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Let $\lambda_1,\lambda_2>1$. How to prove that the dilations $f_i:\mathbb R^n\to \mathbb R^n$, $f_i(x)=\lambda_ix$ for $i=1,2$ are conjugate? That is, how to prove there exists an homeomorphism $h:\mathbb R^n\to \mathbb R^n$ such that $h\circ f_1=f_2\circ h$?

It seems natural to set $h$ to be the identity on the unit sphere and $h(z)=\frac{\lambda_2}{\lambda_1}z$ for the points of norm $\lambda_1$. Then we get $h\circ f_1=f_2\circ h$ on the unit sphere. How to use this idea to define a good $h$ that works for all of $\mathbb R^n$?

Bruno Stonek
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One of the most straightforward way to construct conjugacy is the following. You take fundamental domains for both dilations, for example $H_1 = \lbrace 1 < \vert\vert x \vert \vert_2 \leqslant \lambda_1 \rbrace$ and $H_2 = \lbrace 1 < \vert\vert x \vert \vert_2 \leqslant \lambda_2 \rbrace$ for $f_1$ and $f_2$ respectively. You define homeomorphism $h^{*}$ from $H_1$ to $H_2$ as $x \mapsto \left (\frac{\lambda_1 - \lambda_2}{\lambda_1 -1} + \frac{\lambda_2-1}{\lambda_1 -1} \cdot \vert \vert x \vert \vert \right) \frac{x}{\vert\vert x \vert \vert}$. Then you extend it in following way: for each point $p \in \mathbb{R}^n\setminus{O}$ exists $k \in \mathbb{Z}$ such that $f_1^k(p) \in H_1$ (this is because $H_1$ is a fundamental domain); you map $p$ to a point $f_2^{-k}(h^{*}(f_1^k(p)))$. Then you extend this homeomorphism by continuity to a homeomorphism of $\mathbb{R}^n$ (by setting $O \mapsto O$). It's easy to show that constructed homeomorphism conjugates two dilations.

Evgeny
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  • Thanks! That's really what I was trying to do. Your signs are a bit odd (remember I imposed $\lambda_i>1$, what you wrote works for $\lambda_i<1$). How did you figure out how to define $h^*$? – Bruno Stonek Nov 12 '13 at 17:39
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    Yes, that's my fault, I'll correct it. As for $h^{*}$, this method works really with any homeomorphism from $H_1$ to $H_2$ that maps unit sphere to unit sphere and $\vert \vert x \vert\vert_2 = \lambda_1$ to $\vert\vert x \vert\vert_2 = \lambda_2$. I've chosen the most simple that acts as dilation (as @user72694 said, the problem is rotational invariant). – Evgeny Nov 12 '13 at 17:46
  • Thank you, it's all clear now. It's a nice idea to choose "fundamental domains" (hadn't thought of that in this context), define the homeos there, then extend as you do. I'll keep it in mind. – Bruno Stonek Nov 12 '13 at 18:04
  • You're welcome! It seems to be a powerful idea in this branch of dynamical systems (nonetheless it was got from group theory, I think). – Evgeny Nov 12 '13 at 18:13
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First take logs to reduce the problem to two translations $g_i:R\to R$, where $g_i(t)=t+\mu_i$. to show that these are conjugate, show that each one is conjugate to $g(t)=t+1$. This is done by conjugating by a scaling map with factor $\mu_i$.

Mikhail Katz
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  • Thank you for your answer. Could you elaborate a little bit, maybe expand on what you mean by "take logs to reduce the problem to..."? – Bruno Stonek Nov 12 '13 at 16:47
  • First, the problem is rotationally invariant, so it is really a question about $R^+$ rather than $R^n$. – Mikhail Katz Nov 12 '13 at 17:02
  • Applying log to the relation $f(x)=\lambda x$, we get $\log f = \log \lambda + \log x$. Letting $\mu =\log \lambda$, $t=\log x$, and $g(t)=\log f (e^t)$, we get the relation $g(t)= t+\mu$. – Mikhail Katz Nov 12 '13 at 17:03