# Is the homology of the Moore complex associated to a classifying space $BG$ the group homology of $G$?

Let $$G$$ be a group. We define its standard resolution by setting

$$C_n = \mathbb{Z}G^{n+1}$$

to be the free $$\mathbb{Z}$$-module on $$G^{n+1}$$ for each $$n \geq 0$$, with a $$\mathbb{Z}[G]$$-module structure given by

$$g(g_0,\dots,g_n) := (gg_0,\dots,gg_n)$$

for each $$g,g_0,\dots,g_n \in G$$ and defining the differentials as

$$\partial_n : (g_0, \dots,g_n) \in C_n \mapsto \sum_{i=0}^n(-1)^i(g_0,\dots,\widehat{g_i},\dots,g_n) \in C_{n-1}.$$

This gives a projective resolution $$(C_\bullet,\partial_\bullet) \xrightarrow{\varepsilon} \mathbb{Z} \to 0$$ and so we can compute the group homology of $$G$$ as the homology of tensoring this complex with $$\mathbb{Z}$$,

$$H_\bullet(G) = H_\bullet(\mathbb{Z} \otimes_{\mathbb{Z}G} C_n).$$

From $$G$$ we can also form its classifying space $$BG$$, defined as the simplicial set

$$BG_n = \{(g_0,\dots,g_n) : g_i \in G\}$$

with face and degeneracy operators

$$d_i(g_0,\dots,g_n) = (g_0,\dots,g_ig_{i+1},\dots,g_n),\\ s_i(g_0,\dots,g_n) = (g_0,\dots,g_i,1,g_{i+1},\dots,g_n).$$

From here one can take the Moore complex of $$BG$$

$$\dots \to \mathbb{Z}BG_n \xrightarrow{\partial} \mathbb{Z}BG_{n-1} \to \cdots \to \mathbb{Z}BG_0$$

given by the free $$\mathbb{Z}$$-modules generated by each set $$BG_n$$ and the boundary operator given by $$\partial = \sum_{i=0}^n (-1)^id_i$$, that is,

$$\partial(g_0, \dots,g_n) = \sum_{i=0}^n(-1)^i(g_0,\dots,g_ig_{i+1},\dots,g_n).$$

The homology of $$BG$$ is then defined as the homology of this complex,

$$H_\bullet(BG) := H(\mathbb{Z}BG_\bullet,\partial_\bullet).$$

The constructions look suspiciously similar, so my question is:

What is the relationship between these two? Do we have $$H_\bullet(G) = H_\bullet(BG)$$, and if so, is there an isomorphism between the aforementioned complexes that I am missing?

Yes, the two are isomorphic, there are plenty of ways to see that. I'll begin by sketching a topological argument and then make it more concrete below. If you're not comfortable with algebraic topology you can skip the first part (except the definition of $$EG$$) up to "why yes there is"

If you're comfortable with algebraic topology, here's one way : note that what you wrote $$C_\bullet$$ is actually the singular chain complex of the simplicial set $$EG$$ which is defined in the obvious way, and has a free $$G$$-action. Moreover, this simplicial set $$EG$$ is contractible, as it is the nerve of a contractible groupoid (this groupoid is simply the groupoid that has $$G$$ as objects, and one arrow precisely between any two objects).

It follows that $$C_\bullet$$ is a projective resolution of the $$G$$-module $$\mathbb Z$$, so that what you called $$H_\bullet(G)$$ is in fact $$\mathrm{Tor}^{\mathbb Z[G]}_\bullet(\mathbb Z, \mathbb Z)$$

On the other hand, let's have a look at $$H_\bullet(BG)$$ : $$BG$$ is a simplicial set which is the nerve of the group $$G$$. It is therefore a Kan complex, and one can easily compute its homotopy groups to be $$G$$ for the fundamental group and $$0$$ for higher homotopy groups.

