« first day (2849 days earlier)      last day (554 days later) » 

jdc
jdc
1:05 AM
@TimCampion and @MaximeRamzi, the "H-group" definition I found online that makes the maps I said work is that X be a "group up to homotopy" in the same sense that an H-space is a "unital magma up to homotopy." That is, one asks for a homotopy between the maps on X^3 with values x.(y.z) and (x.y).z and a weak "inverse" operation on X such that the functions on X taking x to i(x).x and x.i(x) are both homotopic to the identity. If X is a loop-space this is automatic from the lemmas in the standard introduction to the fundamental group, unless I'm missing something, but I assume all these noti…
 
 
6 hours later…
Dustin Clausen
Dustin Clausen
6:37 AM
@BryanShih yes, from your questions (which are very legitimate!) it looks like it might have been a good idea for us to include some more details in our arguments...
 
Maxime Ramzi
Maxime Ramzi
7:36 AM
@jdc if (X,m) is homotopy associative, and of the homotopy type of a CW-complex, then the existence of i is equivalent to the condition that (\pi_0(X),\pi_0(m)) be a group !
This is due to the Eckmann-Hilton argument
and the fact that the existence of i is equivalent to the "shear map" being an equivalence, where the shear map is (x,y) \mapsto (xy, y)
(of course I put in "of the homotopy type of a CW complex" just in case we're talking about topological spaces; if we're working in the oo-category of spaces there's obviously no worry to have about that: the shear map will be a weak equivalence, and therefore i will exist up to inverting the relevant maps)
 
 
5 hours later…
Tim Campion
Tim Campion
12:20 PM
@MaximeRamzi For the "shear map" condition to work, maybe we need to assume the H-space is homotopy commutative too?
 
Maxime Ramzi
Maxime Ramzi
12:45 PM
@TimCampion I don't believe so
anything with a multiplication which is associative up to homotopy and has a neutral element (up to homotopy) such that the shear map has a homotopy inverse has an inverse map i
it's just defined as x \mapsto second component of inverse of shear on (e,x)
uhm this formula works for shear(x,y) = (xy,x), not the one I gave earlier
 
Maxime Ramzi
Maxime Ramzi
1:00 PM
It's actually fun to work out : if f = (f_1,f_2) is an inverse to the shear map, then because sf is homotopic to id, you can actually see that f_1 is homotopic to the second projection. It follows that one may assume f_1 = pr_2. Then if you write out what sf homotopic to id means, you'll see that it implies that \mu(y, f_2(e,y)) is homotopic to e, i.e. i(y) = f_2(e,y) is an inverse map
Actually the "hard part" of the argument is working out the details of the Eckmann-Hilton argument: if you don't pay attention, you'll end up proving something only about the component of e
 
Maxime Ramzi
Maxime Ramzi
1:13 PM
Mhm now that I think about it those Eckmann-Hilton/Whitehead details might be a place where people usually use commutativity, but you don't need it. Let me spell it out, just to make sure : the above argument tells you, by Whitehead's theorem, that the shear map is an equivalence on the component of e.
This gives you an inverse map i on this component. Now, for each component C of M, fix, once and for all, a y_C in the component C^{-1}. Then define an i component by component by putting i(z)= y_C i(zy_C) if z is in C. Then to prove that mu(x,i(x)) is homotopic to e, you just need to do that component by component. Fixing a component C, mu(x,i(x)) is mu(z, mu(y_C, i(zy_C)) which is homotopic to mu(zy_C , i(zy_C)), which is homotopic to e
 
 
9 hours later…
Peter Haine
Peter Haine
10:01 PM
@TimCampion Re: that question, I wrote this note to try to understand a bit better what was going on there (in a model-independent way). Maybe it is useful to others.
 
Tim Campion
Tim Campion
@PeterHaine Oh nice, thanks! I'll have to take a look
 
Peter Haine
Peter Haine
What I concluded was that you need something extremely weak for the free loop space fibration to split: a multiplication map with a 1-sided unit such that the shear map is an equivalence. (I called this a "left/right A_{1½}-algebra", but would really love a less silly name.)
 
Tim Campion
Tim Campion
10:17 PM
A_2 is two-sided unital, and A_3 is homotopy associative, right?
Somehow I always feel like the numbering is off by one
ah yes, you recall this in your note
 
Peter Haine
Peter Haine
Yeah, I recalled it because I always get confused by the numbering :) An important example that has this weird structure: the 7-sphere. The multiplication is non-associative, but it is two-sided unital and shear map is an equivalence.
 
jdc
jdc
So homotopy-associativity is not needed? I haven't read the linked note yet, but it seemed necessary for my naive argument and for pi_0 being a group to be equivalent to the existence of a homotopy inverse.
 
Tim Campion
Tim Campion
10:32 PM
In Peter's note he assumes there is a map giving the homotopy inverse
wait, the assumption is that the shear map is an equivalence
sorry
so we're at three different ways of saying "grouplike" and it's not clear (to me) which are equivalent when the multiplication is not associative
 
jdc
jdc
10:45 PM
Ah, OK, thanks for clarifying
 
Peter Haine
Peter Haine
@TimCampion I'm not sure either, though it seems to me that the shear map being an equivalence is the most natural condition. It is also nice that it can be formulated in any ∞-category.
 
Peter Haine
Peter Haine
11:09 PM
Maybe useful to say: the proof of the splitting given in that note basically involves nothing. Once you have the right notions in place, you just draw the diagram
Here 1_C is the terminal object of the ambient ∞-category C, u is the unit of the multiplication on M, and sh is the shear map. The pullback of the top row is M × ΩM, and the pullback of the bottom row is the free loop object LM. So if the shear map is an equivalence, the induced map on pullbacks M × ΩM → LM is an equivalence.
 

« first day (2849 days earlier)      last day (554 days later) »