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.

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    $\begingroup$ It is trivially true for 3-manifolds, because they have exactly one smooth structure. $\endgroup$ – ThiKu Apr 12 '16 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 '16 at 17:21
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    $\begingroup$ This is more-or-less completely answered by this question, given that smoothing theory works in dimension 8. $\endgroup$ – Mike Miller Apr 12 '16 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 '16 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. $\endgroup$ – Mike Miller Apr 12 '16 at 22:39

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.

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    $\begingroup$ $TOP/O$ 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 $PL$. $\endgroup$ – Igor Belegradek Apr 13 '16 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. $\endgroup$ – Oscar Randal-Williams Apr 13 '16 at 14:46
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    $\begingroup$ 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. $\endgroup$ – Igor Belegradek Apr 13 '16 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 '16 at 21:38

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?

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  • $\begingroup$ 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. $\endgroup$ – Igor Belegradek Apr 13 '16 at 15:48

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