I'm interested in the following collection of questions: Let $S^n_k = \sqcup_k S^n$ be a disjoint union of $k$ distinct $n$-dimensional spheres. Write $Emb(S_k^n, S^{n+2})$ for the space of embeddings of these spheres into $S^{n+2}$. Pick your favorite embedding $e: S_k^n \to S^{n+2}$, and let $X_e = S^{n+2} \setminus im(e)$ be the complement of the image of the embedding.

  1. What is $\pi_1(Emb(S_k^n, S^{n+2}), e)$? Since this is probably unknown, what is known?

  2. How is this related to the mapping class group $\pi_0(Diff(X_e))$ of $X_e$?

I ask #2 because in dimension $n=0$, they are the same: the space of embeddings is the configuration space of points in the sphere. Its fundamental group is the (spherical) braid group, which is the same as the mapping class group of the punctured sphere. My guess is that life is not so simple in higher dimensions. Lastly, does any of this simplify out when you get into the range of dimensions where surgery theory starts working well?

  • 1
    $\begingroup$ For n=1, k=1 these fundamental groups based at each component were completely determined by Ryan Budney (who is almost sure to answer this question). $\endgroup$ – Dev Sinha Apr 18 '10 at 17:18

There is a locally-trivial fibre bundle

$$ Diff(S^{n+2}, L) \to Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n, S^{n+2})$$

here $Emb_L(\sqcup_k S^n, S^{n+2})$ is the component of the link $L$ you're interested in the full embedding space $Emb(\sqcup_k S^n,S^{n+2})$ and to simplify technicalities, assume $Diff(S^{n+2})$ is the group of orientation-preserving diffeomorphisms of $S^{n+2}$. The bundle is given by restricting a diffeomorphism of $S^{n+2}$ to $L$. $Diff(S^{n+2}, L)$ is the subgroup of $Diff(S^{n+2})$ which preserves the link $L$.

First observation is that the map $Diff(S^{n+2}) \to Emb_L(\sqcup_k S^n,S^{n+2})$ is null-homotopic. It's a simple argument -- isotope your link $L$ to sit in a hemi-sphere of $S^{n+2}$. Then apply a linearization process to linearize (simultaneously) all the diffeomorphisms of $S^{n+2}$ on that hemi-sphere. What I'm claiming is that $Diff(S^{n+2})$ has as a deformation-retract the subgroup that is linear on a fixed hemi-sphere -- so it gives a product decomposition $Diff(S^{n+2}) \simeq SO_{n+3} \times Diff(D^{n+2})$ (first observed by Morlet, or Cerf, I would guess) among other things.

So now you have a fibration:

$$\Omega Emb_L(\sqcup_k S^n, S^{n+2}) \to Diff(S^{n+2}, L) \to Diff(S^{n+2})$$

where the induced maps on homotopy groups are a short-exact sequence. In particular the fundamental group of your link space injects into $\pi_0 Diff(S^{n+2}, L)$, and its cokernel is precisely $\pi_0 Diff(S^{n+2})$. This group is frequently non-trivial as it is the group of exotic $n+3$-sphere provided $n \geq 3$.

$Diff(S^{n+2}, L)$ is somewhat closely related to $Diff(X_e)$, especially in high dimensions -- the spherical normal bundle to $L$ is particularly symmetric in low dimensions which causes trouble there. In general the sphereical normal bundle is equivalent to a disjoint union of $S^n \times S^1$, so to make the comparison between $Diff(S^{n+2}, L)$ and $Diff(X_e)$ you'd need to ask what kind of automorphisms $Diff(X_e)$ allows on the spherical normal bundle to $L$. There's probably a decent answer to that which doesn't take too much work but the above is a start.

edit: by Cerf's pseudoisotopy theorem, the kernel of the map $\pi_0 Diff(S^{n+2}, L) \to \pi_0 Diff(X_e)$ contains the exotic sphere "part" of $\pi_0 Diff(S^{n+2}, L)$.

  • $\begingroup$ This is perfect. Exactly what I was looking for. Thanks, Ryan. $\endgroup$ – Craig Westerland Apr 18 '10 at 22:10

In dimension $k=1$, i.e. for embeddings $\sqcup_k S^1\hookrightarrow S^3$, and using a totally unlinked embedding as your basepoint $e$, this is what's called the ring group or the loop group. It is closely related to the braid group and has been studied a ton, but two places with references you could follow are Brendle-Hatcher "Configuration spaces of rings and wickets" and Brownstein-Lee "Cohomology of the group of motions of n strings in 3-space".

The fundamental group $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)$ can been identified with McCool's "symmetric automorphism group". This is all the automorphisms of a free group $\langle x_1,\ldots,x_k\rangle$ which take each generator $x_i$ to a conjugate of some generator $x_j$. (A loop around one component of the link has to go to a loop around some component of the link.)

This is the image of $\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ in $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$, but since $S^3\setminus\sqcup_k S^1$ is not aspherical, this doesn't give us that $\pi_1(\text{Emb}(\sqcup_k S^1,S^3),e)=\pi_0(\text{Diff}(S^3\setminus\sqcup_k S^1))$ yet. I would be glad to see an argument that a diffeomorphism acting trivially on $\text{Aut}(\pi_1(S^3\setminus\sqcup_k S^1))$ must be isotopic (or even homotopic) to the identity.

  • $\begingroup$ Tom, do you know anything about the stability of the corresponding group as the dimension of the knots increases? Note that n-spheres are automatically unlinked for n>1. Thanks. $\endgroup$ – Craig Westerland Apr 19 '10 at 0:08
  • $\begingroup$ About your question -- since $S^3 - \sqcup_k S^1$ is a $K(F_k, 1)$ (for the free group $F_k$), every nontrivial self-map is determined by what it induces in $\pi_1$. So if homotopy = isotopy in this dimension (not obvious to me), then $pi_0(Diff(S^3 - \sqcup_k S^1))$ will embed in $Aut(F_k)$. $\endgroup$ – Craig Westerland Apr 21 '10 at 0:24
  • 1
    $\begingroup$ @Craig: I started to write down that argument, but I don't think $S^3-\sqcup_k S^1$ is aspherical (it's not a $K(G,1)$) for unlinked circles. For each circle, add back in all but one point: this gives a map $S^3-\sqcup_k S^1\to S^3-\sqcup_k \ast$. The latter is equivalent to a bouquet of 2-spheres, and this map seems to be surjective on $\pi_2$. $\endgroup$ – Tom Church Apr 21 '10 at 2:40
  • $\begingroup$ Good point! So perhaps it embeds in $$\prod_n Aut(\pi_n(S^3 \setminus \sqcup_k S^1))$$ That will be nonzero for infinitely many values of $n$, but I bet that your argument indicates that it is determined by $n=1, 2$. $\endgroup$ – Craig Westerland Apr 21 '10 at 4:21
  • $\begingroup$ More: I think you can adapt your argument to get a homotopy equivalence from $S^3 \setminus \sqcup_k S^1$ to a bouquet of $k$ circles and $k-1$ 2-spheres. Unfortunately, however, look at the universal covering of this space: it's an infinite $k$-valent tree with $k-1$ $S^2$'s wedged on at every vertex. So $\pi_2$ of this cover (and hence the space itself) is infinitely genereated! $\endgroup$ – Craig Westerland Apr 21 '10 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.