Timeline for Can the Bousfield class of projective space be computed directly?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2022 at 17:48 | comment | added | Neil Strickland | @JamesCameron I think that $F\to S$ is not null in the case $X=\mathbb{C}P^2$ with $p=2$. | |
| Apr 28, 2022 at 17:45 | comment | added | J Cameron | With this strategy, in the triangle $F \to S \to DX \wedge X$ you are showing that the map $F \to S$ is null. In general if $X$ is type $0$, this map is just tensor nilpotent, not necessarily null, right? Could it happen that the map $F \to S$ is not null when $X$ is a complex projective space? | |
| Apr 28, 2022 at 17:34 | history | edited | Neil Strickland | CC BY-SA 4.0 |
deleted 2 characters in body
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| Apr 28, 2022 at 17:22 | history | edited | Neil Strickland | CC BY-SA 4.0 |
Edited to incorporate insights from the comments
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| Apr 28, 2022 at 16:55 | comment | added | Neil Strickland | @MaximeRamzi I think you are right, I went through that analysis previously but miscalculated slightly. So that just leaves the case where $n$ is divisible by $p(p-1)$. | |
| Apr 28, 2022 at 16:17 | comment | added | Maxime Ramzi | Isn't it possible to use a map with Euler class $q$ then ? I think in this case the reduced trace is $\sum_{1\leq i\leq n} q^i = q \frac{q^n-1}{q-1}$. This nonzero (hence invertible) mod $p$ if and only if both $q$ and $q^n-1$ are. So we just pick a $q$ not divisible by $p$, and such that $q^n$ is not $1$ mod $p$. If all nonzero $q$'s satisfy $q^n = 1$ mod $p$, then $(p-1)$ divides $n$. So if $(p-1) \nmid n$ or $p\nmid n$, this can be made to work. | |
| Apr 28, 2022 at 16:03 | comment | added | Neil Strickland | @DylanWilson see my correction to the answer - the relevant trace here is the trace on the reduced homology, which is zero mod $p$. | |
| Apr 28, 2022 at 15:58 | history | edited | Neil Strickland | CC BY-SA 4.0 |
Correct error about reduced vs unreduced Euler characteristic
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| Apr 28, 2022 at 15:27 | comment | added | Dylan Wilson | doesn't the same argument (by considering the Lefschetz trace) prove the general case? (the map $\mathbb{C}P^n \to \mathbb{C}P^n$ with Euler class $p$ times the generator should have trace equal to 1 mod $p$, giving the sphere as a retract of $\mathbb{C}P^n \wedge D\mathbb{C}P^n$.) | |
| Apr 28, 2022 at 15:14 | history | answered | Neil Strickland | CC BY-SA 4.0 |