Choose a arbitrary discrete group $G$. The classifying space $BG$ of $G$ is 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. In other words is it possible to construct a homotopy $h_t: EG \times I \to EG$ which contracts $EG$ to a point explicitly? That is $h_0(EG) =EG, h_1(EG) = \{*\}$. How looks it concretely geometrically?

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 $EG$?


1 Answer 1

  1. As far as I know, this construction, due to Milnor, is the most "conventional" one.

  2. To prove contractibility of Milnor's $EG$, fix the vertex $p=[1]$ of $EG$. By the construction, $[1]$ together with any simplex $s=[g_0,...,g_n]$ in $EG$ span the simplex $\Delta_s=[1,g_0,...,g_n]$ in $EG$ (unless $g_0=1$ in which case we take $\Delta_s=[1,g_1,...,g_n]$). In particular, for every point $q$ in $EG$ we obtain a canonical affine line segment $pq\subset \Delta_s$, where $q$ is in $s$.

Thus, one defines the "straight-line homotopy" of the identity map to the constant map $EG\to \{p\}$ using the above line segments. Since $G$ is discrete, this homotopy is continuous. This is your explicit contraction.

Addendum. In fact, Milnor in his paper also constructs a contractible $G$-complex for an arbitrary topological group $G$ (not necessarily discrete) by taking a countably infinite join of copies of $G$. Milnor proves weak contractibility of his $EG$. For a proof of contractibility, see for instance Proposition 14.4.6 in

Tammo Tom Dieck, "Algebraic Topology," EMS Textbooks in Mathematics, 2008

The contraction is also explicit.


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