23
$\begingroup$

Near the top of the second page of this paper, it is claimed that if two 4-manifolds $X$ and $Y$ are homeomorphic, then their squares $X \times X$ and $Y \times Y$ are diffeomorphic. Why is this true? It is trivially correct for 3-manifolds and lower dimensions. What about higher dimensions?

References would be greatly appreciated as well.

$\endgroup$
5
  • 2
    $\begingroup$ It is trivially true for 3-manifolds, because they have exactly one smooth structure. $\endgroup$
    – ThiKu
    Apr 12, 2016 at 17:20
  • $\begingroup$ Thanks! I learned this seconds after posting, via one of the suggested questions, and edited to reflect this. $\endgroup$
    – kosta
    Apr 12, 2016 at 17:21
  • 2
    $\begingroup$ This is more-or-less completely answered by this question, given that smoothing theory works in dimension 8. $\endgroup$
    – mme
    Apr 12, 2016 at 17:57
  • $\begingroup$ Thanks for the references @MikeMiller. I read through the answers but could not extract out of that, in a form that I understand, an answer to my question. Perhaps you could elaborate a bit? Thanks in advance! $\endgroup$
    – kosta
    Apr 12, 2016 at 20:30
  • $\begingroup$ @kosta My apologies, this is more subtle than I realized. I only see how to write down a proof in the simply connected case, and in a way that has nothing to do with my comment. $\endgroup$
    – mme
    Apr 12, 2016 at 22:39

2 Answers 2

11
$\begingroup$

It follows by smoothing theory. If $h : X \to Y$ is a homeomorphism between smooth 4-manifolds, one obtains two maps $X \to BO$ which become homotopic in $BTOP$. The difference between them is therefore a map $d: X \to TOP/O$.

Now $TOP/O$ is [No it isn't, see comments] 6-connected, so $d$ is nullhomotopic. Therefore the map $$X \times X \overset{d \times d}\to TOP/O \times TOP/O \overset{\oplus}\to TOP/O$$ is also nullhomotopic, but this is the difference construction applied to the homeomorphism $h \times h : X \times X \to Y \times Y$. It being nullhomotopic means that $h \times h$ is homotopic through homeomorphisms to a diffeomorphism.

EDIT: Out of embarrassment about my mistake, let me try to show something like the opposite: there is a smooth 4-manifold $M$ and a homeomorphism $h : M \to M$ such that $h \times h$ is not isotopic to a diffeomorphism (even though its source and target are the same manifold, so definitely diffeomorphic).

Let $M = S^1 \times \mathbb{RP}^3$. To construct $h$ I will instead construct a topological $s$-cobordism $W$ from $M$ to $M$ and then use the fact that $\pi_1(M) = \mathbb{Z} \oplus \mathbb{Z}/2$ is abelian and hence "good" in the sense of Freedman, so $W$ is homeomorphic to $I \times M$, this homeomorphism yielding $h$.

To construct $W$, use the topological surgery exact sequence, in particular the portion $$\mathcal{S}^{TOP}(M \times I, \partial) \overset{\eta}\to [(M \times I, \partial), G/TOP] \overset{\sigma}\to L_5(\mathbb{Z} \oplus \mathbb{Z}/2)$$ which is an exact sequence of abelian groups. The homotopy type of $G/TOP$ is $K(\mathbb{Z}/2,2) \times K(\mathbb{Z},4)$ up to degree 5, so the middle term is identified with $$H^1(M;\mathbb{Z}/2) \oplus H^3(M;\mathbb{Z})$$ as a group. Furthermore, by chasing through this identification we see that if $[H : W \to M \times I] \in \mathcal{S}^{TOP}(M \times I, \partial)$ is an $s$-cobordism (corresponding to a homeomorphism $h : M \to M$) then the projection to $\eta(H)$ to $H^3(M;\mathbb{Z})$ followed by reduction mod 2 to $H^3(M;\mathbb{Z}/2)$ is precisely the obstruction $\kappa(h) \in H^3(M;\mathbb{Z}/2)$ to $h$ being covered by a map of vector bundles (or of PL-microbundles).

For $M = S^1 \times \mathbb{RP}^3$ we have a class $x \in H^3(M;\mathbb{Z})$ which is i) torsion and ii) reduces to $\bar{x} \neq 0 \in H^3(M;\mathbb{Z}/2)$. Looking at page 171 in Wall's book we find that $L_5(\mathbb{Z} \oplus \mathbb{Z}/2) = \mathbb{Z} \oplus \mathbb{Z}$ (when the orientation character is trivial) so we must have $\sigma(x)=0$, as $\sigma$ is a group homomorphism.

Thus there is a $[H : W \to M \times I] \in \mathcal{S}^{TOP}(M \times I, \partial)$ such that $\eta(H) = 0 \oplus x$, and the associated homeomorphism $h : M \to M$ has $\kappa(h) = \bar{x} \neq 0 \in H^3(M;\mathbb{Z}/2)$.

Finally, $\kappa(h \times h) = \bar{x} \times 1 + 1 \otimes \bar{x} \in H^3(M \times M ;\mathbb{Z}/2)$ is still non-zero, so $h$ is not isotopic to a diffeomorphism.

$\endgroup$
4
  • 5
    $\begingroup$ $TOP/O$ is not $6$-connected as $\pi_3(TOP/O)=\mathbb Z_2$. Rather $PL/O$ is $6$-connected and your nice argument works with $TOP$ replaced by $PL$. $\endgroup$ Apr 13, 2016 at 14:17
  • $\begingroup$ So it is. I suppose that breaks the argument, unless there is some reason the associated class in $H^3(X;\mathbb{Z}/2)$ must vanish. $\endgroup$ Apr 13, 2016 at 14:46
  • 2
    $\begingroup$ This is just a note to myself so I do not forget how the class in $H^3$ arises. When we try to homotope $d$ to a constant map the obstruction to building the homotopy on the $k$-skeleton of $X$ is in $H^k(X;\pi_k(Top/O))$. The only obstruction is in $H^3(X;\pi_3(Top/O))$, and if that obstruction vanishes the homotopy can be build. In particular your argument works if $H_1(X:\mathbb Z_2)=0$. I suspect in Ivan Smith's paper he talks about simply-connected $X$, so your argument applies in this case. $\endgroup$ Apr 13, 2016 at 15:03
  • $\begingroup$ Thanks for the answer! I think I will need some time to fully comprehend, so please forgive me that I don't accept yet... $\endgroup$
    – kosta
    Apr 14, 2016 at 21:38
6
$\begingroup$

The original paper of Ivan Smith assumes that the 4-manifolds are simply-connected. So this controls $H^3(X; \mathbb{Z}/2) \cong H_1(X; \mathbb{Z}/2) = 0$.

Possibly this gives an interesting invariant for non-simply-connected 4-manifolds in general?

$\endgroup$
1
  • $\begingroup$ Speaking of an invariant: to get off the ground one has to find a $4$-manifold $X$ such that the map $d: X\to TOP/O$ is not null-homotopic. Then one could look at pairs of $4$-manifolds $X_1$, $X_2$ and see whether the associated map $d_1\times d_2$ stays homotopically non-trivial when projected to $TOP/O$ via the Whitney sum. $\endgroup$ Apr 13, 2016 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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