Aspherical manifolds that cannot be triangulated...Mike Davis (joint with Jim Fowler and Jean Lafont) Aspherical manifolds that cannot be triangulated Introduction Dimension 4 Dimensions

IntroductionDimension 4

Dimensions > 4

Aspherical manifolds that cannot betriangulated

Mike Davis(joint with Jim Fowler and Jean Lafont)

University of Wisconsin, MilwaukeeDecember 6, 2013

Mike Davis (joint with Jim Fowler and Jean Lafont) Aspherical manifolds that cannot be triangulated


Dimensions > 4

ProblemCan every manifold be triangulated? In other words, is everymanifold homeomorphic to a simplicial complex?



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The reason for asking this question was that at the beginning ofthe twentieth century one could only do algebraic topology forsimplicial complexes. After the advent of singular homology thisreason was no longer relevant.



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Kneser’s Questions - 1924Is a polyhedron with the local homology properties ofEuclidean space, locally homeomorphic to Euclideanspace?Is a space which is locally homeomorphic to Euclideanspace, triangulable (homeomorphic to some polyhedron)?(Hauptvermutung). If there are two such triangulations,must they be PL equivalent?

“polyhedron” = space homeo to simplicial cx



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AnswersIn dimensions ≤ 3 all mflds can be triangulated (due toMoise in dim 3). Also, for dim ≤ 3,

(homeo) =⇒ (PL homeo).Conventional wisdom: These questions were answered byKirby-Siebenmann in dimensions≥ 5 and by Freedman indim 4.

Actually Kirby-Siebenmann’s answer was only for certainspecial types of triangulations called “PL manifolds.” A PL mfldis one in which every simplex has a “dual cell.”



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Dual cells

A simplicial cx L is a PL n-manifold if each k -simplex has a dualcell, which is PL homeomorphic to an (n − k)-disk. Orequivalently, if the link of each simplex is PL homeomorphic toSn−k−1. (By the Poincare Conjecture we can drop the phrase“PL” except possibly when n − k − 1 = 4.)



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Theorem (Kirby - Siebenmann 1969)

A topological n-mfld Mn, n ≥ 5, admits a PL structure ⇐⇒ anobstruction ∆ ∈ H4(Mn;Z/2) vanishes.

In other words, TOP/PL is the Eilenberg-MacLane spaceK (Z/2,3).

There is also an obstruction to uniqueness in H3(Mn;Z/2).



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For simplicial cx L to be a topological mfld it is not necessary forit to have dual cells. However, it must be a “homology mfld.”This means that L must have the same local homology groupsas Rn, i.e., H∗(L,L− x) ∼= H∗(Rn,Rn − 0). This entails that thelink of each k -simplex has the same homology as does Sn−k−1.

Example (Poincare)

∃ a 3-dim mfld H3 with same homology as S3 but 6∼= S3

(because π1(H3) 6= 1). The suspension of H3 is a homology4-mfld. (However it’s not a mfld in a nbhd of a suspensionpoint.)



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Question (Milnor - Seattle 1963)

Let H3 be a homology 3-sphere with π1 6= 1. Is the doublesuspension of H3 homeomorphic to S5? (In particular, is thedouble suspension a mfld?)

The answer is yes (Edwards and Cannon ∼1975).

Theorem (The Double Suspension Thm)

The double suspension of any homology sphere Hn is ∼= Sn+2.

Amazing Fact

Since H3 can be triangulated, we get a triangulation of S5

which is not equivalent to a PL triangulation.



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Milnor’s Questions - Seattle 1963

• Let M3 be a homology 3-sphere with π1 6= 1. Is the doublesuspension of M3 homeomorphic to S5?• Is simple homotopy type a topological invariant?• Can rational Pontrjagin classes be defined as topologicalinvariants?• (Hauptvermutung). If two PL-manifolds are homeomorphic,does it follow that they are PL-homeomorphic?• Can topological manifolds be triangulated?• The Poincare hypothesis in dimensions 3, 4.• (The annulus conjecture). Is the region bounded by twolocally flat n-spheres in (n + 1)-space necessarilyhomeomorphic to Sn × [0,1]?



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It follows from Freedman’s work in 1981 that there are 4-mfldswhich do not admit a PL structure. From later work of Casson itfollows that these cannot be triangulated (as homology mflds).

By Kirby - Siebenmann, in dimensions ≥ 5, ∃ mflds which donot admit a PL structures. However, because of the DoubleSuspension Thm, the possibility remained that all mflds of dim≥ 5 could be triangulated. In the late 1970s Galewski - Sternanalyzed the situation. They showed that all mflds of dim ≥ 5could be triangulated ⇐⇒ ∃ a homology 3-sphere with certainproperties.

.

In 2013 this question about homology 3-spheres was resolvedby Manolescu by using gauge theory. He showed no suchhomology 3-spheres exist. So, ∃ nontriangulable Mn for n ≥ 5.Mike Davis (joint with Jim Fowler and Jean Lafont) Aspherical manifolds that cannot be triangulated


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DefinitionA space is aspherical if its universal cover is contractible.

