One funny remark, that \pi_3(S^2) = Z exactly tells you that any proof of Eckman-Hilton is given by repeatedly applying either the standard proof or its inverse. In a sense there are exactly two “good” proofs of EH (the standard one and it’s inverse). In principle it’s not so automatic to see that a given proof is one of the good two, but in practice it’d be hard to give a bad one accidentally. By contrast, put two people in two separate rooms and there’s a good chance they’ll produce the two different good proofs (ie the clockwise proof and the counterclockwise proof). Best, Noah On Mon, Mar 8, 2021 at 10:15 AM Kristina Sojakova < sojakova.kristina@gmail.com> wrote: > Thanks Dan! I think we should have no trouble showing that what I used > is equal to your proof but packaged a bit differently. > > On 3/8/21 4:10 PM, Dan Christensen wrote: > > It's great to see this proved! > > > > As a tangential remark, I mentioned after Jamie's talk that I had a > > very short proof of Eckmann-Hilton, so I thought I should share it. > > Kristina's proof is slightly different and is probably designed to > > make the proof of syllepsis go through more easily, but here is mine. > > > > Dan > > > > > > Definition horizontal_vertical {A : Type} {x : A} {p q : x = x} (h : p = > 1) (k : 1 = q) > > : h @ k = (concat_p1 p)^ @ (h @@ k) @ (concat_1p q). > > Proof. > > by induction k; revert p h; rapply paths_ind_r. > > Defined. > > > > Definition horizontal_vertical' {A : Type} {x : A} {p q : x = x} (h : p > = 1) (k : 1 = q) > > : h @ k = (concat_1p p)^ @ (k @@ h) @ (concat_p1 q). > > Proof. > > by induction k; revert p h; rapply paths_ind_r. > > Defined. > > > > Definition eckmann_hilton' {A : Type} {x : A} (h k : 1 = 1 :> (x = x)) : > h @ k = k @ h > > := (horizontal_vertical h k) @ (horizontal_vertical' k h)^. > > > > > > > > On Mar 8, 2021, Kristina Sojakova wrote: > > > >> Dear all, > >> > >> I formalized my proof of syllepsis in Coq: > >> > https://github.com/kristinas/HoTT/blob/kristina-pushoutalg/theories/Colimits/Syllepsis.v > >> > >> > >> I am looking forward to see how it compares to the argument Egbert has > >> been working on. > >> > >> Best, > >> > >> Kristina > >> > >> On 3/8/2021 2:38 PM, Noah Snyder wrote: > >> > >> The generator of \pi_4(S^3) is the image of the generator of \pi_3 > >> (S^2) under stabilization. This is just the surjective the part > >> of Freudenthal. So to see that this generator has order dividing > >> 2 one needs exactly two things: the syllepsis, and my follow-up > >> question about EH giving the generator of \pi_3(S^2). This is why > >> I asked the follow-up question. > >> > >> Note that putting formalization aside, that EH gives the generator > >> of \pi_4(S^3) and the syllepsis the proof that it has order 2, are > >> well-known among mathematicians via framed bordism theory (already > >> Pontryagin knew these two facts almost a hundred years ago). So > >> informally it’s clear to mathematicians that the syllepsis shows > >> this number is 1 or 2. Formalizing this well-known result remains > >> an interesting question of course. > >> > >> Best, > >> > >> Noah > >> > >> On Mon, Mar 8, 2021 at 3:53 AM Egbert Rijke > >> wrote: > >> > >> Dear Noah, > >> > >> I don't think that your claim that syllepsis gives a proof > >> that Brunerie's number is 1 or 2 is accurate. Syllepsis gives > >> you that a certain element of pi_4(S^3) has order 1 or 2, but > >> it is an entirely different matter to show that this element > >> generates the group. There could be many elements of order 2. > >> > >> Best wishes, > >> Egbert > >> > >> On Mon, Mar 8, 2021 at 9:44 AM Egbert Rijke > >> wrote: > >> > >> Hi Kristina, > >> > >> I've been on it already, because I was in that talk, and > >> while my formalization isn't yet finished, I do have all > >> the pseudocode already. > >> > >> Best wishes, > >> Egbert > >> > >> On Sun, Mar 7, 2021 at 7:00 PM Noah Snyder > >> wrote: > >> > >> On the subject of formalization and the syllepsis, has > >> it ever been formalized that Eckman-Hilton gives the > >> generator of \pi_3(S^2)? That is, we can build a > >> 3-loop for S^2 by refl_refl_base --> surf \circ surf^ > >> {-1} --EH--> surf^{-1} \circ surf --> refl_refl_base, > >> and we want to show that under the equivalence \pi_3 > >> (S^2) --> Z constructed in the book that this 3-loop > >> maps to \pm 1 (which sign you end up getting will > >> depend on conventions). > >> > >> There's another explicit way to construct a generating > >> a 3-loop on S^2, namely refl_refl_base --> surf \circ > >> surf \circ \surf^-1 \circ surf^-1 --EH whiskered refl > >> refl--> surf \circ surf \circ surf^-1 \circ surf^-1 - > >> -> refl_refl_base, where I've suppressed a lot of > >> associators and other details. One could also ask > >> whether this generator is the same as the one in my > >> first paragraph. This should be of comparable > >> difficulty to the syllepsis (perhaps easier), but is > >> another good example of something that's "easy" with > >> string diagrams but a lot of work to translate into > >> formalized HoTT. > >> > >> Best, > >> > >> Noah > >> > >> On Fri, Mar 5, 2021 at 1:27 PM Kristina Sojakova > >> wrote: > >> > >> Dear all, > >> > >> Ali told me that apparently the following problem > >> could be of interest > >> to some people > >> ( > https://www.youtube.com/watch?v=TSCggv_YE7M&t=4350s): > >> > >> > >> Given two higher paths p, q : 1_x = 1_x, > >> Eckmann-Hilton gives us a path > >> EH(p,q) : p @ = q @ p. Show that EH(p,q) @ EH(q,p) > >> = 1_{p@q=q_p}. > >> > >> I just established the above in HoTT and am > >> thinking of formalizing it, > >> unless someone already did it. > >> > >> Thanks, > >> > >> Kristina > >> > >> -- > >> You received this message because you are > >> subscribed to the Google Groups "Homotopy Type > >> Theory" group. > >> To unsubscribe from this group and stop receiving > >> emails from it, send an email to > >> HomotopyTypeTheory+unsubscribe@googlegroups.com. > >> To view this discussion on the web visit > >> > https://groups.google.com/d/msgid/HomotopyTypeTheory/0aa0d354-7588-0516-591f-94ad920e1559%40gmail.com > . > >> > >> > >> -- > >> You received this message because you are subscribed > >> to the Google Groups "Homotopy Type Theory" group. > >> To unsubscribe from this group and stop receiving > >> emails from it, send an email to > >> HomotopyTypeTheory+unsubscribe@googlegroups.com. > >> To view this discussion on the web visit > >> > https://groups.google.com/d/msgid/HomotopyTypeTheory/CAO0tDg7MCVQWLfSf13PvEu%2BUv1mP2A%2BbbNGanKbwHx446g_hYQ%40mail.gmail.com > . > > -- > You received this message because you are subscribed to the Google Groups > "Homotopy Type Theory" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to HomotopyTypeTheory+unsubscribe@googlegroups.com. > To view this discussion on the web visit > https://groups.google.com/d/msgid/HomotopyTypeTheory/1f7bbcb8-ee50-0c24-174e-d3852e52bbee%40gmail.com > . > -- You received this message because you are subscribed to the Google Groups "Homotopy Type Theory" group. 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