# Tag Info

20

In general, a twisted coefficient system on a manifold $M$ (also called a local system by more algero-geometrically minded people) is given by a representation of $\pi_1(M)$ (the holonomy representation, also called the monodromy representation by more algebrao-geometrically minded people). Conversely, any representation of $\pi_1(M)$ gives a twisted ...

14

[At Mariano's suggestion, I'll copy my answer from MathOverflow.] 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 ...

12

Akhil, you're thinking of this the opposite of how I think group cohomology was discovered. The concept of group cohomology originally centered around the questions about the (co)homology of $K(\pi,1)$-spaces, by people like Hopf (he called them aspherical rather than $K(\pi,1)$ spaces, and Hopf preferred homology to cohomology at that point). I think the ...

11

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 ...

10

Often elements of $H^1$ are used to classify objects of a certain kind. E.g. if $k$ is a field and $k^s$ its separable closure, and $G_k = Gal(k^s/k)$, then if $X$ is a variety over $k$, and $Aut(X)$ is the group of automorphisms of $X_{/k^s}$, then the set $H^1(G_k,Aut(X))$ classifies varieties over $k$ that become isomorphic to $X$ when both are regarded ...

9

First of all, your question doesn't make sense if you do not specify what topology you want on your group. A group can be equipped with various topologies that yield non-isomorphic topological groups. There is, however, a nice topology that can be defined on Galois groups called the Krull topology. Let $E/F$ be a Galois extension and $\mathrm{Gal}(E/F)$ its ...

8

Your last question has an easy answer : it is very desirable to learn algebraic topology! It's the most concrete setting for a cohomology theory, and it would be hard to appreciate much (for instance) sheaf cohomology without at least some understanding of the algebraic topology background. With that said, the following theorem of Thom gives what I ...

7

Here are my solutions from 3 years ago when I was studying under Ken Brown as an undergraduate: V.3.5: The symmetric group $S_3$ on three letters has a semi-direct product representation $S_3=\mathbb{Z}_3\rtimes \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{Z}_3$ by conjugation. Thus $H^*(S_3)=H^*(S_3)_{(2)}\oplus H^*(S_3)_{(3)}\cong ... 7 Result If$G$is not perfect (i.e. if$G\neq[G,G]$) , then the$\mathbb Z[G]$-module$\mathbb Z$is not flat. Example If$G\neq 0$is commutative, then the$\mathbb Z[G]$-module$\mathbb Z$is not flat Proof 1) If$I\subset A$is an ideal of a ring and$A/I$is$A$-flat, then$I/I^2=0$Indeed, tensor the injection$0\to I\to A$with$A/I$and ... 6 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 ... 6 Matt E's response is the canonical first answer. As someone (else) who works with nonabelian group (in particular, Galois) cohomology frequently, let me give a second answer. Certain interesting maps between commutative cohomology groups are defined via a non-commutative intermediary. The justification for this is that, though one does not in general have ... 6 Learning algebraic topology, at least to the level of singular homology and cohomology, and Poincare duality, is essential if you plan to use homological techniques in your work with any sort of serious understanding. It is not just that the simplicial techniques involved in singular homology and cohomology are important in themselves (although they ... 6 This is not about groups, but Lie algebras. Please keep in mind that by trade I am a group theorist, which happened to venture in the field of graded (over the positive integers) Lie algebras, discovering that they do not look too different from (residually) nilpotent groups. Anyway, I was studying certain twisted loop algebras, trying too prove they were ... 6 We have$H^2(G,R^\times) \cong {\rm Hom}(M(G),R^\times) \oplus {\rm Ext}(G/G',R^\times)$, where$M(G)$is the Schur multiplier of$G$. If$R^\times$is divisible (which is true, for example, if$R$is algebraically closed), then${\rm Ext}(G/G',R^\times) = 0$, and all extensions arises from a homomorphism from$M(G)$to$R^\times$, and any homomorphism$G ...

6

If you look in the book by Cartan-Eilenberg on Homological Algebra (and surely many others, but each of us has a favorite!) you will find explicit projective resolutions for both free groups and for free abelian groups. From that one can immediately compute homology and cohomology. You can also use technology. For example, it is easy to show that the ...

5

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 ...

5

Take some algebra $\Lambda$, and consider its category of finitely generated left-modules $_\Lambda \operatorname{mod}$. It is folklore that for modules $M,N$, the group $\operatorname{Ext}^1 (M,N)$ controls extensions of $N$ by $M$, i.e. modules obtained by gluing $M$ and $N$ in some way. So suppose we know all the simple modules $S_i$ and all the ...

