Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideal-theoretic arguments). I've seen the proof of the long exact sequence and of basic properties of the Herbrand quotient, and I've started to look through how these are used in the proofs of class field theory.

So far, all I can tell is that the cohomology groups are given by some ad hoc modding out process, then we derive some random properties (like the long exact sequence), and then we compute things like $H^2(\mathrm{Gal}(L/K),I_{L})$, where $I_L$ denotes the group of fractional ideals of a number field $L$, and it just happens to be something interesting for the study of class field theory such as $I_K/\mathrm{N}(I_L)$, where $L/K$ is cyclic and $\mathrm{N}$ denotes the ideal norm. We then find that the cohomology groups are useful for streamlining the computations with various orders of indexes of groups.

What I don't get is what the intuition is behind the definitions of these cohomology groups. I do know what cohomology is in a geometric setting (so I know examples where taking the kernel modulo the image is interesting), but I don't know why we take these particular kernels modulo these particular images. What is the intuition for why they are defined the way they are? Why should we expect that these cohomology groups so-defined have nice properties and help us with algebraic number theory? Right now, I just see theorem after theorem, I see the algebraic manipulation and diagram chasing that proves it, but I don't see a bigger picture.

For context, if $A$ is a $G$-module where $G$ is cyclic and $\sigma$ is a generator of $G$, then we define the endomorphisms $D=1+\sigma+\sigma^2+\cdots+\sigma^{|G|-1}$ and $N=1-\sigma$ of $A$, and then $H^0(G,A)=\mathrm{ker}(N)/\mathrm{im}(D)$ and $H^1(G,A)=\mathrm{ker}(D)/\mathrm{im}(N)$. Note that this is a slight modification of group cohomology, i.e. Tate cohomology, which the cohomology theory primarily used for Class Field Theory. Group cohomology is the same but with $H^0(G,A) = \mathrm{ker}(N)$. The advantage of Tate cohomology is that it is $2$-periodic for $G$ cyclic.

share|improve this question
1  
Some of the answers in this related question deal with group cohomology: mathoverflow.net/questions/640/… –  Qiaochu Yuan Jan 6 '10 at 4:30
6  
I should point out that this is not ordinary group cohomology, but Tate cohomology, which has some minor alterations to make it 2-periodic. –  S. Carnahan Jan 6 '10 at 6:06
1  
A similar question has been discussed at mathoverflow.net/questions/8599/… –  Leonid Positselski Jan 6 '10 at 12:53
    
The best treatment of group cohomology which I have seen is the the book "lectures on algebraic geometry 1" by Gunter Harder. Check it out. –  Steven Gubkin Aug 31 '10 at 17:12
    
@Steven: I agree that that book is quite nice, though as far as I can tell, it doesn't seem to give particularly nice intuition for group cohomology. –  David Corwin Sep 5 '10 at 17:20
show 1 more comment

9 Answers 9

Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology.

Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176).

If moreover $G $ is cyclic with generator $s$, this implies that an element of $K$ has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$.

Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$ the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff $x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$ .

So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$ $$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.

share|improve this answer
4  
Oh, I agree that cohomologists must have known this for ages. But Davidac897's question proves that it might be helpful to have elementary examples explicitly written down to show at an early stage that group cohomology can have applications even before the whole somewhat intimidating machinery is developed. –  Georges Elencwajg Jan 6 '10 at 12:25
5  
I don't like this folklore application because before having done the general proof of $H^1(G,K^*)=0$, where you have to come up with special characters whose linear independancy is used, you have already parameterized the rational circle many times; besides, there are nice geometric approaches to it. –  Martin Brandenburg Jan 6 '10 at 16:28
18  
@Martin: I don't understand your statement about chronology. It is quite possible to prove Theorem 90 without using a parametrization of the rational circle, let alone many of them. Also, I don't see why the existence of a nice geometric proof means one should stop looking for other arguments and points of view. –  S. Carnahan Jan 6 '10 at 16:50
1  
@Elencwajg:This is a very nice observation! I will definitely use it as motivation in my teaching! –  SGP Jun 8 '11 at 2:01
2  
This example was used by Emil Artin in his Hamburg lectures on algebra in I think 1961. –  Franz Lemmermeyer Nov 14 '12 at 5:32
show 3 more comments

I feel like my answer to every question of this kind is "the treatment in Silverman's book on elliptic curves is really nice," but the treatment in Silverman's book on elliptic curves is really nice!

