Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $Top(4)/O(4)$?

share|improve this question
add comment

6 Answers

up vote 19 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
And do you know an example of distinct smooth, stably parallelizable, compact 4-manifolds that are homeomorphic? Also, I'm skeptical that topological 5-manifolds have unique smoothings; I believe smooth structures on a smoothable topological 5-manifold $M$ are distinguished by an invariant in $H^3(M, \mathbb{Z}/2)$. –  John Francis Dec 6 '09 at 23:08
add comment

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.

share|improve this answer
PS. I am still mystified about CW structures: when is a topological 4-manifold a CW complex? Again, if and only if smoothable? –  Andrew Ranicki Sep 15 '10 at 5:59
add comment

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.

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.

share|improve this answer
Do these papers deal with issue of whether the tangent bundles of these distinct smoothings are isomorphic? At first glance, I don't see that point considered. –  John Francis Dec 6 '09 at 22:49
add comment

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.

share|improve this answer
Assuming the $o=ks$ claim (that I, unfortunately, don't know how to show at the moment), this seems to me to give that the plumbing of $E_8\oplus E_8$ has a stable tangent bundle. This might be standard, but how do you go from that to the unstable statement? A priori, it's plausible that a microbundle $\tau_M \oplus \mathrm{R}^k$ could admit a vector bundle structure even if $\tau_M$ doesn't. –  John Francis Dec 12 '09 at 18:13
Fair point. If I read Dold-Whitney (Annals 1959) correctly, I can create an $SO(4)$ vector bundle with $w_2=0$ and $p_1$ any even integer. Altering this over a 4-ball, via a map $S^3\to SO(4)$ that factors through $SU(2)$, I can adjust its Euler class freely. In this way I can cook up a "tangent bundle" that underlies the Spivak fibration, provided that $o=0$. One then ought to check, using $ks$, whether it underlies the microbundle. –  Tim Perutz Dec 12 '09 at 20:11
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.