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Hello Everyone,

I have a question about smooth embeddings of spheres into larger spheres (by sphere I mean the usual, non-exotic kind, in case that makes a difference).

Let $h : S^n \rightarrow S^{n+k}$ be a smooth embedding (with $n,k > 0$). It's always possible to extend $h$ (or any continuous map from $S^n$ to $S^{n+k}$ for that matter) to a continuous map $H : D^{n+1} \rightarrow S^{n+k}$ defined on the closed ball $D^{n+1}$. For example, let $\overline{H} : S^n \times [0,1] \rightarrow S^{n+k}$ be any null-homotopy of $h$ and then let $H$ be the induced map on the quotient $D^{n+1} = S^n \times [0,1] / S^n \times \{ 1 \}$ (sorry if I'm being pedantic).

I'd like to know whether there exists such a continuous extension $H$ having a certain property, namely that

$H(D^{n+1} - S^n) \cap h(S^n) = \emptyset$

That is, the image of the restriction of $H$ to the ball $D^{n+1} - S^n$ should not intersect the image of the original embedding $h$. This is a weakening of the condition that $h(S^n)$ bound an embedded $n+1$-disk in $S^{n+k}$, since the restriction of $H$ to the ball $D^{n+1} - S^n$ needn't be injective.

Such an $H$ exists if $h$ is the standard embedding of $S^n$ into $S^{n+k}$, or more generally if the pair $(S^{n+k}, h(S^n))$ is homeomorphic to the standard pair $(S^{n+k}, S^n)$.

It follows that $H$ exists when $k=1$, since in this case the Generalized Schoenflies theorem provides a homeomorphism between $(S^{n+1}, h(S^n))$ and the standard pair $(S^{n+1},S^n)$ (this is because $h$ is smooth).

I'd like to know if anything can be said about the existence of $H$ when $k>1$. Sorry for the long preamble if this is totally trivial.

Thanks!

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    $\begingroup$ Perhaps take a look at D. Rolfsen, Knots and Links, in particular the intro to Chapter 2. I think such a continuous $H$ always exists for $k>2$, but there may not necessarily exist a smooth $H$. $\endgroup$
    – Mark Grant
    Commented Apr 22, 2012 at 6:32
  • $\begingroup$ Thanks Mark. Knots and Links had everything I needed, except of course for $k=2$. I'm studying a very specific class of embeddings; I'm hoping that $H$ exists for them even when $k=2$ but I guess won't get that for free. $\endgroup$ Commented Apr 25, 2012 at 22:02

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Your condition can be stated equivalently as the knot longitude, as an embedded $S^n$ in the knot complement $C_h = S^{n+k} \setminus h(S^n)$ is null homotopic.

Generally the answer is no when $k=2$:

  • When $n=1$ this follows from the loop theorem which you can find in the book Mark Grant cites.

  • For all $n$ your condition implies that the fundamental group of $C_h$ is infinite cyclic. Conjecturally this implies the knot is smoothly trivial, when $n=2$. In higher dimensions this is not quite enough for the knot to be trivial as the complement $C_h$ can have non-trivial higher homotopy groups. Try Googling "Classification of simple knots" with names like Kearton, Farber and Levine.

When $k>2$ my understanding is generally you can construct your continuous map $H$. For example, see Masamichi Takase The Hopf Invariant of a Haefliger Knot.

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  • $\begingroup$ Thanks Ryan. I should have emphasized that I only need my $H$ to be continuous. $\endgroup$ Commented Apr 25, 2012 at 22:02

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