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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.