compact subsets. If each $\pi_0(X-K)$ is finite, the space of ends is compact.

Now, let G be a finitely generated group. For each finite generating set we can build a Cayley graph C, which has one vertex for each element of G, and one edge for each pair of elements which differ by (right) multiplication by a generator. The graph C is locally finite and connected, and we define the space of ends of G, denoted $\mathcal{E}(G)$, to be just $\mathcal{E}(C )$. It turns out that this does not depend on the choice of a finite generating set, but is really an invariant of the group.

The theory of ends of groups is completely understood, thanks to the work of Stallings:

Theorem (Stallings, ends of groups): Let G be a finitely generated group. Then $\mathcal{E}(G)$ has cardinality 0,1,2 or $\infty$. Moreover,

1. $|\mathcal{E}(G)|=0$ if and only if G is finite;
2. $|\mathcal{E}(G)|=2$ if and only if G is virtually equal to $\mathbb{Z}$; and
3. $|\mathcal{E}(G)|=\infty$ if and only if G splits as a nontrivial amalgam or HNN extension $G = A*_B C$ or $G=A*_B$ where B is finite, and G is not virtually cyclic.

Actually, the only hard part of this theorem is the third bullet; the rest is elementary, and was known to Freudenthal. The third case is equivalent to the existence of a nontrivial action of G on a tree T (which is not a line) with finite edge stabilizers. It follows that groups with infinitely many ends are non-amenable.

Now, let M be a compact Riemannian manifold, and suppose that the fundamental group G has infinitely many ends. This implies that the universal cover $\tilde{M}$ also has infinitely many ends, and we may find a compact subset $K$ of $\tilde{M}$ whose complement has at least two unbounded regions. Define a function f on $\tilde{M}$ which is equal to 0 on some (but not all) of the unbounded regions of $\tilde{M} - K$ and 1 on the rest. Then $df$ has compact support (contained in K) and is therefore $L_2$. On the other hand, if $g$ is any function with $dg=df$ then $g-f$ is a constant, so $g$ is constant and nonzero on some end of $\tilde{M}$, and is therefore not $L_2$. It follows that $[df]$ is nonzero in unreduced $L_2H^1(\tilde{M})$.

Now, on $L_2$ functions f we have an equality $\langle \Delta f,f\rangle = \|df\|_2^2$. The Laplacian is self-adjoint, with non-negative real spectrum. So to prove that $L_2H^1(\tilde{M})$ is equal to $\overline{L_2H}^1(\tilde{M})$ it suffices to establish a spectral gap for $\Delta$; i.e. to prove an estimate of the form

$\inf_f (\int f \Delta f d\text{vol})/(\int f^2 d\text{vol}) =: \lambda_0 > 0$

for all functions f of compact support (which are dense in $L_2$).  In exactly this context one has the following famous theorem of Brooks:

Theorem (Brooks): with notation as above, one has $\lambda_0=0$ if and only if $\pi_1(M)$ is an amenable group.

One can think of the size of $\lambda_0$ as governing the rate of dissipation of the $L_2$ norm of a function f as it evolves by the heat equation $\partial_t f = -\Delta f$. Geometrically it is plausible that heat dissipates at a definite rate when it is concentrated in a region whose boundary is big compared to its volume (since then a definite amount of heat can escape out the boundary). So heat should dissipate at a definite rate unless there are a sequence of compact regions $U_i$ in $\tilde{M}$, exhausting $\tilde{M}$, for which $\text{vol}(\partial U_i)/\text{vol}(U_i) \to 0$. To each such region $U_i$ one can assign a finite subset $G_i$ of G, by looking at which translates of a basepoint are contained in $U_i$; this sequence of subsets is known as a Følner sequence, and the existence of a Følner sequence for a countable group G is one of the definitions of amenability (the equivalence to the other standard definitions is due to Følner). The hard details of Brooks’ argument are to show that one can take subsets $U_i$ whose boundary is regular enough that the comparison between volumes of subsets and their boundaries in the continuous and the discrete world is uniform.

