Finiteness and infiniteness results for Torelli groups of (Hyper)Kähler manifolds
Abstract.
The Torelli group of a closed smooth manifold is the subgroup of the mapping class group consisting of elements which act trivially on the integral cohomology of . In this note we give counterexamples to Theorem 3.4 by Verbitsky [12] which states that the Torelli group of simply connected Kähler manifolds of complex dimension is finite. This is done by constructing under some mild conditions homomorphisms and showing that for certain Kähler manifolds this map is nontrivial. We also give a counterexample to Theorem 3.5 (iv) in [12] where Verbitsky claims that the Torelli group of hyperkähler manifolds are finite. These examples are detected by the action of diffeomorphsims on . Finally we confirm the finiteness result for the special case of the hyperkähler manifold .
1. Introduction
Let be a closed connected oriented manifold. The mapping class group is the group of connected components of the orientation preserving self diffeomorphisms of . The subgroup consisting of diffeomorphisms which act trivially on the integral cohomology ring is called the Torelli group of , denote by . In this note we consider two classes of manifolds, the first is characterized by the following assumptions:
Assumption (*) The manifold has trivial first rational homology and the first Pontrjagin class (considered in rational cohomology) is a linear combination of products of dimensional classes, i.e. there exist classes and rational numbers such that . We denote the set consisting of the classes and of the rational numbers by , like definition data.
We construct a homomorphism
which we consider as a sort of Johnson homomorphism since it is landing in an abelian group and constructed via the mapping torus. The construction is as follows. Consider the mapping torus of . Since acts trivially on the cohomology groups the Wang sequence looks like:
Since this implies that the inclusion induces an ismomorphism and so for each class there is a unique class restricting to . Next we look at . By construction it’s restriction to is zero. Thus there exists a unique class denoted by mapping to this class in the Wang sequence.
Proposition 1.1.
Let fulfil the Assumptions (*), then the map
is a homomorphism.
Next we are looking for Kähler manifolds where this invariant is nontrivial. For this we prove a purely topological result.
Theorem 1.2.
Let be a simply connected manifold with and nontrivial cohomology product . Then the Assumptions (*) are fulfilled and the invariant is independent of any choices and is denoted by . The image of is a lattice in , i.e. a finitely generated abelian group of same rank as the dimension of .
Considering hypersurfaces of degree in we obtain complex dimensional Kähler manifolds with infinite Torelli group, since implies that . Taking the product with copies of one obtains examples in all complex dimensions . Namely if then by definition of we have for and the obvious choice of
where we identify with under the projection map.
Thus we obtain the Corollary contradicting Verbitsky’s theorem:
Corollary 1.3.
For all there are Kähler manifolds of complex dimension with infinite Torelli group, actually containing elements of infinite order. In particular taking for example the quintic in we obtain CalabiYau manifolds with infinite Torelli group.
Remark 1.4.
For simply connected manifolds with we compute in [7] the full mapping class group as well as the Torelli group and give generators of these groups. In particular we prove that an element in the Torelli group has finite order if and only if .
Remark 1.5.
In general the invariant depends on the choice of the data. So, in general there are many homomorphisms depending on the choice of the data. The simplest example is , where for and the generator of the difference of the invariants with these data is nontrivial, as shown in [7].
Now we come to the second class of manifolds which is characterized by the following assumptions:
Assumption (**) The manifolds are simply connected and dimensional, whose Betti numbers satisfy one of the following conditions

, , ;

, , ;

, , and ).
Theorem 1.6.
If satisfies the assumptions (**), then the Torelli group is infinite.
We apply this result to find a counterexample to Theorem 3.5 (iv) in [12]. For this we consider the complex dimensional hyperkähler manifold . For the construction and the following information we refer to [10] (where it is called ). The manifold is simply connected and the Betti numbers are: , and . Thus our theorem implies:
Corollary 1.7.
The Torelli group of the hyperkähler manifold is not finite.
This contradicts Verbitsky’s Theorem 3.5, iv ([12]).
There is another complex dimensional hyperkähler manifold denoted by , where is a surface (for the construction see again for example [10]). In this case we show that the Torelli group is finite. This follows from:
Theorem 1.8.
Let be a simplyconnected closed manifold satisfying the following conditions

isomorphic to the second symmetric power of .

