Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $f$ is a diffeomorphism of $\mathbb R^n$ and $K$ is a compact set in $\mathbb R^n$, can we find another diffeomorphism $\tilde f$ of $\mathbb R^n$ such that:

(1)$f=\tilde f$ on a neighborhood of $K$. (2)There is a bounded set $V$ and $\tilde f=id$ outside $V$?

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

The function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=-x$ and $K=[-1,1]$ provides a counterexample - any continuous map $\tilde{f}$ satisfying properties 1) and 2) would necessarily fail to be injective by the intermediate value theorem.

As Jason DeVito points out below, we can use a similar setup to create a counterexample for any $n$. Letting $f:\mathbb{R}^n\to\mathbb{R}^n$ be defined by $f(x_1,x_2,\ldots,x_n)=(-x_1,x_2,\ldots,x_n)$, we have for all $p\in\mathbb{R}^n$ $$\det(df_p)=\begin{vmatrix} -1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 \end{vmatrix}=-1.$$ Thus, any diffeomorphism $\tilde{f}$ satisfying 1) would also have to have $\det(d\tilde{f}_p)=-1$ for some $p\in\mathbb{R}^n$. Because $\tilde{f}$ is a diffeomorphism, it can't have $\det(d\tilde{f}_p)=0$ for any $p\in\mathbb{R}^n$. Because $\det(d\tilde{f}_p)$ varies continuously with $p$, we must have $\det(d\tilde{f}_p)<0$ for all $p\in\mathbb{R}^n$. But $\det(dI_p)=1>0$ for all $p\in\mathbb{R}^n$, so we can't have $\tilde{f}=I$ anywhere.

share|improve this answer
3  
Your answer can be promoted to $n$ by the same kind of reasoning. Pick your favorite orientation reversing diffeomorphism $f$ and let $K$ be a cube. Then $f:K\rightarrow f(K)\subseteq \mathbb{R}^n$ reverses orientation, so the determinant of the differential is negative. Since the determinant is always nonsingular, it must, thus, be negative everywhere. Hence, $\tilde{f}$ can't be the identity outside $V$. –  Jason DeVito Apr 28 '12 at 3:41
    
+1 Excellent point! You are certainly welcome to post that as a separate answer, or I can incorporate it into my post. Hmm, now I wonder if the statement becomes true if we require $f$ to be orientation-preserving, then. –  Zev Chonoles Apr 28 '12 at 3:45
    
A modified form of the function you gave should show that it doesn't work in $\mathbb{R}^n$, either. I think $f(x',x_n) = (x', -x_n)$, on the box $[-1,1]^n$ is a counterexample. Maybe we could also pick $f(x) = -x$ on $[-1,1]^n$, but I think this requires that $n$ is odd. –  Nicholas Stull Apr 28 '12 at 3:49
    
@Zev: It's all your idea, feel free to edit it in if you wish. –  Jason DeVito Apr 28 '12 at 3:50
    
If $f$ is orientation-preserving and there is a $\epsilon>0$ such that $|f-id|<\epsilon$, then the conclusion holds. –  Hezudao Apr 28 '12 at 12:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.