In particular, for H^1 at least I always find it quite nice to think in terms of twists. Quite generally: if X is a variety over a field K, and L/K is a Galois extension, we say that X'/K is a L-twist of X if there exists an isomorphism between X and X' over L. (If we just mean X' is an L-twist for some L, we just call it a twist of X.)

Anyway, it is a nice exercise to check that L-twists of X yield classes in H^1(Gal(L/K),Aut(X/L)). (And in favorable circumstances, the L-twists are actually in bijection with the Galois cohomology set.) This is the basis of the whole story of principal homogeneous spaces, or torsors, which makes up much of the last chapter of Silverman.

share|improve this answer
add comment

Seconding some remarks already made: despite the (stodgy) tradition of standard math curricula to ignore these things, it is very useful to understand that group (co)homology consists of the "derived functors" of the functors that take G-[co]fixed vectors on representations spaces. (Unfortunately, the two "co"'s get reversed.) Thus, there is an intrinsic definition. These higher derived functors are the universal/right things to "correct" for the non-exactness of the (co)fixed-vector functors. True, this definition does not explain why we would care so much, but, in the end, it is a large part of what (co) homology is, I think.

It is also very useful to know/prove that any pro/in-jective resolution can be used to compute the (co)homology.

Thus, the particular choices (homogeneous/inhomogeneous bar resolution(s), etc.) are not the definition, despite common assertions that this is so.

The point of clever/insightful choice of resolutions is facilitation of computations about specific situations.

C. Weibel's book on homological algebra does some good sample computations in group (co) homology, in the context that it is an example.

share|improve this answer
4  
I think this answer is really good. My point of view is that many things we care about in number theory are acted upon by Galois groups, and often "fixed elements come from below". Somethines, this is sadly false, like for ideals and ideal class group. Group cohomology is then very natural, because you have a tautological exact sequence defining the class group and you would like to compare objects in this sequence (which you care about genuinely because it is tautological) which are fixed by Galois with those coming from below. –  Filippo Alberto Edoardo Nov 14 '12 at 2:28
add comment

I'm not sure if this is what you're looking for, but I always think of group (co)homology in terms of the homology of the classifying space for your group. Assuming $G$ is discrete, then there is a topological space $BG$ with the property that $\pi_1 BG=G$ and the higher homotopy groups vanish. By construction, $BG$ has a contractible cover $EG$ so that $EG/G=BG$.

$H_n(BG)$ is the same as the algebraically defined $H_n(G)$ since, if we take the cellular chain complex of $EG$ we end up with a resolution of the integers by $G$-modules because of the action of $G$ on $EG$ passes to the chain groups. Then tensoring by the integral group ring of $G$ just divides out the $G$ action and we get the cellular chain complex of $BG$.

share|improve this answer
    
What is the difference between BG and the Eilenberg-Maclane space K(G,1) mentioned in Allen Hatcher's book? –  Anweshi Jan 6 '10 at 12:15
    
They are the same thing. –  Andy Putman Jan 6 '10 at 16:10
5  
They are the same thing as long as $G$ has the discrete topology. If $G$ is a topological group, etc, $BG$ won't necessarily be $K(G,1)$. $G \to EG \to BG$ is a fibration so $\pi_n BG = \pi_{n-1} G$. –  jd.r Jan 6 '10 at 18:13
1  
The standard construction is actually in Hatcher section on $K(G,1)$'s on page 89. A more "hands on" construction exists if you have a finite presentation of your group. The only reference I know of is in an appendix in Knudson's "Homology of Linear Groups". I don't have the book handy and google books doesn't offer a preview of it so I can't give a more specific location, sorry. Basically, you construction a reduced CW complex dimension by dimension killing off higher homotopy as you go. –  jd.r Jan 7 '10 at 0:35
1  
A very beautiful construction of EG, BG, etc is in Paolo Salvatore's paper: front.math.ucdavis.edu/9907.5073 A pleasant feature of his construction is that it works for any topological monoid, and the "geometric picture" of an element of EG, BG, is very explicit. –  Ryan Budney Jan 7 '10 at 6:45
show 3 more comments

First, let me say beyond knowing that group cohomology comes from derived functors and all the properties that come with it, I don't have much intuition for general group cohomology. However, in number theory there are a number of places in which you can realize the cohomology groups as parametrizing some other objects you are interested in. To understand these other objects (which you could feasibly have very real intuition about) it then suffices to use the machinery of group cohomology. Below are a few examples. I forget to put continuous subscripts on everything, so beware. Mistakes are mine.