Therefore if you take a universal covering $$\tilde EG\to |BG|$$ (the notation is not a coincidence, I'll explain later - note in the meantime that such a universal covering has $$\tilde EG/G \cong |BG|$$), it has a free $$G$$-action and is contractible; so that its singular chain complex $$C_\bullet(\tilde EG)$$ is a projective resolution of $$\mathbb Z$$ as a $$G$$-module, therefore to compute $$\mathrm{Tor}^{\mathbb Z[G]}_\bullet (\mathbb Z,\mathbb Z)$$, one may use this resolution and take $$\mathbb Z\otimes_{\mathbb Z[G]}C_\bullet(\tilde EG)$$, which, by $$\tilde EG/G\cong |BG|$$ is nothing but $$C_\bullet(|BG|)$$, whose homology is $$H_\bullet(|BG|)$$, and this coincides with simplicial homology, i.e. $$H_\bullet(BG)$$.

So the two homologies are isomorphic, as they are isomorphic to $$\mathrm{Tor}^{\mathbb Z[G]}_\bullet (\mathbb Z,\mathbb Z)$$.

Now this is all very abstract for an actually very concrete thing : take the geometric realization of $$EG$$ as above : it is contractible and has a free action of $$G$$, the quotient $$|EG|/G$$ is therefore a space with exactly one nonzero homotopy group, and it's equal to $$G$$ : if you know a bit about homotopy theory, this tells you that $$|EG|/G$$ is homotopy equivalent to $$|BG|$$. This suggests to maybe look at something earlier in our chain of functors: is there actually something relating $$EG/G$$ and $$BG$$ ?

Why yes, there is : $$EG$$ is the nerve of a certain category, $$BG$$ of another one, can we get a functor between the categories ?

Well send any object $$g\in EG$$ to the single object $$*$$ of $$BG$$ (here I'm seeing them both as categories), and the unique morphism $$g\to h$$ in $$EG$$ can be sent to $$h^{-1}g \in G$$ (I had written $$hg^{-1}$$ initially, it works too, but it won't work later because of the side you chose for the action - this one will fit nicely with the side of the action). One checks easily that this yields a functor $$EG\to BG$$ and so a map of simplicial sets.

Moreover, at the level of categories, one can clearly see that this exhibits $$BG$$ as $$EG/G$$, so this works for simplicial sets too.

Note that the connection between the nerve of $$EG$$ and your construction is as follows: an $$n$$-simplex of said nerve is a composable string of arrows $$g_0\to ...\to g_k$$, but as there is exactly one arrow between any two objects, this amounts to the list $$(g_0,...,g_k)$$ of objects (i.e. elements of $$G$$), and the $$i$$th boundary map, which corresponds to composition for usual nerves, corresponds to just deleting $$g_i$$ from the list in our case : so your $$C_\bullet$$ was indeed the simplicial chain complex of $$EG$$.

Now what does our functor $$EG\to BG$$ look like on the level of nerves ? Well it sends $$(g_0,...,g_k)$$ to $$g_0\to...\to g_k$$ to $$*\overset{g_1^{-1}g_0}\to ... \overset{g_k^{-1}g_{k-1}}\to *$$, that is, the map on chain complexes is, very concretely :

$$(g_0,...,g_k) \mapsto (g_1^{-1}g_0,...,g_k^{-1}g_{k-1})$$

You can now check that this is $$G$$-equivariant (with the trivial action on the right), and that it is a map of chain complexes (simply because it was a map of simplicial sets to begin with and so it commutes with boundaries ! you can also see it more concretely if you wish : if you remove $$g_i$$ on the left, then you can multiply $$g_{i+1}^{-1}g_i$$ and $$g_i^{-1}g_{i-1}$$ on the right, and you get the correct thing)

Since it is $$G$$-equivariant, it factors through $$(C_\bullet)_G \to \mathbb ZBG_\bullet$$, and you can in fact now check that this is an isomorphism (it will come from the fact that $$C_\bullet$$ is a free $$\mathbb Z[G]$$-module, and so taking the coinvariants amounts to taking a quotient)

Finally you have to understand that for any $$\mathbb ZG$$-module $$M$$, $$M_G \cong \mathbb Z\otimes_{\mathbb Z[G]}M$$

tldr: Yes, they are isomorphic, and the isomorphism comes from something much earlier in the creation of these chain complexes : it comes from the categories of whose nerve those complexes are the simplicial chain complexes

• This was very helpful, many thanks for taking the time to answer! – Guido Oct 29 '19 at 3:06