Mapping class groups of exotic tori and actions by
Abstract.
We determine for which exotic tori of dimension the homomorphism from the group of isotopy classes of orientation-preserving diffeomorphisms of to given by the action on the first homology group is split surjective. As part of the proof we compute the mapping class group of all exotic tori that are obtained from the standard torus by a connected sum with an exotic sphere. Moreover, we show that any nontrivial -action on agrees on homology with the standard action, up to an automorphism of . When combined, these results in particular show that many exotic tori do not admit any nontrivial differentiable action by .
A homotopy -torus is a -dimensional smooth manifold that is homotopy equivalent to the standard torus and hence also homeomorphic to it, by a known instance of the Borel conjecture; see [HW69] for , [FQ90, 11.5] for , [Wal68, 6.5] and the Poincaré conjecture for . If is not diffeomorphic to the standard torus , it is called exotic. For instance, given an exotic sphere of dimension , the connected sum is an exotic -torus.
One of the prominent features of the standard torus is that it admits a faithful action by through orientation-preserving diffeomorphisms, induced by the linear action of on . For a general homotopy -torus one might thus wonder:
-
(A)
Is there a faithful action ? If not, is there even any nontrivial action?
As the -action on the standard torus splits the homomorphism induced by the action on the first homology group , it seems natural to approach Question (A) by first considering the following weaker question which is an instance of a high-dimensional version of a Nielsen realisation problem posed by Thurston [Kir97, Prob. 2.6]:
-
(S)
Is the homomorphism given by the action on split surjective?
This homomorphism factors through the mapping class group of isotopy classes of orientation-preserving diffeomorphisms, so one can weaken the question further to:
-
(S0)
Is the homomorphism given by the action on split surjective?
This work establishes several results regarding these three questions. Note that a positive answer to (S) implies positive answers to (A) and (S0). As part of our results, we
-
answer Question (S0) in all dimensions ,
-
conclude that for many exotic tori the answer to all three questions is negative.
In what follows, we describe these results and various extensions of them in more detail.
Splitting the homology action up to isotopy
Our first main result answers (S0) for :
Theorem A.
For a homotopy torus of dimension , the morphism
induced by the action on is split surjective if and only if is diffeomorphic to for a homotopy sphere such that is divisible by in the abelian group .
Here is Kervaire–Milnor’s finite abelian group of homotopy -spheres [KM63] and for is the value of under the Milnor–Munkres–Novikov pairing where is the generator of the first stable homotopy group of spheres (see [Bre67] for more on this pairing). The question whether for a given is divisible by can in most instances be reduced to a problem in stable homotopy theory which can in turn be solved in many cases. This approach is discussed in Section 1.3, but to already illustrate its practicability at this point, we display in Table 1 below the first groups of homotopy spheres together with the subgroups of split spheres, i.e. those for which is divisible by , which is by A equivalent to being split. Note that among the dimensions for which is nontrivial, there are dimensions in which all spheres are split such as , dimensions in which none are split such as , as well as dimensions in which some but not all are split such as . In Section 1.3 we also explain why both cases—the sphere being split or not—occur for exotic spheres in infinitely many dimensions.
| and | ||||||||
|---|---|---|---|---|---|---|---|---|
Actions of on homotopy tori
Our second main result shows that all nontrivial -actions on homotopy tori agree on homology with the standard action up to an automorphism.
Theorem B.
Fix , a homotopy -torus , and an automorphism group
Any homomorphism is either trivial or has the property that its postcomposition
with the action on is an automorphism. Moreover, if also , then the same holds when replacing by the group of isotopy classes.
In particular, given any nontrivial homomorphism , we obtain a splitting of the action on first homology, given by . Applying this to shows that the above questions (S) and (A) are in fact equivalent. Applying it to also shows that (S0) is equivalent to the following isotopy-analogue of (A).
-
(A0)
Is there a faithful action ? If not, is there even any nontrivial action?
Combining these implications with A results in the following corollary which answers all questions (A), (S), (S0), (A0) in the negative for a large class of homotopy tori and partially answers Question 1.4 and Problem 1.5 in work of Bustamante and Tshishiku [BT21].
Corollary C.
Let be a homotopy torus of dimension . If
-
(i)
is not diffeomorphic to a connected sum with , or
-
(ii)
is diffeomorphic to for some such that is not divisible by ,
then every homomorphism from to or to is trivial.
Remark (The Zimmer programme).
One motivation for considering Question (A) stems from the Zimmer programme, part of which studies actions of on manifolds. For instance, it follows from a version of Zimmer’s conjecture, now a theorem due to Brown–Fisher–Hurtado [BFH20], that does not act faithfully on smooth manifolds of dimension . For actions of on -manifolds, there is a conjectural classification by Fisher–Melnick [FM22, Conjecture 3.6] which would imply that if acts faithfully on a homotopy -torus , then is the standard torus. C implies this for a large class of homotopy tori.
Remark (Regularity).
Our results are phrased in terms of the group of -diffeomorphisms, but they also hold for the groups of -diffeomorphisms for finite . For Theorems A and D, this follows from the isomorphism . For B it follows from the observation that the statement for the group also implies the statement for all its subgroups. The deduction of C from Theorems A and B works the same way. In particular, this shows that homotopy tori as in C do not admit any -action by .
Mapping class groups of exotic tori
As an ingredient for the proof of A, we determine the mapping class groups of exotic tori of the form for in all dimensions in terms of the known mapping class group of the standard torus. Note that is trivial when and , so in these cases there is nothing to show. To state the result, we first recall the previously known description of . As mentioned above, the action of on induces a splitting of the action map , so there is a semidirect product decomposition
For , the kernel is abelian and isomorphic to the sum of -modules
| (1) |
where acts through the standard action on , and denotes the coinvariants with respect to the involution induced by multiplication by on (see [Hat78, Theorem 4.1, Remark (3) on p. 9]111[Hat78, Theorem 4.1] asserts that the computation of also holds for . However, this relies on a claim attributed to Igusa (see the middle of p. 7 loc.cit.) for which—to our knowledge—no proof has been provided so far. and [HS76, Theorem 2.5]). In addition to this description of , our identification of involves the aforementioned homotopy sphere and the unique nontrivial central extension
of by ; see Section 2.1. Our result identifies the group as a semidirect product of or acting on a quotient of by a nontrivial subgroup depending on which is contained in the summand of (1) corresponding to the terms .
Theorem D.
For a homotopy sphere of dimension , there is an isomorphism
which is compatible with the homomorphisms to .
In particular, this result shows that the mapping class group for is given by a quotient of by a finite abelian subgroup which is always of order at least and has order precisely if and only if is the standard sphere, so from this mapping class point of view the standard torus admits “the most symmetries”, as one would expect.
