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In my few months of studying group cohomology, I've seen two "standard" complexes that are introduced:

  • We let $X_r$ be the free $\mathbb{Z}[G]$-module on $G^r$ (so, it has as a $\mathbb{Z}[G]$-basis the $r$-tuples $(g_1,\ldots,g_r)$ of elements of $G$). The $G$-module structure of $X_r$ comes by virtue of being a $\mathbb{Z}[G]$-module.
    The boundary maps $\partial_r:X_r\to X_{r-1}$ are


  • We let $E_r$ be the free $\mathbb{Z}$-module on $G^{r+1}$ (so, it has as a $\mathbb{Z}$-basis the $(r+1)$-tuples $(g_0,\ldots,g_r)$ of elements of $G$). The $G$-module structure of $E_r$ is defined by $g(g_0,\ldots,g_r)=(gg_0,\ldots,gg_r)$. The boundary maps $d_r:E_r\to E_{r-1}$ are

We may then proceed to compute the cohomology of $G$ with coefficients in a $G$-module $A$ using $$0\to \text{Hom}_G(X_0,A)\to\text{Hom}_G(X_1,A)\to\cdots$$ or by using $$0\to \text{Hom}_G(E_0,A)\to\text{Hom}_G(E_1,A)\to\cdots$$ Elements of $\text{Hom}_G(X_r,A)$ are "inhomogeneous cochains" and elements of $\text{Hom}_G(E_r,A)$ are "homogeneous cochains". In either case, all that matters is what happens to the basis elements, so really we can say that an "inhomogeneous cochain" is a function $f:G^r\to A$, and that a "homogeneous cochain" is a function $f:G^{r+1}\to A$ that satisfies $f(gg_0,\ldots,gg_r)=g\cdot f(g_0,\ldots,g_r)$.

Lang defines them both in his Topics in cohomology of groups, and says

... we have a $\mathbb{Z}[G]$-isomorphism $X\xrightarrow{\approx}E$ between the non-homogeneous and the homogeneous complex uniquely determined by the value on basis elements such that $$(\sigma_1,\ldots,\sigma_r)\mapsto (e,\sigma_1,\sigma_1\sigma_2,\ldots,\sigma_1\sigma_2\ldots \sigma_r)$$

but Serre defines the only the homogenous cochains in Local Fields and then says that a cochain

... is uniquely determined by its restriction to systems of the form $(1,g_1,g_1g_2,\ldots,g_1\cdots g_i)$. That leads us to interpret the elements of $\text{Hom}_G(E_r,A)$ as "inhomogeneous cochains", i.e. as functions $f(g_1,\ldots,g_i)$ of $i$ arguments, with values in $A$, whose coboundary is given by ...

To put it bluntly, my question is: Why are we doing this? I can think of some possible reasons:

  • Historical - perhaps one way was defined first, now the other is more popular, but the older definition is still included out of tradition.

  • Practical - perhaps there are important computations that are significantly easier to see or do using one or the other approach, or where it is useful to switch between them for some reason.

  • Big picture - perhaps there is a high-level interpretation of one or both approaches that ties in with some other field where (co)homology plays a role.

So, what's the real motivation for defining both "homogeneous" and "inhomogeneous" cochains?

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I don't know what the real answer is, but I'll just point out that the inhomogeneous definition has the advantage that it's defined on a tuple of smaller size (r instead of r+1), while the homogeneous definition has the advantage that the boundary map is "simpler" (omitting a term rather than combining via a product). –  Ted Sep 18 '11 at 17:22
The second complex makes it obvious that we are dealing with the homology of a simplicial set. Cycles and cocycles on the first one, on the other hand, are what we usually find in nature (derivations, factor sets in extensions, etc) –  Mariano Suárez-Alvarez Sep 18 '11 at 18:29
@Mariano: I'm afraid I don't really understand your explanation of the first complex; Serre, for example, talks about factor sets using homogeneous cochains. Could expand a bit more on the places we see inhomogeneous cochains "in nature" (in an answer, if you want)? –  Zev Chonoles Sep 18 '11 at 22:55
By the way, one should keep in mind the fact that the two complexes are in fact isomorphic! –  Mariano Suárez-Alvarez Sep 23 '11 at 3:27
@Mariano: Indeed - that was one of the reasons I found the situation a bit perplexing. –  Zev Chonoles Sep 23 '11 at 3:38
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2 Answers

up vote 5 down vote accepted

Let me give you three examples where nature picks your first complex:

