I started reading the Cohomology theory of groups. But I am not able to get any intuition or motivation behind the following :

It is concerned with the formal definitions of crossed and principal crossed homomorphisms. Crossed homomorphisms are those maps $f:G\to M$ satisfying $f(ab)=f(a)+af(b)$ ( For all $a,b \in G$ ) where as the Principal crossed homomorphisms are given by $f(a)=am-m$ for some $m \in M$. I don't really understand the motivation or the need consider the terms $f(ab)=f(a)+af(b),f(a)=am-m.$ What do they tell us ? . Do they serve as some means for calculating the so called "difference ( Given in above link ). I don't think they exist blindly or randomly. There must be some deep intuition behind that.

I would be very happy to hear if some one posts a detailed explanation. Please name some good articles that will give a good motivation.

Thanks a lot.

  • $\begingroup$ (1) is explained in more detail on the same page, under the heading "Long exact sequence of cohomology". What precisely do you want to know more about? $\endgroup$ – Zhen Lin May 10 '12 at 16:02
  • $\begingroup$ @ZhenLin : How does the functor measures the extent in which the invariants doesn't respect exact sequences is what I am looking for. How is it done ? . $\endgroup$ – IDOK May 10 '12 at 16:04
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    $\begingroup$ @Iyengar: That is measured by the long exact sequence of cohomology groups, as Wikipedia says. Do you understand what an exact sequence is? $\endgroup$ – Zhen Lin May 10 '12 at 16:17
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    $\begingroup$ What more is there to say? If $H^1(G; A) = 0$ then any short exact sequence $0 \to A \to B \to C \to 0$ gives a short exact sequence $0 \to A^G \to B^G \to C^G \to 0$. Otherwise you have to replace the last $0$ with $H^1(G; A)$. $\endgroup$ – Zhen Lin May 10 '12 at 16:23
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    $\begingroup$ Oh, Sorry sir. Thanks a lot, that explanation I was looking for. Really good. I am poor at english , and sometimes I cant understand the context and perspective. @ZhenLin $\endgroup$ – IDOK May 10 '12 at 16:24

There is indeed some nice intuition behind these definitions, and the good news is that not even all that deep. Remember two things: First, that this cohomology all comes by the "fixed by" functor $M\to M^G$, and second, that these crossed homomorphisms come from the definition of cochains, and more directly, the coboundary operator from $n$-chains to $n+1$-chains.

Now, if you didn't have these odd definitions (crossed homomorphisms, etc.) in front of you, how would you have constructed them from scratch? You'd follow your nose from algebraic topology: You start with continuous functions $f:G^n\to M$, and you'd have your coboundary operator be the standard gadget on forms. Namely, $\partial f$ should be the function from $G^{n+1}$ to $M$ given by one of these alternating sums where you omit one index at a time: $$ \partial f(\sigma_0,\ldots,\sigma_n)=\sum_{i=0}^n(-1)^if(\sigma_0,\ldots,\widehat{\sigma_i},\ldots,\sigma_n). $$ There's a natural $G$-action on these functions, where the action of $\sigma$ on $f$ gives the new function $\sigma f$ defined by $$ (\sigma f)(\sigma_0,\ldots,\sigma_n)=\sigma\cdot f(\sigma^{-1}\sigma_0,\ldots,\sigma^{-1}\sigma_n) $$ Now, following the standard recipe, we apply the "fixed by $G$" functor, and take such forms as our cochains.

And now, the best-kep secret in group cohomology -- this works! The groups $C^n(G,M)$ of cocohains and the coboundary operator $\partial$ defined above lead to the standard notions of closed and exact cochains (things which have trivial boundary, and things which are themselves coboundaries, respectively), and voila, cohomology! No crossed homomorphisms, funky coboundary operators, etc.

So why do the less intuitive versions of these things even exist? Because they're better. Or at least more efficient. The point is that once you insist on $G$-invariance, the cochains defined above have a redundant variable in place. Everything that seems scary about the definitions of group cohomology comes from translating the above construction into the ones where you remove the extra degree of freedom. In the literature, this is called moving from homogeneous to inhomogeneous cochains (see, e.g., Chapter 1 of Cohomology of Number Fields.).