Endomorphisms of
As an ingredient for the proof of B, we prove the following classification results for endomorphisms of for :
Theorem E.
Fix . Every nontrivial endomorphism of is an automorphism. Moreover, all automorphisms of agree, up to postcomposition with a conjugation by an element in , with either the identity or the inverse-transpose automorphism.
Remark.
Some comments on E.
-
(i)
The proof is “elementary” in that it does neither rely on Margulis’ superrigidity or normal subgroup theorem, nor on the congruence subgroup property. Using these results, there are likely other proofs. The argument we give was hinted at by Ian Agol in a comment to a question on MathOverflow [Mat17] and sketched by Uri Bader in the case as a response to the question (however this sketch has a small gap; see Remarks 4.3 and 4.8).
-
(ii)
For , the statement of E fails: consider the composition
where the first arrow is abelianisation and the second sends a generator to .
- (iii)
Acknowledgements
We would like to thank Wilberd van der Kallen for helpful comments. AK acknowledges the support of the Natural Sciences and Engineering Research Council of Canada (NSERC) [funding reference number 512156 and 512250]. AK was supported by an Alfred J. Sloan Research Fellowship. BT is supported by NSF grant DMS-2104346. MB is supported by ANID Fondecyt Iniciación en Investigación grant 11220330.
1. Collar twists
As preparation to the proof of Theorems A and D, we collect various results on a certain map defined by twisting a collar of the complement of an embedded -disc in a closed smooth -manifold . After explaining the construction, we discuss how this map behaves under taking products and connected sums, followed by some results on the collar twisting map for specific choices of , first homotopy spheres and then homotopy tori.
1.1. The collar twist
Given a closed connected oriented -dimensional manifold , we write
for the complement of a fixed embedded disc that is compatible with the orientation (which is unique up to isotopy), and we write for the group of diffeomorphisms of that fix a neighbourhood of the boundary sphere pointwise, equipped with the smooth topology. The latter is homotopy equivalent to the larger group of diffeomorphisms of that fix the centre of the disc as well as the tangent space at this point. The group is the fibre of the fibration assigning to a diffeomorphism that fixes its (orientation-preserving) derivative at that point, so after delooping and using the equivalence , there is a homotopy fibration sequence
| (2) |
where is induced by extending a diffeomorphism of to by the identity. The connecting map has the following geometric description: there is a homomorphism which sends a smooth loop that is constant near the endpoints to the self-diffeomorphism of given by mapping to , and a homomorphism induced by a choice of collar of the boundary sphere in . Delooping their composition gives a map
that agrees with the aforementioned connecting map; see e.g. [Kra21, p. 9]. Following Section 3 of loc.cit., we call the collar twisting map of . This map is relevant to the study of the mapping class groups of and , since the sequence (2) induces an exact sequence of groups
| (3) |
so the second morphism in this sequence is an isomorphism if and only if the image
of the standard generator of the leftmost group under the first map is trivial. We call this element the collar twist of . Note that the collar twist lies in the centre of , because the image of the connecting map in the long exact sequence of homotopy groups for any fibration has this property. Alternatively, one could use that the collar twist is supported in a collar and that every diffeomorphism fixing boundary can be isotoped to also fix any chosen collar, thereby having disjoint support from the collar twist.
1.2. Collar twists of products and connected sums
The following proposition shows that collar twisting maps behave well with respect to products and connected sums. Here and in what follows, we identify with the boundary connected sum via the preferred isotopy class of diffeomorphisms between these two manifolds.
Proposition 1.1.
Let and be closed oriented connected manifolds of dimension and .
-
(i)
The compositions
and
are homotopic. In particular,
-
(ii)
If and are of the same dimension , then the map
and the composition
are homotopic after restriction to the subspace . In particular, we have
In order to prove 1.1, it is convenient to view the collar twisting map as the instance of a more general construction for a compact smooth -dimensional manifold equipped with an embedding . First, one extends the latter inclusion to an embedding where is the disc of radius and is obtained by attaching an external collar to . This extension is unique up to isotopy. Given a smooth function and a smooth loop that is constant near the endpoints, consider the diffeomorphism by sending to . In other words, thinking of as foliated by the leaves for and , the diffeomorphism preserves the leaves and acts on the leaf by rotation with the element at time of the loop . If one additionally assumes that
-
(i)
on a neighbourhood of where is the unit disc,
-
(ii)
on a neighbourhood of ,
then agrees with the identity on a neighbourhood of so restricts to a diffeomorphism of the complement. This diffeomorphism of the complement extends via the identity to a diffeomorphism of fixing a neighbourhood of the boundary pointwise, so we obtain a map
which depends continuously on and is a homomorphism with respect to pointwise multiplication on the domain and composition on the target. Since the space of smooth functions satisfying (i) and (ii) is contractible by linear interpolation, the delooping of
is independent of up to homotopy, so only depends on the isotopy class of the embedding . This map generalises the collar twisting map in the following sense.
Lemma 1.2.
For , the maps
are homotopic. Here the embedding is chosen to be compatible with the orientation.
Proof.
It suffices to show that the two maps before delooping are homotopic as maps of topological groups. Going through the construction, one sees that both maps are instances of the following construction applied to smooth loops that are constant near the ends: pick a smooth map which is in a neighbourhood of , and in a neighbourhood of , consider the self-diffeomorphism of sending to , restrict it to a diffeomorphism of , and extend the result to a diffeomorphism of by the identity. As the space of choices for is contractible by linear interpolation, all maps constructed this way are homotopic. ∎
A similar argument also shows the following naturality property of the map .
Lemma 1.3.
Given a compact submanifold of codimension , the map
and the composition
are homotopic. Here the embedding is the restriction of the embedding .
Proof of 1.1.
For part (i), note that the composition is an instance of using the embedding , so its postcomposition with is homotopic to by 1.3, which in turn implies the claim as a result of 1.2. For part (ii), view as being obtained from by gluing on a pair-of-pants bordism . To show the claim, it suffices to show that the maps are homotopic, where simultaneously twists collars of the two incoming boundary spheres and twists a collar of the outgoing boundary sphere. Viewing as for an embedding , the map is given by and the map as the composition of with , so the claim follows from 1.3 applied to , , and . ∎
1.3. Collar twists of exotic spheres
We now turn to the collar twisting map for homotopy spheres , but we actually restrict our attention to the collar twist it induces on fundamental groups. We begin with a recollection of the classification of homotopy spheres.
