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Let $M$ be a smooth oriented manifold, and let $M^E$ be an exotic copy, i.e homeomorphic but not diffeomorphic to $M$.

Is it true that $M\times M$ is diffeomorphic to $M\times M^E$?

I am interested in knowing the answer for closed manifolds.

One example which I can think of is an exotic $\mathbb R^4$. But we know that $\mathbb R^8$ has unique smooth structure. So in that case this is true.

One of the motivation for asking this question is following: I want to see an example of

symplectic manifold $\times$ non-symplectic manifold is symplectic manifold

. There are some manifolds whose some exotic structure doesn't admit symplectic structure. Also we know that Cartesian product of two symplectic manifold always admits symplectic structure. If the answer of my original question is positive, then that will imply the example I am looking for.

EDIT: A potential example of a symplectic manifold with non-symplectic exotic copy could be: Consider $E$ as elliptic surface. $F$ be fiber. Let $N_1= F\times D^2$. Consider $K$non-fibered knot is $S^3$ with non-monic $\Delta_k$. $T= K\times S^1$. And let $N_2=T\times D^2$ $\subset S^3\times S^1$. Them $E^E = (E-N_1)\cup (S3\times S^1-N_2)$ (gluing along meridian, sometimes it is called knot surgery). Then observe that in homology level we are not changing anything. So Freedman implies that it is homeomorphic with $E$. But Siberg-Witten invariant depends on Alexander polynomial of the knot. So thus we can conclude it has no symplectic structure. Similar ideas are written here https://arxiv.org/pdf/dg-ga/9612014.pdf

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  • $\begingroup$ I don't think there is a general theorem that an exotic factor has to disappear after squaring. This kind of matters are analysed by surgery theory. The problem might be computationally feasible for specific manifolds. What is the simplest closed symplectic manifold with an exotic non-symplectic structure? $\endgroup$ Aug 22, 2018 at 16:58
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    $\begingroup$ @IgorBelegradek: the simplest to state (maybe?) is $K3\#\overline{\mathbb{CP}}\vphantom{A}^2$, which is an exotic copy of $3\mathbb{CP}\vphantom{A}^2\#20\overline{\mathbb{CP}}\vphantom{A}^2$; the former is symplectic, the latter isn't. $\endgroup$ Aug 22, 2018 at 17:12
  • $\begingroup$ @IgorBelegradek: According to this paper, $\mathbb{CP}^2\# 2\overline{\mathbb{CP}^2}$ is another such example. $\endgroup$ Aug 22, 2018 at 17:17
  • $\begingroup$ @IgorBelegradek I have edited it. $\endgroup$ Aug 22, 2018 at 17:42

1 Answer 1

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Your question seems to be about simply connected exotic 4-manifolds, in which the answer is yes. That's because $M$ and $M^E$ are h-cobordant (by Wall), say via an h-cobordism W. Then $M \times W$ is an h-cobordism between $M \times M$ and $M\times M^E$, which is trivial by the high-dimensional h-cobordism theorem.

For higher dimensional manifolds, I don't think this is true. An example could come from $M = $ an exotic $\mathbb{C}P^4$. But I'd have to think about this for a bit.

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    $\begingroup$ I wonder if the obstruction classes to making lifts to $PL/O$ homotopic are in any reasonable sense additive under taking products of smoothable $PL$ manifolds (maybe up to filtration?). It seems not implausible that one could get many examples this way from $n$-tori. Of course, this isn't simply connected. $\endgroup$
    – mme
    Aug 22, 2018 at 19:17
  • $\begingroup$ That's a great answer. Which infact giving the answer for my second question as well. $\endgroup$ Aug 22, 2018 at 21:24
  • $\begingroup$ @MikeMiller I agree; the key point to such examples would be product formulas for various surgery-theoretic invariants. For example, if a homotopy $T^n$, say $T^{PL}$ has non-trivial Kirby-Siebenmann invariant then its product with any torus will as well so $T\times T$ and $T\times T^{PL}$ aren't PL homeomorphic. For the OP's purpose, you have to know if $T^{PL}$ is smoothable. $\endgroup$ Aug 23, 2018 at 11:37
  • $\begingroup$ (continued after a peek at Wall's book) Chapter 15.A of Wall's book shows that all PL tori are smoothable. There are many smooth structures $T^E$ lifting $T^{PL}$, but for all of them $T\times T$ and $T\times T^E$ aren't diffeomorphic because they're not even PL homeomorphic. $\endgroup$ Aug 23, 2018 at 11:49

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