Timeline for Adams' theorems on the Hopf-Whitehead J-homomorphism
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2014 at 6:58 | comment | added | Thomas Kragh | Hmm. I thought there were easier ways of proving that but thinking about it I can't figure any out - sorry. | |
| Feb 2, 2014 at 19:25 | comment | added | Johannes Ebert | If I want to show that the $J$-homomorphism is injective on $\pi_{8n}$, I have to show that a homotopy $8n+1$-sphere is stably parallelizable, but unfortunately the only argument known to me uses the fact on the $J$-homomorphism. So it seems to be circular. | |
| Feb 2, 2014 at 18:32 | comment | added | Thomas Kragh | It extends because all homotopy spheres are stably paralizable. So, if you look at any smoothly embedded homotopy disc $X^n \subset \mathbb{R}^n \subset \mathbb{R}^{n+N}$ , which is equal to standard $D^n$ close to the boundary - then the standard framing on $S^{n-1}$ extends to a framing on $X$ (for large $N$). | |
| Feb 2, 2014 at 18:26 | comment | added | Johannes Ebert | Ok, now I understand what you are saying. But I do not see how you turn the disc into a standard disc, i.e. why the ''framing does extend''. | |
| Feb 2, 2014 at 12:33 | comment | added | Thomas Kragh | @Johannes: Assume $f: S^{2n}\to O$ is any map, and that we have a framed null-cobordism of the corresponding framed sphere. By surgery we may assume that this framed null-cobordism is a disc. If this disc is not the standard smooth disc (rel $S^{2n}$) then we may locally cut out a disc at some interior point and replace it with the "inverse" smooth structured disc (framing does extends) and make it the standard disc (rel $S^{2n}$), by isotoping (rel $S^{2n}$) we may assume this disc is the standard embedded disc, this framed standard disc (and the isotopy) now gives a null homotopy of $f$. | |
| Feb 2, 2014 at 10:19 | comment | added | Johannes Ebert | @Allen: is the idea that one defines an invariant similar to the $e$-invariant, but with the Adams operation in place of the Chern character? And the Bott cannibalistic classes instead of the $\hat{A}$-class? | |
| Feb 2, 2014 at 10:17 | comment | added | Johannes Ebert | @Thomas: I do not understand this argument. The framing of $S^{8n}$ gives an element of $\pi_{8n}^{st}$ (this is the geometric description of the $J$-homomorphism). Now one has to show that if the original element in $\pi_{8n}(O)$ is nonzero, then the stable framing is NOT nullbordant. | |
| Feb 2, 2014 at 9:54 | comment | added | Thomas Kragh | Question 1: The case $n=0$ mod 8 can be proved using surgery theory of framed manifolds, and probably also the other case. Indeed, a map $S^{8n} \to O$ defines a stable framing of $S^{8n}$, and this framed manifold represents its image in $\pi_n^{st}$, any framed cobordism showing that it is in the kernel can by surgery be turned into a disc (dim$=8n+1$), and hence a null-homotopy of the map. The odd case takes more work (if possible - which I think it is) since the surgery does not lead to a disc but may have cells of dimension $4n+1$. Is this considered "similarly direct"? | |
| Feb 1, 2014 at 22:41 | comment | added | Allen Hatcher | For question 1, I think there is a similar sort of argument to show injectivity in these cases, though it needs a bit more input, notably Adams operations in real K-theory. I have some handwritten notes on this from 20 years ago that would take some work to decipher after this long a time. My recollection is that I extracted these from Adams' J(X) - IV paper. My plan was, and still is, to include this in that unfinished book mentioned in the question, though as the years pass the chances of this ever happening become increasingly slim. | |
| Feb 1, 2014 at 13:41 | answer | added | Dustin Clausen | timeline score: 18 | |
| Jan 31, 2014 at 22:33 | history | edited | Johannes Ebert | CC BY-SA 3.0 |
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| Jan 31, 2014 at 21:23 | history | edited | Johannes Ebert | CC BY-SA 3.0 |
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| Jan 31, 2014 at 21:16 | comment | added | user43326 | One simple description of the unit map $S\rightarrow KO$ is that at the level of the corresponding infinite loop spaces, it is the limit of the classifying map induced by the regular representation $\Sigma _n\rightarrow O(n)$. | |
| Jan 31, 2014 at 21:14 | comment | added | Neil Strickland | The J-homomorphism is only an infinite loop map with respect to the multiplicative infinite loop structure on $Q_0S^0$, so you get a map of spectra from $\Sigma^{-1}kO$ to $gl_1(S^0)$ rather than $S^0$. Here $\pi_k(gl_1(S^0))=\pi_k(S^0)$ for $k>0$, but other properties of $gl_1(S^0)$ are not too well understood. Also, you need to use the connective spectrum $kO$ rather than the periodic $KO$ here. | |
| Jan 31, 2014 at 20:17 | history | asked | Johannes Ebert | CC BY-SA 3.0 |