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In group cohomology, one defines $H^i(G;A)$ for $G$ a group and $A$ a $G$-module (an abelian group with a $G$-action) as the $i$-th right derived functor of the functor $$(-)^G: G-mod \rightarrow Ab, A \mapsto A^G.$$ Now, I know that one can also define a chain complex $C^*(G,A)=\{C^n(G,A)\}_{n \geq0}$ with differential $$ (df)(g_1,\ldots,g_{n+1}) =g_1 \cdot f(g_2,\ldots,g_{n+1})+ \sum_{i=1}^{n}(-1)^if(g_1,\ldots,g_i g_{i+1},\ldots,g_{n+1})+ (-1)^{n+1}f(g_1,\ldots,g_n),$$ and that $H^i(C^*(G,A)) = H^i(G;A)$. My question is, are there more differentials on $C^*(G,A)$ that give the same cohomology groups ? For example, I've found various sources where one defines derivations (normally, the kernel of the map $d: C^1(G,A) \rightarrow C^2(G,A)$) as functions satisfying the condition $f(g_1 g_2) = g_2\cdot f(g_1)+f(g_2)$, which aren't in the kernel of $d$, since the kernel of $d$ are the functions that satisfy $f(g_1 g_2) = g_1\cdot f(g_2)+f(g_1)$ (this happens, for instance, in "the Arithmetic of Elliptic curves" by Silverman). If there are more, why do they give the same cohomology groups ?

More particular, let PDer be the principal derivations (functions for which there is an $a \in A$ so that $f(g) = g\cdot a - a$, let LDer be the functions which satisfy $f(g_1 g_2) = g_1\cdot f(g_2)+f(g_1)$ and RDer be the functions which satisfy $f(g_1 g_2) = g_2\cdot f(g_1)+f(g_2)$. Is there an isomorphism

$$LDer/PDer \cong RDer/PDer$$ and if so, why ?

As always, help would be appreciated.

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Do you mean different differentials on the same sequence of groups, or do you allow chain complexes with different underlying groups? – Zhen Lin May 14 '12 at 8:43
Different differentials, the chain complex stays fixed (so $G$ and $A$ are fixed) – KevinDL May 14 '12 at 8:45
up vote 3 down vote accepted

Strictly speaking, a chain complex is a sequence of groups equipped with a differential operator, so two different differential operators gives rise to two different chain complexes.

One sufficient condition is the existence of a chain homotopy equivalence. Let's write $C^\bullet$ and $C'^\bullet$ for the two complexes. Then, we need to find chain homomorphisms $f^\bullet : C^\bullet \to C'^\bullet$ and $h^\bullet : C'^\bullet \to C^\bullet$ and a sequence of homomorphisms $\alpha^n : C^n \to C^{n-1}$, $\beta^n : C'^n \to C'^{n-1}$ such that \begin{align} d' \circ f & = f \circ d \\ d \circ h & = h \circ d' \\ d \circ \alpha + \alpha \circ d & = h \circ f - \textrm{id} \\ d' \circ \beta + \beta \circ d' & = f \circ h - \textrm{id} \end{align} It is a standard exercise to show that the homomorphisms in cohomology induced by $f^\bullet$ and $h^\bullet$ are isomorphisms.

In your chain complex, there is an obvious symmetry that we can exploit to construct "alternative" differential operators. Let $C^n = C'^n$ be the group of all functions $f : G^n \to A$, and equip $C^\bullet$ with the standard differential. For each natural number $n$, choose a permutation $\sigma_n \in S_n$. We have an obvious automorphism $\sigma_n^* : C'^n \to C'^n$ and we define $d' : C'^n \to C'^{n+1}$ by the formula below: $$d' = \sigma_{n+1}^* \circ d \circ (\sigma_n^*)^{-1}$$ It is easy to check that this makes $C'^\bullet$ into a chain complex, and $\sigma_\bullet^* : C^\bullet \to C'^\bullet$ is a chain isomorphism. Thus $C^\bullet$ and $C'^\bullet$ have the same cohomology.

One thing you might be wondering about is why the standard resolution even gives the right answer in the first place: after all, the right derived functors are supposed to be defined by injective resolutions, and this isn't one. Secretly, we are using the natural isomorphism $$H^n(G; A) \cong \textrm{Ext}^n_{\mathbb{Z} G}(\mathbb{Z}, A)$$ where $\mathbb{Z}$ is considered as a $G$-module with trivial $G$-action, and the fact that $\textrm{Ext}^\bullet$ can be computed by a projective resolution of the first variable. There is a canonical free resolution of $\mathbb{Z}$, called the bar resolution, and it is constructed as follows. Let $P_n$ be the free $\mathbb{Z} G$ module on $G^{n}$-many generators. (I am assuming $G$ is a discrete but not necessarily finite group here.) If we write the generators of $P_n$ as $[ g_1, \ldots, g_n ]$, then we can define a surjective $G$-equivariant homomorphism $p : P_0 \to \mathbb{Z}$ by $$[] \mapsto 1$$ and clearly $\ker p$ consists of those sums $\sum_i a_i g_i []$ such that $\sum_i a_i = 0$. Notice that it is generated as a $\mathbb{Z}$-module by $\{ (g - 1)[] : g \in G \}$. Let $d : P_{n+1} \to P_n$ be the $G$-equivariant homomorphism defined by $$[g_0, \ldots, g_n] \mapsto g_0 [g_1, \ldots, g_n] + \sum_{i=1}^{n-1} (-1)^i [g_0, \ldots, g_{i-1} g_i, \ldots, g_n] + (-1)^n [g_0, \ldots, g_{n-1}]$$ A standard calculation shows that $P_\bullet$ is a chain complex under this differential operator and that it is exact in positive degrees, with $\operatorname{im} (d : P_1 \to P_0) = \ker p$, so it is indeed a free resolution of $\mathbb{Z}$. Now, taking $C^\bullet = \operatorname{Hom}_{\mathbb{Z} G}(P_\bullet, A)$ gives the standard chain complex for calculating the group cohomology of $A$.

But there's another canonical free resolution $Q_\bullet$, called the normalised bar resolution, where $Q_n$ is the free $\mathbb{Z} G$-module on $(G \setminus \{ e \})^n$-many generators, and the differential $d' : P'_{n+1} \to P'_n$ is constructed essentially the same way. Notice that the chain complex $Q_\bullet$ is naturally a quotient of $P_\bullet$, in such a way that $P_n$ splits as $K_n \oplus Q_n$ as $\mathbb{Z} G$-modules, where $K_\bullet$ is the kernel of the projection $P_\bullet \to Q_\bullet$. Thus, we can construct a chain complex $K_\bullet \oplus Q_\bullet$ which is degreewise isomorphic to $P_\bullet$, but with a subtly different differential. This is another way of getting a non-standard differential on the standard chain complex.

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