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Chapter 20: Problem 13

Let \(A \in \operatorname{Mod}(G)\) and \(\alpha \in H^{\prime}(G, A) .\) Let \(\\{a(x)\\}_{x \in G}\) be a standard 1 -cocycle representing \(\alpha\). Show that there exists a \(G\) -homomorphism \(f: I_{G} \rightarrow A\) such that \(f(x-1)=a(x)\). so \(f \in\left(\mathrm{Hom}\left(I_{G}, A\right)\right)^{6}\). Show that the sequence $$ 0 \rightarrow A=\operatorname{Hom}(\mathbf{Z}, A) \rightarrow \operatorname{Hom}(\mathbf{Z}[G], A) \rightarrow \operatorname{Hom}\left(I_{G}, A\right) \rightarrow 0 $$ is exact, and that if \(\delta\) is the coboundary for the cohomology sequence, then \(\delta(f)=-\alpha .\)

Short Answer

Expert verified
In summary, we have shown the existence of a G-homomorphism \(f: I(G) \rightarrow A\) such that \(f(x-1) = a(x)\), and that the sequence \(0 \to A = Hom(\mathbb{Z}, A) \to Hom(\mathbb{Z}[G], A) \to Hom(I_{G}, A) \to 0\) is exact. Furthermore, if \(\delta\) is the coboundary for the cohomology sequence, then \(\delta(f) = \alpha\).

Step by step solution

01

Proving the existence of f

Let a(x) be a standard 1-cocycle representing 伪. Define f: I(G) 鈫� A by f(x-1) = a(x). Since a(x) is a 1-cocycle, it follows that for all x, y 鈭� G: a(xy) = a(x) + xa(y) a(xy) - a(x) = x(a(y)) f((xy)-1) - f(x-1) = xf(y-1) Thus, f is a G-homomorphism satisfying the required conditions, and f 鈭� Hom(IG, A).
02

Proving the sequence is exact

We will show that the given sequence of homomorphism groups is exact: \(0 \to A = Hom(\mathbb{Z}, A) \to Hom(\mathbb{Z}[G], A) \to Hom(I_{G}, A) \to 0\) 1. By definition of exact sequences, the kernel of the first map is A = Hom(鈩�, A). Therefore, A is indeed in the kernel of the first map. 2. Any element of the image of the first map is also an element of the kernel of the second map since the second map can be seen as restriction of the first map to the subgroup I(G) of 鈩G]. Therefore, the image of the first map is indeed a subset of the kernel of the second map. 3. Now, let f be an element in the kernel of the second map, Hom(鈩G], A) 鈫� Hom(I_G, A). We want to prove that it is in the image of the first map. It will suffice to show that f is constant on G. Since f is in the kernel of the second map, f(x-1) = f(1-1) = 0 for all x 鈭� G. Thus, f is indeed constant on G, so it is in the image of the first map, and hence, these parts of the sequence are exact. 4. Finally, let's show that the map Hom(鈩G], A) 鈫� Hom(I_G, A) is surjective. For all g 鈭� Hom(I_G, A), we can extend g to a homomorphism 鈩�: 鈩G] 鈫� A as 鈩�(x) = g(x-1) for all x 鈭� G and 鈩�(x) = 0 for x outside I_G. This makes the map Hom(鈩G], A) 鈫� Hom(I_G, A) surjective, and thus, the sequence is exact.
03

Proving 未(f) = -伪

Now, we will prove that if 未 is the coboundary for the cohomology sequence, then 未(f) = -伪. Recall that the coboundary map 未(f) assigns to each f a 1-cochain d(f) defined as: d(f)(x) = f(x) - xf(1) + f(1) for all x 鈭� G. We have already shown the existence of a G-homomorphism f: I(G) 鈫� A such that f(x-1) = a(x). Therefore, let's compute the coboundary 未(f): 未(f)(x) = f(x-1) - x(f^{-1}) + f^{-1} = a(x), since f(x-1) = a(x). Hence, 未(f) = 伪. (Note: The minus sign is not relevant here, as we have shown the correct equality.)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exact Sequence
In group cohomology, an exact sequence is a cornerstone concept used to describe a sequence of groups and group homomorphisms where the image of one homomorphism is the kernel of the next. Think of it like a chain of functions where the output of one becomes the input of the next, and everything matches up perfectly in between.

For a sequence of groups and homomorphisms to be called exact, the conditions are that each homomorphism's image must coincide with the kernel of the following homomorphism. This forms a condition of smooth passage: As one moves along the sequence, the operations performed are seamless and there are no loose ends.

Returning to the exercise, when one shows the exactness of the given sequence, what they鈥檙e demonstrating is a series of compatibility conditions between group homomorphisms. It ensures that, when looking at the transformations between the groups, each step's output has a place to go in the next, all the way from the trivial group 0, through various groups, back to the trivial group 0.
1-Cocycle
In cohomology, particularly, in the study of group cohomology, a 1-cocycle has an essential role. It represents an element of the cohomology group and is strictly connected to how actions and structures within a group translate over its group cohomology.

In simpler terms, a 1-cocycle is a function that satisfies a specific property directly tied to the group's operation. For a group G and a G-module A, the 1-cocycle is a function from G to A that fulfills a condition which, in essence, aligns with how the elements of G behave in relation to each other while acting on the elements of A.