1.3.1. Classification of homotopy spheres
Recall (e.g. from [Lev85, p. 90-91]) that Kervaire–Milnor’s finite abelian group of homotopy -spheres [KM63] fits for into an exact sequence
| (4) |
where is a certain cyclic subgroup and is the cokernel of the stable -homomorphism from the homotopy groups of the stable orthogonal group (which are known by Bott periodicity) to the stable homotopy groups of spheres. The order of the cyclic subgroup is known in all cases except (combine [Lev85, Corollaries 2.2, 3.20, Theorem 4.9] with [HHR16, Theorem 1.3]):
The map in the sequence (4) is known to be trivial as long as for . It is known to be nontrivial for , but the case (i.e. ) is still open (see [HHR16, Theorem 1.4]). The question whether or and the question whether is surjective or not (these questions turn out to be equivalent; see [Lev70, p. 88]) is the last remaining case of the Kervaire invariant one problem. The upshot of this discussion is that apart from the two problematic dimensions , the group is described in terms of the group up to extension problems. In most cases, also these extension problems have been resolved:
-
For even, vanishes and the map to is an isomorphism as long as for , so in these cases there are no extension problems.
-
For with , we have an exact sequence which admits a splitting since in these dimensions is known to be annihilated by (see e.g. the table [IWX20, Table 1]), so . For the question whether the map is split surjective (rather than just surjective which is open too; see above) is known as the strong Kervaire invariant one problem.
1.3.2. Collar twists of homotopy spheres and the Milnor–Munkres–Novikov pairing
We begin the discussion of collar twists of homotopy spheres with a general observation: if is a closed oriented manifold of dimension and is a homotopy sphere, then writing for the inverse sphere obtained by reversing the orientation, the maps
are inverse homotopy equivalences, so in particular induce an isomorphism on fundamental groups. For , combining the latter with the usual isomorphism given by gluing together two copies along their boundary via diffeomorphisms of supported on a hemisphere results in a chain of isomorphisms
We write
for the image of the collar twist under these isomorphisms. This defines a set-theoretical function which can be rephrased (see 1.5 below) in terms of a well-known construction in the study of homotopy spheres, namely the bilinear Milnor–Munkres–Novikov pairing (see e.g. [Bre67]) for . The latter is related to the multiplication in the stable homotopy groups of spheres by a commutative diagram
| (5) |
with bottom horizontal map induced by the multiplication on the stable stems, using that products of elements in and contained in if (see p. 442 of loc.cit.).
Proposition 1.5 (Kreck, Levine).
We have where is the generator.
Proof.
Levine writes for [Lev70, p. 245-246] and Kreck writes for it [Kre79, p. 646]. For even , the claim is [Kre79, Lemma 3 c)]. For odd , the subgroup is trivial, so it suffices to show the claimed equality after passing to (see Section 1.3.1). The latter follows from [Lev70, Corollary 4] using that Levine’s subgroup is generated by by definition; see p. 246 loc.cit.. ∎
Remark 1.6.
1.5 has immediate consequences for collar twists of homotopy spheres. For example, since is -torsion and the Milnor–Munkres–Novikov pairing is bilinear, the sphere is trivial if has odd order, so the collar twist of is in these cases trivial too.
The combination of 1.5, the classification of homotopy spheres as recalled in Section 1.3.1, and the diagram (5) allows one to reduce most questions on collar twists of exotic spheres to questions in stable homotopy theory. As an example of this principle, we rephrase the condition featuring in the statements of Theorem A and D (whether is divisible by or not) in most cases in terms of the cokernel of the stable -homomorphism:
Lemma 1.7.
If is divisible by , then so is . The converse holds
-
(i)
for ,
-
(ii)
for for , and
-
(iii)
for for with .
Proof.
By commutativity of (5), the class is the image of under the morphism , so if the latter is divisible by , then so is the former. To prove the partial converse, we distinguish some cases and make frequent use of the classification of homotopy spheres as recalled in Section 1.3.1, without further reference.
-
For , the map is split surjective, so . Since has order two and is cyclic of order divisible by , the -component of the order element has to be divisible by , so the full element is divisible by if and only if its image is divisible by .
-
For , the map is also split surjective, so . In this case the -component of turns out to vanish, which implies the result. The reason for this vanishing is that is contained in the subgroup of homotopy spheres that bound a spin manifold [Law73, §4 + Diagram (6)] and on this subgroup the -component with respect to the can be computed as the image of the -invariant from [Bru69, §3] which vanishes for by [Law73, Proposition 4.1] (this uses that the pairings denoted and in loc.cit. are compatible, by diagram (B) on p. 835 of loc.cit.).
-
For and for , and for with for we have and there is nothing to show.
-
For with for we have , so an element in is divisible by if and only if this holds for its image in . ∎
Remark 1.8.
To extend 1.7 to for , it would suffice to show that the -component of for under the splitting recalled in Section 1.3.1 is trivial. We do not know whether this is the case.
In view of 1.7, the question whether is divisible by can in many dimensions be analysed with inputs from stable homotopy theory. The following two remarks contain some applications in this direction:
Remark 1.9.
As has order two, whether is divisible by or not can be tested -locally. At the prime , the groups and multiplication by on them have been computed up to dimensions about . The result is summarised in [IWX20, Figure 1] where every dot represents a nontrivial element, the diagonal and vertical lines indicate that two elements are related by multiplication with or , respectively, and the image of consists of the blue dots, apart from the blue dots in degrees . Combining this with 1.7 and the classification of homotopy spheres recalled in Section 1.3.1, one can in most dimensions up to about determine the groups and the subgroups of those such that is divisible by . The result of this analysis for is recorded in Table 1 of the introduction.
Remark 1.10.
There are also many infinite families of homotopy spheres for which one can decide whether is divisible by or not. We again rely on Section 1.3.1.
-
As an infinite family of nontrivial in odd dimensions such that is divisible by , one may for instance take any for that lies in the nontrivial subgroup . This is because is trivial as a result of (5), so it is in particular divisible by . There are also examples in even dimensions: as , the class in of Adams’ element (which is nontrivial in as ) lifts uniquely to a homotopy sphere . As since is known to be contained in , it follows from 1.7 that is divisible by .
-
As an infinite family of nontrivial in odd dimensions such that is not divisible by , one may take any that maps to the class in represented by Adams’ element which is known to have the property that is not divisible by . In even dimensions, one may use the families of nontrivial homotopy spheres for from [Kra21, Proposition 2.11 (i)] which have the property that is nontrivial and detected in the spectrum of topological modular forms (see the proof of the cited proposition). Moreover, in these dimensions is known to be annihilated by (see e.g. [Beh20, Figure 1.2]), so is not divisible by in and hence neither in .
