3
$\begingroup$

(Migrated from Math Stack Exchange)

A smooth embedding $f : D \to \mathbb{R}^3$ can be isotoped to a canonical inclusion $D \hookrightarrow \mathbb{R}^3$. (This is part of a proof that only the unknot has the disk as a Seifert surface.) A consequence of this is that two disjoint circles $C_1, C_2 \subset D$ cannot link in their image, that is, $f(C_1)$ is not linked with $f(C_2)$.

However if $f$ is an immersion then linking is enabled near a double curve: find two small disks inside $D$ whose images intersect along a double curve, then perturb them until the boundaries of those disks form a Hopf link.

The enablement of linking via an immersion seems interesting but is there any established theory on this that you could refer me to?

$\endgroup$
  • 1
    $\begingroup$ Here is a simple comment: each two-component link can be realized in this way. In fact, each knot in $S^3$ bounds an immersed disk with some clasp singularities. You can choose one clasp disk for each component respectively, and perturb them into general position. By taking a band sum of them you will obtain the desired immersed disk. For links with components more than 2 this is also correct. $\endgroup$ – Zhiyun Cheng Mar 20 '16 at 4:46
  • $\begingroup$ That sounds very interesting, do you have a reference for this construction? $\endgroup$ – Herng Yi Apr 4 '16 at 17:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.