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Unit ball diffeomorphism

Jakob

New member
Jan 11, 2014
23
Hello.

I have problems with the following exercise:

Let \(\displaystyle f\) be a diffeomorphism of a unit ball in \(\displaystyle \mathbb{R}^n\) such that \(\displaystyle f^2 = id\), and \(\displaystyle f=id\) on a certain neighbourhood of zero. Is \(\displaystyle f=id\) ?

I know it is true for \(\displaystyle n=1\). Then we deal with \(\displaystyle f: [-1,1] \rightarrow [-1,1]\) and \(\displaystyle f\) must send endpoints to endpoints and in fact it must fix them, because it is the identity on a neighbourhood of \(\displaystyle 0\). That and the fact that continuous functions map connected sets to connected sets imply that $f=id$

However, I don't know how to prove the result for higher dimensions.

Could you help me with that?

Thank you.
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Hello.

I have problems with the following exercise:

Let \(\displaystyle f\) be a diffeomorphism of a unit ball in \(\displaystyle \mathbb{R}^n\) such that \(\displaystyle f^2 = id\), and \(\displaystyle f=id\) on a certain neighbourhood of zero. Is \(\displaystyle f=id\) ?

I know it is true for \(\displaystyle n=1\). Then we deal with \(\displaystyle f: [0,1] \rightarrow [0,1]\) and \(\displaystyle f\) must send endpoints to endpoints and in fact it must fix them, because it is the identity on a neighbourhood of \(\displaystyle 0\). That and the fact that continuous functions map connected sets to connected sets imply that $f=id$

However, I don't know how to prove the result for higher dimensions.

Could you help me with that?

Thank you.
If $f^2 = id.$, then the (higher-dimensional) chain rule says that the derivative $Df$ satisfies $(Df)^2 = I$.

If a matrix $A$ satisfies $A^2=I$ then $0 = A^2-I = (A+I)(A-I)$. If $x$ is a nonzero vector in the range of $A-I$ it follows that $(A+I)x = 0$ and therefore $(A-I)x = 2x$. Thus $\|A-I\| \geqslant 2$. What that tells you is that if $A^2=I$ but $A\ne I$ then $A$ is not close to $I$. In other words, $I$ is an isolated point in the space of all matrices whose square is $I$.

Now come back to the problem about $f$. If $f = id.$ in a neighbourhood of $0$ then $Df=I$ in that neighbourhood. But $Df$ is a continuous function, and since $I$ is an isolated point in its range, it can never jump from $I$ to some other value. Therefore $Df=I$ throughout the unit ball and it follows that $f= id.$ everywhere.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Correction: After thinking about it some more, I don't believe that the above argument works. What the chain rule says is that $(Df)(f(x))(Df)(x) = I.$ In other words, the derivative is being evaluated at two distinct points. Of course, if $f(x)=x$ then $((Df)(x))^2 = id.$, but that is no help at all because if $f(x)=x$ then we already know that $(Df)(x) = id.$
 

Jakob

New member
Jan 11, 2014
23
I see. Maybe the result isn't true for higher dimensions?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I know it is true for \(\displaystyle n=1\). Then we deal with \(\displaystyle f: [0,1] \rightarrow [0,1]\) and \(\displaystyle f\) must send endpoints to endpoints and in fact it must fix them, because it is the identity on a neighbourhood of \(\displaystyle 0\). That and the fact that continuous functions map connected sets to connected sets imply that $f=id$
For n=1 the unit ball is actually [-1,1].
What is your reason to conclude that f must send endpoints to endpoints?
 

Jakob

New member
Jan 11, 2014
23
For n=1 the unit ball is actually [-1,1].
What is your reason to conclude that f must send endpoints to endpoints?
Thank you for pointing that out. I'll correct my first post.

If $f(-1) = \delta \in (-1,1)$, then $(-1, 1]$ would be mapped to $[-1, \delta) \cup (\delta, 1]$ which is impossible because continuous functions map connected sets to connected sets.

Is that correct?
 
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