Timeline for Does a self map from the wedge sum of two spheres have either a fixed point or a point of period 2?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 9, 2013 at 9:13 | answer | added | nsrt | timeline score: 12 | |
| Jul 9, 2013 at 7:43 | comment | added | Neil Strickland | @PedroPerez Will's argument says that if $f$ is a self-map such that neither $f$ nor $f^2$ has a fixed point, then $f$ must act with determinant one on $H^2(X)\simeq\mathbb{Z}^2$. In particular, $f$ must be a homotopy equivalence. I don't think that Will is making any claim about whether such a map $f$ exists. | |
| Jul 9, 2013 at 7:12 | comment | added | Pedro Perez | @Will I do not know what are you saying, yes or no? | |
| Jul 9, 2013 at 6:45 | comment | added | Will Sawin | By the Lefschetz trace formula, if $M$ is the $2 \times 2$ matrix that gives the action of $f$ on $H^2$, then $tr(M)=-1$ and $tr(M^2)=-1$, so $tr(M)^2-2 det (M) = -1$, so $det(M)=1$. | |
| Jul 9, 2013 at 6:37 | comment | added | Will Sawin | @Noam: Yes. On one sphere, send each point to its antipode. Send everything else to the antipode of the base point. | |
| Jul 9, 2013 at 6:33 | comment | added | Noam D. Elkies | Is there an example where $f$ has no fixed point? | |
| Jul 9, 2013 at 6:28 | review | First posts | |||
| Jul 9, 2013 at 6:29 | |||||
| Jul 9, 2013 at 6:10 | history | asked | Pedro Perez | CC BY-SA 3.0 |