# Group cohomology topologically with simplicial sets

I have a question about the usual formula for the differential in the usual projective resolution of $$\mathbb{Z}$$ as a $$G$$-module for a finite group $$G$$

Recall that for a $$G$$-module $$A$$, $$C^i(G,A)$$ is defined as the abelian group of functions $$G^i\to A$$ and the differential is $$d(f)(g_0,...,g_i)=$$ $$g_0\cdot f(g_1,...,g_i) + \displaystyle\sum_{j=1}^i(-1)^jf(g_0,...,g_{j-1}g_j, ..., g_i) + (-1)^{i+1}f(g_0,...,g_{i-1})$$

I was trying to find an interpretation for this formula, and for this I found out that there was a nice topological point of view on group cohomology. Pick a $$K(G,1)$$ space $$X$$, and let $$p:\tilde{X}\to X$$ be its universal covering space.

Then $$\pi_1(X)=G$$ acts on $$\tilde{X}$$ and thus on the singular complex $$C_*(\tilde{X})$$, making $$C_*(\tilde{X})\to \mathbb{Z}$$ a projective (actually free since the action on $$C_*(\tilde{X})$$ comes from an action on the simplices and the action of $$G$$ on $$\tilde{X}$$ is free) resolution of $$\mathbb{Z}$$ as a trivial $$G$$-module ( the complex $$C_*(\tilde{X})\to \mathbb{Z}$$ is exact because $$\tilde{X}$$ is simply connected and has zero higher homotopy groups (because so does $$X$$), thus is contractible, because it's a CW-complex)

Now we also know that $$|BG|$$ is a $$K(G,1)$$ space, where $$BG$$ is a the nerve of $$G$$ (seen as a category), and we already know its universal cover, it's $$|EG|$$ (I don't know if it's standard notation, so let me make it clear) where $$(EG)_n = G^{n+1}$$ and $$d_i(g_0,...,g_n) = (g_0,...,\widehat{g_i},...,g_n)$$ and $$s_i(g_0,...,g_n) = (g_0,...,g_i,g_i,...,g_n)$$; which gives another explicit cochain complex whose cohomology is $$H^*(G,\mathbb{Z})$$. I had already seen this complex and now this gives me a topological interpretation for it.

But my trouble is with the first one I introduced. Indeed the formula for $$d$$ seems like a twisted version of $$\displaystyle\sum_{j=0}^i(-1)^j d_j$$ where $$d_j$$ is the boundary of $$BG$$; twisted by the action of $$G$$ but only on the first summand. I'd like to understand this connection with BG (and not $$|BG|$$, that I understand, I think) more precisely than "it looks a bit similar":

How do we get this formula for $$d$$ from the nerve $$BG$$ ? What is the topological interpretation of this formula ?

• Do you know about cohomology with local coefficients? (Or, better, sheaf cohomology?) Jan 6, 2019 at 17:03
• @MoisheCohen I know the definition of sheaf cohomology and some of its applications but that's about it; and the same for cohomology with local coefficients. But since I'm only looking for some interpretation of these formulas, some motivation, maybe it's enough ? Jan 6, 2019 at 17:07
• The answer to your question lies in the cohomology with local coefficients $H^*(BG, V)$, where $V$ is the flat bundle (local system) associated with the $G$-module $A$. You have to think about simplicial cohomology of this local system. When you write down the cochain complex for this, you will discover the coboundary formula in the group cohomology. The "twisting" comes from the fact that the local system is (in general) nontrivial. Jan 6, 2019 at 17:11
• @MoisheCohen : when you write $BG$ you mean what I called $|BG|$ or is there some notion of cohomology with local coefficients for simplicial sets (that I don't know about ) ? And if yes, what's the flat bundle associated with $A$ ? Is it just "$A$ on objects and identifying (in a fixed way) $\hom(x,y)$ with $G$ for all $x,y$, we get the action on maps" ? * Jan 6, 2019 at 17:34
• @Max Since $A$ is a $G$-module, there is a natural map $G \to \text{End}(A)$. As $G = \pi_1 BG$, this gives a representation $\pi_1 BG \to \text{End}(A)$. Consider the locally constant $A$-valued sheaf on $BG$ corresponding to this representation - that's the associated flat bundle. You can alternatively describe this as $(EG \times A)/G \to BG$ where $G$ acts diagonally on $EG \times A$. Jan 6, 2019 at 20:21

Consider the simplical set $$EG$$. It has a free $$G$$-action and is contractible, so that $$C_*(\mathbb Z[EG])$$ is a complex of free $$\mathbb Z[G]$$-modules with a quasi-isomorphism $$C_*(\mathbb Z[EG]) \to \mathbb Z[0]$$.

$$EG$$ in degree $$n$$ is $$G^{n+1}$$ so this chain complex is $$\mathbb Z[G^{n+1}]$$ in degree $$n$$, and the differential is $$\sum_i (-1)^i d_i$$.