,

.
Then is finite.
Corollary 1.9.
The Torelli group of , where is a surface, is finite.
Proof.
Remark 1.10.
This implies that Verbitsky’s Theorem 3.5 is true for and so his main result about the moduli space of hyperkähler manifolds holds for .
We would like to thank Daniel Huybrechts for bringing Verbitsky’s paper [12] to our attention when we told him about our paper [7] where we give a complete computation of the mapping class group of certain manifolds. Since counterexamples to Verbitsky’s theorem might be of separate interest we wrote this note. We have sent this note to several people and received a paper by Richard Hain [4] confirming our result that Theorem 3.4 in [12] is incorrect by showing that the Torelli group of certain complex dimensional Kähler manifolds have an abelian quotient of infinite rank. These examples are different from our examples, they are detected by the induced map , whereas for complex dimensional complete intersections , and so the Torelli groups acts trivial on .
2. Proofs
Proof of Proposition 1.1.
To prove that is a homomorphism we first note that if is a fibre in then we have an exact sequence with rational coefficients
and the term on the left is by suspension isomorphism isomorphic to and under this isomorphism corresponds to the class in which restricts to .
Now we construct from by cutting along and regluing it via from the left to the right. Then we consider the two fibres over and and denote them by and . We consider the restriction map (with rational coefficients)
We have an isomorphism
If we identify with , the homomorphism
corresponds to the homomorphism given by the sum, and the element
is mapped to . This implies that .
∎
Next we consider the manifolds in Proposition 1.2. Let be a generator, by Poincaré duality and thus equals to times a generator of . The condition that the cohomology product is nontrivial is equivalent to saying that is nonzero. It’s shown in [13] and [5] that for every there exist such . Also there is a connected sum decomposition , where has the same homology as .
We introduce a construction of diffeomorphisms, which is a generalization of Dehn twists. Let , , choose a smooth map representing , such that a neighborhood of is mapped to the identity. Define by
We call a Dehn twist in with parameter .
Proof of Theorem 1.2.
The main ingredients of Theorem 1.2 already appeared in [6]. The proof of a slightly different situation will be given in [7]. For convenience of the reader we repeat the proof.
By the assumptions the rational Pontrjagin class is a multiple of , say, , . Therefore our invariant is the pre image of in .
In the decomposition we number the standard embedded spheres
They represent a symplectic basis of .
Let be a generator, be its image induced by the inclusion . Then the dimensional vector bundle over corresponding to has trivial Euler class and equals times a generator of (see [9, Lemma 20.10]).
Now let be the Dehn twist in with parameter . Since the Euler class of is trivial, it’s easy to see that acts trivially on homology, hence . We claim that equals times the Poincaré dual of in . This implies that the image of is a subgroup of of the same rank, hence it is a lattice in .
It is enough to show the claim for . By the adjunction formula this is equivalent to
Since is the identity, we have embeddings . Let , then is a preiamge of in the short exact sequence
There is a decomposition , where denotes the fiber connected sum along . Then and . This implies . Therefore we have
Now notice that is the identity on a tubular neighborhood of ’s for , therefore the normal bundle of in is trivial, and we have for . Let
be the quotient map, from the geometric construction it’s easy to see that the normal bundle of in equals . Therefore
∎
The proofs of Theorem 1.6 and Theorem 1.8 are based on modified surgery theory [8]. We recall some basic definitions here. The normal type of a smooth oriented manifold is a fibration , which is characterized by the assumptions that there is a lift of the normal Gauss map by a equivalence and that the homotopy groups of the homotopy fibre are trivial in degree (for details see [8]).
To show Theorem 1.6 we will construct infinitely many mapping classes in , whose actions on are pairwisely distinct. Let be the normal type of , denote the homotopy fiber of by , and the kernel of the Hurewicz homomorphism by .
Lemma 2.1.
Under the assumptions (**) we have

the image of the Hurewicz map has rank ;

