Timeline for how to prove the $n$-times self-product of a map is null-homotopic
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2015 at 11:41 | review | Suggested edits | |||
| Nov 16, 2015 at 11:51 | |||||
| Nov 16, 2015 at 11:40 | comment | added | Quan | @AchimKrause Thanks! could you explain more in detail for the last paragraph? Thanks! | |
| Nov 16, 2015 at 5:39 | comment | added | Quan | @TomGoodwillie Thanks Prof. Goodwillie. In your comment, I do not understand the first two steps: "The Chern character from $\tilde{KO}^0(K)$ to the product of the rational cohomology groups $H^4j(K)$ is rationally an isomorphism if $K$ is a finite complex. Therefore in order for a vector bundle on $K$ to be such that some multiple of it is trivial it is sufficient if the rational Pontryagin classes vanish." Could you explain? Thanks! | |
| Nov 15, 2015 at 15:57 | comment | added | Achim Krause | It is $1+kw_1+\frac{k(k-1)}{2}w_1^2$, which is $1$ for $k=4n$ or $k=4n+1$. | |
| Nov 15, 2015 at 12:55 | comment | added | user_11437 | Sorry, I'm confused, maybe you find my mistake. If you take the inclusion $\mathbb{R} P^2\rightarrow B\Sigma_2$, then the associated bundle is $E = \gamma_{\mathbb{R} P^2} \oplus \mathbb{R}$ and $w(kE) = (1+w_1)^k = 1 + kw_1 + (k-1)w_1^2$ which is never zero. | |
| Nov 14, 2015 at 15:01 | vote | accept | Quan | ||
| Nov 14, 2015 at 14:53 | vote | accept | Quan | ||
| Nov 14, 2015 at 14:55 | |||||
| Nov 14, 2015 at 14:09 | comment | added | Tom Goodwillie | By the way, for finite $G$ the group $\tilde{ko}^0(BG)$ is indeed not a torsion group; in fact it is torsion-free, isn't it? According to the Atiyah-Segal completion theorem it is a completion of the real representation ring of $G$. | |
| Nov 14, 2015 at 14:06 | comment | added | Tom Goodwillie | An equivalent argument goes like this: The Chern character from $\tilde{ko}^0(K)$ to the product of the rational cohomology groups $H^{4j}(K)$ is rationally an isomorphism if $K$ is a finite complex. Therefore in order for a vector bundle on $K$ to be such that some multiple of it is trivial it is sufficient if the rational Pontryagin classes vanish. Of course they vanish if the bundle is pulled back from a bundle on $B\Sigma_k$ (or any space having trivial rational cohomology). | |
| Nov 14, 2015 at 10:12 | history | answered | Achim Krause | CC BY-SA 3.0 |