We consider smooth embeddings of $S^1\sqcup S^2$ inside $S^4$. Let $f$ be such an embedding with the property that $f|_{S^2}$ is isotopic to the standard embedding, say in the first 3 coordinates. Let $\textrm{Emb}(S^1\sqcup S^2,S^4)_\textrm{std}$ denote the isotopy class of such embeddings as above. This is also an abelian group via embedded connect sum operation. [The notion of isotopy and ambient isotopy may not coincide as the codimension is $2$. Hence, the questions I am asking below are applicable for both notions, if they differ.]

(1) Is it true that $\textrm{Emb}(S^1\sqcup S^2,S^4)_\textrm{std}$ is infinite cyclic?

Remark: While reading Habegger's 1986 paper Knots & links in codimension greater than 2 (available here), it seems that one can deduce (1) via Theorem 1.2 although this is contradicted when one uses Corollary 1.3. This may be due to some typo's in the formula or lack of my understanding.

(2) If we remove the assumption of the embedding being standard on $S^2$, then we are led to consider $\textrm{Emb}(S^1\sqcup S^2,S^4)$. How much of the group structure of $\textrm{Emb}(S^1\sqcup S^2,S^4)$ is known in terms of $\textrm{Emb}(S^2,S^4)$?

(3) More generally, is it known that $\textrm{Emb}(S^1\sqcup S^{n-2},S^n)_\textrm{std}$ is infinite cyclic? [It seems to be the case based on Habegger's paper but I am stuck at the same confusion/typo as pointed above.] How much of $\textrm{Emb}(S^1\sqcup S^{n-2},S^n)$ can be captured by $\textrm{Emb}(S^{n-2},S^n)$?

Edit: As pointed out by D. Ruberman, $\textrm{Emb}(S^1\sqcup S^{n-2},S^n)$ need not be a group. However, even for $n=2$, I'm interested in seeking a characterization of the embeddings in terms of $\textrm{Emb}(S^2,S^4)$ and the linking number of $S^1$ and $S^2$ in $S^4$.


In dimensions at least 4, one has that homotopy of circles implies isotopy of circles. So if you fix an embedding of $S^2$ (eg the standard embedding), then you are really asking about embeddings of the circle in the complement of that fixed $2$-sphere, i.e. about the fundamental group of the complement. If you fix the standard embedding, then the fundamental group is $\mathbb{Z}$. (By the way, these groups are typically not abelian, so you have to fix a base point in order to get a group.) The same comments would apply if $4$ is replaced by $n$ and $n-2$ by $2$.

I don't really understand what you mean more generally about the group structure on $\operatorname{Emb}(S^1 \sqcup S^2,S^4)$. Beyond the need for a base point to use in multiplying circles by ‘connected sum’, there doesn't seem to be an inverse with respect to connected sum of $2$-spheres. It's different in Habegger's situation, because the codimension is bigger than two.

  • $\begingroup$ Thanks for clarifying the group structure part. Please have a look at my edit (in the question). The relevant parts of questions 2 & 3 should make sense now. $\endgroup$ Jun 19 at 17:28
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    $\begingroup$ As I wrote, there is in general no a priori group structure on this set. The best I can see is that for each isotopy class A of knotted $2$-spheres, the set of isotopy classes of links is in 1-1 correspondence with the set of free homotopy classes in $\pi_1(S^4 - A)$, which is not in any natural way a group. If the $S^2$ is unknotted, then this is the same as $\pi_1(S^4 - A) = \mathbb{Z}$. $\endgroup$ Jun 19 at 19:01

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