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I am given a 2-disc $D^2$ embedded into $\Bbb R^4$, that is, I have an injective continuous map $\phi:D^2\to\Bbb R^4$. I want to "double" this disc in the sense that I am looking for a second embedded disc $\smash{\psi:D^2\to\Bbb R^4}$ that agrees with $\phi$ on the boundary $\smash{\partial D^2 = S^1}$ but whose image is otherwise disjoint from the image of $\phi$.

Question: Can I always find such a second embedded disc?

If it helps, we can assume that $\phi$ is piece-wise linear, but then $\psi$ should be as well (in fact, Will's comment shows that we should probably work in a category that does not contain an equivalent of Alexander's horned sphere).

I also believe this is equivalent to asking whether every embedding of the northern hemisphere $\subset S^2$ extend to an embedding of the full sphere.

If $\phi$ were differentiable ...

... (at least in the interior of $D^2$) then I believe we can choose a continuously varying normal vector $n:\mathrm{int}(D^2)\to\Bbb R^4$ at each interior point of the disc and define

$$\psi(x):=\phi(x)+\epsilon(x)n(x),$$

where $\epsilon(x)$ is positive but sufficiently small on $\mathrm{int}(D^2)$ and tends to zero as $x$ approaches $\partial D^2$ (so I don't care what $n(x)$ is on the boundary).

But I do not want to assume differentiability and so I have no idea for how to choose the normal vector at each point.

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  • $\begingroup$ I suppose the answer is no in $\mathbb R^3$: Start with a disc, and at two distinct points in the interior, cut out a small hole and glue on a long tube (one pointing toward the "heads" side of the disc, one toward the "tails" side), then wrap the tubes around the edge of the disc until you can grow them together in the style of the Alexander horned sphere. (Sorry if this makes no sense at all.) You asked about $\mathbb R^4$, so this doesn't answer your question. But have you looked at generalizations of the horned sphere to dimension $4$? $\endgroup$
    – Will Brian
    Oct 22, 2021 at 20:12
  • $\begingroup$ Do you want the images of $\phi$ and $\psi$ to be close? $\endgroup$
    – Shijie Gu
    Oct 22, 2021 at 21:14
  • $\begingroup$ If you assume $\phi$ is locally flat you can use the argument in the smooth case by the existence of normal microbundles (for example proven in Freedman-Quinn's Topology of 4-manifolds). $\endgroup$
    – skupers
    Oct 22, 2021 at 22:41
  • $\begingroup$ @WillBrian My visualization skills are failing me - how do you wrap the tubes around the edge of the disc without breaking injectivity? $\endgroup$
    – Steven Stadnicki
    Oct 22, 2021 at 22:50
  • $\begingroup$ @WillBrian Because I am worried about such pathologies, let's say we stay in PL (as far as I know the horned sphere does not work there). $\endgroup$
    – M. Winter
    Oct 22, 2021 at 23:02

2 Answers 2

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Without PL structure and smoothness, the counterexamples are constructed by Bob Daverman. That is, there exist wildly embedded 2-disks in $\mathbb{R}^n$ ($n\geq 4)$. In fact, those disks have non-simply connected complement in $\mathbb{R}^n$. See

  1. On the absence of tame disks in certain wild cells. Geometric topology (Proc. Conf., Park City, Utah, 1974), pp. 142–155. Lecture Notes in Math., Vol. 438, Springer, Berlin, 1975. MR0400236
  2. On the scarcity of tame disks in certain wild cells. Fund. Math. 79 (1973), no. 1, 63–77. MR0326742
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    $\begingroup$ I'm probably being thick here, but I don't understand if the disks constructed by Daverman are so wild that no other disk can coexist with them. $\endgroup$
    – PseudoNeo
    Oct 23, 2021 at 7:32
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    $\begingroup$ @PseudoNeo: I think Daverman's technique allows one to create wildness on the boundary of the disk. One may use certain sticky Cantor sets so that it cannot be pushed off itself via a small ambient isotopy. If you are worried about the area of the wildness is not large enough, one may use suspension to produce everywhere wild 2-disk in S^n$ ($n\geq 4$). That is, using a (wild) arc A on the Bing sling (an everywhere wild 1-sphere), the suspension of it yields an everywhere wild 2-disk. This utilizes the shrinking theorem of J. J. Andrews and M. L. Curtis. $\endgroup$
    – Shijie Gu
    Oct 23, 2021 at 16:55
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Yes, let us assume that $D$ is PL.

Let us move all inner vertices of $D$ randomly a bit. Denote by $D'$ the new disc; it might intersect $D$ at a collection of isolated points $p_1,\dots,p_n$; we can assume that no $p_i$ belongs to 1-skeleton. For each $p_i$ choose a polygonal path $\gamma_i$ to $\partial D$ in $D$; we may assume that $\gamma_i$ do not intersect each other and they do not pass thru the vertices.

Note that one can remove the intersection point $p_i$ by modifying $D'$ in a small neighborhood of $\gamma_i$ (a simplified version of the Whitney trick).

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  • $\begingroup$ Thank you very much for the this answer! Do you know where to read about how this can be made rigorous? Also, why are the $p_i$ better not in the 1-skeleton and why do the $\gamma_i$ better not pass through vertices? $\endgroup$
    – M. Winter
    Oct 25, 2021 at 22:49
  • $\begingroup$ @M.Winter edges do not create a problem, but vertices might be problematic and generic condition kills both. No, I do not know a directly related text, but it should be easy to follow once you understand Whitney trick. $\endgroup$
    – Anton Petrunin
    Oct 26, 2021 at 4:35
  • $\begingroup$ I checked some sources regarding the Whitney trick, and they seem to require a smooth embedding and dimension $\ge 5$. Is this not necessary here for some reason? $\endgroup$
    – M. Winter
    Oct 28, 2021 at 20:59
  • $\begingroup$ @M.Winter not here --- we are in a much simpler situation --- mainly an embedded disc comes for free. $\endgroup$
    – Anton Petrunin
    Oct 28, 2021 at 23:21
  • $\begingroup$ Sorry to come back to this so late, but after thinking a bit I have doubt that "an embedded disc comes for free". At least it is not obvious to me. For example, what if the 2-disc is a cone over a (non-slice) PL-knot in $S^3$? Can you elaborate a bit? $\endgroup$
    – M. Winter
    Mar 4, 2022 at 19:37

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