Failure of smoothing theory for topological 4-manifolds - MathOverflow

Failure of smoothing theory for topological 4-manifolds

2009-12-05T23:10:28Z

Smoothing theory fails for topological 4-manifolds, in that a smooth structure on a topological 4-manifold $M$ is not equivalent to a vector bundle structure on the tangent microbundle of $M$. Is there an explicit compact counterexample, i.e., are there two compact smooth 4-manifolds which are homeomorphic, have isomorphic tangent bundles, but are not diffeomorphic? (The uncountably many smooth structures on $\mathbb{R}^4$ should give a noncompact counterexample, since $Top(4)/O(4)$ does not have uncountably many components.)

Addendum to question, added 12/11/09:

I'm also interested in the other type of counterexample, of a nonsmoothable topological 4-manifold whose tangent microbundle does admit a vector bundle structure. Does someone know such an example? Tim Perutz's answer to my first question, below, says that homeomorphic smooth 4-manifolds have isomorphic tangent bundles. If it's not true that all topological 4-manifolds have vector bundle refinements of their tangent microbundle, what is the obstruction in the homotopy of $Top(4)/O(4)$?

Answer by Igor Belegradek for Failure of smoothing theory for topological 4-manifolds

2009-12-06T05:00:13Z

Take any two closed simply-connected homeomorphic smooth closed 4-manifolds that are not diffeomorphic. Then their products with $\mathbb R$ are diffeomorphic because the smooth structure on a such a product is unique. (Indeed, since PL/O is 6-connected, it is enough to show that the associated PL structure is unique, but the set of PL-structures on a PL-manifold $M$ of dimension $\ge 5$ is bijective to the set of homotopy classes of maps from $M$ to $TOP/PL$, and the latter space is $K(\mathbb Z_2, 3)$, so the set of PL structures on $M$ is bijective to $H^3(M,\mathbb Z_2)$, which vanishes by Poncare duality if $M$ is homotopy equivalent to a simply-connected $4$-manifold; in fact the argument shows that all we need is $H_1(M;\mathbb Z_2)=0$).

It follows that the original closed simply-connected $4$-manifolds are tangentially homotopy equivalent, i.e. there is a homotopy equivalence that pulls stable tangent bundles to each other.

Answer by Spinorbundle for Failure of smoothing theory for topological 4-manifolds

2009-12-06T18:03:52Z

I think you are searching for the following:
An exotic {4}-manifold by Selman Akbulut

We construct two compact smooth 4-manifolds $Q_1, Q_2$ which are homeomorphic but not diffeomorphic to each other. In particular no diffeomorphism $\partial Q_1 \rightarrow \partial Q_2$ can extend to a diffeomorphism $Q_1 \rightarrow Q_2$

Alternatively the boundary case
An exotic orientable 4-manifold by Robert E. Gompf

In the present paper, we exhibit two compact orientable manifolds (with boundary), $M_1$ and $M_2$, which are homeomorphic, but not diffeomorphic.

The minimal symplectic case.
http://www.msp.warwick.ac.uk/gt/2008/12-02/p019.xhtml

Finally you will perhaps like the following notes by David Gay

This paper will outline in an informal way the construction of a family of 4manifolds which are homeomorphic but not diffeomorphic.

The first section of the paper (after the introduction, so it is section 2 in the paper), describes the usual construction "of an infinite family of diffeomorphism classes of 4manifolds in two homeomorphism classes".
(Roughly speaking, the basic examples of non diffeomorphic but homeomorphic 4-manifolds are constructed as follows : Let $E(1)$ be the algebraic surface, obtained by blowing up 9 points in $\mathbb{C}P^2$. This is an elliptic surface. Let $E(2)$ be the sum of two copies of $E(1)$ (how this is done, is explained in section 2). Define inductively $E(n)$ as the fiber sum of $E(n-1)$ and $E(1)$. By logarithmic transformations you can build from these $E(n)$'s the elliptic surfaces $E(n, m_1,...m_n)$, where $m_1,...,m_n$ are the orders of the transformation. The basic examples of not diffeomorphic but homeomorphic 4-manifolds are such $E(n,p,q)$'s where $p,q$ are relativly prime.)
Since you asked for compact examples, this doesn't answer your question. Nevertheless I think (hope) that this last link is useful, since it provides a short overview and introduction to non diffeomorphic but homeomorphic 4-manifolds.

Answer by Tim Perutz for Failure of smoothing theory for topological 4-manifolds

2009-12-07T00:46:22Z

For a pair of smooth, simply connected, compact, oriented 4-manifolds $X$ and $Y$,

Any isomorphism of the intersection lattices $H^2(X)\to H^2(Y)$ comes from an oriented homotopy equivalence $Y\to X$ (Milnor, 1958).
Any oriented homotopy equivalence is a tangential homotopy equivalence (Milnor, Hirzebruch-Hopf 1958).
Any oriented homotopy equivalence comes from an h-cobordism (Wall 1964).
Any oriented homotopy equivalence comes from a homeomorphism (Freedman).
It need not be the case that $X$ and $Y$ are diffeomorphic (Donaldson). Many examples are now known: e.g., Fintushel-Stern knot surgery on a K3 surface gives a family of exotic K3's parametrized by the Alexander polynomials of knots.

