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This question might not have a good answer. It was something that occurred to me yesterday when I found myself in a pub, needing to do an explicit calculation with 2-cocycles but with no references handy (!).

Review of group cohomology.

Let $G$ be a group acting (on the left) on an abelian group $M$. Then $H^0(G,M)=M^G=Hom_{\mathbf{Z}[G]}(\mathbf{Z},M)$ and hence $$H^i(G,M)=Ext^i_{\mathbf{Z}[G]}(\mathbf{Z},M).$$ Now $Ext$s can be computed using a projective resolution of the first variable, so we're going to get a formula for group cohomology in terms of "cocycles over coboundaries" if we write down a projective resolution of $\mathbf{Z}$ as a $\mathbf{Z}[G]$-module. There's a very natural projective resolution of $\mathbf{Z}$: let $P_i=\mathbf{Z}[G^{i+1}]$ for $i\geq0$ (with $G$ acting via left multiplication on $G^{i+1}$) and let $d:P_i\to P_{i-1}$ be the map that so often shows up in this sort of thing: $$d(g_0,\ldots,g_i)=\sum_{0\leq j\leq i}(-1)^j(g_0,\ldots,\widehat{g_j},\ldots,g_i).$$ Check: this is indeed a resolution of $\mathbf{Z}$. So now we have a "formula" for group cohomology.

But it's not the usual formula because I need to do one more trick yet. Currently, the formula looks something like this: the $i$th cohomology group is $G$-equivariant maps $G^{i+1}\to M$ which are killed by $d$ (that is, which satisfy some axiom involving an alternating sum), modulo the image under $d$ of the $G$-equivariant maps $G^i\to M$.

The standard way to proceed from this point.

The "formula" for group cohomology derived above is essentially thought-free (which was why I could get this far in a noisy pub with no sources). But we want something more useful and it was at this point I got stuck. I remembered the nature of the trick: instead of a $G$-equivariant map $G^{i+1}\to M$ we simply "dropped one of the variables", and considered arbitrary maps of sets $G^i\to M$. So we need to give a dictionary between the set-theoretic maps $G^i\to M$ and the $G$-equivariant maps $G^{i+1}\to M$. I could see several choices. For example, given $f:G^{i+1}\to M$ I could define $c:G^i\to M$ by $c(g_1,g_2,\ldots,g_i)=f(1,g_1,g_2,\ldots,g_i)$, or $f(g_1,g_2,\ldots,g_i,1)$, or pretty much anything else of this nature. The point is that given $c$ there's a unique $G$-equivariant $f$ giving rise to it. Which choice of dictionary do we use. Each one will give a definition of group cohomology as "cocycles" over "coboundaries". But which one will give the "usual" definition? Well---in some sense, who cares! But at some point maybe someone somewhere made a decision as to what the convention from moving from $f$ to $c$ was, and now we all stick with it. The standard decision was the rather clunky $$c(g_1,g_2,\ldots,g_i)=f(1,g_1,g_1g_2,g_1g_2g_3,\ldots,g_1g_2g_3\ldots g_i).$$

Why is the standard definition ubiquitous?

Actually though, I bet that no-one really made that decision. I bet that the notion of a 1-cocycle and a 2-cocycle preceded general homological nonsense, and the dictionary between $f$s and $c$s was worked out so that it agreed with the definitions which were already standard in low degree.

But this got me thinking: if, as it seems to me, there is no "natural" way of moving from $f$s to $c$s, then why do the 1-cocycles and 2-cocycles that naturally show up in mathematics all satisfy the same axioms??. Why doesn't someone do a calculation, and end up with $c:G^2\to M$ satisfying some random axiom which happens not to be the "standard" 2-cocycle axiom but which is an axiom which, under a non-standard association of $c$s with $f$s, becomes the canonical cocycle axiom for $f$ that we derived without moving our brains? Does this ever occur in mathematics? I don't think I've ever seen a single example. In some sense it's even a surprise to me that there is a uniform choice which specialises to the standard choices in degrees 1 and 2.

Examples of cocycles in group theory.

1) Imagine you have a 2-dimensional upper-triangular representation of a group $G$, so it sends $g$ to the $2\times 2$ matrix $(\chi_1(g),c(g);0,\chi_2(g))$. Here $\chi_1$ and $\chi_2$ are group homomorphisms. What is $c$? Well, bash it out and see that $c$ is precisely a 1-cocycle in the sense that everyone means when they're talking about 1-cocycles. So we must have used the dictionary $c(g)=f(1,g)$ when moving between $c$s and $f$s above. Why didn't we use $c(g)=f(g,1)$?

