Timeline for Why does the Steenrod algebra act faithfully on $H^\ast(BC_p)$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2021 at 3:17 | vote | accept | Tim Campion | ||
| Dec 13, 2019 at 7:41 | answer | added | user43326 | timeline score: 11 | |
| Dec 12, 2019 at 22:40 | comment | added | Tim Campion | @ConnorMalin I mean that for every nonzero $\phi \in A^\ast$, there exists $\alpha \in H^\ast(BC_p)$ such that $\phi(\alpha) \neq 0$. So the map $A^\ast \to Hom(H^\ast(BC_p),H^\ast(BC_p))$ is injective. | |
| Dec 12, 2019 at 22:39 | comment | added | Connor Malin | Tim, could you clarify for me what you mean by acting faithfully? For example, how can $\mathscr{A}$ be said to act faithfully on $H^*(K(\mathbb{F}_2,1),\mathbb{F}_2)=\mathbb{F}_2 [x]$ when $sq^1 (x^2)=0$. | |
| Dec 12, 2019 at 17:54 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Dec 12, 2019 at 17:49 | comment | added | Tyler Lawson | You're correct. But in characteristic p, a divided power algebra on x turns into a tensor of truncated polynomial algebras generated by x^{p^k}/(p^k)!, because if y is an element, then y^p = p! (y^p/p!) == 0. And thanks to the magic of characteristic two, these truncated polynomial algebras are also exterior algebras. | |
| Dec 12, 2019 at 17:40 | comment | added | Tim Campion | @TylerLawson Thanks! But I thought that Thm 8.11 of Wilson's BP sampler was saying that $H_\ast(K(F_p, n))$ is the tensor product of an exterior algebra and a truncated polynomial algebra, rather than a divided power algebra? Admittedly, I'm very unsure of how to read results involving Hopf rings... | |
| Dec 12, 2019 at 17:33 | comment | added | Tyler Lawson | Unfortunately it's not the case that $H_* (K(\Bbb F_p, n))$ is generated under the Pontrjagin product by suspended classes from $H_*(K(\Bbb F_p, n-1))$. For example, $H^*(K(\Bbb F_2, 2))$ is a polynomial algebra on elements $x, Sq^1(x), Sq^2 Sq^1(x), \dots$ that are all primitive under the coproduct. When you take duals, you get a divided power algebra, which is an exterior algebra on classes dual to $x^{2^k}, (Sq^1(x))^{2^k}, (Sq^2 Sq^1(x))^{2^k}, \dots$ -- the suspended classes only cover the $x^{2^k}$. | |
| Dec 12, 2019 at 17:21 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Dec 12, 2019 at 17:08 | history | asked | Tim Campion | CC BY-SA 4.0 |