1.4. Collar twists of tori
The next class of manifolds for which we establish some results on their collar twisting maps are homotopy tori. For this class of manifolds, it is convenient to study the fibre sequence (2) involving the collar twisting maps by comparing it to an analogous sequence for block-homeomorphisms (see e.g. [HLLRW21, Section 2] for a discussion of block-automorphisms suitable for our needs) via a map of fibre sequences
| (6) |
The bottom row of this diagram deserves an explanation. To construct it, first consider the forgetful map from the space of orientation-preserving block-homeomorphisms of to the space of orientation-preserving homotopy self-equivalences. The space is defined as the homotopy pullback of this map along the inclusion map of those orientation-preserving self-equivalences of that preserve the chosen point . The delooping of the latter map is the universal -fibration, so is by construction equivalent to the total space of the universal oriented -block-bundle. The right-hand map in the bottom sequence is the delooping of the map that takes the stable topological derivative of a block-homeomorphism of at , or equivalently, it classifies the stable vertical topological tangent bundle of the universal oriented -block-bundle (see Section 2 loc.cit.), similarly to how the right-hand map of the upper sequence classifies the vertical tangent bundle of the universal oriented smooth -bundle. The rightmost vertical map classifies the underlying stable Euclidean bundle of an oriented -dimensional vector bundle and the middle vertical map is induced by the forgetful map . The left-hand map of the bottom sequence is defined as the homotopy fibre inclusion of the right-hand map, or equivalently, as the delooping of the derivative map.
Note that the bottom row only depends on the underlying topological manifold of , so in particular agrees for homotopy tori with the corresponding sequence of the standard torus . For the latter, the middle space has a very simple description:
Lemma 1.11.
For any , the map
induced by the action on is an equivalence.
Proof.
As , the analogous map from the space of orientation homotopy self-equivalences of is an equivalence, so it suffices to show that the forgetful map is an equivalence. We will do so by proving that the right-hand map in the map of homotopy fibre sequences
comparing the universal -block-bundle with the universal -fibration, is an equivalence. Using the action of on and the action of on itself, a diagram chase in the ladder of long exact sequences induced by this map of fibre sequences shows that the middle arrow is surjective on all homotopy groups. Injectivity on homotopy groups is equivalent to the claim that for , any self-homeomorphism of fixing on the boundary that is homotopic to the identity relative to the boundary is also concordant to the identity relative to the boundary. For , this from the fact that the topological structure sets in the sense of surgery theory are trivial as long as [KS77, p. 205, Theorem C.2], but there is also a more direct proof in all dimensions [Law76]. ∎
Corollary 1.12.
Let be a homotopy torus of dimension . The collar twisting map
is injective on for . In particular, is nontrivial for .
Proof.
From the map of long exact sequences induced by the map (6) together with the fact that the higher homotopy groups of vanish as a result of 1.11, we see that the map in question is injective on if the map is injective. This maps factors as the stabilisation map followed by the forgetful map . The latter is injective on all homotopy groups (combine [Bru68] with [KS77, p. 246, 5.0.(1)]) and the former for by stability, so the claim follows. ∎
Remark 1.13.
The collar twist is also nontrivial for which follows for instance from 2.2 below. Moreover, for the map induces an isomorphism on all higher homotopy groups since the component of the identity is contractible. This is well-known for and follows for from a combination of [Hat83] and [Hat76] using that is Haken.
Remark 1.14.
Replacing with , the statement of 1.11 (and thus also that of 1.12) holds for many other closed aspherical manifolds , in particular for those of dimension whose fundamental group satisfies the Farrell–Jones conjecture and also for those of dimension if the fundamental group is good in the sense of [FQ90, p. 99] (see e.g. [HLLRW21, Proposition 5.1.1] for an explanation of this).
2. Mapping class groups of exotic tori and the proof of D
Equipped with the results on collar twists from the previous section, we turn towards studying the mapping class groups of homotopy tori of the form .
2.1. Central extensions of special linear groups
The strategy will be to relate the mapping class groups of homotopy tori to well-known central extensions of special linear groups. We first recall these extensions and discuss some of their properties. The universal cover of the stable special linear group over the reals gives a central extension
which we may pull back along the lattice inclusion to a central extension
| (7) |
for . For , also a different central extension will play a role, namely the pullback
| (8) |
along the inclusion of the universal cover central extension
Since the inclusion map is surjective on fundamental groups for , the extension (7) agrees with the pushout of (8) along the quotient map . Everything we need to know about these extensions, together with some useful information on the low-degree (co)homology of is summarised in the following lemma.
Lemma 2.1.
Proof.
That the abelianisation is cyclic of order for can be read off from the standard presentation of (see e.g. [Mil71, Corollary 10.5]), and the fact that vanishes for follows for instance from [Mil71, Corollary 10.3] together with the observation that for every elementary square matrix can be written as a commutator of two elementary matrices. The computation of for follows from the isomorphism [Ser03, 1.5.3] and the Mayer–Vietoris sequence for group homology of amalgamated products [Bro94, Corollary II.7.7] (c.f. Exercise 3 on p. 52 of loc.cit.), for from [vdK75], and for from [Mil71, Corollary 5.8, Remark on p. 48, Theorem 10.1]. Using these calculations, the computations and for implicitly claimed in (iii) and (iv) follow from the universal coefficient theorem. The latter also implies the part of (ii) for once we show (iv). The claim on the stabilising map for can be proved via the arguments in [vdK75].
To prove (iii), we use the general fact that for a given central extension , the Serre spectral sequence induces an exact sequence . The identity map induces a preferred class in and its image in is the class that classifies the given extension. Applying this to the extension (8), we see that in order to show that this extension generates it suffices to show that vanishes. Now agrees up to isomorphism with the braid group on three strands (see e.g. [Mil71, p. 83]), so the claim follows from the universal coefficient theorem and the facts that and [Arn14, p. 32].
For (iv), we use the universal coefficient theorem to see that is surjected upon by the map induced by reduction modulo , so it follows from (iii) that the extension (7) is nontrivial for and hence also for all higher values of since the former is the pullback of the latter along the inclusion . As has only a single nontrivial element for , this gives (iv).∎
2.2. Mapping class groups of homotopy tori and D
We now determine the mapping class groups of homotopy tori of the form . The argument has three steps.
-
Step \raisebox{-.9pt} {\arabic{subsubsection}}⃝
Determine in terms of .
-
Step \raisebox{-.9pt} {\arabic{subsubsection}}⃝
Determine .
-
Step \raisebox{-.9pt} {\arabic{subsubsection}}⃝
Determine .
Throughout this section, we abbreviate , fix a basis of , and use the bases for the first homology groups of , , and that are induced by the chosen basis of .