the abelian group has rank .
Proof.
We use Sullivan’s rational homotopy theory [11] to show the assumptions (**) imply the conclusions in this lemma. For this we construct the minimal model from information of the cohomology ring.
Let be free generators of the minimal model in degree with differential . Let be a part of the free generators in degree , which generate . They have trivial differentials. The elements generate a space of dimension , on which the differential is trivial. Since by Poincaré duality the th Betti number is , there must be at least linearly independent indecomposable elements in degree , which by the differential are mapped injectively to a subspace of the vector space generated by the . In degree the dimension of the subspace generated by () is , whereas the th Betti number is , there are additional (to the ) free generators (where ) with trivial differential in degree . There might be more generators in degree depending on the product . But alone from this information the two conclusions in the lemma follow. Namely is isomorphic to the degree subspace modulo the decomposable elements and so contains the subspace with basis , . The dual of the rational Hurewicz homomorphism maps injectively. Dualizing we see that the dual of the are mapped injectively showing (1) and the dual of the are mapped to in rational homology, so the rank of is at least .
Consider the following commutative diagram
where is an isomorphism if and is injective with a codimension image if , and is injective (by the LeraySerre spectral sequence of the fibration ). A little diagram chasing show that
This show that under the assumption . ∎
Let be the group of cocycles with coefficients in , be the group of fiber homotopy classes of fiber homotopy equivalences of . We will first define a map
as follows.
Let be the skeleton of , be the cells, then the dimensional cellular chain group is a free abelian group generated by (). Let be the pinch map, where we pinch each cell to . Let be a cocycle. For each , by abuse of notation, we view it as a map . Let be the inclusion, define to be the composition
By the construction is compatible with the fiber projection , i. e. the following diagram commutes up to homotopy
Now we show that we can extend to a map on the skeleton of compatible with . Let be a cell, with attaching map . By BlakersMassey theorem we have an isomorphism
under this isomorphism the image of is clearly . Since is a cocyle, , therefore extends to a map , compatible with the fiber projection . (The extension may not be unique, we choose one extension.) Since for , we can extend further to a fiber map .
Lemma 2.2.
is a fiber homotopy equivalence. is the identity on for . is the identity on for . is the identity on .
Proof.
It’s clear from the construction that is the identity on for and is the identity on for . On the chain level is
where is a boundary by definition. Therefore is the identity on . By the Hurewicz theorem we see that is an isomorphism for and is a surjection for . Notice that is a finitely generated abelian group, a surjective endomorphism of a finitely generated abelian group must also be injective, therefore is an isomorphism on . Since is coconnected, induces isomorphisms on for all . Therefore is a fiber homotopy equivalence.
Since for , by rational homotopy theory, there are no indecomposable cohomology classes in for . Therefore elements in are linear combinations of products of cohomology classes of degree . Since is the identity on for , is the identity on for and hence also the identity on . Thus is the identity on . By the LeraySerre spectral sequence of the fibration , is the identity on . ∎
Proof of Theorem 1.6.
By Lemma 2.1 the image of has rank . Let be homology classes in the image which generate a free abelian subgroup of rank in , be a preimage of , represented by a map . Then we may take of the form
where is a complex, and for .
Let be a homomorphism, since there are surjective homomorphisms
we may pick a preimage of , say . Let be the image of under . Consider the action of on , especially the image of under : let be the inclusion of , clearly the homotopy class of the composition
equals .
By Lemma 2.1 there are infinitely many such that are pairwisely distinct. Therefore we have constructed infinitely many with properties in Lemma 2.2.
Let be the dimensional bordism group. It’s isomorphic to the dimensional stable homotopy group of the corresponding Thom spectrum. By the AtiyahHirzebruch spectral sequence there is an isomorphism
where the image of a bordism class is the image of the fundamental class .
Now we fix a normal smoothing . Consider the normal smoothing , where is constructed as above. Since is the identity on , we have
Therefore there are infinitely many (indexed by ) such that
Let be a bordism between and , by the theory of modified surgery, the obstruction to changing by surgery to an cobordism is in . Therefore there is a diffeomorphism such that . Since is a equivalence, we see that is the identity on for , and is the identity on for . Therefore by Poincaré duality, is the identity on and hence . But on are pairwisely distinct. Therefore we have constructed infinitely many elements in . ∎
Proof of Theorem 1.8.
Let be the normal type of , be the homotopy fiber of . Fix a normal smoothing .
We claim that and are finite. For this we consider the minimal model of . It has generators in degree , where is the second Betti number. By assumption for is a basis of . This and the fact that implies that there are no generators in degree and hence is finite. This implies that there are no decomposable elements in degree . Thus the differential on elements of degree is zero. But then there are no indecomposable elements in degree 4, since they would produce indecomposable cohomology classes. Thus is finite. Finally we conclude that also and are finite. This follows from the homotopy sequence of the fibration since for and implying that is nontrivial.
Let be a selfdiffeomorphism, then is also a normal smoothing. Let be the subset of consisting of selfdiffeomorphisms such that and are homotopic as liftings of .
Lemma 2.3.
Under our conditions is a finite index subgroup of .
Proof.
It suffices to show that there are finitely many homotopy classes of over , which in turn follows if there are finitely many lifts of the Gauss map over . This follows by induction over the skeleta and the Puppe sequence and the fact that induces the identity on from the fact that the homotopy groups and are finite and for . ∎
We finish the proof of the theorem by showing that is finite. Given , choosing a homotopy between and we obtain a normal structure , where is the mapping torus. This represents an element in . Again by the AtiyahHirzebruch spectral sequence we have .
Lemma 2.4.
.
Proof.
By the LeraySerre spectral sequence of the fibration , it suffices to show that . But since there are no indecomposable classes in for , it is enough to show that . Again by the LeraySerre spectral sequence there is an exact sequence
where can be identified with the composition
If , then is surjective. This implies . ∎
Therefore the finitely generated abelian group is finite. There are finitely many , together with homotopies such that for any and homotopy , for some . This implies that there is a bordism between and , where the two boundary components are identified by and respectively. The modified surgery obstruction to changing to a relative cobordism is . If , there is a relative cobordism between and . Such an cobrodism gives rise to a diffeomorphism which is a pseudoisotopy between and . Therefore by Cerf’s pseudoisotopy theorem [3] is isotopic to . If , then for any with this property, we may glue and and get a bordism between and where the two boundary components are identified by and respectively. Now the surgery obstruction , therefore and are isotopic. This shows that there are at most elements in . ∎
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