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00:00
@SaalHardali I hope to be proved wrong, but I do not think that this has anything to do with any higher algebra. After all, they are determined by the subcategory of the homotopy category spanned by wedges of spheres
A slightly more interesting perspective might be that they come from the fact that the Goodwillie derivative of the identity form the Lie operad
hmmm it is interesting.
Whitehead products aren't preserved by maps in any sense though right?
What do you mean? The Whitehead product is a class in π_{2n-1}(S^n∨S^n), so of course is preserved
So the if I have a map between spaces I have an induced map on graded (quasi-)lie algebras?
This paper describes all the algebraic structure you can put on unstable homotopy groups
But I believe that the short answer is yes
00:28
Auch I put the wrong link this is the paper I meant.
And apparently that paper has disappeared..
01:12
@SaalHardali @DenisNardin it might be something related to some kind of Lie coalgebra structure on the spheres?
Perhaps the spheres form a coalgebra for some graded co-quasi-Lie operad in Top.
01:29
@JonBeardsley Something like that is certainly true (at least in the homotopy category). In fact it is literally true rationally (or more generally K(n)-locally). I think that generally it will boil down to understanding properly the Goodwillie spectral sequence
 
6 hours later…
07:09
@DenisNardin which paper did you mean? I'm interested
 
5 hours later…
12:38
sanity check:
If I have an open covering U_i, of a space X and on each U_i a sheaf F_i, then if I have isomorphisms (omitting restriction maps) \phi_{ij}: F_i \to F_j on the intersections that satisfy the cocycle conditions on the triple intersection I can glue this data to get a sheaf F on X. Suppose I had the following situation instead:
I have an open covering U_i as before, but now instead of having isomorphisms \phi_{ij} on U_i \cap U_j I have an open covering \{W_{ijk}\} of U_i \cap U_j together with isos between F_i and F_j restricted to W_{ijk} such that these isomorphisms satisfy a
I would guess yes, and that this is some statement of descent along hypercoverings (?)
13:27
@BrunoStonek I meant Hilton's On the homotopy groups of a union of spheres but it seems to have disappeared from the internet together with the Journal of the London Mathematical Society
There are probably modern accounts of the result (almost certainly in Neisendorf unstable homotopy theory book) but I was fond of Hilton's paper
how odd
13:42
Or must one add more advanced conditions on the triple intersections for the gluing to work? It just seems odd to me that we only mentioning how to glue along cech coverings, but maybe there's something that goes wrong otherwise
@Dedalus Can't you glue this isomorphisms to an isomorphism on U_i ∩ U_j using the sheaf condition for F_i and F_j?
14:12
@DenisNardin That's what I thought, but I'm now a bit confused. Suppose we just have two opens, $U_1$ and $U_2$ and $F_1$ and $F_2.$ In such a case, we have an open covering $\{ W_{12k} \}$ of $U_1 \cap U_2$ and on each $W_{12k}$ an isomorphism $\phi_{12k} : F_1 \to F_2.$ To glue these isos $\phi_{12k}$ to an isomorphism on $U_1 \cap U_2,$ we need that on the intersection $ W_{12k} \cap W_{12l}$ both $\phi_{12k}$ and $\phi_{12l}$ must be equal. But is this guaranteed by cocycle conditions?
Depends on what you call cocycle conditions I guess.
Precisely thanks! Although apparently MIT has no access to that publisher...
I just wanted to recommend the paper, not read it but it feels weird that the paper evaporated out of reach
15:56
is the Dold-Kan correspondence over Q (say, the normalized chain complex functor from simplicial Q-vector spaces to non-negative Q-chain complexes) strong monoidal?
 
2 hours later…
17:50
@BrunoStonek no, but close. See here: ncatlab.org/nlab/show/…
@UrsSchreiber thanks, I'm aware of that for the general situation, but I'm asking about when we're over $\mathbb{Q}$
I thought it might be different for the following reason: the fact that AW is not -symmetric- colax (over Z) allows you to build the Steenrod algebra. But the Steenrod algebra over Q is trivial...
18:12
but the rational cochains are not a strictly commutative dg-algebra, which it would be if AW were symmetric. hmm.
18:58
ok, I think I get it. the fact that the rational Steenrod algebra is trivial "implies" that we have hope in finding an honestly commutative model for the rational cochains -- even if the rational cochains are not, themselves, commutative
right. and we have that (the PL de Rham complex)
yes. hooray

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