Here's a sketch of why homotopy equivalences preserve tangent bundles: $X$ and $Y$ have three characteristic classes: $w_2$, $p_1$ and $e$. However, $e[X]$ is the Euler characteristic, and $p_1[X]$ three times the signature. By the Wu formula, $w_2$ is the mod 2 reduction of the coset of $2H^2(X)$ in $H^2(X)$ given by the characteristic vectors, hence is determined by the lattice. In trying to construct an isomorphism of tangent bundles over a given homotopy equivalence, the obstructions one encounters are in $H^2(X;\pi_1 SO(4))=H^2(X;Z/2)$ and in $H^4(X;\pi_3 SO(4))=Z\oplus Z$, and these can be matched up with the three characteristic classes.

Answer by Tim Perutz for Failure of smoothing theory for topological 4-manifolds

2009-12-12T05:18:31Z

This is in response to John's addendum. As I understand it, one has the following hierarchy:

Any Poincare complex $X$ has a Spivak normal spherical fibration $S$.
If $X$ carries a topological manifold structure then $S$ has a microbundle reduction.
If $X$ carries a smooth manifold structure then $S$ has a vector bundle reduction refining the microbundle reduction.

I'm going to concentrate on simply connected Poincare 4-complexes $X$ with even intersection form. These have Kirby-Siebenmann smoothing obstruction $ks\in H^4(X;\mathbb{Z}/2)=\mathbb{Z}/2$ equal to $\sigma(X)/8$ mod 2, where $\sigma$ is the signature. This is just the obstruction coming from Rochlin's theorem: $\sigma$ is divisible by $16$ if $X$ is smoothable.

Freedman tells us that $X$ has a unique topological manifold structure, and hence $S$ has a canonical microbundle structure. So, to ask whether there is a vector bundle reduction of the microbundle is the same as asking whether $S$ has a vector bundle reduction.

Let $BG$ be the classifying space for stable spherical fibrations. To solve the obstruction-theory problem of lifting $X\to BG$ to a map $X\to BO$, we need to know the low-dimensional homotopy groups of $BO$ and $BG$ - specifically, whether $\pi_i(BO)\to \pi_i(BG)$ is surjective. I read off from a table in Ranicki's book "Algebraic and geometric surgery" that this is so for $i=1$ and $2$, but that $\pi_3(BG)=\mathbb{Z}/2$ whereas $\pi_3(BO)=0$. So there is an obstruction $o\in H^4(X;\mathbb{Z}/2)$ to finding a vector bundle reduction.

I'm a bit nervous of $ks$ due to my ignorance of topological manifold theory, but I think it should then be the case that $o=ks$ (they seem to be similar beasts; I'm thinking of $ks$ as coming from $\pi_3 (BTOP)$, where $o$ comes from $\pi_3(BG)$). What I actually want to use is the corollary, which if true should have a direct proof - that $o=\sigma/8$ mod 2. Anyone?

Given any unimodular matrix $Q$, I can build a Poincare 4-complex with $Q$ as its intersection matrix (plumb together disc-bundles over $S^2$ according to $Q$, cone off the homology 3-sphere boundary). If it's correct that $o=\sigma/8$, then when $Q=E_8$, I get a complex with no tangent bundle, whereas when $Q=E_8\oplus E_8$ I get a complex which has a tangent bundle but which is not smoothable by Donaldson's diagonalizability theorem.

Answer by Peter Teichner for Failure of smoothing theory for topological 4-manifolds

2010-03-18T07:09:57Z

John, if you look at chapter 8 of Freedman-Quinn's book on topological 4-manifolds, you'll find the following computation of the homotopy groups of Top(4)/O(4):

$\pi_3 = Z/2$ and $\pi_i = 0$ for $i=0,1,2,4$.

This implies that

a topological 4-manifold has a linear reduction of its tangent bundle if and only if the Kirby-Siebenmann invariant vanishes
if it exists, the reduction is unique.

Donaldson's and Freedman's results imply lots of examples of non-smoothable 4-manifolds with trivial Kirby-Siebenmann invariant: any unimodular intersection form arises from a closed simply connected topological 4-manifold, and in the even case the Kirby-Siebenmann invariant is the signature/8 mod 2. If the form is definite, it cannot arise from a smooth manifold. Furuta even showed that Euler characteristic/signature must be $\geq 10/8$ in order to be realized smoothly. The conjectured bound is 11/8 and is realized by the Kummer surface.

Answer by Andrew Ranicki for Failure of smoothing theory for topological 4-manifolds

2010-09-14T21:23:10Z

I have always been mystified about handlebody structures on topological 4-manifolds. Already in 1970 Kirby and Siebenmann had established that topological n-manifolds have a handlebody structure for n>5 (see Essay III.2 in the 1976 K-S book), and Quinn proved this for n=5 in Ends of Maps III (1982). Finally I just sent an email to Kirby, who gave a simple argument that a topological 4-manifold has a handlebody structure if and only if it is smoothable. I have posted his email on the surgery pages of the Manifold Atlas Project.