Step \raisebox{-.9pt} {\arabic{subsubsection}}⃝. Fixing a disc or a point
We first determine the group in terms of the group . This step works for general homotopy tori , not just those of the form .
Lemma 2.2.
For a homotopy -torus , there are pullback squares
| for and for |
Proof.
If , then is the standard -torus for which the claimed square is well-known (for a reference, compare the standard presentations of and e.g. in [Mil71, p.82–83] and [Kor02, Section 5]). For , we consider the map of central extensions
| (9) |
induced by (6) for , using that the bottom sequence only depends on the underlying topological manifold. Exactness at follows from the fact that by 1.11 and exactness at follows from exactness at . 1.11 also shows that the homology action map is an isomorphism, so we are left to show that the bottom extension is isomorphic to . It suffices to show this for large enough , since the bottom extension in dimension maps by taking products with to the corresponding extension in dimension (which have both kernel ), so the extension for is the pullback of the extension for along the inclusion . We may thus assume in which case there is a single nontrivial central extension of by (see 2.1), so we only need to exclude that the bottom extension in (9) is trivial. To show this, we consider (9) for and extend it to the top as
where the top middle vertical map is induced by taking products with followed by restriction, commutativity of the left upper square follows by an application of 1.1 (i) to and , and the right upper square is induced by the commutativity of the left upper square. We will show that the bottom extension is nontrivial by showing that its pullback along the composition is nontrivial. This composition is isomorphic to the inclusion and the composition to the quotient map , so it follows that the pullback in question is isomorphic to the mod reduction of the extension , i.e. the extension . The latter is nontrivial by 2.1 (iv).∎
Step \raisebox{-.9pt} {\arabic{subsubsection}}⃝. The pointed mapping class group of
Next, we determine the group . For the evaluation fibration whose fibre is has a splitting given by the standard action of on itself, so the long exact sequence in homotopy groups induces the first out of two isomorphisms
the second isomorphism was explained in the introduction. Combining this with 2.2 for , we obtain an isomorphism
| (10) |
Now recall that the collar twist generates the kernel of the map to so it corresponds under the isomorphism (10) to the element where is the central element that generates the kernel of the map to . The composition
| (11) |
which we abbreviate by , can be identified in terms of (10) as follows:
Lemma 2.3.
Proof.
By an application of 2.2, this would follow from showing the analogous statement for instead of for once we know that the postcomposition of (11) with the isomorphism (10) has image in . That the statement holds in follows from [Hat83, p. 9, Remark (5)], so it suffices to show that (11) lands in the subgroup . To show that, note that diffeomorphisms in the image of (11) are topologically isotopic to the identity since is contractible by the Alexander trick. In particular, (11) lands in the kernel of the forgetful map which agrees via the isomorphism precisely with the subgroup (see the proof of 2.2), so the claim follows. ∎
Given a homotopy sphere and , we have the isomorphism discussed in Section 1.3
| (12) |
so from the exact sequence (3), we see that is isomorphic to the quotient of by the central subgroup generated by the preimage of the collar twist under (12).
Lemma 2.4.
Proof.
We already explained how the second part follows from the first. To prove the first, we use the relation in ensured by 1.1 (ii), using which we express the element in question in as
Here we used the equality from 1.5 and the definition of from Section 1.3. By the discussion above, and corresponds under the isomorphism (10) to the elements and in , so the element we are looking for is indeed .∎
The quotient of appearing in 2.4 can be further simplified:
Lemma 2.5.
There is an isomorphism of groups
that is compatible with the homomorphisms to .
Proof.
Since the element of the finite abelian group is of order , it is not divisible by if and only if it generates a direct -summand. We first assume that this is the case, so . Writing for the -invariant subgroup complementary to the central summand in (1), we have . The latter admits an epimorphism to given by sending to . This is well-defined since the central element has order and acts trivially on since the action factors by construction through . The kernel of this epimorphism is the subgroup generated by , so we obtain an isomorphism between and , as claimed.
Now assume that does not generate a direct summand, so is divisible by . We have a (non-central) extension
| (14) |
As has order , the -equivariant map from into the kernel of (14) induced by inclusion is an isomorphism, so in order to show that is isomorphic to it suffices to show that (14) splits. Writing as in the previous case, the extension (14) is by construction the sum of the trivial -extension by the -module with the central extension classified by the image of the unique nontrivial element in under the composition induced by the inclusion and quotient maps of coefficients, so it suffices to show that this image is trivial. From the universal coefficient theorem and the computations in 2.1, we see that for any abelian group, the map induced by reducing modulo is an isomorphism, so to show that the class in question is trivial, it suffices to do so after reducing modulo . The latter follows by noting that the composition of -modules is trivial after passing to since vanishes in by assumption. ∎
Step \raisebox{-.9pt} {\arabic{subsubsection}}⃝. Fixing a point or not
Using the description of from the previous step, we are now in the position to determine . In view of the fibration sequence
| (15) |
this amounts to understanding the image of the “point-pushing” homomorphism . For , the image is trivial a result of the action of on itself (see the beginning of Step \raisebox{-.9pt} {\arabic{subsubsection}}⃝), but for any other homotopy sphere such an action is not available and it in fact follows from the following lemma that the image is never trivial.
Lemma 2.6.
Proof.
It suffices to show that the two compositions agree on the first standard basis vector , since both compositions are -equivariant (the action on the source is through and the action on the target is by conjugation after extending diffeomorphisms from to by the identity) and the orbit of under the -action spans .
We work with the following model of : view as , choose an orientation-preserving embedding disjoint from the origin and a representative of , extend by the identity to a diffeomorphism supported in , and form the mapping torus . We parametrise this quotient by in the evident way, use as base point, and we view as the complement of an embedded disc that contains the part where the nontrivial gluing happened (i.e. the image of in the quotient) and is disjoint from the image of in the quotient. The latter is so that the loop in is contained in . We chose a basis for such that this loop represents .
The first claim we show is that the image of under the map is represented by the diffeomorphism given by using on the image of in , and extending it to all of by the identity. This is because being the connecting map in the long exact sequence induced by the evaluation fibration with fibre , the point-pushing map sends to the isotopy class of any diffeomorphism that arises as the value at time of a path in with and . A possible choice of such path is given by for and for , which indeed agrees with at time .
The second claim we make is that the image of under the second composition in the statement is given by the diffeomorphism obtained by choosing an orientation-preserving embedding such that the image of in is contained in and avoids the origin, using on the image of in and extending it to a diffeomorphism of by the identity. This would imply the result, since the image of under both maps in consideration arises from the following construction: choose an embedding that represents (which is unique up to isotopy as ), use on this image, and extend by the identity.