So in conclusion, if G is a group with infinitely many ends, then reduced and ordinary $L_2$ cohomology agree in dimension 1, and we can construct a nontrivial class $[df]$ as above. Putting this together we deduce the following:

Corollary (Gromov): A Kähler group is either finite, or has 1 end.

Proof: A group with two ends is virtually equal to $\mathbb{Z}$, which is not Kähler because it has $b^1$ odd. A group with infinitely many ends has nontrivial reduced $L_2$-cohomology in dimension one. But for a Kähler group, this implies the group is commensurable with the fundamental group of a closed surface of genus at least 2; such groups have only 1 end after all. qed

6. Ends and extensions

The arguments of Gromov can be generalized considerably. It should be remarked from the outset that at very few points in the proof of Gromov’s theorem did we use the fact that the manifold $\tilde{M}$ was the universal cover of M.

The following is proved by Arapura-Bressler-Ramachandran:

Theorem (Arapura-Bressler-Ramachandran):  Let M be a complete Kähler manifold with bounded geometry, and suppose that $\mathcal{H}^1_{(2),\text{ex}}(M)$ has dimension at least 2. Then there is a hyperbolic Riemann surface S and a proper holomorphic map $M \to S$ with connected fibers. Moreover, the fibers of the map are permuted by the holomorphic automorphisms of M, and the map induces an isomorphism from $\mathcal{H}^1_{(2),\text{ex}}(S)$ to $\mathcal{H}^1_{(2),\text{ex}}(M)$.

Here the subscript “ex” means the harmonic $L_2$ 1-forms which are exact (as ordinary forms). Given an exact harmonic $L_2$ form $\alpha$ we can take the holomorphic (1,0) part $\varphi$ which is $L_2$ and closed. But we cannot assume it is exact if $H^1(M)$ is nontrivial. If we only have one $\alpha$, then we are more or less stuck. But if we have at least two such forms, then the following remarkable Lemma (due originally to Gromov) applies:

Lemma (cup product): Let M be a complete Kähler manifold with bounded geometry, and let $\alpha_1,\alpha_2$ be real, harmonic exact $L_2$ 1-forms. Let $\varphi_i$ be their (1,0)-components. Then $\varphi_1\wedge\varphi_2=0$ pointwise.

Proof: The first remark to make is that on a complete Kähler manifold with bounded geometry, any harmonic $L_2$ form is actually bounded. Equivalently, since harmonic forms are smooth, there is no sequence of points $p_i$ going off to infinity such that the operator norms $|\alpha|_{p_i}$ diverge. Since the manifold has bounded geometry, we can integrate the square of $\alpha$ on disjoint balls of definite radius centered at such points, and the claim will therefore follow if we show the integral of the square of a harmonic form on a ball of definite radius is controlled by below by its value at the center. Assume we are in flat space; then this claim is obviously true for a linear form. But a harmonic form satisfies an elliptic 2nd order equation, which shows that the higher derivatives can be controlled in terms of the first derivative; the claim follows.

Now let $\alpha$ be an exact $L_2$ harmonic form, and write $\alpha = df$. Suppose $\psi$ is a closed $L_2$ form. Then $\alpha \wedge \psi$ is in $L_2$ because $\alpha$ is bounded (as above). If we define $f_t$ to be equal to f where $|f| and locally constant elsewhere, then $df_t$ is equal to $df$ where $|f| and vanishes elsewhere. But now $f_t$ is bounded, so $f_t\psi$ is in $L_2$, whereas $d(f_t\psi) = df_t\wedge \psi \to \alpha \wedge \psi$ in $L_2$. We deduce that $\alpha \wedge \psi$ is zero in reduced cohomology.