For example, when $n=0$, it's easy to see that the $G$-invariance of a function $f:G\to M$ implies that it is in fact constant, determined by $f(1)$. And this holds true in higher dimensions as well. The next one up is your specific questions: "homogeneous" 1-chains are functions $f:G^2\to M$ satisfying $G$-invariance and the coboundary condition $df(\sigma_0,\sigma_1,\sigma_2)=f(\sigma_1,\sigma_2)-f(\sigma_0,\sigma_2)+f(\sigma_0,\sigma_1)=0$ for all $\sigma_0,\sigma_1,\sigma_2\in G$. In other words, cochains satisfy $G$-invariance and the identity. $$ f(\sigma_0,\sigma_2)=f(\sigma_1,\sigma_2)+f(\sigma_0,\sigma_1) $$ Not so bad, at first glance. Certainly easy to remember But now your cocycle condition is both an equation with one extra variable and an extra condition ($G$-invariance) that's not built into the equation. But! If you translate this into the language of *in*homogeneous cochains, you get precisely the $f(\sigma_1,\sigma_2)=f(\sigma_1)+\sigma_1*f(\sigma_2)$ condition -- now you have a single condition defining coboundary-ness. The translation between homogeneous and inhomogeneous forms is an easy but notationally-heavy exercise. Your other question is about the completely analogous translation of what makes a 1-chain a coboundary.

So in the end, it just so happens that for both computational aspects and theoretical development, its significantly handier to work with functions of one fewer variable, even if it comes at the cost of a little intuition. You get used to it. :)

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    $\begingroup$ What analogy to algebraic topology are you alluding to when you say the cochains should be functions $G^n \to M$? $\endgroup$ – Zhen Lin May 14 '12 at 7:33
  • $\begingroup$ +1, good answer, but you missed telling why should one consider such form of equation like $f(ab)=f(a)+a.f(b)$ why not something else. I think there is some reason why one should consider them. Please do explain me that sir. $\endgroup$ – IDOK May 14 '12 at 13:49
  • $\begingroup$ @ZhenLin: Nothing fancy; among other options, singular homology (and then dualizing.) $\endgroup$ – Cam McLeman May 14 '12 at 14:39
  • $\begingroup$ @Iyengar: Ah, but I did! The weirdness of that equation comes from taking the completely ordinary-looking cocycle condition for homogeneous forms, and removing the redundancy by traslating into the inhomogeneous forms formulation. I've updated the answer to say this more explicitly, but writing down the details is a very very important exercise to do. The book I referenced (available on google books) contains everything you need. $\endgroup$ – Cam McLeman May 14 '12 at 14:53
  • $\begingroup$ @CamMcLeman: Ah, so you think of $G^n$ as being the set of "singular $n$-cubes" in $G$? That's an interesting way of looking at it, and it's probably even valid if we identify $G$ with the Eilenberg–MacLane space $K(G, 1)$. $\endgroup$ – Zhen Lin May 14 '12 at 15:02

Group cohomology was initially observed in nature: people noticed that the homology of certain spaces with trivial higher homotopy groups depends only on the fundamental group, and with considerable ingenuity managed to predict purely algebraically what that homology is from the fundamental group itself.

In a sense, then, people found that $H^1$ can be described in terms of crossed homomorphisms by computing the group explicitly and there is no a priori intuition of why it should be given in that way. Moreover, one can provide many equivalent descriptions of $H^1$, some of which do not involve crossed homomorphisms in any way—and, in particular, saying that crossed homomorphisms are an "ingredient" of $H^1$ is a misunderstanding: they are just one of its several avatars only...

On the other hand, people had already encountered crossed homomorphisms and principal crossed homomorphisms in nature before, independently of cohomology. I don't know who exactly made the connection, but surely someone did see that the $H^1$ that topologists found was the same one that was implicit in, say, Hilbert's Theorem 90.

  • $\begingroup$ I honestly have no idea what you meant by that. $\endgroup$ – Mariano Suárez-Álvarez May 18 '12 at 14:21
  • $\begingroup$ Answer is good sir, thank you for helping. +1 $\endgroup$ – IDOK May 19 '12 at 17:45

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