Timeline for Relative version of Browder's theorem on H-spaces
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7, 2022 at 19:57 | comment | added | Danny Ruberman | @JohnKlein I'm not sure that I know how to get the stronger assumption. I'd be glad to continue the conversation off-line and tell you more than makes sense to try to put in the comments. I think you might know where to find me... | |
| Dec 7, 2022 at 15:13 | comment | added | John Klein | @DannyRuberman If $X$ and $Y$ are topological groups and $f$ is a homomorphism. Then the homotopy fiber $F$ of $f$ at the identity element of $Y$ has the homotopy type of a topological group. If you knew that $H_\ast(F)$ is totally finitely generated then it follows that $F$ has the homotopy type of a homotopy finite space. In particular, $F$ is a homotopy finite $H$-space, so $\pi_2(F) = 0$. However, you are assuming less than I am: you are assuming that the kernel of $H_*(f)$ is totally finitely generated. Can you get away with my stricter assumption? | |
| Dec 7, 2022 at 2:12 | comment | added | Danny Ruberman | @JohnKlein I don't think so. Since I'm easily confused by logic, let me state it in the contrapositive setting in which the question arises. I have two H-spaces X and Y (actually groups) and a map, and I know that the kernel on $\pi_2$ is non-zero. So I know that $H_*(X)$ is infinitely generated, but I'd like to know that the kernel is infinitely generated. | |
| Dec 6, 2022 at 20:32 | comment | added | John Klein | @DannyRuberman you also want to assume in the second paragraph of your post that $H_*(X)$ and $H_*(Y)$ are both finitely generated. Right? | |
| Dec 2, 2022 at 15:42 | comment | added | Danny Ruberman | @AchimKrause Good question; I don't know an example. I'm far from an expert on this so I was hoping that someone would tell me how to find that `magic'. | |
| Dec 1, 2022 at 0:30 | comment | added | Achim Krause | Do you have an example of an H-space map with finitely generated (non-trivial) kernel on homology, which doesn't just have $H_*(X)$ finitely generated? Feels weird to me, shouldn't there be some Hopf algebra magic that makes this impossible? | |
| Nov 30, 2022 at 23:48 | history | edited | Danny Ruberman | CC BY-SA 4.0 |
added 46 characters in body
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| Nov 30, 2022 at 23:47 | comment | added | Danny Ruberman | @archipelago Sorry for the confusion; I've edited to clarify "finitely generated". | |
| Nov 30, 2022 at 23:27 | comment | added | archipelago | Makes sense. Thanks. | |
| Nov 30, 2022 at 23:10 | comment | added | Oscar Randal-Williams | @archipelago: "finitely generated" refers to the totality of the integral homology groups. | |
| Nov 30, 2022 at 22:44 | comment | added | archipelago | $X=K(\mathbb{Z},2)$ is a counterexample to your first paragraph. Did you forget an assumption? | |
| Nov 30, 2022 at 21:42 | history | asked | Danny Ruberman | CC BY-SA 4.0 |