Given a short exact sequence for some finite groups $A,B,C$, $$1\to A \to B \to C \to 1,$$

how could we construct an exact sequence of their cohomology group out of it?

One version of the story I found is here,

$$1\to H^0(G,A) \to H^0(G,B) \to H^0(G,C)\to H^1(G,A) \to H^1(G,B) \to H^1(G,C)$$

but I am looking for a relation of an exact sequence relating the following cohomology groups by group homomorphism: $H^d(A,\mathbb{R}/\mathbb{Z})$, $H^d(B,\mathbb{R}/\mathbb{Z})$, $H^d(C,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(A,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(B,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(C,\mathbb{R}/\mathbb{Z})$.

Is there a relation like

$$\to H^d(A,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z}) \to H^d(C,\mathbb{R}/\mathbb{Z})\to ?$$

or the other way around

$$\to H^d(C,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z}) \to H^d(A,\mathbb{R}/\mathbb{Z})\to ?$$

and how do we proceed the sequence?

  • 1
    $\begingroup$ Perhaps the Lyndon-Hochschild-Serre spectral sequence is what you're looking for? In low degrees you get a short-ish exact sequence, which is usually called the inflation-restriction sequence. $\endgroup$ – JHF May 1 '17 at 16:48
  • $\begingroup$ Thanks JHF, +1, It looks to me that in LHS $1\to H^1(C,\mathbb{R}/\mathbb{Z}) \to H^1(B,\mathbb{R}/\mathbb{Z}) \to H^1(A,\mathbb{R}/\mathbb{Z})\to H^2(C,\mathbb{R}/\mathbb{Z}) \to H^2(B,\mathbb{R}/\mathbb{Z})$, but does this continue go on for $H^d$? $\endgroup$ – wonderich May 1 '17 at 17:34
  • $\begingroup$ In general, no. You only get an exact sequence at the beginning because they aren't that many terms in the associated graded of the $E_\infty$-page of the spectral sequence. However, if you know that there is a vanishing region in the spectral spectral for low degrees, you may get a corresponding exact sequence for the first nonvanishing degrees. Second, I want to point out that in the inflation-restriction sequence, the coefficient modules may change depending on the group action, so you should be careful with the sequence you've written . $\endgroup$ – JHF May 1 '17 at 18:04
  • $\begingroup$ +1, JHF, Thanks, can you write a much complete answer? For example, when do I have $H^d(C,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z})$ and $H^{d-1}(C,\mathbb{R}/\mathbb{Z}) \to H^{d-1}(B,\mathbb{R}/\mathbb{Z})$? $\endgroup$ – wonderich May 1 '17 at 18:09

As requested, I am turning my comments into an answer.

First, given a short exact sequence $$1 \to A \to B \to C \to 1$$ of finite discrete groups, one can think of them as a sequence of topological spaces $$K(A,1) \to K(B,1) \to K(C,1).$$ If this were a cofiber sequence, then we would get a long exact sequence in cohomology as usual, but unfortunately this is not the case. Instead, this is a fibration, so we have the (Lyndon-Hochschild-)Serre spectral sequence. More precisely, suppose we are given compatible systems of local coefficients, which in this case amounts to a $B$-module $M$. Then we have a spectral sequence $$E_2^{p,q} = H^p(C, H^q(A, M)) \Rightarrow H^{p+q}(B,M).$$

This is a first-quadrant spectral sequence, and some of the terms at the lower left corner of he $E_2$ page are unaffected by diffentials and hence survives intact to the $E_\infty$ page. Then, when reconstructing $H^*(B, M)$ from the associated graded on the $E_\infty$ page, we find that we get a short "long" exact sequence $$0 \to H^1(C, M^A) \to H^1(B, M) \to H^1(A, M)^C \xrightarrow{\delta} H^2(C, M^A) \to H^2(B, M).$$ The map $\delta$ is given by the $d_2$ differential $E_2^{0,1} \to E_2^{2,0}$. In group cohomology, this sequence is called the inflation-restriction sequence. However, this sequence can be extended only for a couple more terms, but even then one of the terms would be less explicit. For example, the next term in the sequence would be $H^1(C, H^1(A,M))$.

On the other hand, if you can show that $E_2^{p,q} = 0$ for $0 < p \leq d$ and $0 < q < d$, then you would have a five-term exact sequence starting in a higher degree $$0 \to H^d(C, M^A) \to H^d(B, M) \to H^d(A, M)^C \to H^{d+1}(C, M^A) \to H^{d+1}(B,M).$$

Finally, you asked when you have maps $H^d(C, M) \to H^d(B, M)$. The answer is always, and this map is called the restriction. (There is also sometimes a wrong-way map $H^d(B,M) \to H^d(C,M)$ called the transfer.) The problem is that they don't knit together to form an exact sequence, and there certainly isn't going to be a "connecting homomorphism" in general, as we see from the spectral sequence.

  • $\begingroup$ thanks, JHF, +1, looks a very serious one. $\endgroup$ – wonderich May 1 '17 at 19:16
  • $\begingroup$ I think there is always a group homomorphism map $H^d(A,M) \to H^d(B,M)$ is that correct? $\endgroup$ – annie heart May 1 '17 at 19:22
  • $\begingroup$ @annieheart In this case, yes, since $A$ and $B$ are finite. $\endgroup$ – JHF May 1 '17 at 19:24
  • $\begingroup$ How do you call the name of the space $K(G,n)$ and how is it defined? $\endgroup$ – annie heart May 14 '17 at 1:04
  • $\begingroup$ $K(G,n)$ are called Eilenberg-Mac Lane spaces. Their homotopy type is defined by the property that they only have one nontrivial homotopy group $G$ in degree $n$. There are various constructions for point-set models for them, for example via the bar construction. Of course, any two such CW models of them will be homotopy equivalent. $\endgroup$ – JHF May 14 '17 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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