Timeline for Is the $H$-space structure on $S^7$ associative up to homotopy?
Current License: CC BY-SA 3.0
13 events
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| Jul 25, 2019 at 18:03 | comment | added | Tyrone | @ConnorMalin, correct. You need a multiplication $\mu$ to begin with, then you get a new one by defining $\mu'=\mu+\alpha\circ q$, (as elements of the algebraic loop $[X\times X,X]$) where $\alpha:X\wedge X\rightarrow X$ and $q$ is the quotient to the smash. Since the only spheres that admit (integral) multiplications are $S^1,S^3,S^7$, among spheres the statement is quite special to these chaps. | |
| Jul 25, 2019 at 17:40 | comment | added | Connor Malin | @Tyrone Oh so this is special to $S^7$ since it is an H-space? That makes more sense. | |
| Jul 25, 2019 at 16:25 | comment | added | Tyrone | Recall that any multiplication defined on $X\times X$ needs to restrict to $X\vee X$ as the folding map up to homotopy. Now use the cofibration sequence $X\vee X\rightarrow X\times X\rightarrow X\wedge X$, which splits after suspension, and the fact that the homotopy suspension is monic for any H-space $X$ (since it retracts of $\Omega\Sigma X$). | |
| Jul 25, 2019 at 16:23 | comment | added | Tyrone | @ConnorMalin, the statement is that the number of multiplications on a given H-space $(X,\mu)$ is in one-to-one correspondence with the homotopy set $[X,\wedge X,X]$. See for instance Zabrodsky, pg 26. | |
| Jul 25, 2019 at 15:09 | comment | added | Connor Malin | @Tyrone Are you sure multiplications are in correspondence with $\pi_{14}S^7$? Of course you are using the fact that $S^n \wedge S^m=S^{n+m}$ but shouldn't maps out of the smash product $S^n \wedge S^n$ classify multiplications with a two sided "absorbing element", since maps out of the smash product correspond to maps where the slices are basepointed maps? Multiplication maps should just be classified by $S^n \times S^n$. | |
| Dec 6, 2016 at 8:38 | history | edited | YCor | CC BY-SA 3.0 |
linked to abstract page of AMS + wrote reference
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| Dec 6, 2016 at 7:25 | comment | added | მამუკა ჯიბლაძე | @NajibIdrissi Great, thanks. I should remember that. I now wonder existence of a factorization map of $\mu$ from what kind of quotient of $S^7\times S^7$ to $S^7$ does this imply... | |
| Dec 5, 2016 at 8:42 | comment | added | Najib Idrissi | @მამუკაჯიბლაძე The octonions form an alternative algebra, i.e. $x(xy) = (xx)y$ and $(yx)x = y(xx)$ (and $S^7$ is the unit sphere of the algebra of octonions as far as I can tell). | |
| Dec 5, 2016 at 5:47 | comment | added | მამუკა ჯიბლაძე | @Tyrone A question arises which I wanted to formulate rigorously but failed - is it still "homotopy something"? That is, is the existing multiplication completely generic in some sense? | |
| Dec 4, 2016 at 15:59 | comment | added | Tyrone | Multiplications on $S^7$ are in 1-1 correspondence with the elements of $\pi_{14}S^7=\mathbb{Z}_{8}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{5}$. Given a product $\mu:S^7\times S^7\rightarrow S^7$, the obstruction it being homotopy associative lies in $\pi_{21}S^7=\mathbb{Z}_{8}\oplus\mathbb{Z}_{4}\oplus\mathbb{Z}_{3}$. As Jon says, there are no homotopy associative multiplications on $S^7$. In fact $S^7$ has no 2-local or 3-local associative products. One must invert both the primes 2 and 3 before $S^7$ admits a homotopy associative multiplication (and then any multiplication will be HA). | |
| Dec 4, 2016 at 14:21 | vote | accept | SashaP | ||
| Dec 4, 2016 at 3:25 | comment | added | Theo Johnson-Freyd | Or n=0, of course. | |
| Dec 3, 2016 at 23:00 | history | answered | Jonathan Beardsley | CC BY-SA 3.0 |