# In a Simplicial group the Degenerate elements don't intersect the kernel of the face maps.

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 :

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

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 $Z_2\cap D_2$.
• 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). Sep 24, 2014 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? Sep 25, 2014 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.) Sep 25, 2014 at 12:51
• Further thought, Whitehead's original papers on crossed complexes introduced the $\Gamma_n$ functors and in the 1991 paper of Carrasco and Cegarra they prove that the vanishing of the $\Gamma_n(X)$ is equivalent to the homotopy type of $X$ being completely modelled by its crossed complex. This can also be derived from results in the work of Ronnie Brown and Phil Higgins, as well as being generalised by Baues. The condition of $M\cap D=1$ and its interpretation still could do with exploration however. Aug 19, 2016 at 8:11