I just seen De Michelis and Freedman's paper Uncountably many exotic $\mathbf{R}^4$'s in standard 4-space, J. Differential Geometry 35 (1992) pp 219-254, doi:10.4310/jdg/1214447810.

If I understand correctly, they said there are many small exotic $\mathbb{R}^4_\xi$ manifolds which can embedding into a standard $\mathbb{R}^4$.

I can not catch all the ideas in their paper, but I would very like to know the question:

Can two small exotic $\mathbb{R}^4_\xi$ and $\mathbb{R}^4_{\xi'}$ manifolds which embedding in a standard $\mathbb{R}^4$ be combined (annihilated) each other and become a global standard $\mathbb{R}^4$ manifold?

Or can a global standard $\mathbb{R}^4$ manifold be continuously deformed to create two small exotic smooth $\mathbb{R}^4_\xi$ and $\mathbb{R}^4_{\xi'}$ manifolds?

  • $\begingroup$ If you can embed in $\mathbb R^4$, then can’t you embed in a half space, and glue two half spaces together? Is this what you want or something else? $\endgroup$
    – Will Sawin
    Commented Oct 21, 2019 at 13:23
  • $\begingroup$ Thanks for your help. Let me use other way (maybe imprecisely) to present my question. Do exist $\mathbb{R}^4_\xi$ and $\overline{\mathbb{R}^4_\xi}$ such that $\mathbb{R}^4_\xi \cup \overline{\mathbb{R}^4_\xi} = B^4 \hookrightarrow \mathbb{R}^4$ ? ( Where the $B^4$ with standard smooth structure. ) $\endgroup$ Commented Oct 21, 2019 at 14:12
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    $\begingroup$ I think the OP means 'can the connect sum of two exotic R^4 be diffeomorphic to R^4?'. Certainly the same is true of exotic 7-spheres, which form an order-28 Abelian group under connect sum (with S^7 being the identity, of course). $\endgroup$ Commented Oct 21, 2019 at 14:28
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    $\begingroup$ You should probably be informed that these exotic $\mathbb{R}^4$ look like open sets with very fractal-like boundary. So, it is hard to say what is ever continuously deforming such a set. Your first question might make sense - you could probably consider subsets of $\mathbb{R}^4$ which admit non-empty smooth part of the boundary along which it is smoothly isomorphic to the half space. We could glue them together along (small open disks in) these boundaries which will be roughly the operation of connective sum you want. There are probably some subtle details though. $\endgroup$ Commented Oct 21, 2019 at 15:31

1 Answer 1


If the "combining" you're referring to is the end-sum, it turns out that no such exotic $\mathbb{R}^4$'s exist. This was actually shown by Gompf in the appendix to "An infinite set of exotic $\mathbb{R}^4$'s" (Journal of Differential Geometry 1983). The basic idea is to suppose $R_1 \natural R_2 = \mathbb{R}^4$ and use the Eilenberg swindle to get

$$R_1 = R_1 \natural (\natural_{i=1}^\infty \mathbb{R}^4) = R_1 \natural (\natural_{i=1}^\infty (R_2 \natural R_1)) = R_1 \natural R_2 \natural R_1 \natural R_2 \dots = \mathbb{R}^4 \natural \mathbb{R}^4 \natural \dots = \mathbb{R}^4 $$

[Edit: old note] I think any such exotic $\mathbb{R}^4$ would have to be standard at infinity. To see this, suppose two exotic $\mathbb{R}^4$'s $R_1$ and $R_{2}$ have end sum $R_1 \natural R_2$ diffeomorphic to the standard $\mathbb{R}^4$. The end sum $R_1 \natural R_2$ is constructed by taking smoothly properly embeddings of rays $\gamma_i: [0, \infty) \rightarrow R_i$. We then take tubular neighborhoods of ray which will be diffeomorphic to $[0,\infty) \times \mathbb{R}^3$. Delete each of these tubular neighborhoods to get $U_i \subset R_i$. We then glue $U_1$ and $U_2$ together along the new boundary to get $R_1 \natural R_2$. When Gompf introduced this in his aforementioned paper, he showed that this produces a well defined smooth manifold homeomorphic to $\mathbb{R}^4$.

If $R_1 \natural R_2$ is diffeomorphic to $\mathbb{R}^4$, then there is a neighborhood of infinity $V \subset R_1 \natural R_2$ that is diffeomorphic to a neighborhood of infinity of $\mathbb{R}^4$, namely $S^3 \times \mathbb{R}$. This induces a diffeomorphism with a neighborhood of infinity of each $U_i$ with a neighborhood of infinity of $\mathbb{R^3} \times (-\infty,0]$. We can then glue the neighborhood of the rays $\gamma_i$ back in. This will induce a diffeomorphism of the neighborhoods of infinity of $R_i$ with a neighborhood of infinity of $\mathbb{R^3} \times (-\infty,0]$ glued with $[0,\infty) \times \mathbb{R}^3$. This will be the standard $\mathbb{R}^4$ and so $R_i$ is diffeomorphic to $\mathbb{R}^4$ at infinity.

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    $\begingroup$ Your implication "exotic S4 to exotic R4" is not correct, it's a subtle issue that reduces to the smooth Schoenflies conjecture (I've chatted with Gompf about precisely this for one of my research projects). That is, there could possibly exist a standard R4 with an exotic collar at infinity. In your proposed argument, the collar is "standard" from the perspective of the disk but this would not be isotopic to the "standard" collar if we are diffeomorphic to standard R4 (so if we have an exotic S4 it could be possible that we get a standard open 4-ball whose closure is an exotic 4-ball). $\endgroup$ Commented Oct 22, 2019 at 3:01
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    $\begingroup$ Also your proof of the reverse implication is not quite correct (but the statement is correct): You're not really plugging in a point, you're gluing on a 4-ball. Then instead of defining 𝜙′ by removing a point, you need to remove that 4-ball, and this is where you then need to argue that 𝜙′(4-ball) is ambient-isotopic to a standard 4-ball in S4 (which in turn is a theorem of Palais-Cerf). $\endgroup$ Commented Oct 22, 2019 at 4:44
  • $\begingroup$ Thanks all! I think this almost answers my question. It seems not easy to combine two exotic's into one standard. I optimistically hope there is other operation can do it. I will still think about it. $\endgroup$ Commented Oct 22, 2019 at 13:11

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