2) Imagine you're trying to construct the boundary map $H^0(C)\to H^1(A)$. Follow your nose. Claim: your nose-following will lead you to the standard cocycle representing the cohomology class. Why did it not lead to a non-standard notion? Why do the notions (1) and (2) agree?

3) Imagine we have a group hom $G\to P/M$ where $M$ is an abelian normal subgroup of the group $P$. Can we lift it to a group hom $G\to P$? Well, let's take a random set-theoretic lifting $L:G\to P$. What is the "error"? A completely natural thing to write down is the map $(g,h)\mapsto L(g)L(h)L(gh)^{-1}$ because this will be $M$-valued. This is a 2-cocycle in the standard sense of the word. Why did the natural thing to do come out to be the standard notion of a 2-cocycle? Aah, you say: there are other natural things that one can try. For example we could have sent $(g,h)$ to $L(gh)^{-1}L(g)L(h)$. For a start this "looks slightly less natural to me" (why does it look less natural? That's somehow the question!). Secondly, this is really just applying a canonical involution to everything: we're inverting on $G$ and inverting on $M$, which is something that we'll always be able to do. It certainly does not correspond to the "more natural" dictionary that I confess I tried down the pub, namely $c(g,h)=f(1,g,h)$, which gave much messier answers. Why does my "obvious" choice of dictionary lead me to 20 minutes of wasted calculations? Why is it "wrong"?

The real question.

Has anyone ever found themselves in a situation where a natural cocycle-like construction is staring them in the face, and they make the construction, and find themselves with a non-standard cocycle? That is, a cocycle which will induce an element of a cohomology group but only after a non-standard dictionary is applied to move from $f$s to $c$s?

Edit: Here is what I hope is a clarification of the question. To turn the cocycles from $G$-equivariant maps $G^{i+1}\to M$ to maps of sets $G^i\to M$ we need to choose a transversal for the action of $G$ on $G^{i+1}$ and identify this with $G^i$. There seems to be one, and only one, way to do this that gives rise to the "standard" definition of n-cocycle that I (possibly incorrectly) percieve to be ubiquitous in mathematics. I call this "the clunky way" because the map seems odd to me. Why is it so clunky? And why, whenever a 2-cocycle falls out of the sky, does it always seem to satisfy the axioms induced by this clunky method? Why aren't there people popping up in this thread saying "well here's a completely natural "cocycle" $c(g,h)$ coming from theory X that I study, and it doesn't satisfy the usual cocycle axioms, we have to modify it to be $c'(g,h)=c(g,gh)$ before it does, and this boils down to the fact that in theory X we would have been better off if history had chosen a non-clunky identification?"

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5 Answers 5

up vote 4 down vote accepted

The difference between f(1,g) and f(g,1) is generally an issue of whether mathematicians give preference to "domains" or "ranges" of maps.

Here is one way that you could think of this. I can write EG for a category whose objects are objects are elements of G, and where each pair of objects has a unique map between them. This category has an action of G on it, and you can ask about G-equivariant functors from this category to another category what has a G-action on it.

To define such a functor on the level of objects it suffices to define F(1), where 1 is the unit; equivariance forces us to define F(g) = g F(1). On the level of morphisms, however, we have to make a choice. The unique morphism g→h becomes a morphism F(g)→F(h), and to make such maps compatible with the G-action it suffices to make one of the following sets of choices:

  • We could define maps fh:F(1)→F(h) for all h, and get all the other maps as g fh:F(g)→F(gh). To be a functor, we need this to satisfy the cocycle condition fgh = (g fh) fg.
  • We could define maps dh:F(h)→F(1) for all h, and get all the other maps as g dh:F(gh)→F(g). To be a functor, we need this to satisfy the cocycle condition dgh = dg (g dh).

In group cohomology, H1(G,M) classifies splittings in the semidirect product of G with M, and the cocycle condition we get comes from our convention of writing this group as pairs (m,g) (which is in the same order as the exact sequence it fits into) and not (g,m). Similarly for H2(G,M).

I would say that I've hit nonstandard cocycle definitions several times because I've been too lazy to come up with sensible conventions about when I'm thinking about domains and ranges or trying to sweep it under the rug, especially when dealing with Hopf algebroids and cohomological calculations there.

