Timeline for About a generalization of the Borsuk-Ulam theorem
Current License: CC BY-SA 4.0
9 events
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Dec 23, 2019 at 19:37 | comment | added | Paul Cusson | Yes I meant $+2\pi /n$. And thanks for clarifying about the ping, I wasn't sure how that worked exactly. By you are right, mean about my counterexample, right? Thanks for the input. | |
Dec 23, 2019 at 19:32 | vote | accept | Paul Cusson | ||
Dec 23, 2019 at 19:28 | comment | added | Mark Grant | PS I got pinged anyway, the author of any question or answer always gets notified of any comments on their post. | |
Dec 23, 2019 at 19:27 | comment | added | Mark Grant | About your second question, there has been some work done on extending Borsuk-Ulam to homology spheres, see the references in the linked thesis (Conner-Floyd, Munkholm,...) | |
Dec 23, 2019 at 19:26 | comment | added | Mark Grant | @PaulCusson: You are right, I think, and your example above has $A(g)$ empty whilst having points for which $g(x)=g(f(x))$. (I suppose you meant to write $f(\theta)=\theta + 2\pi/n$ or something similar?) There doesn't seem to be much in the literature about your precise question. | |
Dec 23, 2019 at 0:25 | comment | added | Paul Cusson | Sorry, @Mark I forgot to ping you | |
Dec 22, 2019 at 21:08 | comment | added | Paul Cusson | In particular, $g : S^1 \to \mathbb{R}$ such that $g(\theta) = \cos(\theta)$, and $f:S^1 \to S^1$ such that $f(\theta) = \theta + 1/n$ for some natural $n$ large enough. The map $f$ is such that $f^n$ is the identity on $S^1$, and we should get $A(g)$ empty, but $g(S^1)$ and $g(f(S^1))$ have two points of intersections, so two such "fixed points". | |
Dec 21, 2019 at 6:02 | comment | added | Paul Cusson | This is interesting. But what if I don't want other powers of $f$ to fix the value, and just want $g(x)= g(f(x))$? We could still have $A(g)$ empty in this case, if for every $x$ some other power doesn't fix the value. | |
Dec 20, 2019 at 12:08 | history | answered | Mark Grant | CC BY-SA 4.0 |