Diffeomorphisms of $(-1,1)^n$ sending fixed point to origin

Question

Consider $I^n\equiv(-1,1)^n:=\{(x_1,\dots,x_n)\in \mathbb{R}^n: |x_i|<1 \ \text{for each} \ 1\leq i\leq n\}.$ Let's fix a point $a=(a_1,\dots,a_n)\in I^n$.

Question: How to construct a $C^1$-diffeomorphism $\Phi:I^n\to I^n$ such that $\Phi(a)=\vec{0}$?

I believe it suffices to construct a diffeomorphism $f:I\to I$ such that $f(a)=0$, where $a$ is some fixed point in $I$.

I tried to come up with some constructions but they did not work out. I know usually one needs to show some efforts but in this case I don't have anything to show.

Answer 1

Hint Can you find a diffeomorphism $\Psi : I^n \to \Bbb R^n$? If so, let $T : \Bbb R^n \to \Bbb R^n$ be, e.g., the translation $x \mapsto x + \Psi(0) - \Psi(a)$. Then, $\Phi := \Psi^{-1} \circ T \circ \Psi$ is a diffeomorphism $I^n \to I^n$, and by construction $\Phi(a) = 0$.

Remark By the way, like you suggest it's enough to solve this problem for $n = 1$ and deal with each coordinate separately: Given $a = (a_1, \ldots, a_n) \in I^n$, if we can find diffeomorphisms $\Phi_{a_i} : I \to I$, $i = 1, \ldots, n$, respectively satisfying $\Phi_{a_i}(a_i) = 0$, then $$\Phi(x_1, \ldots, x_n) := (\Phi_{a_1}(x_1), \ldots, \Phi_{a_n}(x_n))$$ is a diffeomorphism $I^n \to I^n$ satisfying $\Phi(a) = 0$.

Answer 2

A very simple way to do this is just by polynomial interpolation. It's a little more convenient to look for $f:I\to I$ with $f(0)=a$ rather than $f(a)=0$ (to get $f(a)=0$, then, you can just take the inverse). To get $f:I\to I$ with $f(0)=a$, you can find a polynomial $f$ with $f(-1)=-1$, $f(1)=1$, and $f(0)=a$. As long as $f'>0$ on $I$, this will be a diffeomorphism (not just $C^1$ but $C^\infty$, or in fact analytic).

The simplest case to try is a quadratic. Note that the required values for $f$ mean that $1$ and $-1$ must be the roots of $f(x)-x$, so $f(x)$ will be $-a(x^2-1)+x$. The derivative is then $f'(x)=-2ax+1$, which is positive on $I$ as long as $|a|\leq 1/2$.

So, this solves the problem for $|a|\leq 1/2$. For $|a|>1/2$, you can use a trick: just compose $f$ with itself to make the value at $0$ get further and further away from $0$. Let $f_c(x)=-c(x^2-1)+x$ and consider the $n$-fold composition $f_c^n$, which will still be a diffeomorphism $I\to I$ if $|c|\leq 1/2$. If $c>0$, then $f_c(x)>x$ for all $x$, so $f_c^n(0)$ is an increasing sequence. The limit of this sequence will be a fixed point of $f_c$ by continuity, which can only be $1$. So in particular, $f_{1/2}^n(0)$ gets arbitrarily close to $1$ as $n\to\infty$ if $c>0$. On the other hand, $f_0(x)=x$ so $f_0^n(0)=0$ for all $n$. Since $f_c$ is continuous in $c$, this means that the values of $f_c^n(0)$ cover every value between $0$ and $1$ as $n\to\infty$ and as $c$ varies between $0$ and $1/2$. So, this solves the problem for all positive $a$, and negative $a$ can be handled similarly by taking $f_c^n$ for $c<0$.