Timeline for What is the intuition for higher homotopy groups not vanishing?

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15 events

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Jul 21, 2019 at 22:32	comment	added	Noah Snyder		Right, that’s the right thing to do for $S^2$. What I was writing down gives $2 \in \mathbb{Z}$ instead of 1. But note it doesn’t give 0. To work out what a given construction does you should work out the corresponding Pontryagin-Thom diagram.
Jul 21, 2019 at 22:22	comment	added	Simon Henry		Ah I see ! Now I understand what you meant. The reason I was confuse is because what I do is a bit different: I Use $EHclock_{x,x}$ to get a $3$-cell $x^2 \rightarrow x^2$ that I then compose with the inverse of $x^2$ to bring it back to a $3$-cell $1 \rightarrow 1$. And $EHcounterclock_{x,x}$ is the inverse of $EHclock_{x,x}$.
Jul 21, 2019 at 22:16	comment	added	Noah Snyder		Though in a sense I think the nonvanishing of $\pi_3(S^2 \vee S^2)$ more directly illustrates why just the group structure fails to capture higher loop spaces. The free grouplike $E_2$ algebra generated by two 2-loops has an interesting 3-loop given by $xy \rightarrow yx \rightarrow xy$ because there are two distinct proofs of Eckman-Hilton, the clockwise one and the counterclockwise one.
Jul 21, 2019 at 22:08	comment	added	Noah Snyder		That said, the error wasn't what you're pointing out. If $EHclock_{x,y}$ denotes one of the two 3-paths xy = yx and $EHcounter_{x,y}$ the other, then you can build a full twist either by composing $EHclock_{x,x^{-1}}$ with $EHcounter_{x,x^{-1}}^{-1}$ or with $EHclock_{x^{-1},x}$. I had in mind the former, but you had in mind the latter. But actually (fun exercise!) both of those 3-paths are themselves connected by a 4-path.
Jul 21, 2019 at 22:04	comment	added	Noah Snyder		@SimonHenry: Sorry I realized my comment had an error and so deleted it, I was writing down the interesting new generator for $\pi_3(S^2 \vee S^2)$ instead of $\pi_3(S^2)$.
Jul 21, 2019 at 21:36	comment	added	Simon Henry		@NoahSnyder : I think I'm misunderstanding your comment: isn't "counterclockwise" Eckman-Hilton the inverse of the clockwise Eckman-Hilton ? so if you compose them you should get something trivial no ?
Jul 21, 2019 at 14:59	history	edited	Simon Henry	CC BY-SA 4.0	added 101 characters in body
Jul 21, 2019 at 13:26	history	edited	Simon Henry	CC BY-SA 4.0	added 69 characters in body
Jul 21, 2019 at 13:19	history	edited	Simon Henry	CC BY-SA 4.0	added 69 characters in body
Jul 20, 2019 at 21:51	history	edited	Simon Henry	CC BY-SA 4.0	added 967 characters in body
Jul 20, 2019 at 21:48	comment	added	Simon Henry		@KevinCarlson : I've edited to show how to get a non trivial element in $\pi_3(\mathbb{S^2})$. In theory you can get all elements of $\pi_n(\mathbb{S}^m)$ this way, though it is going to be very non automatic and I have no idea how you could show that you have found all elements, so I wouldn't call that a way to compute $\pi_n(\mathbb{S}^m)$, to me it is more a way of 'naming' its elements.
Jul 20, 2019 at 21:45	history	edited	Simon Henry	CC BY-SA 4.0	added 967 characters in body
Jul 20, 2019 at 20:59	comment	added	Kevin Carlson		Is it actually possible to do some computations with one of these perspectives, ideally the second one? I've thought a bit about it but I don't know how to find the Hopf fibration as a generator of the endomorphism group of the 2-identity in a weak 3-groupoid with one object freely generated by a 2-endomorphism of the 1-identity. Maybe its double, via Whitehead products, but even in HoTT to get the generator they seem to construct the fibration and use the long exact sequence, which I think of as a fundamentally geometric argument.
Jul 20, 2019 at 20:19	history	edited	Simon Henry	CC BY-SA 4.0	added 16 characters in body
Jul 20, 2019 at 20:04	history	answered	Simon Henry	CC BY-SA 4.0