Let $S$ be the suspension operator. Let $K1$ and $K_2$ be two knots in $S^3$ which are not equivalent. Does this imply that their suspensions in 4 sphere are not equivalent in the sense that their complements are not homeomorphic?
$\begingroup$
$\endgroup$
7
-
$\begingroup$ What do you mean by "suspension"? $\endgroup$– Ryan BudneyCommented Aug 24 at 7:36
-
$\begingroup$ He means “double cone”, I think. $\endgroup$– Sam NeadCommented Aug 24 at 7:37
-
2$\begingroup$ The answer is “the complement of the suspension in the four-sphere is homotopy equivalent to the complement of the knot in the three-sphere”. So the fundamental group (and a bit of discussion of peripheral subgroups) distinguishes. $\endgroup$– Sam NeadCommented Aug 24 at 7:41
-
2$\begingroup$ The title of the question is not great… you would avoid down votes (I think) if you made the title something like “Does suspension preserve the inequivalence of knots?” $\endgroup$– Sam NeadCommented Aug 24 at 7:42
-
1$\begingroup$ With this version of suspension you can recover the original knot by looking at a link of the cone point. $\endgroup$– Ryan BudneyCommented Aug 24 at 19:36
|
Show 2 more comments