Let $M^3$ be a compact $3$-manifold with $\partial M=N$ a connected surface. Suppose one has a smooth embedding of $N$ into the interior of $M$ and $N$ bounds a domain $D$ in $M$. Can we show that $D$ is homeomorphic to $M$?

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    $\begingroup$ This is certainly not true. One class of examples is given by the complements of satellite knots: en.wikipedia.org/wiki/Satellite_knot. These contain tori that are not parallel to the boundary, and indeed these tori bound a "deeper" knot complement. $\endgroup$
    – HJRW
    Jun 10, 2021 at 10:23
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    $\begingroup$ If you drop the dimensions down to $2$, it is also not true. There is however a way to phrase it for curves. $\endgroup$ Jun 10, 2021 at 15:23
  • $\begingroup$ This is true only if M is the 3-ball. $\endgroup$ Jun 11, 2021 at 15:21
  • $\begingroup$ @BrunoMartelli Why not for other handlebodies? $\endgroup$
    – Will Sawin
    Jun 20, 2021 at 23:57
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    $\begingroup$ You can embed a surface of genus >= 1 inside a ball in a strange way, so that it does not bound a handlebody in it. $\endgroup$ Jun 21, 2021 at 14:02

1 Answer 1


No, this need not hold.

Here is a concrete example. Suppose that $K \subset S^3$ is the trefoil knot in the three-sphere. It is an old result (certainly known to Alexander) that $K$ is not isotopic to the unknot. Let $N(K)$ be a small tubular neighbourhood of $K$, taken in $S^3$. So $N(K)$ is a solid torus. Let $n(K)$ be the interior of $N(K)$. We define $M = S^3 - n(K)$. This is a (compact) knot exterior. By the "old result" $M$ is not homeomorphic to a solid torus. Note that $T = \partial M = \partial N(K)$ is a two-torus.

Now choose an embedded loop $\alpha$ in $M$. Then $N(\alpha)$, a small tubular neighbourhood of $\alpha$, is another solid torus. Again $T' = \partial N(\alpha)$ is a two-torus. However, $N(\alpha)$ is not homeomorphic to $M$.

The comments above give many other examples. In general, the ways that a fixed surface embeds into a fixed three-manifold is an interesting question.


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