First: Let $G$ be a group, let $A$ be a $G$-module, and let $A\rtimes G$ be the direct product (as a set, this is $A\times G$, and it becomes a group with multiplication such that $(a,g)\cdot(b,h)=(a+g\cdot b,gh)$ for all $a$, $b\in A$ and all $g$, $h\in G$) Consider the projection map $p:A\rtimes G\to G$, which is a group homomorphism. A section of $p$ is a group homomorphism $s:G\to A\rtimes G$ such that $f\circ s=\mathrm{id}_G$. It is immediate to check that a section determines uniquely and is determined uniquely by a function $\sigma:G\to A$ such that $$g\cdot\sigma(h)+\sigma(g)=\sigma(gh)$$ for all $g$, $h\in G$; indeed, the relation between $s$ and $\sigma $ is that $s(g)=(\sigma(g),g)$ for all $g\in G$. The function $\sigma$ is then a $1$-cocycle defined on your first complex.

Second: Consider an extension

          i      f
0 ---> A ---> E ---> G ---> 1

of a group $G$ by an abelian group $A$ (whose operation I'll write $+$). Let $\sigma:G\to E$ be a set-theoretic section of $f$. For all $g$, $h\in G$ we have $f(\sigma(g)\sigma(h))=f(\sigma(g))f(\sigma(h))=gh=f(\sigma(gh))$, so that there exists a unique element $\alpha(g,h)\in A$ such that $$\sigma(g)\sigma(h)=\iota(\alpha(g,h))\sigma(gh).$$

There is an action of $G$ on $A$ such that $$\iota(g\cdot a)=\sigma(g)\iota(a)\sigma(g)^{-1}$$ for all $g\in G$ and all $a\in A$. It is eeasy to check that this is indeed an action of $G$ on $A$ by group automorphisms (here it is where we need that $A$ be abelian) In other words, $A$ is a $G$-module.

Now, whenever $g$, $h$, $k$ are in $G$ we have $$(\sigma(g)\sigma(h))\sigma(k)=\iota(\alpha(g,h))\sigma(gh)\sigma(k)=\iota(\alpha(g,h)+\alpha(gh,k))\sigma(ghk)$$ and $$\sigma(g)(\sigma(h)\sigma(k))=\sigma(g)\iota(\alpha(h,k))\sigma(hk)=\iota(g\cdot\alpha(h,k))\sigma(g)\sigma(hk)=\iota(g\cdot\alpha(h,k)+\alpha(g,hk))\sigma(ghk).$$ Since multiplication in $G$ is associative, the left-hand sides in these last two equations are equal, so so are their right hand sides---and since $\iota$ is injective, we see that $$g\cdot\alpha(h,k)+\alpha(g,hk)=\alpha(g,h)+\alpha(gh,k)$$ or, equivalently, that $$g\cdot\alpha(h,k)-\alpha(gh,k)+\alpha(g,hk)-\alpha(g,h)=0.$$ This means that $\alpha$ determines a $2$-cocyle on your first the complex.

Third: If $G$ is a group, the category of $G$-modules over a field $k$ is a monoidal category $\mathscr M_G$ with respect to the tensor product of representations. If $\alpha:G\times G\times G\to k^\times$ is a $3$-cocycle defined on your first complex and with values in the multiplicative group of $k$, then one can "twist" the associativity isomorphisms of $\mathscr M_G$ using $\alpha$ to obtain a different, slightly more fun monoidal category $\mathscr M_G(\alpha)$, and if you work this out in detail, you will see that again the cocycle condition with respect to your first complex is precisely the pentagon condition for a monoidal structure.

These are just three instances where nature picks inhomopgenous cochains.

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The inhomogeneous cochain construction is a standard free resolution of $\mathbb Z$ (the trivial $G$-module) as a $\mathbb Z[G]$-module, and it is explicitly constructed to be such. Since taking $G$-invariants is the same as forming $Hom_G(\mathbb Z, A)$, this is what is needed to compute derived functors of this operation (which is what group cohomology is, from a derived functor point of view).

On the other hand, homogenous cochains are what you get if you compute the cohomology of local systems (twisted coefficients) on the classifying space for $G$, which is how group cohomology first arose (explicitly --- there were implicit examples of group cohomology classes much earlier) in the literature. The "homogeneity" reflects the fact that we are computing with a certain $G$-equivariant simplicial complex.

Loosely, and roughly, speaking, the inhomogeneous picture is more algebraic, and the homogeneous picture is more topological.

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The homogeneous cocycles are very similar to the forming of a boundary of a topological CW-complex. It therefore gives intuition that the homogeneous approach is more useful in algebraic topology. In fact, the homogeneous cocyles formula is the same as Cech cohomology formula when considering cohomology of sheaves. –  LinAlgMan Aug 5 '13 at 13:08
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