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I'm trying to understand the Hochschild-Serre spectral sequence by an example. Consider the short exact sequence of groups: $1\to N\to G\to G/N\to 1$ where $G\cong \mathbb{Z}_4$, $N\cong\mathbb{Z}_2$. Assume $G$ acts on the module $\mathbb{C}^*$ by conjugation (the generator of $G$ sends $a+bi$ to $a-bi$). Note that $N$ acts trivially on the module. The inflation-restriction exact sequence that comes from the Hochschild-Serre spectral sequence is:

$1\to H^1(G/N,\mathbb{C}^*)\to H^1(G,\mathbb{C}^*)\to H^1(N,\mathbb{C}^*)^{G/N}\to H^2(G/N,\mathbb{C}^*)\to H^2(G,\mathbb{C}^*)$

I computed these groups explicitly and got that they are:

$1\to 1\to 1\to \mathbb{Z}_2\to \mathbb{Z}_2\to H^2(G,\mathbb{C}^*)$

and the last morphism is trivial. This means that $d_2:H^1(N,\mathbb{C}^*)^{G/N}\to H^2(G/N,\mathbb{C}^*)$ is nontrivial. But I fail to compute it explicitly.

I begin with a cocycle $\alpha\in H^1(N,\mathbb{C}^*)^{G/N}$. Observe that $d_1(\alpha)$ should be trivial as an element of $E_1^{1,1}$. Therefore $d_1(\alpha)$ is in the image of $d_0:E_0^{1,0}\to E_0^{1,1}$ (a coboundary). Deriving a preimage of this coboundary should give me the desired element of $H^2(G/N,\mathbb{C}^*)$. But when I try to follow these steps with a specific cocycle (the one that sends the nontrivial element of $N$ to $-1$), I see that $d_1(\alpha)$ is trivial already in $E_0^{1,1}$ because $\alpha=\bar{\alpha}$. Moreover, $d_0:E_0^{1,0}\to E_0^{1,1}$ is trivial since $N$ acts trivially on the module. I conclude that $d_2$ is trivial, contradicting the inflation-restriction exact sequence.

What am I doing wrong?

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The difficulty with the procedure that you describe is that you have to be extremely careful with the structure of the $E_1$-page of the spectral sequence, and in particular you need a very concrete model. I am not sure what model you are using, and it is difficult to say without more information. Here is an example of one method by which this can be carried out.

To compute this differential, you start with a cocycle $f: N \to \Bbb C^*$. You must then choose some way to extend it to a function $G: G \to \Bbb C^\times$ in the following way: choose arbitrary values $f(a)$ for $a$ representatives of the cosets of $G/N$, and then for a general $g = a \cdot n$ in $G$ define $f(g) = f(a) \cdot {}^af(n)$. Then compute the coboundary $\delta f$, which will be a 2-cocycle and will be well-defined on $G/N$.

Your cocycle has $f(0) = 1$, $f(2) = -1$. We could e.g. define $f(1) = i$ and then find $f(3) = f(1 + 2) = f(1) \cdot \overline{f(2)} = -i$, so $f(x) = i^k$. Then $$ \begin{align*} \delta f(k,l) &= f(k) \cdot f(k+l)^{-1} \cdot {}^kf(l)\\ &= \begin{cases} i^k i^{-k-l} i^l &\text{if $k$ is even,}\\ i^k i^{-k-l} (\bar{i})^{l} &\text{if $k$ is odd.} \end{cases} \\ &= \begin{cases} 1 &\text{if $k$ is even},\\ (-1)^l &\text{if $k$ is odd.} \end{cases}\\ &= (-1)^{kl}. \end{align*} $$ This represents the nontrivial element of $H^2(\Bbb Z/2, \Bbb C^\times)$. If we choose a different value of $f(1)$ we alter this by a coboundary.

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  • $\begingroup$ Thank you very much. This is very helpful. Can you extend this method for computing $d_2:H^p(G/N,H^1(N,\mathbb{C}^*))\to H^{p+2}(G/N,\mathbb{C}^*)$ for all $p$? Your method reminds me of a snake lemma. Can you explain where it comes from? $\endgroup$ – David Levit-Gurevich Feb 6 '19 at 23:07

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