Why aspherical mflds?

At one point the only examples of closed aspherical mflds camefrom differential geometry or from Lie groups; hence, weresmooth mflds (and hence, could be triangulated). In 1991Januszkiewicz and I applied Gromov’s hyperbolizationtechnique to Freedman’s 4-mflds to show the existence ofnontriangulable aspherical 4-mflds. The product of one of thesewith a torus also does not have a PL structure. (However, bythe Double Suspension Thm, any such product with a positivedim’l torus can be triangulated.)



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Gromov’s hyperbolization procedure showed that you couldconvert simplicial complexes (hence, triangulated mflds) intoaspherical ones. To get nontriangulable aspherical 4-mfldsJanuszkiewicz and I had to use Freedman’s work. To convertthe Galewski - Stern mflds into nontriangulable aspherical mfldsof dim ≥ 5 you need a further trick: relative hyperbolization.Using this, we show that these Galewski - Stern mflds can bechosen to be aspherical, at least for dim ≥ 6.



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1 Introduction

2 Dimension 4Rokhlin’s Theorem and the µ-invariantFreedman’s E8-manifoldHyperbolization

3 Dimensions > 4Kirby - SiebenmannGalewski - Stern + ManolescuRelative hyperbolization



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Polyhedral homology mflds

DefinitionsA simplicial cx Ln is a polyhedral homology n-mfld (a PHM forshort) if for each k -simplex σ, Lk(σ, L) has the same homologyas Sn−k−1. Ln is a PL mfld if ∀σ ∈ L, Lk(σ, L) is PLhomeomorphic to Sn−dimσ−1.



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Rokhlin’s Theorem and the µ-invariantFreedman’s E8-manifoldHyperbolization

For any closed orientable mfld M4k , cup product defines anondegenerate symmetric bilinear form on

H2k (M4k )/torsion.

In dimension 4 this form is even if w2 = 0.

FactIf B is an even, nondegenerate, symmetric bilinear form over Z,then its signature, σ(B), is divisible by 8.

Theorem (Rokhlin 1952)

If M4 is a closed PL 4-mfld, with w1 = 0 and w2 = 0, then

σ(M4) ≡ 0 mod 16.



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Fact

If H3 is a homology 3-sphere, then H3 = ∂W 4, where W 4 is aPL mfld with even intersection form.

The µ-invariantDefine

µ(H3) =σ(W 4)

8∈ Z/2.

This defines a homomorphism µ : ΘH3 → Z/2, where ΘH

3 is thegroup of homology cobordism classes of homology 3-spheres.



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Associated to the Dynkin diagram E8 there is a nondegenerateeven, symmetric form which is positive definite of rank 8(hence, σ = 8).

By a process calle “plumbing,” one can use the E8 diagram to tobuild a smooth 4-mfld with bdry and intersection form given byE8.

Q(E8) := the E8 plumbing.

∂Q(E8) = H3, Poincare’s homology 3-sphere. σ(Q(E8) = 8. LetX 4 := Q(E8) ∪ c(H3). It is a PHM of signature 8.



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Theorem (Freedman 1982)

H3 = ∂C4, where C4 is a top contractible mfld. PutM4 = Q(E8) ∪ C4, the “E8-manifold”.

By Rokhlin’s Thm, M4 does not have a PL structure.

FactAny triangulation of a 4-mfld is automatically PL. (Pf: By thePoincare Conj, the link of any vertex is PL homeomorphic toS3.) So, Freedman’s M4 is not triangulable.



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Hyperbolization (Gromov)A hyperbolization procedure is a functor h from{simplicial complexes} to {locally CAT(0) spaces} togetherwith a map f : h(K )→ K with the following properties:

h preseves local structure: ∀σ ∈ K , Lk(h(σ)) ∼= Lk(σ)(Lk means “link”.) In particular, if K is a mfld (or a PHM),then so is h(K ).f ∗ is a split injection on cohomology.When K is a mfld, f pulls back stable tangent bundle tostable tangent bundle. So, f ∗ pulls back characteristicclasses of K to those of h(K ).



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(D - Januszkiewicz)

Apply h to the E8 homology mfld X 4.Resolve it to N4 = (h(X 4)− nbhd of cone pt) ∪ C4.Then N4 is aspherical and not triangulable.

Theorem (DJ 1991)∃ closed aspherical 4-mflds that cannot be triangulated. Forn ≥ 5, ∃ closed aspherical n-mflds which are not homotopyequivalent to PL mflds.

Proof of 2nd sentence.

N4 × T k is not PL.



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Remark

By Double Suspension Thm, for k > 0, N4 × T k ∼= X 4 × T k

(where X 4 is the PHM). So, N4 × T k can be triangulated.