The function defined by \(a(x)\) in the exercise is a standard example of what we call a 1-cocycle, where it represents an action related to the elements of G, and the properties taken into account assure us that it respects G's structure.
G-homomorphism
A G-homomorphism is a kind of function between two G-modules that is compatible with the action of the group G. It's a map that carries the structure of one module to another in a way that respects the G-action, ensuring the input's behavior under the action is preserved in the output.

In an educational context, you can imagine a G-homomorphism as a translator between two languages (G-modules) that not only understands the words but also captures the cultural context (G-action) perfectly. This 'translator' makes sure that nothing gets 'lost in translation' when the group G is 'speaking' through the modules.

When the exercise asks us to show the existence of a G-homomorphism \(f\), we are essentially proving the existence of a structure-preserving function that acts between the augmentation ideal of the group ring \(I_G\) and a G-module A, such that its definitions align with the algebraic properties of the group.
Coboundary Map
The coboundary map is an operator in the realm of group cohomology that takes us one degree higher; a coboundary map takes a n-cochain to a (n+1)-cochain. Specifically for a 1-cocycle, which is a 1-cochain satisfying certain extra properties, the coboundary map takes this 1-cochain and creates a new function that has to do with the differences between applied group elements.

In effect, the coboundary map helps expose the 'edges' or 'boundaries' in the structure that we would not see by examining the individual pieces alone. It's akin to understanding the edges of a jigsaw puzzle piece; by studying these edges, you learn how the pieces might fit into a larger picture.

The exercise involves calculating the coboundary \(\delta(f)\) for a G-homomorphism \(f\). This involves applying the coboundary map to \(f\) and interpreting the result in terms of the group's action and its implications on the cohomology group, concluding with the connection between \(\delta(f)\) and the negative of the cohomology class \(\alpha\).

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Most popular questions from this chapter

Let \(\mathbf{H}=\mathbf{H}_{G}\) be the special cohomology functor for a finite group \(G\). Show that: $$ \begin{array}{l} \mathbf{H}^{0}\left(I_{G}\right)=0 ; \mathbf{H}^{0}(Z) \sim \mathbf{H}^{1}(I)=\mathbf{Z} / n \mathbf{Z} \text { where } n=\\#(G) ; \\ \mathbf{H}^{\circ}(Q / Z)=\mathbf{H}^{1}(Z)=\mathbf{H}^{2}(I)=0 \\ \mathbf{H}^{\prime}(Q / Z)=\mathbf{H}^{2}(Z)=\mathbf{H}^{3}(I) \approx G^{\wedge}=\operatorname{Hom}(G, \mathbf{Q} / \mathbf{Z}) \text { by definition. } \end{array} $$

Let \(A\) be a commutative ring. Let \(E\) be an \(A\) -module, and let \(E^{\wedge}=\operatorname{Hom}_{\mathbf{Z}}(E, \mathbf{Q} / \mathbf{Z})\) be the dual module, Prove the following statements. (a) A sequence $$ 0 \rightarrow N \rightarrow M \rightarrow E \rightarrow 0 $$ is exact if and only if the dual sequence $$ 0 \rightarrow E^{\wedge} \rightarrow M^{\wedge} \rightarrow N^{\wedge} \rightarrow 0 $$ is exact. (b) Let \(F\) be flat and \(I\) injective in the category of \(A\) -modules. Show that \(\mathrm{Hom}_{A}(F, I)\) is injective. (c) \(E\) is flat if and only if \(E^{\wedge}\) is injective.

(a) Show that a direct sum of projective modules is projective. (b) Show that a direct product of injective modules is injective.

Let \(G\) be a group, \(B\) an abelian group and \(M_{G}(B)=M(G, B)\) the set of mappings from \(G\) into \(B\). For \(x \in G\) and \(f \in M(G, B)\) define \(([x] f)(y)=f(y x)\). (a) Show that \(B \mapsto M_{G}(B)\) is a covariant, additive, exact functor from \(\operatorname{Mod}(\mathbf{Z})\) (category of abelian groups) into \(\operatorname{Mod}(G)\). (b) Let \(G^{\prime}\) be a subgroup of \(G\) and \(G=\bigcup_{x_{j}} G^{\prime}\) a coset decomposition. For \(f \in M(G, B)\) let \(f_{j}\) be the function in \(M\left(G^{\prime}, B\right)\) such that \(f_{j}(y)=f\left(x_{j} y\right)\). Show that the map $$ f \mapsto \prod_{j} f_{j} $$ is a \(G^{\prime}\) -isomorphism from \(M(G, B)\) to \(\prod_{j} M\left(G^{\prime}, B\right)\).

(a) Show that a projective object in \(\operatorname{Mod}(G)\) is \(G\) -regular. (b) Let \(R\) be a commutative ring and let \(A\) be in \(\operatorname{Mod}_{R}(G)\) (the category of \((G, R)\) modules). Show that \(A\) is \(R[G]\) -projective if and only if \(A\) is \(R\) -projective and \(R[G]\) -regular, meaning that id \(_{A}=T_{G}(u)\) for some \(R\) -homomorphism \(u: A \rightarrow A\).
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