First, in the realm of (edit:local) class field theory we have the Brauer group $Br(K)$ which is the abelian group (under the tensor operation) of central simple algebras over $K$ up to the equivalence $A \sim M_n(A)$. It turns out that there is a natural isomorphism between $Br(K)$ and $H^2(G_K, \overline{K}^{\times})$. This isomorphism provides two worlds in which to make important calculations. For instance, the statement that every central simple algebra over $K$ splits over an unramified extension of $K$ may be proved explicitly using the Brauer group or it may be proven by checking that $H^2(G(K^{nr}/K),\overline{K}^\times) = H^2(G_K,\overline{K}^\times)$. From either proof you are able to obtain a proof of the other. Perhaps, also in the realm of class field theory, the local Artin map arises as a map on Tate groups given by a certain cup product but that would take a bit longer to explain. You should look at Milne's notes on CFT for all of this (Ch III and Ch IV for what I have said).

Here is another. Suppose that $B$ is a topological ring and $G$ is a topological group and $G$ acts continuously on $B$. Then, we consider finite free $B$-modules $X$ equipped with a semi-linear action, i.e. $g(bx) = g(b)g(x)$ for all $b \in B$ and $x \in X$. It turns out that all such objects are parametrized by $H^1(G,GL_d(B))$. (Warning: this is non-abelian cohomology, so this is only a pointed set. The point corresponds to the trivial semi-linear representation $B^d$ with the diagonal action.) This comes up very early in the part of $p$-adic Galois representations where one studies the period rings $B_{dR}, B_{HT}$, etc.

Finally, consider the situation where one has a representation $\overline{\rho}:G_{\mathbb Q,S} \rightarrow GL_n(\mathbb F_p)$ and one wants to know whether it has litings $\rho: G_{\mathbb Q,S} \rightarrow GL_n(R)$ where $R$ is some complete DVR with residue field $\mathbb F_p$. If we already have a lifting to $GL_n(R')$ and $R$ and $R'$ are nice enough (there is a surjection $R \rightarrow R'$ whose kernel $I$ is killed by the maximal ideal in $R$) then the obstruction to lifting further lies in the cohomology group $H^2(G_{\mathbb Q,S},I\otimes Ad(\overline{\rho}))$ where $Ad(\overline{\rho})$ is the vectorspace $M_n(\mathbb F_p)$ together with conjugate action by $\overline{\rho}$. This (I've specialized a few things) is written down in Mazur's paper "Deforming Galois Representations".

To finish, I will just say what my adviser told me when I asked him a similar question as you are asking: "Just wait and you will see how much clearer your thought can become with group cohomology."

share|improve this answer
add comment

The online notes of J.S. Milne on Class Field Theory contain a chapter on group cohomology including Tate cohomology that is easily accessible (and exactly what you need to understand class field theory).

For algebraic intuition the keyword is derived functors. Group cohomology (and homology) are examples of derived functors, which can be considered as a reason why the definitions as are they are, and why they are interesting/useful.

share|improve this answer
add comment

Topology and Logic as a Source of Algebra by Mac Lane in the AMS Bull. January 1976 may be helpful.

share|improve this answer
add comment

See also this Math.SE post I wrote for some more motivation: http://math.stackexchange.com/a/270266/873. Recall that $\mathrm H^*(G,M)=\mathrm{Ext}^*(\mathbb{Z},M)$.

After learning some more math, I've come across the following example of a use of group cohomology which sheds some light on its geometric meaning. (If you want to see a somewhat more concrete explanation of how group cohomology naturally arises, skip the next paragraph.)

