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Section 3.4 Cohomology of cyclic groups

Reference.

[37], IV.7; [33], IX.1.

We next specialize attention to the case of a finite cyclic group, which will play a key role in many of our calculations. In this case, the cohomology, homology, and Tate groups satisfy a key periodicity property (Theorem 3.4.1) which allows us to define and manipulate a sort of “Euler characteristic”, the Herbrand quotient (Definition 3.4.4).

Subsection The periodicity theorem

In general, for any given \(G\) and \(M\text{,}\) it is at worst a tedious exercise to compute \(H^i_T(G,M)\) for any single value of \(i\text{,}\) but try to compute all of these at once and you discover that they exhibit very little evident structure. Thankfully, there is an exception to that dreary rule when \(G\) is cyclic.

Choose a generator \(g\) of \(G\text{.}\) We start with the four-term exact sequence of \(G\)-modules

\begin{equation*} 0 \to \ZZ \to \ZZ[G] \to \ZZ[G] \to \ZZ \to 0 \end{equation*}

in which the first map is \(1 \mapsto \sum_{g \in G} [g]\text{,}\) the second map is \([h] \mapsto [hg] - [h]\text{,}\) and the third map is \([h] \mapsto 1\text{.}\) Since everything in sight is a free abelian group, we can tensor over \(\ZZ\) with \(M\) and get another exact sequence:

\begin{equation*} 0 \to M \to M \otimes_\ZZ \ZZ[G] \to M \otimes_\ZZ \ZZ[G] \to M \to 0\text{.} \end{equation*}

The terms in the middle are just \(\Ind^G_{1} M\text{,}\) where we first restrict \(M\) to a module for the trivial group and then induce back up. Thus their Tate groups are all zero. The desired result now follows from the following general fact: if

\begin{equation*} 0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \stackrel{h}{\to} D \to 0 \end{equation*}

is exact and \(B\) and \(C\) have all Tate groups zero, then there is a canonical isomorphism \(H^{i+2}_T(G, A) \to H^i_T(G, D)\text{.}\) To see this, apply the long exact sequence to the short exact sequences

\begin{gather*} 0 \to A \to B \to B/\im(f) \to 0\\ 0 \to B/\ker(g) \to C \to D \to 0 \end{gather*}

to get

\begin{equation*} H^{i+2}(G, A) \cong H^{i+1}(G, B/\im(f)) = H^{i+1}(G, B/\ker(g)) \cong H^i(G,D). \end{equation*}

Remark 3.4.2.

In particular, when \(G\) is cyclic, the long exact sequence of a short exact sequence \(0 \to M' \to M \to M'' \to 0\) of \(G\)-modules curls up into an exact hexagon as in Figure 3.4.3.

Figure 3.4.3.

Subsection Herbrand quotients

Definition 3.4.4.

Let \(G\) be a finite cyclic group and let \(M\) be a \(G\)-module. If the groups \(H^i_T(G, M)\) are finite, we define the Herbrand quotient as the ratio

\begin{equation*} h(M) = \#H^0_T(G,M) / \#H^{-1}_T(G, M). \end{equation*}

From the exactness of the hexagon in Figure 3.4.3, we see that if \(M', M, M''\) all have Herbrand quotients, then

\begin{equation*} h(M) = h(M') h(M''). \end{equation*}

Remark 3.4.5.

Expanding on the previous point, if two of \(M', M, M''\) have Herbrand quotients, then so does the third. For example, if \(M'\) and \(M''\) have Herbrand quotients, i.e., their Tate groups are finite, then we have an exact sequence

\begin{equation*} H^{-1}_T(G, M') \to H^{-1}_T(G, M) \to H^{-1}_T(G, M'') \end{equation*}

and the outer groups are all finite. In particular, the first map is out of a finite group and so has finite image, and modulo that image, \(H^{-1}_T(G,M)\) injects into another finite group. So it's also finite, and so on.

Remark 3.4.6.

In practice, it will often be much easier to compute the Herbrand quotient of a \(G\)-module than to compute either of its Tate groups directly. The Herbrand quotient will then do half of the work for free: once one group is computed directly, at least the order of the other will be automatically known.

Remark 3.4.7.

If \(M\) is finite, then \(h(M) = 1\text{.}\) To wit, the sequences

\begin{gather*} 0 \to M^G \to M \to M \to M_G \to 0\\ 0 \to H^{-1}_T(G,M) \to M_G \stackrel{\Norm_G}{\to} M^G \to H^0_T(G,M) \to 0 \end{gather*}

are exact, where \(M \to M\) is the map \(m \mapsto m^g - m\text{;}\) thus \(M_G\) and \(M^G\) have the same order, as do \(H^{-1}_T\) and \(H^0_T\text{.}\)

Exercises Exercises

1.

The periodicity of the Tate groups for \(G\) cyclic means that there is a canonical (up to the choice of a generator of \(G\)) isomorphism between \(H^{-1}_T(G, M)\) and \(H^1_T(G,M)\text{,}\) i.e., between \(\ker(\Norm_G)/MI_G\) and the set of equivalence classes of 1-cocycles. What is this isomorphism explicitly? In other words, given an element of \(\ker(\Norm_G)/MI_G\text{,}\) what is the corresponding 1-cocycle?

2.

Put \(K = \QQ_p(\sqrt{p})\text{.}\) Compute the Herbrand quotient of \(K^*\) as a \(G\)-module for \(G = \Gal(\QQ_p(\sqrt{p})/\QQ_p)\text{.}\)
Hint.
Use the exact sequence \(1 \to \gotho_K^* \to K^* \to \ZZ \to 1\text{.}\)

3.

Let \(G = S_3\) (the symmetric group on three letters), let \(M = \ZZ^3\) with the natural \(G\)-action permuting the factors, and let \(N = M^G\text{.}\) Compute \(H^i(G, M/N)\) for \(i=1,2\) however you want: you can explicitly compute cochains, use the alternate interpretations given above, or use the exact sequence \(0 \to N \to M \to M/N \to 0\text{.}\) Better yet, use more than one method and make sure that you get the same answer.
Hint.

Part of the point of this exercise is that even in this relatively simple-looking situation, it is not all that easy to do the computation. One approach that minimizes the computational complexity is to use the Hochschild-Serre spectral sequence (see [36], Remark II.1.35>) to reduce to working with the cyclic groups \(H = A_3 \cong \ZZ/3\ZZ\) and \(G/H = S_3/A_3 \cong \ZZ/2\ZZ\text{,}\) for which periodicity is applicable.