I keep running across this statement, but how does one show that the only knot that bounds a disk is the trivial knot?
EDIT: I am seeking a more rigorous proof.
Asked
Active
Viewed 480 times
7
-
2Take a small isotopy to ensure that the knot is a smooth embedding, then take an isotopy that looks like $f_t(x) = f(tx)/t$, using smoothness to extend it across the entire disk. – anomaly Feb 12 '18 at 04:11
-
1@anomaly Is $f$ describing the isotopy of the smoothly embedded knot to the standard unknot in $S^{3}$? – Michael Shumate Feb 12 '18 at 14:38
-
Yes.$\phantom{}$ – anomaly Feb 12 '18 at 16:31
-
1How does smoothness allow the appropriate extension? – Michael Shumate Feb 12 '18 at 16:45
-
What happens at time $t = 0$? – anomaly Feb 12 '18 at 17:31
-
1@anomaly Why is $f_t$ a knot? – PVAL-inactive Feb 12 '18 at 20:12
-
Why is $f_t$ an ambient isotopy? Just write out each $f_t:S^1 \to S^3$ for fixed $t$. – anomaly Feb 12 '18 at 20:54
-
@MichaelShumate The proof of the more general statement is a three page paper by Palais available here ( http://www.ams.org/journals/proc/1960-011-02/S0002-9939-1960-0117741-0/S0002-9939-1960-0117741-0.pdf ). If you are interested I suggest you read it. – PVAL-inactive Feb 12 '18 at 21:18
-
1@anomaly I don't understand. If we have the convention $f(1)=f(0)$ for instance , I don't think $f_t$ is ever closed for $t<1$. – PVAL-inactive Feb 12 '18 at 21:23
-
1I'm being lazy about writing out the details of $f$. The point is that this just Alexander's trick, once you manage to get the embedding to be $C^1$. – anomaly Feb 12 '18 at 22:35