I am having a hard time following this isotopy put forth by Milnor in On the Total Curvature of Knots

For each $c$ and $p$ in $\mathbb{R}^{n-1}$ such that $\|c-p\| > < r$, there is an isotopy, $f_u^{c\; > p} (\gamma), 0 \le u \le 1$, of $\mathbb{R}^{n-1}$ onto itself which transforms $c$ into $p$ and leaves fixed all points of $\mathbb{R}^{n-1}$ outside the $(n-2)$-sphere of radius $r$

This concept makes sense, but the paper continues saying:

For example,

$f_u^{c\; p}(\gamma) = \gamma - u ( 1 - \frac{\|\gamma - c\|}{r} )(p -c)$ for $\|\gamma - c\| \le r$

$f_u^{c\; p}(\gamma) = \gamma$ for $\|\gamma - c\| \ge r$

If you plug the value $\gamma = c$ into this where $u = 1$, you get

$c - (1)(1)(p-c) = 2c - p$

I thought the idea was to transform c into p?

It seems weird that you have the vector $p-c$ with some appropriate magnitude being subtracted instead of added.

Any thoughts on this would be really helpful,


  • 3
    $\begingroup$ I suspect there is simply a typo, and that the defintion of $f^{c p}_u(\gamma)$ should be $\gamma + u(\text{etc. ...})$. $\endgroup$
    – Emerton
    Dec 7 '10 at 5:56
  • $\begingroup$ I hope you're right $\endgroup$
    – Ori
    Dec 7 '10 at 6:04
  • $\begingroup$ Dear Ori, If you change the $-$ to a $+$ instead, doesn't everything then make sense? If so, then I'm right! $\endgroup$
    – Emerton
    Dec 7 '10 at 6:15
  • 1
    $\begingroup$ Emerton, you're right. In order to understand this isotopy, I strongly suggest drawing a picture of the function $\varphi(\gamma) = ( 1 - \frac{\|\gamma - c\|}{r} )$... $\endgroup$ Dec 7 '10 at 8:17

[I'm entering my comment above as an answer here, so as to stop this question being bumped to the front page in the future.]

There is simply a typo: the definition of $f_u^{cp}(\gamma)$ should be $\gamma + u$(etc. ...).


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