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Jun 15, 2021 at 23:37 comment added Piotr Hajlasz I hope you do not mind I asked this question: mathoverflow.net/q/395418/121665. I hope I did correctly state the result. Correct me if I am wrong.
Nov 19, 2018 at 19:57 vote accept Fedya
Nov 14, 2018 at 11:36 answer added Tyler Lawson timeline score: 9
Nov 14, 2018 at 5:37 history edited Fedya CC BY-SA 4.0
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Nov 14, 2018 at 0:24 comment added Fedya Ah, gotcha. Yeah, I'm curious about Whitehead products not of the identity map, but thanks!
Nov 14, 2018 at 0:13 comment added skd Yes, that was a typo. The symbol $\iota_n \in \pi_n(S^n)$ is the class of the identity map. In particular, $[\iota_{k+1}, \iota_{k+1}] \in \pi_{2k+1}(S^{k+1})$ where $k$ is as in my previous comment. The statement I claimed is Corollary 3.2 of Barratt-Jones-Mahowald, "The Kervaire invariant problem".
Nov 13, 2018 at 22:46 comment added Fedya @skd I assume you mean the Kervaire element exists if and only if $[\iota_{k+1},\iota_{k+1}]$ is divisible by 2? What do you mean by $\iota_{k+1}$ (a link is fine)?
Nov 13, 2018 at 22:21 comment added skd One family of examples containing both nontrivial and trivial Whitehead products comes from the Kervaire classes. If $k = 2^{j+1}-2$, then the Kervaire element $\theta_j\in \pi_{2^{j+1}-2}(\mathbb{S})$ if and only if $[\iota_{k+1},\iota_{k+1}]$ is divisible by 2.
Nov 13, 2018 at 22:12 comment added Fedya This is why I specified $m>n$.
Nov 13, 2018 at 22:12 comment added Tyrone Also, it is not true in general that $[id_{S^n},id_{S^n}]$ has Hopf invariant $2$.
Nov 13, 2018 at 22:10 comment added Tyrone For your specific question take $\alpha=id_{S^1}$. Then $[id_{S^2},id_{S^2}]=-2\eta$.
Nov 13, 2018 at 21:21 history asked Fedya CC BY-SA 4.0