3
$\begingroup$

We know that $S^p$ and $S^q$ can be linked in $S^d$ if $p+q<d$. Let us consider the simplest case where both $S^p$ and $S^q$ are un-knotted spheres.

I am looking for a surgery to unlink $S^p$ and $S^q$ in $S^d$.

Let us start with an example,

(0) The simples process can be done to unlink a Hopf link of a $S^1$ and a $S^1$ linked in $S^3$. In that case, we cut out a $D^2 \times S^1$ solid torus bounding one of the $S^1$ unknot. And we glue the $D^2 \times S^1$ with the remaining $D^2 \times S^1$ to form a $S^2 \times S^1$, Therefore, we successfully unlink two unknots $S^1$ and $S^1$ in the new 3-manifold $S^2 \times S^1$.

(I) Next, say $S^1$ and a $S^2$ are linked in $S^4$. Can I unlink the $S^1$ and $S^2$ by doing some surgery (with certain constraints described below, and let us say both $S^1$ and $S^2$ maintain unknotted)?

Can I do the surgery to first like the above case, say (1) cut out a thin tube $X$ around $S^1$, and then (2) modify this $X$ under some transformation $\eta$ (like any diffeomorphism), and then (3) re-glue it back to make it form a new manifold $M_a$ such that $M_a$ has no boundary: $\partial M_a =0$, and such that $S^1$ and a $S^2$ are eventually unlinked in $M_a$? How to do that? What is $X$, $\eta$ and $M_a$?

Here $X \cup (S^4-X)=S^4$, and $X \cup_{\eta} (S^4-X)=M_a$. Here $\eta$ means possible transformation when gluing two manifolds together. The $X$ may be $D^3 \times S^1$ or $D^2 \times S^1 \times S^1$, or whatever contains the $S^1$ in the interior of $X$.

(II) How about more general cases? To unlink $S^p$ and $S^q$ in $S^d$ by a similar surgery?

p.s. knowing the answer for (I) is good enough to count an answer. p.s. this is a generalized version of a question, rooted in the post from ME with no feedback.

EDIT: Following Kevin Walker suggestion, let me tentatively define,

$\bullet$ $S^p$ and $S^q$ are linked in $M^d$, if and only if their filled ${p+1}$-ball $D^{p+1}$ and ${q+1}$-ball $D^{q+1}$ (bounded by the surface of $S^p$ and $S^q$ respectively) always intersect with each other in $M^d$.

-

$\bullet$ $S^p$ and $S^q$ are not linked in $M^d$, if and only if there exists some filled ${p+1}$-ball $D^{p+1}$ and ${q+1}$-ball $D^{q+1}$ (bounded by the surface of $S^p$ and $S^q$ respectively) such that $D^{p+1}$ and $D^{q+1}$ do not intersect with each other in $M^d$.

$\endgroup$
11
  • 1
    $\begingroup$ What do you mean when you say two spheres in an ambient manifold are "unlinked"? $\endgroup$ Apr 24, 2015 at 19:45
  • $\begingroup$ Dear Ryan: I mean the case just analogous to the $S^1$ and the $S^1$ becomes unlinked in $S^2 \times S^1$. Start from two $S^1$ linked in $S^3$, by cutting out a solid torus $D^2 \times S^1$ bounded one of $S^1$, and reglue this $D^2 \times S^1$ back via SL$(2,\mathbb{Z})$ transformation $\phi$, this $(D^2 \times S^1) \cup_\phi (S^1 \times D^2)=S^2 \times S^1$ will make the two $S^1$ unlinked. In the original $(D^2 \times S^1) \cup (S^1 \times D^2)=S^3$, the two $S^1$ were set to be linked. Now the two $S^1$ are unlinked. Is that clear? (I am sure what I am talking about.) Thanks. $\endgroup$
    – miss-tery
    Apr 24, 2015 at 20:33
  • 1
    $\begingroup$ I suspect your use of the terms "linked" and "unlinked" is non-standard. You should probably say very carefully and clearly what you mean by these two terms. i.e. "Definition: Two submanifolds of $M$ are unlinked if and only if ...". $\endgroup$ Apr 24, 2015 at 22:30
  • $\begingroup$ Dear Kevin Walker: I have no better way to define "linked" and "unlinked," other than saying that it is the same definition as the previous example for $S^3$. But let me give it a try... $\endgroup$
    – miss-tery
    Apr 25, 2015 at 0:51
  • 1
    $\begingroup$ If I understand your example (0) correctly, then the two copies of $S^1$ inside $S^1 \times S^2$ (the result of the surgery on $S^3$) do not bound disks, and therefore are not unlinked according to your definition. $\endgroup$ Apr 25, 2015 at 1:14

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.