24
$\begingroup$

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.

Improve this question
$\endgroup$
9
  • $\begingroup$ So 'automorphism' here doesn't mean anything about algebraic structure? That is, it's purely a topological question? $\endgroup$
    – LSpice
    Commented Aug 17, 2017 at 21:34
  • $\begingroup$ I mean topologically, yes. Maybe my vocabularies are too algebraic. $\endgroup$
    – Wille Liu
    Commented Aug 17, 2017 at 21:35
  • 3
    $\begingroup$ [Correcting here previous comment] Here's some info from the review of [HKMS] Haynes, Kwasik, Mast, Schultz, Periodic maps on R7 without fixed points; Math Proc Camb Phil Soc, 132, p. 131-136, 2002. by Elmonds. ams.org/mathscinet-getitem?mr=1866329 (continued) $\endgroup$
    – YCor
    Commented Aug 17, 2017 at 22:02
  • 5
    $\begingroup$ Consider $f$ on $R^n$ with $f^m=id$. (1) "classical results of Smith": if $f$ is smooth, and if ($n\le 6$ or $m$ is prime-power) then answer is yes. (2) Kister, 1961-63 relying on Conner-Floyd 1959: construction of fixed-point-free smooth periodic maps on $R^8$. (3) [HKMS] for any $m$ non-prime-power and $n\ge 7$, existence of such smooth $f$. (4) Oliver 1979 fixed-point-free self-homeo on $R^n$ for $n\ge 6$ when $m$ is divisible by at least 3 primes. (5) (no ref): answer is yes for $n\le 4$ without smoothness. Open when $n=5$. $\endgroup$
    – YCor
    Commented Aug 17, 2017 at 22:03
  • 2
    $\begingroup$ In the polynomial case, the fixed point set of any power of $f$ is always homotopically equivalent to a finite CW-complex. So by Verdier's trace formula, $f$ always has a fixed point. $\endgroup$
    – Wille Liu
    Commented Aug 17, 2017 at 22:11

1 Answer 1

Reset to default
24
$\begingroup$

There is, naturally, a huge history regarding a basic question like this. This particular problem was figured out between 1930 and the early 1960's. The main names are P.A. Smith, Conner, Floyd. Here is a math review to get you going:

MR0130929 (24 #A783) Reviewed Kister, J. M. Examples of periodic maps on Euclidean spaces without fixed points. Bull. Amer. Math. Soc. 67 1961 471–474. 54.80 (57.47)

Let T be a homeomorphism of period r on Euclidean space En. The classical result of P. A. Smith is that T has a fixed point if r is prime [Ann. of Math. (2) 35 (1934), 572–578] or a prime power [Amer. J. Math. 63 (1941), 1–8; MR0003199]. The author settles a question of long standing with the following theorem: If r is not a prime power, then there exists a triangulation μ of $E^{9r}$ and a homeomorphism T of period r on $E^{9r}$ without a fixed point, such that T is simplicial over μ. The construction is a modification of an example due to Conner and Floyd [Proc. Amer. Math. Soc. 10 (1959), 354–360; MR0105115].

Improve this answer
$\endgroup$
1
  • 25
    $\begingroup$ The words "naturally", "basic" and "Do some reading!" suggest that the OP ought to know better, which is unfortunate -- if he were expert here, he wouldn't be asking. The question strikes me as being MO-appropriate. $\endgroup$
    – Todd Trimble
    Commented Aug 18, 2017 at 2:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .