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Let $M^n$ be a smooth closed, oriented $n$-manifold. Let $S_0,S_1\subset M^n$ two connected, compact and (positively) oriented submanifolds of $M$ of codimension $k$ diffeomorphic to $S$.

  1. Suppose $k=1$, $\partial S_i = \emptyset$, under which assumptions $S_0$ is isotopic to $S_1$? Is there a set of complete invariants, e.g. refining the homotopy class $[S_i] \in [S, M]$ ?

I am aware this is a general question. However, this should be easier than the usual knot-theoretical setting (i.e. $k=2$). For example, if $\dim M = 3$, and $\pi_1 S_0\hookrightarrow \pi_1 M$ then it's enough for $S_1$ to be homotopic to $S_1$ for being isotopic see Ian Agol's answer. References are welcome too.

  1. Suppose $k = 0$, $\partial S_i \neq \emptyset$. Under which assumptions $S_0$ is isotopic to $S_1$? What is an example when they are not isotopic? Is this case easier or more difficult than the $k=1$ case?

Here again, my hopes for a classification stems from Palais' theorem which asserts that if $S_i\simeq \mathbb{D}^n$ then $S_1$ is always isotopic to $S_0$. So in some degree of generality we have a nice answer.

I'm particularly interested in the following weaker case:

  1. Suppose $k=0$, and $S_i$ have diffeomorphic complement, i.e. $M\setminus \mathrm{int}(S_0)\simeq M\setminus \mathrm{int}(S_1)$, can we conclude that $S_1$ is isotopic to $S_0$?

I expect this extra hypothesis to be strong similarly to the case of knots in $\mathbb{S}^3$.

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  • $\begingroup$ The relevant codimension for studying embeddings is not the geometric dimension but the handle dimension (the minimum over all handle decompositions of the maximum of the indices of the handles). E.g. from the point of view of understanding embeddings of $D^n$, it behaves as if it were 0-dimensional. $\endgroup$
    – skupers
    Commented Dec 7, 2020 at 22:15
  • $\begingroup$ Thank you @skupers. I'm a bit confused though, for example, why we have that surfaces (handle dimension 2) in three manifolds are more easily understood up to isotopy than knots (handle dimension 1)? Also, if the codimension is big enough there is only one isotopy class regardless the handle dimension, (I'm thinking to the fact that all embedding of n-manifolds in $\mathbb{R}^{2n+1}$ are isotopic for $n>1$). That's why it seems to me that the codimension matters. How can I explain these with just the handle dimension? $\endgroup$
    – Roberto Ladu
    Commented Dec 7, 2020 at 22:29
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    $\begingroup$ @RobertoLadu The assumption of $\pi_1$-injectivity is doing a lot for you here. Embeddings of surfaces in $S^3$ are strictly more complicated than knot theory, as every embedding $T^2 \to S^3$ bounds a thickened knot on one side, and isotopic embeddings give rise to isotopic knots. Note that embeddings of genus 2 surfaces are more complicated than embeddings of genus 2 handlebodies: there are genus 2 surfaces in $S^3$ so that neither side of the complement is a handlebody $\endgroup$
    – mme
    Commented Dec 7, 2020 at 22:37
  • $\begingroup$ I am not saying that the only relevant information for deciding whether a space of embeddings is easy to understand is whether the handle dimension of the domain is small. However, it is the criterion for deciding whether embedding calculus applies and hence whether you can apply for example Haefliger's work on embeddings in the metastable range. In low codimensions you're probably better off using surgery (e.g. Chapter 11 of Wall's Surgery on compact manifolds). I think you should expect the cases where the target has low dimension to be different from the high-dimensional case. $\endgroup$
    – skupers
    Commented Dec 7, 2020 at 22:40
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    $\begingroup$ There are some cases when $n=3$ and isotopy holds without the incompressibility assumption, for instance when $M=S^3$ and complements to both $\partial S, \partial S'$ are handlebodies (this is Waldhausen's uniqueness theorem for Heegaard splittings). $\endgroup$
    – Moishe Kohan
    Commented Dec 8, 2020 at 1:22

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