5

This is almost never the case. Here are two counterexamples: 1) Infinite groups: Let $G$ be a free group on $n$ generators, and let $H$ be a subgroup of $G$ isomorphic to a free group on $m$ generators, with $m\neq n$. Then one has $\mathrm{H}^1(G,\mathbb{Z})\cong \mathbb{Z}^n$ and $\mathrm{H}^1(H,\mathbb{Z})\cong \mathbb{Z}^m$. 2) finite groups: Let $G$ ...

5

As an abelian group, $I_G$ is free on $\{(g-1): g \in G\}$, so we may define a homomorphism $I_G \to G/G'$ by $g-1 \mapsto gG'$. The kernel contains $I_G^2$ because of the identity $(g-1)(h-1) = (gh-1)-(g-1)-(h-1)$, so this descends to a map $I_G/I_G^2\to G/G'$ which is inverse to your map $G/G' \to I_G/I_G^2$. Both maps are therefore ...

4

As $(\sigma-1)(c_0+c_1\sigma+\dots c_{n-1}\sigma^{n-1})=(c_n-c_0)+(c_0-c_1)\sigma+\dots (c_{n-2}-c_{n-1})\sigma^{n-1}$, the element $a=c_0+c_1\sigma+\dots c_{n-1}\sigma^{n-1}$ is in the kernel of $\sigma-1$ iff all $c_i$'s are equal, i.e. iff $a=Nc$ for some $c\in\mathbb{Z}$. Similarly, $Na=(\sum c_i)N$, so here the kernel is given by the condition $\sum ... 4 For any module$M$the natural quotient map$A \to H_0(\Gamma,A)$induces an isomorphism$Hom(H_0(\Gamma,A),M) \cong H^0(\Gamma,Hom(A,M)).$(Here we regard$Hom(A,M)$as a$\Gamma$-module by the contragredient action.) Thus, in order to get a relationship between$Hom(H_0(\Gamma,A),\mathbb Z)$and$H^0(\Gamma,\mathbb Z)$, you would need an isomorphism of ... 4 I believe there's a very simple motivation for the cocycle condition: Suppose$f:X \to R$is a function on a space$X$with values in a ring$R$(if you wish,$X$is a manifold, and$R$is$\mathbb{R}$or$\mathbb{C}$, and the function$f$is differentiable or even holomorphic). Suppose that$G$is a group acting on$X$(say on the left). Then for the ... 4$F_n$is a finitely-generated$G$-module, so it is also finitely-generated as a$\mathbf Z$-module because$\mathbf Z[G]$is finitely generated as a$\mathbf Z$-module (since$G$is finite). Now$F_n^G$is a subgroup of a finitely-generated abelian group, so it is also finitely-generated. Thus$H^n(F^G)$is a subquotient of a finitely-generated abelian ... 4 Well, you'll probably want a more conceptual proof, but one thing you can do is check they are computed by the same chain complex: for$K(G,1)$take the simplicial construction of the classifying space$BG$and compute its cohomology in the usual way for simplicial sets (using the dual to the complex of formal linear combinations of simplices); for the group ... 4 Yes.$H^1(G,M) = \operatorname{Ext}_{kG}^1(k,M)$describes how many ways$M$can be a submodule of$N$with$N/M \cong k$.$P(k)$can only have composition factors that are “connected” to$k$by paths of nonzero$\operatorname{Ext}_{kG}^1$. We prove this by induction on the Loewy length, that is, by considering the semisimple modules$J^n P(k) / J^{n+1} ...

4

My standard way to think about $H^1(G,A)$ is as functions $G \to A$ such that $f(gh)= g \cdot f(h) + f(g)$ ("cocycles"), modulo functions of the form $g \mapsto (g-1)\cdot a$ for some $a \in A$ ("coboundaries"). This is what you get by working with the bar resolution. In answer to your question about sections, it's group homomorphisms that you need. You ...

4

Acyclic and exact are not the same. As Akhil says in his answer, the long exact for group cohomology is indeed exact, but a projective resolution of a module is acyclic because it is not exact in degree zero. Originally, one used to say that a projective resolution $P_\bullet$ of a module $M$ was "acyclic over $M$", and that means that there is a map ...

4

A non-zero element of $H^2(H,\mathbb Z)$ is the class $\alpha$ of the non-split extension of $H$ by $\mathbb Z$ $$0\to\mathbb Z\to\mathbb Z\xrightarrow f\mathbb Z/3\to0$$ with $f(1)=1+3\mathbb Z$. Using esssentially the first description of the action (that is, a pullback construction), you can check that the transpositions of $S_3$ act on this extension by ...

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