# Does every embedded 2-sphere in $\mathbb{R}^n$ bound an embedded ball?

Fix $n \geq 3$ and let $S \subset \mathbb{R}^n$ be a smoothly embedded $2$-sphere. Must there exist a smoothly embedded $3$-ball $B \subset \mathbb{R}^n$ such that $\partial B = S$? This is true for $n=3$ by the Schoenflies theorem. Also, it is true for $n \geq 7$; indeed, since $\mathbb{R}^n$ is contractible we can find a continuous map $B \rightarrow \mathbb{R}^7$ taking $\partial B$ homeomorphically onto $S$, and by Whitney's theorem we can jiggle this continuous map a little to make it an embedding. This might also work for $n=6$, but the homotopy to make $B$ embedded might be complicated and mess up $S$ (I don't know the details of Whitney's strong embedding theorem), so I'm not sure.

What about $4 \leq n \leq 6$?

• Doesn't the existence of non-trivial $2$-knots imply the answer is no for $n =4$? – Mark Grant Jan 9 '15 at 6:59
• Try looking at Kawauchi's "A survey of knot theory" or Rolfsen's "Knots and links". Any book on knot theory that doesn't restrict to three dimensions. Hillman's book "Algebraic Invariants of Links" would also suffice. Perhaps math.stackexchange.com would be a better venue for this question. – Ryan Budney Jan 9 '15 at 20:49

Haefliger proved that $S^k$ in $\mathbb{R}^n$ is unknotted if $2n > 3(k+1)$. Therefore, any 2-sphere embedded in $\mathbb{R}^n$ with $n\ge 5$ is unknotted. On the other hand, as Mark Grant points out in his comment, there are knotted 2-spheres in $\mathbb{R}^4$ (see this paper by Andrews and Curtis, for example).