I don't have a good answer for higher cocycle conditions other than saying that writing 2-cochains using f(g,1,h) is somehow more unusual than either of the other 2 choices because it's somehow derived from focusing on the "middle" object in a double composite of maps.

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Two quick comments: (1) if you really have seen nonstandard cocycle definitions, then somehow this answers the question: it really does mean that in some sense our definition is arbitrary rather than "the right one". (2) you mention f(g,1,h) and say it's more unusual than the other two choices. But in fact none of those three choices are used in practice! f(1,g,gh) is the one used. This is exactly the slip I made. Why f(1,g,gh) rather than f(1,g,h)? –  Kevin Buzzard Nov 27 '09 at 14:33
    
I guess I would call (2) the consequence of the fact that you've chosen f(1,h) to represent 1-cochains, so it's more natural to talk about f(g,gh) and e.g. the coboundary has a more direct expression in terms of (1,g,gh). –  Tyler Lawson Nov 27 '09 at 17:51
    
I think that perhaps it's this comment (f(1,h) induces f(g,gh) induces (1,g,gh)) that is the best I'm going to get, so I'll accept this answer. –  Kevin Buzzard Dec 2 '09 at 7:30
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Let me just beat a dead horse a little. As Mariano and DC have mentioned (and you mentioned in your original question), there are multiple equivalent ways to compute group cohomology, or Ext in general, because you can pick any resolution you want. However, if you want a systematic resolution that always works then you need to work harder.

The resolution that you wrote down in terms of G acting on Gi+1 works perfectly well but it depends on a special fact about the group algebra: it's a Hopf algebra, and as a result it has a sensible way to act on a tensor product of copies of itself. Not all rings have such properties.

The "clunky" definition works for more general rings. Given any ring R and an R-module M, there is always an exact sequence

$$\cdots \to R \otimes R \otimes M \to R \otimes M \to M \to 0$$

called the bar resolution of R over M, and it arises from a simplicial construction via some natural adjoint-functor considerations - namely, there is an adjoint pair of the forgetful functor from R-modules to ℤ-modules and its left adjoint, the free R-module functor, and together they construct a simplicial object which you convert into a chain complex etc etc. If you specialize this construction to R = ℤ[G], you recover the "clunky" resolution and its specific formula for the boundary and coboundary operators. So somehow the definition that is less intuitive comes from forgetting that we're working in a grouplike context and being systematic there.

Because I can't help myself, I should mention that the above bar resolution is only a free resolution if R and M are free over ℤ (or if you replace ℤ and tensor with some ground ring over which they are both free). If not, then in order to get a canonical resolution you have to dip all the way down to the forgetful functor from R-modules to sets and its adjoint, the free R-module on a set, and this gives you an absolutely nightmarish but always-valid resolution that is even less intuitive.

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A keyword to search for when looking for information about Tyler's last paragraph is "MacLane homology". –  Mariano Suárez-Alvarez Dec 3 '09 at 15:36
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Hey Kevin -

You write: "The standard decision was the rather clunky $$c(g_1,g_2, \cdots g_i)=f(1,g_1,g_1g_2,g_1g_2g_3,\cdots, g_1g_2g_3\cdots g_i)$$

A topologist thinks of group cocycles as operating on "simplices labeled by elements of a group". Since orderings matter, we have "the" standard $n$-simplex, which comes with a labeling of its vertices by the numbers $0,1,2,\cdots,n$. The way the group comes in is that each edge of the simplex from vertex $i-1$ to vertex $i$ is labeled by an element of the group, i.e. $g_i$ is the label going from vertex $i-1$ to vertex $i$. This is because the simplex is "really" part of a $K(G,1)$ with one vertex, and one edge for each element of the group, one triangle for each element of the multiplication table of the group (i.e. for each expression $g \times h = gh$) and so on. These $g_i$ are the terms on the LHS of your equation.

The universal cover of this $K(G,1)$ is a contractible simplicial complex whose vertices are now labeled by elements of the group (because the group "is" the deck group of the cover, and acts simply transitively on vertices). Your original simplex has a unique lift to the cover taking the vertex $0$ to some specific vertex, which might as well be the one labeled by the identity. The vertex $i$ then lifts to the composition $g_1g_2\cdots g_i$. So the terms on the RHS of your equation are the labels on the vertices.

I am really only repeating what Mariano wrote in very slightly more topological language; you didn't seem happy with his answer, so maybe you won't be happy with mine either.

