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Suppose $U \subseteq \mathbb{R}^m$ is some connected open subset and let $f: U \to U$ be a differentiable map with $f^n = \text{id}$ for some $n\in\mathbb{N}$. Let $\text{Fix}(f)=\{x\in U: f(x) = x\}$. Suppose that $\text{Fix}(f)$ contains some non-empty open set. Is it true, that $f$ has to be the identity map?

If $U$ is any topological space and $f$ only continuous this would be false in general. For example take a plus $+$ as topological space and reflect it across the horizontal axis. Under which conditions on $X$ would this be maybe true for topological spaces?

psl2Z
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  • Hint: https://math.stackexchange.com/questions/55376/fixed-points-set-of-an-isometry – Moishe Kohan Apr 23 '23 at 23:48
  • @MoisheKohan If we assume $f$ is $C^{\infty}$, maybe, we could introduce a Riemannian metric $g := \sum_{k=1}^n (f^k)^h$, where $h$ is some metric on $U$. Then $f^g=g$, because $f^n=\text{id}$, so $f$ is an isometry. Then a connected component $V$ of $\text{Fix}(f)$ is a closed, totally geodesic submanifold without boundary, and contains w.l.o.g. by assumption an open subset of $U$. Therefore $V$ is open (submanifold without boundary, $\dim V = n$) and closed and since $U$ is connected: $V=U$. So $U=V\subseteq\text{Fix}(f) = U$ and $f = \text{id}$. – psl2Z Apr 24 '23 at 09:45
  • @MoisheKohan Is this then also true for arbitrary differentiable maps, maybe $C^1$ or less? Or maybe even for continuous maps between connected open subsets of $\mathbb{R}^n$, or some other "good enough" topological space? – psl2Z Apr 24 '23 at 09:48
  • Right, that's the standard proof. The result is also true for periodic homeomorphisms but is much less trivial, Newman's theorem. – Moishe Kohan Apr 24 '23 at 13:18
  • @MoisheKohan Thank you for the hints. I posted an answer. – psl2Z Apr 24 '23 at 23:24

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Based on Moishe Kohan's hints:

Assume f is smooth. Introduce a Riemannian metric $g:=\sum_{k=1}^n(f^k)^*h$, where $h$ is some metric on $U$. Then $f^∗g=g$, because $f^n=\text{id}$, so $f$ is an isometry. Then a connected component $V$ of $\text{Fix}(f)$ is a closed, totally geodesic submanifold without boundary (Fixed Points Set of an Isometry), and contains w.l.o.g. by assumption an open subset of $U$. Therefore $V$ is open (as a submanifold without boundary of $\dim V=n=\dim U$) and closed. Since $U$ is connected, $V=U$. So $U=V\subseteq \text{Fix}(f)=U$ and $f=\text{id}$.

If $X$ is a connected topological manifold and $f$ only continuous this still holds: $\mathbb{Z}_n \to \text{Homeo}(X), 1 \mapsto f$ is an effective action, if $n$ is minimal with $f^n = \text{id}$. By one of Newman's theorems on transformation groups $N = \{x\in X: (\mathbb{Z}_n)_x=1\}$ is open and dense in $X$. Therefore it exists some $x \in \text{Fix}(f) \cap N \neq \emptyset$ and therefore $1=(\mathbb{Z}_n)_x = \mathbb{Z}_n$, so $n=1$ and $f=\text{id}$.

psl2Z
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Newman, M. H. A., A theorem on periodic transformations of spaces, Q. J. Math., Oxf. Ser. 2, 1-8 (1931). ZBL0001.22703.

For a modern proof see Theorem 9.5 in

Bredon, Glen E., Introduction to compact transformation groups, Pure and Applied Mathematics, 46. New York-London: Academic Press. XIII,459 p. (1972). ZBL0246.57017.

The result also holds for periodic groups of homeomorphisms of homology manifolds.

Moishe Kohan
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