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Kirby - SiebenmannGalewski - Stern + ManolescuRelative hyperbolization

Theorem (Kirby - Siebenmann 1969)A top n-mfld, n ≥ 5, admits a PL structure ⇐⇒ an obstruction∆ ∈ H4(Mn;Z/2) vanishes.



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Polyhedral Mfld Characterization Theorem

Theorem (Edwards 1978 + Perelman)A PHM (of dim > 2) is a top mfld ⇐⇒ the link of each vertex issimply connected.

ExampleThe double suspension of a homology sphere, with π1 6= 1.

Such triangulations are not PL.

In the early seventies such considerations led Siebenmann toask if all mflds could actually be triangulated (before the DoubleSuspension Thm or Freedman’s E8 4-mfld were known).



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I will describe highlights of a theory worked out in the 1970 s byseveral people, Siebenmann, Matumoto, most notably Galewski- Stern (with important contributions by others, eg, Cohen,Sullivan, Martin, Maunder).

Suppose X is a PHM. Let λ ∈ H4(X ; ΘH3 ) be the cohomology

class which associates to the “dual cell” (actually dual cone) ofa codim 4 simplex σ, the class of Lk(σ) in ΘH

3 (where ΘH3 is the

group of homology cobordism classes of homology 3-spheres).λ is the obstruction to finding an “acyclic resolution” of X by aPL manifold.



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Consider the coefficient sequence:

0→ Kerµ−→ΘH3

µ−→Z/2→ 0.

Fact 1

When X is a top mfld, µ∗ takes λ ∈ H4(X ; ΘH3 ) to the

Kirby-Siebenmann obstruction ∆ ∈ H4(X ;Z/2).

Fact 2If M is a top mfld, then the obstruction to triangulation isβ(∆) ∈ H5(M; Kerµ), where β is the Bockstein associated tothe above coefficient sequence.



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Theorem (Galewski-Stern ∼ 1980)

In dim n > 4, ∃ nontriangulable Mn iff the sequence0→ Kerµ→ ΘH

3 → Z/2→ 0 does not split, ie, ⇐⇒ 6 ∃ ahomology 3-sphere H3 with µ(H3) 6= 0 and H3#H3 = 0 in ΘH

3 .

Theorem (Manolescu 2013)The sequence does not split.



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Galewski-Stern mflds

It suffices to consider the Bockstein associated to

0→ Z/2 ×2−→ Z/4−→Z/2→ 0.

This Bockstein is Sq1 (the first Steenrod square).

∃ a (nonorientable) PHM P5 with bdry stint P5 is a top mfld (this uses Edwards’ Thm).∆(P5) = µ∗(λ(P5)) 6= 0 in H4(P5;Z/2) and∆(∂P5) = 0.Sq1(∆) 6= 0 in H5(P5, ∂P5;Z/2).∃ a PHM bordism U from ∂P5 to a PL mfld V 4, and V 4 isbdry of PL mfld W 5.M5 := P5 ∪ U ∪W 5 is not triangulable.



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Relative hyperbolization (D - Januszkiewicz - Weinberger, 2001)

Let (M, ∂M) be a triangulated mfld with bdry. Put

H(M, ∂M) := h(M ∪ c(∂M))− (nbhd of cone point)

Key properties

H(M, ∂M) is mfld with bdry; its bdry is ∂M.π1(∂M)→ π1(H(M, ∂M)) is injective.H(M, ∂M) is aspherical iff ∂M is aspherical.

Corollary (DJW)If an aspherical mfld bounds a triangulable mfld, then it boundsan aspherical mfld.



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Corollary (DJW)If an aspherical mfld V bounds a triangulable mfld W, then itbounds an aspherical mfld W ′.

Proof.W ′ = H(W ,V ).



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GS mflds in dimensions ≥ 6

Put P6 := P5 × S1. Since ∆(∂P6) = 0, ∂P6 admits a PLstructure (by Kirby-Siebenmann).

Put M6 = P6 ∪ U ∪W , where U is the mapping cylinder of a(necessarily non-PL) homeomorphism from ∂P6 to a PL mfldV 5 and W is a PL 6-mfld bounded by V 5.



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Theorem (D-Fowler-Lafont)

In each dim n ≥ 6, ∃ an aspherical mfld Nn that cannot betriangulated.



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Proof.

Start with h(P6). Then h(∂P6) is homeomorphic to a PL mfldV 5. Let U be the mapping cylinder of a homeomorphismV 5 → h(∂P6). V 5 is bdry of a PL 6-mfld W . Put

N6 := h(P6) ∪ U ∪H(W ,V ).

We check immediately thatN6 is aspherical.∆(N6) 6= 0 and Sq1(∆(N6)) 6= 0.

So, N6 cannot be triangulated.



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Thank you.


Documents

Aspherical manifolds that cannot be triangulated...Mike Davis (joint with Jim Fowler and Jean Lafont) Aspherical manifolds that cannot be triangulated Introduction Dimension 4 Dimensions