For a discrete group $G$ I have the following two definitions, which I think are correct:

  • The nerve of $G$ is $NG$, a simplicial set whose $n$-simplices are $G^n$ ($G^0$ being the trivial group $\{1\}$) and whose face maps are $d_0(g_1,...,g_n)=(g_2,...,g_n)$, $d_n(g_1,...,g_n)=(g_1,...,g_{n-1})$, and $d_i(g_1,...,g_n)=(g_1,...,g_i g_{i+1},...,g_n$ for all $0<i<n$. The degeneracy maps are $s_0(g_1,...,g_n)=(1,g_1,...,g_n)$ and $s_i(g_1,...,g_n)=(g_1,...,g_i,1,g_{i+1},...,g_n)$ for $i>0$. The geometric realization of $NG$ is a space which may be viewed as a CW complex with $n$-cells that can be identified with the nondegenerate simplices of $NG$, that is, those that don't have $1$ as any of the $g_i$ in $(g_1,...,g_n)$. Generally this space is considered to be the (a?) classifying space for $G$.
  • The bar resolution of $G$ is the sequence $0\leftarrow \mathbb{Z}\leftarrow G^0 \leftarrow G^1 \leftarrow...$ with boundary maps $\partial_n(g_1,...,g_n)$$=g_1*(g_2,...,g_n)$$+\overset{n-1}{\underset{i=1}{\sum}}(-1)^i(g_1,...,g_i g_{i+1},...,g_n)$$+(-1)^n(g_1,...,g_{n-1})$.

I'm looking for a way to understand the cohomology groups of $G$, as defined using the bar resolution, in terms of the cellular cohomology groups of $BG$. It seems like there should be a very direct relationship between the face maps $d_i$ which can be thought of as determining the cellular structure of $BG$ and the bar resolution's boundary map $\partial_n$; in fact $\partial_n$ looks almost exactly like $\overset{n}{\underset{i=0}{\sum}}(-1)^i d_i(g_1,...,g_n)$, except that the first term is $g_1*(g_2,...,g_n)$ instead of just $(g_2,...,g_n)$. Essentially, my question is where this "extra multiplication by $g_1$" comes from. I also wonder if my definitions are missing something or my understanding of them has a gap. I'd like to understand this in a way that doesn't appeal to the "other" usual construction of $BG$ as a quotient of $EG$ if at all possible.

  • 1
    $\begingroup$ The free abelian group on the nerve is a simplicial abelian group; applying the Dold-Kan correspondence to it produces the bar resolution: ncatlab.org/nlab/show/Dold-Kan+correspondence $\endgroup$ Sep 27, 2020 at 21:29
  • $\begingroup$ I'm sure it's a matter of my own ignorance, and not a problem with your comment, but I honestly can't really grasp the relevance of your comment to my question. I generally have a lot of trouble understanding anything at all on nlab's pages. $\endgroup$
    – Xindaris
    Sep 28, 2020 at 15:26
  • $\begingroup$ Dold-Kan tells you how the face maps of the nerve produce the boundary maps of the bar resolution. Admittedly I haven’t worked this out in detail myself. $\endgroup$ Sep 28, 2020 at 16:12

1 Answer 1


I believe I have settled this issue for myself, and the key ingredient is indeed the Dold-Kan correspondence. However, I find a synthesis of these two expositions to be much more comprehensible than anything nlab has to offer: (1), (2) They also both refer to the same book reference, namely this one, which is also helpful in understanding the result.

The basic idea is that any chain complex corresponds to a simplicial abelian group, and any simplicial abelian group $A$ has a corresponding chain complex $A_*$. The image of the degenerate simplices under this correspondence forms a subcomplex $DA_*$, and there is another subcomplex $NA_*$ which turns out to have the property that $NA_*\oplus DA_*\cong A_*$. In other words, $A_*/DA_*\cong NA_*$. The Dold-Kan correspondence theorem itself says that $A\to NA_*$ gives an equivalence of categories between chain complexes and simplicial abelian groups, which then gives a way to "normalize" certain chain complexes, including the chain complex defining the simplicial homology of a topological space.

In order to apply this to my situation, which involves cohomology rather than homology, it is necessary to "dualize" this result. The key insight, I think, is that actually any simplicial set $X$ can be "upgraded" into an abelian simplicial group $\mathbb Z (X)$, whose $n$-simplices $\mathbb Z(X)[n]$ are $\mathbb Z (X[n])$, which consists of formal sums $\sum_{x\in X}k_x x$ such that $k_x\in \mathbb Z$ and only finitely many of them are nonzero. Then any cochain complex $C^n(X,A)$ defined as $\lbrace f:X[n]\to A \rbrace$ can instead be thought of as $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$, and if $C_0^n(X,A)$ is the functions which are zero on any nondegenerate $n$-simplex, then $C^n(X,A)/C^0_n(X,A)\cong Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$ and the Dold-Kan correspondence gives a chain equivalence between $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$ and $ Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$, which makes the natural projection $C^n(X,A)\to C^n(X,A)/C^0_n(X,A)$ into a chain equivalence also.

After this, I found that it does appear necessary, or at least convenient, to go through $\mathcal E G$, the simplicial set with $\mathcal E G[n]=G^{n+1}$ and $d_i$, $s_i$ given by deletion and repetition of the $i$th element respectively, and whose geometric realization is $EG$. Even though this (or the bar resolution) may be viewed as a simplicial group, $G$ may not be abelian, so that isn't particularly helpful. It's better to think of it as a simplicial set and use a slightly modified version of the result above (essentially, the same formulation can be done for subcomplexes of $C^n(X,A)$ to "normalize" them). Then (in broad strokes) if $C^n(G,A)=\lbrace f:G^{n+1}\to A\text{ such that }gf=fg\text{ for all }g\in G \rbrace$, there is an injective map $C^n(G,A)/C^n_0(G,A)\to C^n_\text{cell}(EG,A)$, and an injective map $C^n_\text{cell}(BG,A)\to C^n_\text{cell}(EG,A)$; if $g$ acts trivially on $A$ then both of which have the same subcomplex as their image. The isomorphism follows from that, and if $G$ doesn't act trivially on $A$ then that first injective map can still be used to get an isomorphism to the cohomology with local coefficients instead.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .