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Let $G_{\bullet}$ be a simplical group. I denote by $D_n \subset G_n$ the subgroup generated by all the degeneracy maps $s_i:G_{n-1} \to G_n$. I also denote $$M_n = \bigcap_{i>0} \mathrm{Ker} d_i \subset G_n$$ and $$Z_n = \bigcap_{i\geq 0} \mathrm{Ker} d_i \subset G_n$$ When $G_{\bullet}$ is a simplicial abelian group it is well known and there many references to the fact that

$$M_n \cap D_n = \{e\}$$

I know how to prove it for the non-abelian case, but I believe there should be a reference somewhere. In fact I'm actually interested in the weaker claim that :

$$Z_n \cap D_n = \{e\}$$

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Tomer: The claim you make about $M_n\cap D_n$ being trivial in the non abelian case is not true. In fact that condition is equivalent to the Moore complex of $G$ being a crossed complex in the sense of Brown and Higgins. This seems to be first proved in the thesis of Ashley (N. Ashley, Simplicial T-Complexes: a non abelian version of a theorem of Dold-Kan, Dissertations Math., 165, (1989), 11 – 58). It is also a lemma in one of the two papers by Brown and Loday that if $G_2=D_2$, so there are no new non-degenerate elements in $G_2$, just ones there because of $G_1$, then $\partial(M_2\cap D_2)=[Ker d_0,Ker d_1]$. This condition implies that $M_1\to M_0$ is a crossed module. The kernel of that crossed module will not be trivial in general but is $Z_2\cap D_2$.

I think you have missposed the question (in its present form) and that there is some extra condition that you have omitted. (It is also possible that I have misunderstood exactly what you are asking!) To me it seems that the homotopy types that you can model with simplicial groups having your first condition would have vanishing Whitehead products and thus be products of Eilenberg-MacLane spaces.

As to a reference for this stuff, apart from the original sources, discussions of this area can be found in two of the draft texts to be found via my n-Lab pages (Menagerie and Profinite algebraic homotopy). They are also in papers of Pilar Carrasco and Antonio Cegarra (1991 JPAA).

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Minor correction: having vanishing Whitehead products does not imply a space is a product of Eilenberg-MacLane spaces. Consider the space $X = P_3 Q S^2$. Since it is a 3-type, the only Whitehead product to worry about is $\pi_2 \times \pi_2 \to \pi_3$, which vanishes since it agrees with the one for $QS^2$, which is an infinite loop space. On the other hand $X$ is not equivalent to $K(\pi_2X,2) \times K(\pi_3X,3)$ because the operation $\pi_2 \to \pi_3$ given by precomposing with the generator of $\pi_3(S^2)$ is non-zero on $X$ (but vanishes for the product of Eilenberg-MacLane spaces). –  Omar Antolín-Camarena Sep 24 at 15:16
    
Thanks for the clarification. Can you add what is $Q$ to help `the reader'? Has anyone calculated the algebraic models (crossed square, or whatever) for this 3-type? –  Tim Porter Sep 25 at 6:12
    
$Q=\Omega^\infty\Sigma^\infty$ so that $QS^2 = \mathrm{colim}_{n\to\infty} \Omega^n S^{n+2}$. I'd like to know about algebraic models for its 3-type too. (I'm sadly ignorant about that sort of thing, all I know is that this one is not modeled by a strict 3-groupoid.) –  Omar Antolín-Camarena Sep 25 at 12:51

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