Timeline for Is the J homomorphism compatible with the EHP sequence?
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22 events
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| Sep 6, 2016 at 17:39 | comment | added | Gregory Arone | @JohnKlein You are right. Not sure how I could miss it. I edited the question. | |
| Sep 6, 2016 at 17:38 | history | edited | Gregory Arone | CC BY-SA 3.0 |
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| Sep 6, 2016 at 2:23 | comment | added | John Klein | @GregoryArone : I just checked your link to Ravenel's book (it's the second edition). On page 30, he only claims that the diagram commutes after looping once. He doesn't assert that the diagram itself commutes. On page 56 of the first edition he doesn't say anything at all about whether the diagram commutes––it's inferred. Obviously, he corrected his misstatement in edition 2. | |
| Sep 4, 2016 at 21:32 | answer | added | John Klein | timeline score: 5 | |
| Sep 2, 2016 at 11:48 | comment | added | John Klein | @GregoryArone: According to Bill Richter: "The statement [Question 1]" is false, and I proved that the right square does not commute. Dan Kahn showed me that James had a simple proof that the right square homotopy commutes after a loop, so I didn't publish my proof that I was so proud of." | |
| Sep 2, 2016 at 6:39 | vote | accept | Gregory Arone | ||
| Sep 1, 2016 at 16:16 | answer | added | John Klein | timeline score: 12 | |
| Aug 23, 2016 at 23:59 | comment | added | John Klein | @GregoryArone: here's what I mean by my last sentence: we know that there's a homotopy fiber sequence $S^n \to J(S^n) \to J(S^{2n})$, in the $(3n)$-range, where $J=$ James construction. If we loop this $n$-times and take suitable path components we get a (2n)-approximate fiber sequence $F_n \to F_{n+1} \to \Omega^nJ(S^{2n})$ that maps into the fiber sequence $F_n \to F_{n+1} \to F_{n+1}/F_n$. The 5-lemma implies the map $\Omega^n J(S^{2n})^{(2n)} \to F_{n+1}/F_n$ is ~ $2n$-connected. Finally, it's clear that the evident map $S^n \to \Omega^n J(S^{2n})^{(2n)} $ is an isomorphism on $\pi_n$. | |
| Aug 23, 2016 at 23:48 | comment | added | John Klein | @GregoryArone: it means that it's a $F_{n+1}/F_n$ is $(n-1)$-connected and the map $S^n \to F_{n+1}/F_n$ is an isomorphism on the $n$-th homotopy group. The argument that $\pi_n(S^n) \to \pi_n( F_{n+1}/F_n) \cong \Bbb Z$ is a consequence of the EHP sequence, right? | |
| Aug 23, 2016 at 20:24 | comment | added | Gregory Arone | @JohnKlein I would like to understand it better. What exactly does it mean that the map $S^n \to F_{n+1}/F_n$ is picking up the bottom cell, and how do I know it? Thanks. | |
| Aug 23, 2016 at 2:43 | comment | added | John Klein | Greg and Tom: doesn't the first question follow from the commutative square which at the top is $BO(n) \to BO(n+1)$ and the bottom is $BF_n \to BF_{n+1}$? The map of horizontal fibers is $S^n \to F_{n+1}/F_n$ and this is picking up the bottom cell. One only needs to show that the map $F_{n+1} \to \Omega^{n+1}S^{2n+1}$ defined by $n$-fold looping of the Hopf invariant factors through a map $F_{n+1}/F_n \to \Omega^{n+1} S^{2n+1}$ which is about $2n$-connected (then the composite $S^n \to F_{n+1}/F_n \to \Omega^{n+1} S^{2n+1}$ is the $\pm E$ up to sign). But it seems to me that it does. | |
| Aug 22, 2016 at 17:14 | comment | added | Tom Goodwillie | What I should have said is that I know that $F(V)$ has linear coeff $S^{-1}$, and I can see that if the given map $O(V)\to F(V)$ induces an equivalence of linear coefficients then also $G(V)$ has coefficient $S^{-1}$ (and the given map ... ). | |
| Aug 22, 2016 at 16:57 | comment | added | Gregory Arone | How do you know that $G(V)$ has linear coefficient $S^{-1}$? I think that the fact that the map $G(V)\to G(V+R)$ factors through $F(V)$, together with the fact that $O(V+R)/O(V)\to G(V+R)/F(V)$ is an equivalence, implies that the map $O(V)\to G(V)$ induces a split monomorphism on linear coefficients. So if I understood why $d_1 G$ is abstractly equivalent to $S^{-1}$, this might finish the job. | |
| Aug 22, 2016 at 12:56 | comment | added | Tom Goodwillie | I had never looked into this before. It seems funny to prove this fact about $G_n/O_n$ using embeddings instead of more directly. Since your question, I've been thinking about a direct proof. In terms of orthogonal calculus, the question can be stated like this. There are functors of $V$, $O(V)\to G(V)\to F(V)$, where $G(V)$ is self-equivalences of the unit sphere and $F(V)$ is based self-equivalences of the suspended sphere $V\cup\infty$. All three functors have linear coefficient $S^{-1}$. The problem is to show that the maps $O(V)\to G(V)\to F(V)$ induce equivalences of linear parts. | |
| Aug 21, 2016 at 8:17 | comment | added | Gregory Arone | OK, Haefliger's main theorem identifies the group of concordance classes of embeddings $S^i\hookrightarrow S^{n+i}$ with $\pi_i$ of the homotopy fiber of the map $G_n/O(n)\to G/O$ (assuming $n\ge 3$). The Sphere Unknotting Theorem says that all such embeddings are trivial roughly for $i<2n-3$, which means that the map $G_n/O(n)\to G/O$ is approximately $2n$ connected. Is this about right, and if yes is this the easiest way to see the homotopy theoretic fact about $G_n/O(n)$? | |
| Aug 20, 2016 at 15:45 | comment | added | Gregory Arone | I went through Haefliger's paper. While there clearly are similar looking statements there, I had trouble locating exactly the one I am looking for. If someone can give me chapter and verse, I will be grateful. | |
| Aug 20, 2016 at 7:51 | history | edited | Gregory Arone | CC BY-SA 3.0 |
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| Aug 19, 2016 at 21:36 | comment | added | Tom Goodwillie | I want to understand this. I've been curious about it for years. Your question is prompting me to think about it. I'll let you know if I think of anything sensible to say. | |
| Aug 19, 2016 at 17:57 | comment | added | Tom Goodwillie | The closely related fact that $G_n/O_n\to G/O$ is approximately $2n$-connected is stated on page 124 of Walls "Surgery on Compact Manifolds", and he gives as reference Haefliger, A., Differentiable embeddings of $S^n$ in $S^{n+q}$ for $q>2$. Ann. of Math. 83 (1966), 402–436. | |
| Aug 19, 2016 at 16:14 | history | edited | Gregory Arone | CC BY-SA 3.0 |
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| Aug 19, 2016 at 16:08 | history | edited | Gregory Arone | CC BY-SA 3.0 |
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| Aug 19, 2016 at 14:59 | history | asked | Gregory Arone | CC BY-SA 3.0 |