Timeline for Is Cohen immersion conjecture (theorem) known for vector bundles?

7 events

when toggle format	what		by	license	comment
May 21, 2019 at 22:37	comment	added	Qfwfq		Yes, the only cases that escape are $r=1$, which doesn't give anything useful; and $n=1$, which formally is better than Whitney by $1$, but equal to the improvement by Haefliger-Hirsch that I can see in wikipedia; and anyway not better than Cohen for any $r$ (And, by the way, there aren't many interesting vector bundles on the circle, are there?...).
May 21, 2019 at 20:33	comment	added	Luis A. Florit		I'm not sure if I understood you correctly, but since a(n)<n, then 2n−a(n)+r(n+1) > 2(n+r) (except maybe for r=1), which is already Whitney's bound for E. So no better. (Sorry, I just saw Mike's answer to yours).
May 21, 2019 at 16:45	comment	added	mme		Because $a(n) \leq \log_2(n)$, roughly your bounds are $(2+r)n - O(\log_2 n)$; the factor of $r$ is pretty noxious. Call the optimal bound $D(n,r)$; Cohen gives $D(n,0) = 2n - a(n)$. A slight improvement on the Whitney embedding theorem (see here) gives $D(n,r) \leq 2n+r$, which is a stronger bound for essentially every $n$. I don't see any clear way to improve the linked argument, even knowing Cohen's result.
May 21, 2019 at 14:41	history	edited	Qfwfq	CC BY-SA 4.0	added 75 characters in body
May 21, 2019 at 14:10	history	edited	Qfwfq	CC BY-SA 4.0	added 336 characters in body
May 21, 2019 at 13:52	history	edited	Qfwfq	CC BY-SA 4.0	added 426 characters in body
May 21, 2019 at 13:33	history	answered	Qfwfq	CC BY-SA 4.0