Skip to main content

Finite order-order self-homeomorphisms of $\mathbf{R}^n$

Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is said to be of finite order if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ hashave a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed point-point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the self-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed point-point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite order-order self-homeomorphisms.

Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that a certain power of it has its fixed point-point set very complicated.

Finite order self-homeomorphisms of $\mathbf{R}^n$

Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is said to be of finite order if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ has a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the self-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite order self-homeomorphisms.

Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that certain power of it has fixed point set very complicated.

Finite-order self-homeomorphisms of $\mathbf{R}^n$

Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is said to be of finite order if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ have a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed-point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the self-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed-point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite-order self-homeomorphisms.

Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that a certain power of it has its fixed-point set very complicated.

Update vocabularies
Source Link
Wille Liu
  • 1.1k
  • 5
  • 17

Finite order automorphismsself-homeomorphisms of $\mathbf{R}^n$

Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A continuous mapself-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is called a finite order automorphismsaid to be of finite order if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order continuous automorphismself-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ has a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the automorphismsself-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite order automorphismsself-homeomorphisms.

Unfortunately, I don't know if there can be some automorphismself-homeomorphism of non-prime order such that certain power of it has fixed point set very complicated.

Finite order automorphisms of $\mathbf{R}^n$

Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A continuous map $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is called a finite order automorphism if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order continuous automorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ has a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the automorphisms of $\mathbf{R}^n$ of prime order has a fixed point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula finite order automorphisms.

Unfortunately, I don't know if there can be some automorphism of non-prime order such that certain power of it has fixed point set very complicated.

Finite order self-homeomorphisms of $\mathbf{R}^n$

Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is said to be of finite order if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ has a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the self-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite order self-homeomorphisms.

Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that certain power of it has fixed point set very complicated.

Source Link
Wille Liu
  • 1.1k
  • 5
  • 17

Finite order automorphisms of $\mathbf{R}^n$

Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A continuous map $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is called a finite order automorphism if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

Question: Does every finite order continuous automorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ has a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the automorphisms of $\mathbf{R}^n$ of prime order has a fixed point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula finite order automorphisms.

Unfortunately, I don't know if there can be some automorphism of non-prime order such that certain power of it has fixed point set very complicated.