Timeline for A question on fixed point theory
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 16, 2014 at 14:06 | vote | accept | Ali Taghavi | ||
| Dec 16, 2014 at 2:21 | answer | added | Eric Wofsey | timeline score: 4 | |
| Jul 18, 2014 at 4:09 | comment | added | Eric Wofsey | More generally, $n$ needs to be large enough compared to $k$ to ensure the existence of an integer $d$ as in the argument above (I haven't checked carefully, but I think $n\geq (k+1)(2k+1)$ should suffice). The condition that either $k$ is odd or $n$ is even is just so that it is impossible for $1+d^{k+1}+\dots+d^n$ to be zero. | |
| Jul 18, 2014 at 4:08 | comment | added | Eric Wofsey | Consider, for instance, $\mathbb{C}P^6/\mathbb{C}P^1$. Its cohomology is the subring of $\mathbb{Z}[x]/x^7$ generated by $s=x^2$ and $t=x^3$. We have a relation $s^3=t^2$, and this forces any endomorphism of the ring to be of the form $s\mapsto d^2s$, $t\mapsto d^3 t$ for some $d\in\mathbb{Z}$. Since $1+d^2+d^3+\dots+d^6$ can never be zero, every map has a fixed point by the Lefschetz fixed point theorem. | |
| Jul 18, 2014 at 0:32 | comment | added | Michael | @EricWofsey: could you elaborate on your answer? | |
| Jul 17, 2014 at 23:54 | answer | added | user56137 | timeline score: 0 | |
| Jul 14, 2014 at 7:37 | comment | added | Eric Wofsey | An slight variation of the usual cohomological argument should give an easy positive answer when $n\gg k$ and $k$ is odd or $n$ is even. For instance, for $k=1$ and $n\geq 6$ every endomorphism of the cohomology ring of $\mathbb{C}P^n/\mathbb{C}P^1$ extends to the cohomology ring of $\mathbb{C}P^n$, and so the fixed point property follows by an easy computation. | |
| Jul 14, 2014 at 5:54 | history | asked | Ali Taghavi | CC BY-SA 3.0 |