6
$\begingroup$

Let $X\in \mathrm{sSet}$ and $FX$ be the Milnor's construction (model for $\Omega\Sigma |X|$) - in each dimension $n$ this is the free group on $X_n$ with one relation $*=1$. I'm interested in $\mathbb Z FX$ - the level-wise group ring on this construction - which homotopy groups gives the integral homology of $FX$. From the classical Bott-Samelson theorem we know that (over the field) homology of $\Omega\Sigma |X|$ as Hopf algebra is isomorphic to the tensor algebra on reduced homology of $X$ (as far as I understand, restriction to field here is given only to ensure Hopf algebra structure on homology). I'm trying to trace this fact back to simplicial level. Note that $\mathbb Z FX$ is also a Hopf algebra (comultiplication is given by diagonal) and homotopy groups of the abelianization of Milnor's construction are exactly reduced homology of $X$. So my question here: is there a (canonical!) morphism of simplicial Hopf algebras $T(FX_{ab}) \to \mathbb Z FX$ (or in another direction) which is homotopy equivalence on the underlying simplicial sets?

Few observations here:

  1. By Magnus-Witt, associated graded $\bigoplus I^k/I^{k+1}$ of $\mathbb Z FX$ is isomorphic to the tensor algebra $T(FX_{ab})$ as algebras (here $I^k$ are powers of augmentation ideal). Also this augmentation ideal filtration give rise to the cobar spectral sequence $E^1_{p,q}=\pi_p(I^q/I^{q+1})\Rightarrow \pi_{p+q}\mathbb Z FX$. Curtis used the same spectral sequence for Kan's construction $GX$ to compute homology of loop spaces. So, if my question have positive answer, the corresponding spectral sequence should collapse on first page and all extensions should be trivial. Not sure, how to see this and how to extract homotopy equivalence from this spectral sequence. Also, Hopf algebra structure here is a mystery for me. Advantage of this approach is that we would have combinatorial control over elements like $g^{-1}$ in group ring and their image in tensor algebra, since $g^{-1}-1=-g-1 \ \mathrm{mod} \ I^2$.

  2. From the short exact sequence of free abelian groups $I\to \mathbb Z FX\to \mathbb Z$ we can choose a splitting (even naturally with respect to $X$ since it serves as a basis) which will give the isomorphism of abelian groups $\mathbb Z FX\cong \mathbb Z\oplus I$. We can continue this process and get an isomorphism of abelian groups between the group ring and it's associated graded. Not sure if homotopy magic will help to turn this "map" to isomorphism of Hopf algebras.

  3. Another point of view is an isomorphism of simplicial algebras $\mathbb Z JX\cong T(FX_{ab})$, here $JX$ denote free simplicial monoid on $X$ (aka James construction). Now, by theorem of Quillen, group completion $JX\to FX$ is a homotopy equivalence, so looks like we done. But in this approach the inverse of homotopy equivalence is untrackable, so one really can not understand images in tensor algebra of elements like $g^{-1}$ from group ring.

So, all of these observations are different pieces of same puzzle which I can not put together. Will appreciate any help!

$\endgroup$
  • $\begingroup$ What does the notation "$X_{ab}$" mean? $\endgroup$ – Charles Rezk Feb 13 '16 at 15:14
  • $\begingroup$ @Charles, by $FX_{ab}$ I meant $(F[X])_{ab}$ - abelianization of Milnor's construction - this is a free abelian group on simplicial set $X$, quotient out by subgroup $\mathbb Z [*]$, here $*$ is distinct point. $\endgroup$ – ghknbhxbfhdby Feb 14 '16 at 2:17
  • $\begingroup$ Your homotopy equivalences are only homotopy equivalences when X is connected, so it's unlikely you can find a natural homotopy inverse $\endgroup$ – Charles Rezk Feb 14 '16 at 2:24
  • $\begingroup$ I see. What if we restrict to reduced simplicial sets? Aside from describing homotopy inverse, it will be interesting to understand relation between congruence modulo some power of $I$ and homotopy equivalence (something like $a=b \ \mathrm{mod} \ I^n \Rightarrow a\sim b$- I doubt that this is true in this form, but maybe something similar). As far as I understand, this is exactly what cobar spectral sequence codify $\endgroup$ – ghknbhxbfhdby Feb 14 '16 at 2:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.