To show this claim, we prove more generally that the composition is given by sending to the diffeomorphism obtained by representing by an embedding and by a diffeomorphism , using on and extending it to by the identity. By the argument from the proof of 2.3, it suffices to show that the described diffeomorphism considered as a diffeomorphism of agrees with the image of under the inclusion . This follows from [Hat78, p. 9, Remark (5)]. ∎
Corollary 2.7.
For and there is an isomorphism
which is compatible with the homomorphisms to .
3. Splitting the homology action and the proof of A
To deduce A from D, we first determine for which homotopy tori the map is surjective. The following was stated in [BT21, p. 4] without proof.
Lemma 3.1.
For a homotopy torus of dimension , the map is surjective if and only if is diffeomorphic to for some .
Proof.
The direction is easy: if , then is surjective because we can precompose it with the map and use that is surjective which holds for instance as a result of 2.2.
For the direction , we may assume since for any torus is diffeomorphic to the standard torus . This allows us to use smoothing theory [KS77, Essay V] which we briefly recall in a form suitable for our purposes: given a closed topological manifold of dimension , the set of concordance classes of smooth structures on is the set of equivalence classes of pairs of a smooth manifold together with a homeomorphism , where two pairs and are equivalent if there is a diffeomorphism such that the homeomorphisms and are concordant. The group of concordance classes of homeomorphisms acts on by postcomposition and the set of orbits is in bijection with the set of diffeomorphism classes of smooth manifolds homeomorphic to , induced by sending to . There is a map
to the set of isomorphism classes of pairs of a stable vector bundle over together with an isomorphism of the underlying stable Euclidean bundle with the stable topological tangent bundle of . The map is given by assigning a pair to the pullback of the stable tangent bundle of along , together with the isomorphism induced by the stable topological derivative of . The map turns out to be a bijection, by one of the main results of smoothing theory. Unwrapping definitions, one sees that the action of on is induced by pulling back the bundle along and postcomposing the isomorphism with the stable topological derivative of . The set is a torsor for the group of stable vector bundles on together with a trivialisation of the underlying stable Euclidean bundle; the group structure and the action are induced by taking direct sums. Thus, if comes already equipped with a smooth structure then we obtain a bijection , postcomposition with which gives a bijection
Going through the definition, the action of on translates to .
We now specialise to . Given a homotopy torus of dimension , a choice of homeomorphism induces a class and a morphism by conjugation with . This agrees with the map to when precomposed with the action map . The latter is an isomorphism as a result of 1.11 and the isomorphism (use the action of on itself to see this), so it suffices to show that is not surjective unless is diffeomorphic to for some . The image of is contained in the stabiliser of , so it is enough to show that is not contained in the invariants of this action unless for some . Since any is isotopic to a diffeomorphism of , the terms in the above description of the -action vanishes, and thus the action is simply by precomposition. In particular, it is an action by group homomorphisms if we equip with the group structure induced by the infinite loop space structure on . Using this infinite loop space structure and the fact that stably splits into a wedge of spheres we also get a direct sum decomposition of -modules . We will show below that the invariants of this action are given by the subgroup . This will imply the claim, since the subgroup corresponds to the classes of the pairs where is the unique homeomorphism up to isotopy that fixes the disc where the connected sum is taken, so in particular is not contained in this subgroup unless is diffeomorphic to for some .
To finish the proof, it thus suffices to show that for a finitely generated abelian group , the -action on by precomposition with the inverse has no invariants for . This is isomorphic to the standard action on up to the automorphism of given by taking inverse transpose, so we may equivalently show that has no invariants . Without loss of generality we may assume that is cyclic. In this case, has a basis as a -module indexed by subsets of cardinality , where the basis vector corresponding to is for and where is the standard -basis of . Now observe that an elementary matrix for acts by sending to if and , and to itself otherwise. On a general element , the matrix thus acts by
so if is an invariant, then for all with and . But since were arbitrary and , these are in fact all coefficients, so all invariants are zero. ∎
3.1. Proof of A
We conclude this section with the proof of A, which says that the map given by the action on admits a splitting if and only if for such that divisible by 2.
Proof of A.
We distinguish the cases whether a given homotopy torus of dimension is diffeomorphic to for some or not. If it is not, then the map is not surjective by 3.1, so it is in particular not split surjective. If it is, then by D the group is isomorphic, compatibly with the map to , to a semidirect product of or depending on whether is divisible by or not. In the first case, the map to visibly admits a splitting. In the second case, a hypothetical splitting would in particular induce a splitting of the projection , which does not exist since this extension is nontrivial. This finishes the proof. ∎
4. Endomorphisms of and the proofs of Theorems B and E
This section serves to deduce B from the classification result for endomorphisms of stated as E, and to prove the latter.
4.1. Proof of B assuming E
Assuming E, we prove B. We first assume . Given a nontrivial homomorphism for , the composition with the action on homology is by E either trivial or an isomorphism, so we have to exclude the former. If it were trivial, then would have image in . Suppose for contradiction that is nontrivial. Its kernel is a normal subgroup, so by [Men65, Corollary 1, p. 36] it is either (a) contained in the centre , which is trivial or depending on the parity of , or (b) of finite index. In either case, the image of contains a nonabelian finite group : in case (a) it contains or , so in particular a nonabelian finite group , and in case (b) the image of is finite itself, and also nonabelian since otherwise would be trivial since is perfect for (see 2.1).
To make use of the nonabelian finite subgroup , following [LR81], we consider the extension whose middle group is the normalizer of considered as a subgroup of the homeomorphism group of the universal cover. Note that the induced action of by agrees by construction with the action on the fundamental group. The pullback of this extension along is, by the Corollary on p. 256 of loc.cit. admissible in the sense of p. 256 loc.cit.. The proof of Proposition 2 loc.cit. then shows that the centraliser of in is abelian. But since acts trivially on we have , so is abelian and thus the same holds for which cannot be true by the choice of , so has to be trivial.
The case follows from the case by postcomposing a given homomorphism with the inclusion , so we are left to prove the addendum concerning homomorphisms from into or under the additional assumption . By the same argument as before, it suffices to show that all homomorphisms from into or are trivial. 4.1 below says that the latter two groups are abelian, so such morphisms factor over the (trivial) abelianisation of (see 2.1) and are therefore trivial, as claimed.
Lemma 4.1.
For a homotopy torus of dimension , the kernels of the homology actions
are both abelian.
Proof.