We define an elliptic curve to be $E=\mathbb{C}/L$ for a two-dimensional lattice $L$. Note that the first homology group of this elliptic curve is isomorphic to $L$ precisely because it is a quotient of the universal cover $\mathbb{C}$ by $L$. A theta function is a section of a line bundle on an elliptic curve. Since any line bundle can be lifted to $\mathbb{C}$, the universal cover, and any line bundle over a contractible space is trivial, the line bundle is a quotient of the trivial line bundle over $\mathbb{C}$. We can define a function $j(\omega,z):L \times \mathbb{C} \to \mathbb{C} \setminus \{0\}$. Then we identify $(z,w) \in \mathbb{C}^2$ (i.e. the line bundle over $\mathbb{C}$) with $(z+\omega,j(\omega,z)w)$. For this equivalence relation to give a well-defined bundle over $\mathbb{C}/L$, we need the following: Suppose $\omega_1,\omega_2 \in L$. Then $(z,w)$ is identified with $(z+\omega_1+\omega_2,j(\omega_1+\omega_2,z)w$. But $(z,w)$ is identified with $(z+\omega_1,j(\omega_1,z)w)$, which is identified with $(z+\omega_1+\omega_2,j(\omega_2,z+\omega_1)j(\omega_1,z)w)$. In other words, this forces $j(\omega_1+\omega_2,z) = j(\omega_2,z+\omega_1)j(\omega_1,z)$. This means that, if we view $j$ as a function from $L$ to the set of non-vanishing holomorphic functions $\mathbb{C} \to \mathbb{C}$, with (right) L-action on this set defined by $(\omega f)(z) \mapsto f(z+\omega)$, then $j$ is in fact a $1$-cocyle in the language of group cohomology. Thus $H^1(L,\mathcal{O}(\mathbb{C}))$, where $\mathcal{O}(\mathbb{C})$ denotes the (additive) $L$-module of holomorphic functions on $\mathbb{C}$, classifies line bundles over $\mathbb{C}/L$. What's more is that this set is also classified by the sheaf cohomology $H^1(E,\mathcal{O}(E)^{\times})$ (where $\mathcal{O}(E)$ is the sheaf of holomorphic functions on $E$, and the $\times$ indicates the group of units of the ring of holomorphic functions). That is, we can compute the sheaf cohomology of a space by considering the group cohomology of the action of the homology group on the universal cover! In addition, the $0$th group cohomology (this time of the meromorphic functions, not just the holomorphic ones) is the invariant elements under $L$, i.e. the elliptic functions, and similarly the $0$th sheaf cohomology is the global sections, again the elliptic functions.

More concretely, a theta function is a meromorphic function such that $\theta(z+\omega)=j(\omega,z)\theta(z)$ for all $z \in \mathbb{C}$, $\omega \in L$. (It is easy to see that $\theta$ then gives a well-defined section of the line bundle on $E$ given by $j(\omega,z)$ described above.) Then, note that $\theta(z+\omega_1+\omega_1)=j(\omega_1+\omega_2,z)\theta(z) = j(\omega_2,z+\omega_1)j(\omega_1,z) \theta(z)$, meaning that $j$ must satisfy the cocycle condition! More generally, if $X$ is a contractible Riemann surface, and $\Gamma$ is a group which acts on $X$ under sufficiently nice conditions, consider meromorphic functions $f$ on $X$ such that $f(\gamma z)=j(\gamma,z)f(z)$ for $z \in X$, $\gamma \in \Gamma$, where $j: \Gamma \times X \to \mathbb{C}$ is holomorphic for fixed $\gamma$. Then one can similarly check that for $f$ to be well-defined, $j$ must be a $1$-cocyle in $H^1(\Gamma,\mathcal{O}(X)^\times)$! (I.e. with $\Gamma$ acting by precomposition on $\mathcal{O}(X)^\times$, the group of units of the ring of holomorphic functions on $X$.) Thus the cocycle condition arises from a very simple and natural definition (that of a function which transforms according to a function $j$ under the action of a group). A basic example is a modular form such as $G_{2k}(z)$, which satisfies $G_{2k}(\gamma z) = (cz+d)^{2k} G_{2k}(z)$, where $\gamma = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \in SL_2(\mathbb{Z})$ acts as a fractional linear transformation. It follows automatically that something as simple as $(cz+d)^{2k}$ is a cocycle in group cohomology, since $G_{2k}$ is, for example, nonzero.

share|improve this answer
add comment

In my opinion, the best you can do is to see group cohomology in the context of homological algebra. For example, the book by Hilton and Stammbach on homological algebra should be a good introduction.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.