($$d_i$$ is the one I mentioned in the question, and the $$G$$-action is coordinate-wise by translation)

The point is that the writing $$\mathbb Z[G^{n+1}]$$ doesn't make the free-ness apparent, and one way to do so will introduce the $$G$$-action on $$d_0$$.

Namely, if $$X$$ is an arbitrary $$G$$-set, then $$G\times X^{triv}\cong G\times X$$ via $$(g,x)\mapsto (g,gx)$$ (this is obviously a bijection, and it is a simple computation to check that it is $$G$$-equivariant). In particular, $$G^{n+1}\cong G\times (G^n)^{triv}$$ via $$\varphi: (g,x)\mapsto (g, g^{-1}x)$$. One issue with this is that it doesn't respect the simplicial structure, but of course, it shouldn't, and can't ! $$EG$$ is not $$G\times -$$ as a $$G$$-simplicial set. A bigger issue is that the formula for $$d_0$$ we get is not that good if we do this. But luckily, in this case, we can trivialize the $$G^n$$ part in several steps.

Indeed, $$G^{n+1} = G^{n-1}\times G\times G\cong G^{n-1}\times G\times G^{triv}$$, and then you can again remove one $$G$$ by doing $$G\times G\times G^{triv}\cong G\times (G\times G^{triv})^{triv}$$, and so on and so forth. If you work this out, you'll see that the isomorphism is given, on $$G^{n+1}$$, by $$G\times (G^n)^{triv}\cong G^{n+1}, (g_0,...,g_n)\mapsto (g_0,g_0g_1,g_0g_1g_2,...., g_0...g_n)$$ (you can check a posteriori that this is a $$G$$-map, and a bijection). Call this $$\varphi^{-1}$$.

If we compute what $$d_i$$ looks like with this presentation, we see that for $$0\leq i< n$$, $$\varphi \circ d_i \circ \varphi^{-1} (g_0,...,g_n) =\varphi\circ d_i (g_0, g_0g_1,...,g_0...g_n) = \varphi (g_0,g_0g_1, ...,\widehat{g_0...g_i} ,..., g_0...g_n) = (g_0,...,g_ig_{i+1},...,g_n)$$

For $$i=n$$, $$\varphi \circ d_n \circ \varphi^{-1} (g_0,...,g_n)= \varphi\circ d_n(g_0,g_0g_1,...,g_0...g_n) = \varphi(g_0,g_0g_1,...,g_0...g_{n-1}) = (g_0,...,g_{n-1})$$.

The important point is to observe that for $$i=0$$, this does not send $$\{e\}\times G^n$$ to $$\{e\}\times G^{n-1}$$

So if we write the $$n$$-term of our chain complex as $$\mathbb Z[G]\otimes_\mathbb Z\mathbb Z[(G^n)^{triv}]$$ to witness that it is indeed a free $$G$$-abelian group, and compute the differential on $$\hom_G(\mathbb Z[G]\otimes_\mathbb Z[G^n],A)$$, we see that if we view elements of this as set maps $$G^n\to A$$, then for all differentials except for $$d_0$$, we have nothing to do, but precomposition by $$d_0$$ makes us go out of $$G^n$$ so there is something to say.

Precomposition by $$d_0$$ looks like this : if $$f: G^n\to A$$, then we can view $$f$$ as a $$G$$-map $$\tilde f: G\times G^n\to A$$ by sending $$(g_0,...,g_n)$$ to $$g_0\cdot f(g_1,..,g_n)$$.

So now, the map $$G^{n+1}\to A$$ associated to $$\tilde f\circ d_0$$ is defined by $$\tilde f\circ d_0 (e,...,g_{n+1})$$, and this is $$\tilde f(g_1,..., g_{n+1}) = g_1\tilde f(e,g_2,...,g_{n+1}) =g_1 f(g_2,...,g_{n+1})$$.

So this gives us the desired $$d_0$$, where the other $$d_i$$'s preserve $$\{e\}\times G^n$$, so there is nothing to say.