For , the homotopy torus is diffeomorphic to the standard torus and both kernels and are trivial, so in particular abelian. To show the claim for , note that because is homeomorphic to , so is abelian since we have by [Hat78, Theorem 4.1]. To show that is abelian, note that as we may view the map as the induced map on path components of the map to the space of orientation-preserving homotopy equivalences, so receives an epimorphism from . Replacing by in the argument for (3) on page 8 of [Hat78] and using that is homeomorphic to , we get that is isomorphic to the abelian group , so the claim follows (the final step can also be proved via smoothing theory). ∎
4.2. Proof of E
In the remainder of this section, we prove E. The proof makes use of the subgroup of unipotent upper triangular matrices which in particular contains the elementary matrices for ; these have on the diagonal and at the th entry, and at all other entries. It is well-known that is an -step nilpotent group whose centre is generated by the elementary matrix , which is an iterated commutator of length , namely . An important ingredient in the proof of E is the following lemma on complex representations of . In its statement and in all that follows, we write
for the automorphism of given by taking inverse-transpose.
Lemma 4.2.
Fix and a homomorphism with .
-
(i)
Assume . If , or if and is not a scalar, then is unipotent.
-
(ii)
If and is unipotent for each , then .
-
(iii)
If and is unipotent for each and , then has rank .
-
(iv)
If is unipotent and has rank for all , then after possibly precomposing with , the matrices all have the same fixed set.
Remark 4.3.
The argument in Bader’s MathOverflow post [Mat17] contains the claim that for any representation the matrix is unipotent. This is incorrect: 4.4 (iii) gives a representation for which is a nontrivial scalar. Also 4.2 (i) fails for (the case (ii) of 4.4 below involves representations for which is a non-scalar semisimple matrix).
Lemma 4.4.
Fix a homomorphism . If , then either
-
(i)
is unipotent, or
-
(ii)
there are and so that after postcomposing with conjugation by ,
If , then either
-
(i)
is unipotent,
-
(ii)
, where , up to conjugation, or
-
(iii)
there are and so that is a nontrivial cube root of , and after postcomposing with conjugation by ,
We omit the proof of 4.4 since it is based on similar (and easier) analysis as the base case in the proof of 4.2 (i) which we explain now.
Proof of Lemma 4.2 (i).
We do an induction on . To simplify the notation we set .
Base case. We treat the case by hand. To show that is unipotent, it suffices to prove that all its eigenvalues equal . Let be the -eigenspace for . Since is central in , restricting to gives a homomorphism whose image of we denote by . Next we distinguish cases depending on the dimension of . By the assumption that is not a scalar when , we know . If , then since is abelian, we have because is a commutator, so . If , we consider the subgroup generated by the images of , , and in . By assumption . Let be an eigenvector for with eigenvalue . Using the relation we conclude that is an eigenvector for with eigenvalue . Since , this forces because eigenvectors with different eigenvalues are linearly independent, and thus . Suppose for a contradiction that . Then has two distinct eigenvalues and . Since is central in , we deduce that and are simultaneously diagonalisable; in particular they commute. But since , this implies , which is a contradiction, so has to be . Finally, suppose that . In this case the argument is very similar to the preceding case: by assumption , and the relation implies that acts freely on the eigenvalues of which implies . If , then has distinct eigenvalues , , and for some . Using the fact that is both central and a commutator in , we reach a contradiction.
Induction step. Fix an eigenvalue for , and let be the corresponding eigenspace. We have to show . As in the base case, since is not a scalar if , so and since is central in , the representation restricts to . As before we write for the image of under this homomorphism. Consider the subgroup of . Let be an eigenvalue of and let be the corresponding eigenspace. As above, is also an eigenvalue for for each . Consider the subgroup of . Since is central in this copy of , there is an induced map , . If , then is a proper subspace of (since the eigenspaces for and are linearly independent). Then , so the induction hypothesis implies that is unipotent, so . Since the same argument applies for each eigenspace of , we conclude that , so as claimed. ∎
Proof of Lemma 4.2 (ii).
Fixing such that is unipotent for all , we want to show . As before we write . Note that the special case implies the case , because if then we may restrict to the subgroup to conclude from the special case, so using we get . To prove the special case , we do an induction on the dimension .
Base case. To settle the case , suppose for a contradiction that is not the identity. Since it is unipotent by assumption, it has up to conjugation the form , so by postcomposing with this conjugation we may assume that equals this matrix. Since is central in , the image of is contained in the centraliser of which consists of matrices of the form . This is an abelian subgroup, so is identity, a contradiction.
Induction step. For the induction step, we fix and suppose for a contradiction that . Consider the subspaces where . Writing we have since is unipotent, since , and (one way to see this is to consider the Jordan normal form). Note that since is central in , the image of preserves so we obtain a morphism by restriction. We write for its image of . Setting , we choose a basis for that extends a basis for and that has the property that
| (16) |
in this basis. To see that such a basis exists, it is again helpful to use the Jordan normal form. Since is central in , the morphism lands in the centraliser of (16) which are the matrices of the form
| (17) |
We claim that and have the form
| (18) |
for some , and . Assuming this claim for now, we observe that the matrices (18) commute, so is the identity. If then we are done since this contradicts (16). If then the relation shows that is the identity, which again contradicts (16). This leaves us with showing (18). We first treat . Since has image in (17), we may postcompose it with
to obtain two homomorphisms . We may apply the induction hypothesis to the restriction of these to the subgroup to conclude that the image of under these two homomorphism is the identity, so has the claimed form.
To deal with the second matrix we argue similarly: postcompose with the restriction to to obtain a morphism , restrict them to the subgroup in , and apply the induction hypothesis. ∎
Proof of 4.2 (iii).
Fix such that is unipotent and nontrivial. The subspace is nontrivial, preserved by the image of , each acts on it by a nontrivial unipotent, and acts nontrivially on it, so 4.2 (ii) implies . Arguing as in the proof of 4.2 (ii), up to changing basis (corresponding to postcomposing with a conjugation), we can assume that (16) holds and by the same argument as in the previous proof has image in matrices of the form (17) and has the form (18). We are left to show since then has rank in view of (16). Assuming for a contradiction that , then , so . Written out in matrices this equation reads as
which implies , but this is a contradiction because the trace of is , whereas that of is nonzero since is unipotent because so is , by assumption. ∎
Before proving Lemma 4.2 (iv), we discuss some properties of rank-1 operators. Given subspaces with , , there is a rank-1 operator with kernel and image , which is unique up to a unit, namely the composition . In what follows, it will be convenient to consider rank-1 operators up to scalars; abusing notation, we will use to denote either this equivalence class of rank-1 operator with kernel and image . In terms of equivalence classes, the composition behaves as
The operator (which is well-defined up to scaling by a unit) is unipotent if and only if (otherwise is diagonalisable and nontrivial). In this case the fixed set of is and its inverse is which is another representative of . Fixing two such equivalence classes of unipotent operators and , we have the commutator relation
| (19) |
If and , then the commutator is not unipotent.