Finally, if we let $\varphi_j = df_j + i\beta_j$ be the decomposition of the (1,0) forms into real and imaginary parts, then we compute

$\varphi_1\wedge \varphi_2 = (df_1\wedge df_2 - \beta_1\wedge \beta_2) + i(df_1\wedge \beta_2 + \beta_1\wedge df_2)$

Now, the imaginary part of this is harmonic and $L_2$; on the other hand, we have just shown it is trivial in reduced $L_2$ cohomology. Thus it must vanish identically. But then $\varphi_1\wedge \varphi_2$ must vanish identically too, proving the lemma. qed

It follows that the space $\mathcal{H}^1_{(2),\text{ex}}(M)$ determines (by taking holomorphic parts) an isotropic subspace of $\mathcal{H}^{1,0}_{(2)}(M)$, which by hypothesis has dimension at least 2. Each form in this space determines a complex codimension 1 foliation whose leaves are tangent to the kernel, and because the forms are closed and the space is isotropic, this foliation $\mathcal{F}$ is independent of the choice of form. Furthermore, on any open subset where two such holomorphic 1-forms $\varphi_1,\varphi_2$ do not both vanish, the ratio $\varphi_1/\varphi_2$ defines a holomorphic map to $\mathbb{C}$.

At this point the following fact is extremely handy:

Proposition: Let M be a connected complex manifold (not assumed to be compact!) and $\varphi_1$ and $\varphi_2$ linearly independent closed holomorphic 1-forms with $\varphi_1\wedge \varphi_2=0$. Then $\varphi_1/\varphi_2$ has no indeterminacy; i.e. it defines a holomorphic map from M to $\mathbb{P}^1$.

This Proposition is Lemma 2.2 in a paper of Napier-Ramachandran, where they seem to suggest that the fact is standard, but give an elementary proof. Since the argument is local, one can write $\varphi_1 = df_1$ and $\varphi_2 = df_2$ and then one observes that the functions $f_1,f_2$ are locally constant on the fiber over each point $\zeta \in \mathbb{P}^1$; the argument then follows essentially from a (co)dimension count.

Anyway, once this proposition is proved, it follows that the components of the level sets of this function agree with the leaves of $\mathcal{F}$, which can be taken to be the points of a Riemann surface S. An argument similar to the one above (using a pair of real harmonic functions instead of a single holomorphic function in the coarea formula) shows that some, and therefore every, leaf is compact of bounded volume. Pulling back an $L_2$ form on S gives something $L_2$ by uniform boundedness of the volume of the fibers; conversely, exact harmonic $L_2$ forms on M descend to S because they are constant on the leaves of the foliation. This proves the theorem.

Corollary: Let G be a Kähler group, and suppose there is an exact sequence

$0 \to K \to G \to H \to 0$

where $\overline{\ell_2H}^1(H)\ne 0$ and $b^1(K)<\infty$. Then H is commensurable to the fundamental group of a compact Riemann surface of genus at least 2.

Proof: Let M be a compact Kähler manifold with fundamental group G, and let N be the cover with fundamental group K. Then H acts on N cocompactly, and it follows that $\overline{L_2H}^1(N)\ne 0$. An unbounded sequence of deck transformations must push most of the mass of an $L_2$ harmonic form off to infinity, so necessarily the space  $\overline{L_2H}^1(N)$ is infinite dimensional; since $H^1(N) = H^1(K)$ is finite dimensional, there is an infinite dimensional space of exact forms. Thus N fibers over S as above. Since the map from N to S is surjective on fundamental groups, it follows that S is of finite type (because $H^1(N)$ is finite dimensional). But the deck group H acts on S discretely and cocompactly by holomorphic automorphisms (which are isometries in the hyperbolic metric), so actually S is the disk. qed

Example (Arapura): The pure braid group $P_n$ surjects onto a (virtually) free group, with finitely generated kernel, and therefore it is never Kähler. Note that $b^1(P_n) = n(n-1)/2$ so these groups can’t always be ruled out as Kähler groups on the oddness of $b^1$ alone. On the other hand, pure braid groups are fundamental groups of hyperplane complements: the group $P_n$ is the fundamental group of the space of ordered distinct n-tuples of points in $\mathbb{C}$, which is the complement of a hyperplane arrangement in $\mathbb{C}^n$. So it follows that this quasiprojective variety can’t be compactified in such a way as that the compactifying locus has big codimension (or one could apply the Lefschetz hyperplane theorem).

(Updated November 26: added references)

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