Intuition for Group Cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T07:42:53Z http://mathoverflow.net/feeds/question/10879 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10879/intuition-for-group-cohomology Intuition for Group Cohomology David Corwin 2010-01-06T04:19:01Z 2013-01-06T05:54:12Z <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/10880#10880 Answer by Mariano Suárez-Alvarez for Intuition for Group Cohomology Mariano Suárez-Alvarez 2010-01-06T04:23:15Z 2010-01-06T04:23:15Z <p>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.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/10887#10887 Answer by Josh Roberts for Intuition for Group Cohomology Josh Roberts 2010-01-06T05:33:51Z 2010-01-06T05:33:51Z <p>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 $EG/G=BG$. </p> <p>$H_n(BG)$ 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 $BG$.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/10894#10894 Answer by Peter McNamara for Intuition for Group Cohomology Peter McNamara 2010-01-06T06:56:35Z 2010-01-06T06:56:35Z <p>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).</p> <p>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.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/10903#10903 Answer by Georges Elencwajg for Intuition for Group Cohomology Georges Elencwajg 2010-01-06T09:46:50Z 2010-01-06T09:46:50Z <p>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.</p> <p>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).</p> <p>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 $a\in K$.</p> <p>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 $u+iv\in \mathbb Q (i)$ . </p> <p>So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$ $$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/10956#10956 Answer by tkr for Intuition for Group Cohomology tkr 2010-01-06T20:12:20Z 2010-01-07T17:32:00Z <p>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.</p> <p>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).</p> <p>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.</p> <p>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".</p> <p>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."</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/11094#11094 Answer by JSE for Intuition for Group Cohomology JSE 2010-01-08T03:09:47Z 2010-01-08T03:09:47Z <p>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!</p> <p>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 <em>L-twist</em> 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 <em>twist</em> of X.)</p> <p>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. </p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/33649#33649 Answer by David Corwin for Intuition for Group Cohomology David Corwin 2010-07-28T12:46:00Z 2013-01-06T05:54:12Z <p>See also this Math.SE post I wrote for some more motivation: <a href="http://math.stackexchange.com/a/270266/873" rel="nofollow">http://math.stackexchange.com/a/270266/873</a>. Recall that <code>$\mathrm H^*(G,M)=\mathrm{Ext}^*(\mathbb{Z},M)$</code>.</p> <p>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.)</p> <p>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.</p> <p>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 <em>the cocycle condition arises from a very simple and natural definition</em> (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 $\gamma = \left(\begin{array}{cc} a &amp; b \ c &amp; d \end{array}\right) \in SL_2(\mathbb{Z})$ 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.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/33663#33663 Answer by Roy for Intuition for Group Cohomology Roy 2010-07-28T14:57:17Z 2010-07-28T14:57:17Z <p>Topology and Logic as a Source of Algebra by Mac Lane in the AMS Bull. January 1976 may be helpful.</p> http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/67194#67194 Answer by paul garrett for Intuition for Group Cohomology paul garrett 2011-06-07T23:32:07Z 2011-06-07T23:32:07Z <p>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 <em>is</em>, I think. </p> <p>It is also very useful to know/prove that <em>any</em> pro/in-jective resolution can be used to <em>compute</em> the (co)homology.</p> <p>Thus, the particular choices (homogeneous/inhomogeneous bar resolution(s), etc.) are not the definition, despite common assertions that this is so.</p> <p>The point of clever/insightful choice of resolutions is facilitation of computations about specific situations.</p> <p>C. Weibel's book on homological algebra does some good sample computations in group (co) homology, in the context that it is an <em>example</em>.</p>