Best,

DC

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Hey DC, long time no see! I was walking past our old place on google street thingy the other day. I only accepted something because MO kept telling me "have you considered accepting an answer to your question??". I interpreted the one I accepted as saying this: "the definition is basically forced on you in degree 0, and now in higher degrees if you explicitly write down the operator H^{n-1}(C)->H^n(A) you'll get the definition people use in higher degrees. –  Kevin Buzzard Dec 3 '09 at 8:59
    
...So I accepted "an algebra answer". But on the other hand I think you and Mariano are offering an alternative reason which I think is probably just as satisfactory, if not more, because now I understand that you're actually able to justify the un-normalisation of the resolution on some conceptual level, which the "algebra" answer I accepted does not really do. As other people have pointed out, sometimes you want to accept multiple answers here, but we can't do that, so you'll just have to make do with my enthuasiam for your answer. Thanks! –  Kevin Buzzard Dec 3 '09 at 9:03
    
Perhaps I should also add that somehow the key to Mariano's answer seems to be in the last para, and there weren't enough details for me to fully understand it until you chipped in. –  Kevin Buzzard Dec 3 '09 at 9:04
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The "trick" you are referring to, of replacing $G$-linear morphisms $f:\mathbb ZG^{i+1}\to M$ by functions of sets $c:G^i\to M$, is not a trick: it is just the observation that $G$-linearity implies that, since $\mathbb{Z} G^{i+1}=\mathbb ZG\otimes\mathbb Z{G^i}$ as a left $G$-module, there is a natural isomorphism $$\hom\_G(\mathbb ZG^{i+1},M)=\hom\_G(\mathbb ZG\otimes\mathbb Z^{G^i},M)=\hom_{\mathbb Z}(\mathbb Z G^i,M)$$ by standard properties of the tensor product. Now $\mathbb ZG^i$ is free abelian on the set $G^i$, so there is a natural isomorphism $$\hom_{\mathbb Z}(\mathbb Z G^i,M)\cong\hom_{\mathrm{Set}}(G^i,M).$$ The composition of theses maps gives the identification between $f$'s and $c$'s. You can write them down explicitly, and compute their inverses: in both cases a one on the left" is involved and this is simply a reflection that we are using left modules.


On clunkiness

The form you call 'clunky' has the following origin. The usual bar resolution for a group $G$ has in degree $n$ the $\mathbb ZG$-module $P\_n=\mathbb ZG^{n+1}$. We can think of it as a free left $G$-module generated by the set of symbols symbols $[g\_1,\dots,g\_n]$. The differential of the complex is then left $G$-linear and for example, $$d[g\_1,g\_2,g\_3] = g\_1[g\_2,g\_3]-[g\_1g\_2,g\_3]+[g\_1,g\_2g\_3]-[g\_1,g\_2],$$ and this plainly involves the multiplication of the group. This is the so-called inhomogeneous description.

We can also think of $P\_n$ in another, homogenous, way: define $$(g\_0,\dots,g\_n)=g\_0[g\_0^{-1}g\_1,\dots,g\_0^{-1}g\_n].$$ The set of all such symbols is now a basis of $P\_n$ over $\mathbb Z$, with the left action of $G$ given by $$g\cdot(g\_0,\dots,g\_n)=(gg\_0,\dots,gg\_n),$$ which is a bit more annoying that before. The good thing is that now the boundary is given by $$d(g\_0,\dots,g\_n)=\sum_{i=0}^n(-1)^i(g\_0,\dots,\hat g\_i,\dots g\_n),$$ which does not involve the product in $G$ at all and, in fact, is precisely the same formula as the one giving the boundary in a simplicial complex.

The clunkiness you refer to is nothing but the translation formula between these two descriptions of the complex, as $$[g\_1,\dots,g\_n]=(1,g\_1,g\_1g\_2,g\_1g\_2g\_3,\dots,g\_1\cdots g\_n).$$

The original description of the complex is the inhomogeneous one, because it was found topologically. The two can be seen as a description of the classifying space for $G$ either by labelling paths in the category by the objects you go through, or by the arrows you follow, or something...

