Timeline for Finite-order self-homeomorphisms of $\mathbf{R}^n$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 18, 2017 at 20:32 | comment | added | Wille Liu | @YCor With Verdier's formula, if we take the trace of $f$, the left hand side will give $\pm 1$, since the only compactly supported cohomology of $X=\mathbf{R}^n$ is $H^{n}_c(X, \mathbf{Q}) \cong \mathbf{Q}$. On the right hand side, only the subset $X_G$ will contribute, so the Euler characteristic of the fixed point set must be $\pm 1$, depending on whether $f$ preserves or inverses the orientation. However, it doesn't say anything about the connectedness for any power of $f$. | |
| Aug 18, 2017 at 20:18 | vote | accept | Wille Liu | ||
| Aug 18, 2017 at 18:37 | comment | added | YCor | @WilleLiou I'm not sure I get the argument in the polynomial case. Say, $f^6=id$. The fixed-point set of both $f^2$ and $f^3$ are non-empty, but I don't see why they are, say, connected. But I'm still far from a correct understanding of Verdier's formula. | |
| S Aug 18, 2017 at 16:33 | history | suggested | Peter Heinig | CC BY-SA 3.0 |
Minor grammatical corrections.
|
| Aug 18, 2017 at 15:19 | review | Suggested edits | |||
| S Aug 18, 2017 at 16:33 | |||||
| Aug 18, 2017 at 1:13 | answer | added | Nicholas Kuhn | timeline score: 24 | |
| Aug 17, 2017 at 22:11 | comment | added | Wille Liu | 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. | |
| Aug 17, 2017 at 22:06 | comment | added | YCor | I can't find any discussion about the case when $f$ is assumed real analytic, or polynomial. | |
| Aug 17, 2017 at 22:03 | comment | added | YCor | 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 $n=5$. | |
| Aug 17, 2017 at 22:02 | comment | added | YCor | [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) | |
| Aug 17, 2017 at 21:45 | comment | added | Wille Liu | I updated the term. By the way, if it was real analytic, woudn't the fixed point set be of finite topological type? | |
| Aug 17, 2017 at 21:43 | history | edited | Wille Liu | CC BY-SA 3.0 |
Update vocabularies
|
| Aug 17, 2017 at 21:35 | comment | added | Wille Liu | I mean topologically, yes. Maybe my vocabularies are too algebraic. | |
| Aug 17, 2017 at 21:34 | comment | added | LSpice | So 'automorphism' here doesn't mean anything about algebraic structure? That is, it's purely a topological question? | |
| Aug 17, 2017 at 21:27 | history | asked | Wille Liu | CC BY-SA 3.0 |