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Kopell's lemma states the following: suppose f and g are commuting $C^2$ diffeomorphisms of $[0, \infty)$ such that $f(x) < x$ for every $x > 0$ and $g(p) = p$ for some $p > 0$. Then $g$ is the identity. The idea goes like this: $g'(0) = 1$ since fixed points of $g$ accumulate on $0$. If $g$ were not the identity on the interval $(f(p), p)$, then in order for $g$ to be $C^1$ as $x \to 0$, $f$ would have to "stretch" the intervals $(f^{n + 1}(p), f^n(p))$ in such a way that $f'$ would have unbounded variation -- impossible since $f$ is $C^2$. (See Navas's book, Groups of Circle Diffeomorphisms.)

I am curious whether something similar holds in dimension two. For example, let $f, g\colon \mathbb{R}^2 \to \mathbb{R}^2$ be commuting $C^2$ diffeomorphisms such that $f(0) = g(0) = 0$. Suppose $f$ is a contraction near $0$; that is, there is an open neighborhood $U \ni 0$ such that $\overline{f(U)} \subset U$ and $\cap_{n \geq 0} f^n(U) = \{0\}$. Suppose there is a simple closed curve $c_0 \subset U$ around $0$ such that $c_0$ is fixed pointwise by $g$. Thus we have concentric topological circles $c_i = f^i(c_0)$ fixed pointwise by $g$. Since the $c_i$ approach $0$, we immediately see that $Dg_0 = Id$. What can we say about the behavior of $g$ on the annulus $A$ lying between $c_0$ and $c_1$?

Definitely, since $Dg_0 = Id$, $g|_A$ cannot rotate points around the annulus; the rotation number of every point will be $0$. This is simply because if some point $x$ did rotate around $A$ under $g$, then $f^n(x)$ would rotate around $f^n(A)$ for every $n$, but we know that $f^n(x) \to 0$.

Is it possible that there's some circle $c$ around $0$ lying between $c_0$ and $c_1$ such that $g(c)$ is inside $c$? I suspect that, as in the 1D case, this is ruled out by the assumption that $g$ is $C^1$ and $f$ is $C^2$, but I don't know how to prove it. Any other observations would be welcome -- thanks!

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