# Classifying space BG and contractable space EG

This question is probably not research level that's why I asked it previously on MSE a week ago. Unfortunately it doesn't get much attention there and I thought I would try it here.

Choose a arbitrary discrete group $$G$$. The classifying space $$BG$$ of $$G$$ is classically constructed by forming a certain contractable $$\Delta$$-complex $$EG$$ (on concrete construction of $$EG$$: see below) endowed with an action by $$G$$. $$BG$$ is obtained by taking quotient $$BG:= EG/G$$.

The concrete construction of the $$\Delta$$-complex $$EG$$ works as follows: The $$n$$-simplices of $$EG$$ are the ordered tuples

$$[g_0,g_1,...,g_n] \cong \Delta_n =\left\{x\in \mathbb {R} ^{n}:x=\sum _{i=0}^{n}t_{i}v_{i}\ {\text{with}}\ 0\leq t_{i}\leq 1\ {\text{and}}\ \sum _{i=0}^{n}t_{i}=1 \right\}$$

with $$g_i \in G$$. The $$v_i$$ are spanning $$\Delta_n$$. We obtain a $$\Delta$$-complex by attaching $$n$$-simplices to the $$(n − 1)$$ simplices $$[g_0,g_1,..., \hat{g}_i,...,g_n]$$ in standard way as a standard simplex attaches to its faces. Here $$\hat{g}_i$$ means that this vertex is deleted.

Question. Does there exist a constructive way to show that $$EG$$ is contractable. By constructive I mean how to construct an explicite homotopy $$h_t: EG \times I \to EG$$ which contracts $$EG$$ to a point. That is $$h_0(EG) =EG, h_1(EG) = \{*\}$$. How looks it concretely geometrically?

I know some abstract arguments like Whitehead's theorem that also provide a reason for contrability of $$EG$$ but this is not what I'm looking for.

I tried to define such homotopy as follows: let $$p \in [g_0,...,g_n]$$. Then $$h_t$$ "slides" step by step $$p$$ along $$(n+1)$$-simplex $$[e,g_0,...,g_n]$$ to $$[e]$$ ($$e \in G$$ the identity element). What I mean by "step by step along $$[e,g_0,...,g_n]$$"?

If we use again the identification $$[g_0,g_1,...,g_n] \cong \Delta_n$$ then as long as we sitting "inside" $$\Delta_n$$ we can interpret the $$g_i$$ as spanning vectors $$v_i$$. Let $$p = \sum _{i=0}^{n}t_{i}g_{i}$$. As $$[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$$ we can interpret $$[g_0,g_1,...,g_n]$$ as $$x = t_{-1}e + \sum _{i=0}^{n}t_{i}g_{i} \in \Delta_{n+1}$$ with $$t_{-1}=0$$. That is the "point" $$p_e:= 1 \cdot e \in [e, g_0,g_1,...,g_n]$$ is not contained in $$[g_0,g_1,...,g_n]$$ and we can define a unique line $$l_pe$$ which contains $$p_e$$ and is perpendicular to $$[g_0,g_1,...,g_n]$$ in our geometric picture $$[g_0,g_1,...,g_n] \subset [e, g_0,g_1,...,g_n]$$ corresponding to $$\Delta_n \subset \Delta_{n+1}$$.

This line uniquely intersects $$[g_0,g_1,...,g_n]$$ in a unique point $$p_l$$. Then we say that our homotopy slide $$p$$ along the unique line through $$p$$ and $$p_l$$ up to the first contact with a boundary of $$[g_0,g_1,...,g_n]$$.

Let this boundary be $$[g_0,..., \hat{g}_i,...,g_n]$$. Then we play the same game with $$[g_0,..., \hat{g}_i,...,g_n]$$ and $$[e, g_0,..., \hat{g}_i,...,g_n]$$.

Does this approach work? And is there known a more conventional "textbook" (that I still haven't found) way for the construction of $$EG$$?

• The abstract nonsense'' explanation can be found here: ncatlab.org/nlab/show/decalage – Bertram Arnold Feb 3 at 13:05
• @BertramArnold: so intuitively decalage "cleans" the obstuctions (like degenerations) of a simplicial set (more precisely it's geom realization) to be contractable? – MortyPB Feb 3 at 16:35

• Could you lose few words on your last argument that the composition G//G→*→G//G is homotopic to identity "via the unique choices of morphisms in G//G". I not fully understand what you mean. By constuction between every $g,h \in G//G$ there exist exactly one map. Why does this imply the desired claim? – MortyPB Feb 3 at 16:39