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I am confused by this answer. I am saying "there are lots of ways to go between f's and c's. One funny way seems to be very good in applications". Your comment at the top says "it's clear that one can go naturally between f's and c's". The whole crux of the matter is the explicit dictionary. You have not said how to identify Z[G] tensor Z[G^i] with Z[G^{i+1}]. I claim that if you asked any reasonable person how to do this, they would give you a natural isomorphism which would not be the one that induces the standard definition of an n-cocycle for n>=2. –  Kevin Buzzard Nov 27 '09 at 17:09
    
As for the rest of the answer, my question becomes this. You say [g_1,g_2,...,g_n]=(1,g_1,g_1g_2,\ldots). So now you seem to be saying "if we do things this way we get the standard definition". This I know. My question becomes "why [g_1,g_2,...,g_n]=(1,g_1,g_1g_2,...)? Why not [g_1,g_2,...,g_n]=(g_1,g_2,..,g_n,1)? This works just as well for the homological algebra arguments, and we get a definition of H^i(G,M)=Z/B, Z="cocycles" etc, but the answers one gets are for some reason not the cocycles that come up naturally in applications. Why is this? –  Kevin Buzzard Nov 27 '09 at 17:14
    
Well, I guess you are calling me unreasonable! :) For me, there is one obvious isomorphism between ZG\otimes ZG^i and ZG^{i+1}, which is the one which is the identity on the basis given by elementary tensors of group elements. Notice this comes from considering G^{i+1} as a G-set with action on the first coordinate only, not with diagonal action. –  Mariano Suárez-Alvarez Nov 27 '09 at 17:45
    
I guess I must be :-) Not intentionally, I might add! In terms of groups, we're asking for a natural map G x G^i -> G^{i+1}. I would choose one which when restricted to 1 x G^i sends (g1,g2,...,gi) to (1,g1,g2,...,gi). You are thinking about things in a different way which I am struggling to fit into my worldview. I know that what I'm supposed to be seeing is a canonical argument telling me that I'm supposed to send it to (1,g1,g1g2,...) but this is the bit that's failing to fit into my head. –  Kevin Buzzard Nov 27 '09 at 18:42
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Group cohomology of a group $G$ looks at homotopy classes of maps $\mathbf{B}G \to \mathbf{B}^n \mathbb{Z}$ from an oo-groupoid $\mathbf{B}G$ that is a delooping of $G$ to, in the simplest case, some Eilenberg-MacLane oo-groupoid $\mathbf{B}^n \mathbb{Z}$.

So $\mathbf{B}G$ here, like everything else in the game, is specified only up to equivalence. There are many ways to realize this.

One is to realize it as the action groupoid $ {*} // G$ of $G$ acting trivially on the point. That gives the second formula that you mention.

But we can take bigger models of the point. One popular one is $\mathbf{E}G = G//G$, the action groupoid of $G$ acting on itself. This is an equivalent model of the point. So the cohomology of $\mathbf{E}G//G$ also computes the group cohomology of $G$. But in components maps out of $\mathbf{E}G/G$ are just maps out of copies of $G$, with one more copy than we had before, that respect the $G$ action. This is the first formual you have.

There are infinitely more, and infinitely more complicated ones than you listed. One for each connected oo-groupoid who looping gives $G$, up to equivalence.

Generally, cohomology is something that lives in a homotopical context, where everything only depends on everything else up to weak equivalence. That's why there are, generally, so many different models for cohomologies. Not just group cohomology.

Some details on this are at nLab: group cohomology

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Right! So we're both convinced that there are many ways of thinking about H^n(G,M). But that's not my question. My question is this: whenever I goof around with an obstruction and construct it by explicitly writing down a natural map G^2-->M, it satisfies the axioms of a 2-cocycle. So why is "the construction that everyone uses" the same as "the one which seems to come up again and again in practice"? You seem to be saying "there are loads of constructions". I'm saying "yes, but one seems to be better than the others for no good reason." –  Kevin Buzzard Nov 27 '09 at 17:37
    
I would say: because of all the many models for BG, one is the smallest, in components. That's the usual bar-resulution *//G. SO that's the one seen most often in applications. Just because *//G happens to tbe smalles model for BG: it doesn't get smaller than the point. –  Urs Schreiber Nov 27 '09 at 17:42
    
But maybe I still don't really understand the question. I have a bit of trouble distinguishing between "the construction that everyone uses" and "the construction which comes up in practice". No matter what construction we are talking about, the one that comes up most in practice will be the one that most people use! :-) (?) –  Urs Schreiber Nov 27 '09 at 17:45
    
Yeah, I don't think I've explained the question very well. I'll add a para at the end. –  Kevin Buzzard Nov 27 '09 at 18:39
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Maybe my question to you is "why is there one definition that always comes up in practice, where we can clearly see that there are many choices of doing things, all but one of which seem to lead to definitions that never come up in practice?" –  Kevin Buzzard Nov 27 '09 at 19:16
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