The following observation will play a role in the proof of 4.2 (iv): Fixing unipotent operators as above for and assuming firstly that commutes with for and secondly that , we may use the commutator formula from above to conclude that for and that or .
Proof of Lemma 4.2 (iv).
Since has rank for , the operators are for of the form as discussed above where is the kernel of , i.e. the fixed set of . We claim that either or . This would imply the result, because the two cases are interchanged when precomposing with . To show this claim, we use that commutes with for . Since , it follows from the discussion after (19) that either or . In the first case, we also have for all , using and the fact that preserves since it commutes with . Similarly, in the second case we also have for all using and that commutes with . ∎
We illustrate the utility of Lemma 4.2 to study representations of by the following two corollaries, which will both play a role in the proof of Theorem E.
Corollary 4.5.
For and , all homomorphisms are trivial.
Under the additional assumption that factors through , this corollary is proved in [Wei97, Lemma 3] using superrigidity and the congruence subgroup property.
Proof of 4.5.
If then is unipotent by 4.2 (i) and since the are conjugate in so are all . We then apply 4.2 (ii) to see that is trivial, so also the conjugates are. As the generate the result follows. For we use that is perfect (see 2.1) and is abelian. For we apply the first part of 4.4: in case (i) we proceed as for and the case (ii) is ruled out because the images of and are not conjugate. ∎
Corollary 4.6.
Fix and a nontrivial homomorphism .
-
(i)
If , then for all the matrix is unipotent and has rank . Moreover, after possibly precomposing with , the matrices all have the same fixed set.
-
(ii)
If , then the same conclusion holds under the additional assumption .
Proof.
We begin with two observations based on the fact that is for all conjugate to . Firstly, to show the first claim of (i) and (ii), it suffices to consider . Secondly, is nontrivial since otherwise were trivial as is generated by the .
In the case , it suffices to prove that is not a scalar, for then everything follows from 4.2, using that is conjugate in to for any . If were a scalar, then all are scalars, so would have image in scalar matrices because the generate . But since is a commutator and scalar matrices commute, this would imply , which is not the case.
Next we consider the case . To show the case , for which we imposed the additional assumption . It suffices by 4.2 to prove that the nontrivial matrix is unipotent which we prove by contradiction. We consider the restriction of to and consult the classification in 4.4. Since we assumed that is not unipotent, we do not need to consider the case (i). Cases (ii) and (iii) of 4.4 can be excluded by showing that for these representations the matrices are not all conjugate in . In almost all cases this can be seen considering their eigenvalues, except in the case
Also these matrices are not conjugate in which one can see by reducing modulo . ∎
Theorem 4.7.
Fix and a nontrivial homomorphism . There exist linearly independent vectors so that, after possibly after precomposing with , the image of preserves the lattice and for all the matrix of the restriction with respect to the basis is .
Remark 4.8.
One might suspect that given there exists a basis for so that the same conclusion of 4.7 holds (this is claimed in the MathOverflow post mentioned in Remark 4.3). This is not the case. For example, there is a nontrivial representation with finite image, constructed by setting
and then defining and . One can then check directly that this extends to a morphism by checking that these matrices satisfy the relations in the standard presentation of in terms of (see [Mil71, Corollary 10.3]). This peculiar representation has finite image because for each , the matrix has order 2, and the subgroup generated by has finite index in by a general theorem of Tits [Tit76] (see also [Mei17, Theorem 3]).
Proof of Theorem 4.7.
Fix a nontrivial homomorphism . We write , considered as a matrix in . After possibly precomposing with , we know from 4.6, that for , the matrix is unipotent and has rank , and that where be the fixed set of the matrix for . Note that each is -dimensional, since has rank 1. Using the fact that for each fixed , the matrices (skipping ) are simultaneously conjugate to , we find that also the hyperplanes (skipping ) all agree. We abbreviate this hyperplane by . Next we claim that the intersection of hyperplanes for are all lines. For this it suffices to show that is trivial. Assume by contradiction that this intersection is nontrivial. By construction, it is the common fixed set for the for all , so it is in fact fixed by the whole image of since the generate the image because the generate . Moreover, since the are defined over , also is nontrivial, so the free abelian group has rank . Combining this with 4.5, we see that the morphism induced by is trivial, so factors over the additive group . The latter is abelian, so must be trivial since is perfect (see 2.1). This contradicts our choice of .
Claim. The image of is .
Proof of Claim. For definiteness, we prove the statement for . Since has rank and is -dimensional, it suffices to show that the image of is contained in for all . Recall that is the fixed set of . Since commutes with , the matrix preserves the image of , but since this image is only one dimensional, it is an eigenspace for , which implies since is unipotent. This proves the claim.
Now we construct the basis . Fix a nonzero vector which we may choose to be an integer vector as is defined over since has image in . Now define inductively . Note that the are integer vectors as . Moreover, each is nonzero: if were trivial then would be contained in which we saw above is trivial, so we get and inductively which is not true. Now we examine what properties the vectors have. First observe that they form a basis for , by the general fact that if are hyperplanes of with trivial intersection, then a choice of nonzero vector from each of the lines gives a basis for . By construction, with respect to the basis , the matrix of is . Now using the commutator relations in , we conclude that after this change of basis the restriction of to upper triangular matrices is the inclusion, so to finish the proof suffices to show the same for the lower triangular matrices since is generated by upper and lower triangular matrices. As for upper triangular matrices, it suffices to consider for every . By construction, has fixed set and for some scalar , so we are left to show . This follows from the braid relation . ∎
Proof of Theorem E.
Fix a nontrivial homomorphism and let be the linearly independent vectors promised by 4.7, so that possibly after precomposing with , the matrix for preserves the lattice , and the restriction is represented by the matrix when written in the basis . In particular, this has as consequence that every orientation-preserving automorphism of extends to an orientation-preserving automorphism of . We claim that this in turn implies for some . Dividing the basis by , this would show that we can choose to form a basis of , so is given by conjugation by an element of . That for some follows from two facts: (a) for every non-characteristic subgroup of full rank, there exists an (orientation-preserving) automorphism of that does not extend to , so has to be characteristic, and (b) every characteristic subgroup of full rank has the form for some . To see these two facts, we fix a subgroup of full rank. By the elementary divisor theorem, there is a basis of and natural numbers such that is a basis of . If is non-characteristic, then for some and (since is clearly characteristic), so the automorphism of that interchanges and does not extend to (by interchanging a second pair of basis vectors we also find an orientation-preserving example of such an automorphism). This shows (a). Moreover, if we assume for some and , then the automorphism of that interchanges and does not restrict to , so cannot be characteristic. This shows (b). ∎
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