Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question from Lang's Algebra (chapter twenty ex 6d). I think the vagueness confuses me as I am not even sure where to start

If $H$ is a normal subgroup of $G$, we have the cohomology groups $H^i (H,A)$.

The question asks to describe how we can get an operation of $G$ on the cohomology function $H_G$ "by conjugation" and functorality.

Any advice or direction on what this operation would be or how to look for it would be greatly appreciated.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Let $C^i(H,A)$ be the group of $i$-cochains, i.e., functions $H^i\rightarrow A$, which define the complex that computes the cohomology $H^i(H,A)$. For $g\in G$, you have the automorphism $\varphi_g:h\mapsto ghg^{-1}:H\rightarrow H$ because $H$ is normal in $G$, and you also have the abelian group automorphism $\psi_g:a\mapsto g^{-1}a:A\rightarrow A$, which is not usually $G$-equivariant, but is compatible with $\varphi_g$ in the sense that $\psi_g(\varphi_g(h)a)=h\psi_g(a)$. These maps give rise to maps $f\mapsto\psi_g\circ f\circ\varphi_g^i:C^i(H,A)\rightarrow C^i(H,A)$ which are compatible with the coboundary maps (here $\varphi_g^i:H^i\rightarrow H^i$ is the $i$-fold product of the map $\varphi_g$ with itself). So these maps descend to maps on cohomology $g^*:H^i(H,A)\rightarrow H^i(H,A)$ for all $i$. Then you get an action of $G$ on $H^i(H,A)$ by $g\cdot\kappa:=g^*(\kappa)$ (I may have the signs switched for a left action, i.e., you might need $\varphi_{g^{-1}}$ and $\psi_{g^{-1}}$ instead, but that works the same way). For example, on $H^0(H,A)=A^H$, this action is just given by $a\mapsto g^{-1}a$, so in particular, $H$ acts trivially on $H^0(H,A)$. Using an inductive argument with long exact sequences, it can be shown that $H$ acts trivially on $H^i(H,A)$ for all $i$, so you really have an action of $G/H$ on $H^i(H,A)$. This is important for defining the Hochschild-Serre spectral sequence $H^p(G/H,H^q(H,A))\Rightarrow H^{p+q}(G,A)$.

share|improve this answer
    
That makes sense. What does being "compatible" imply or I suppose "how" does it imply? Thank you –  user45150 Nov 15 '12 at 1:40
1  
I'm just using the term ``compatible" to mean the equation I wrote down. The importance of this is that you can use it to show that the maps $\alpha^i:C^i(H,A)\rightarrow C^i(H,A)$ defined in my answer give rise to a map of complexes $C^\bullet(H,A)\rightarrow C^\bullet(H,A)$, i.e., that $\partial^{i-1}\circ\alpha^{i-1}=\alpha^i\circ\partial^{i-1}$, where $\partial^{i-1}:C^{i-1}(H,A)\rightarrow C^i(H,A)$ is the coboundary map. –  Keenan Kidwell Nov 15 '12 at 1:52
1  
This construction is a special case of a morphism of pairs $(G,A)\rightarrow(G^\prime,A^\prime)$, where $G,G^\prime$ are groups and $A$ (resp. $A^\prime$) is a $G$-module (resp. $G^\prime$-module). This is defined as the data of a map $\varphi:G^\prime\rightarrow G$ and $\psi:A\rightarrow A^\prime$ such that $\psi(\varphi(g^\prime)a)=g^\prime\psi(a)$ (this is exactly the compatibility condition, although in the special case of my answer, $G^\prime=H=G$ and $A=A^\prime$). Any morphism of pairs gives rise to a map of complexes $C^\bullet(G,A)\rightarrow C^\bullet(G^\prime,A^\prime)$ and thus –  Keenan Kidwell Nov 15 '12 at 1:55
    
maps on cohomology $H^\bullet(G,A)\rightarrow H^\bullet(G^\prime,A^\prime)$. –  Keenan Kidwell Nov 15 '12 at 1:55
    
Thank you. This was very helpful and much appreciated. –  user45150 Nov 15 '12 at 17:26

This is explained succinctly in Ken Brown's classic text Cohomology of Groups, pg48.

You have your projective resolution $F\to\mathbb{Z}$ for your group in question, and you just need to check that the chain map $\tau:F\to F$ given by $\tau(x)=g\cdot x$ is $G$-equivariant and gives you your action (fixed $g\in G$). Then it induces a well-defined map $\tau:F_G\to F_G$ which induces the desired map on homology.

share|improve this answer
    
If $M$ is a $G$-module, then $x\mapsto gx:M\rightarrow M$ is not usually $G$-equivariant if $G$ isn't abelian, right? –  Keenan Kidwell Nov 15 '12 at 1:28
    
No, what is implied by $G$-equivariance here is $\tau(g_1x)=c(g_1)\tau(x)$ for a homomorphism $c:G\to G$, which in our case is conjugation, $c(g_1)=gg_1g^{-1}$. –  Chris Gerig Nov 15 '12 at 1:37
1  
In particular, $G$ acts trivially on $H_*(G)$, and for $H\triangleleft G$ we get an induced $G/H$-action on $H_*(H)$. –  Chris Gerig Nov 15 '12 at 1:38

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.