hello,


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,


Thanks



isotopy doesn't make sense (Milnor)
[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. ...).


