Homology and Tate groups

Section 3.3 Homology and Tate groups

Reference.

[36], II.2.

You may not be surprised to learn that there is a “dual” theory to the theory of group cohomology, namely group homology. What you may be surprised to learn is that one can actually fit the two together, so that in a sense the homology groups become cohomology groups with negative indices. (Since the arguments are similar to those for cohomology, I’m going to skip some details.)

Subsection Homology

Definition 3.3.1.

Let \(M_G\) denote the maximal quotient of \(M\) on which \(G\) acts trivially. In other words, \(M_G\) is the quotient of \(M\) by the submodule spanned by \(m^g-m\) for all \(m \in M\) and \(g \in G\text{.}\) In yet other words, \(M_G = M/M I_G\text{,}\) where \(I_G\) is the augmentation ideal of the group algebra \(\ZZ[G]\text{:}\)

\begin{equation*} I_G = \left\{\sum_{g \in G} z_g[g]: \sum_g z_g = 0\right\}. \end{equation*}

Or if you like, \(M_G = M \otimes_{\ZZ[G]} \ZZ\text{.}\) Since \(M^G\) is the group of \(G\)-invariants, we call \(M_G\) the group of \(G\)-coinvariants.

The functor \(M \to M^G\) is right exact but not left exact: if \(0 \to M' \to M \to M'' \to 0\text{,}\) then \(M'_G \to M_G \to M''_G \to 0\) is exact but the map on the left is not injective. Again, we can fill in the exact sequence by defining homology groups.

Definition 3.3.2.

A \(G\)-module \(M\) is projective if for any surjection \(N \to N'\) of \(G\)-modules and any map \(\phi: M \to N'\text{,}\) there exists a map \(\psi: M \to N\) lifting \(\phi\text{.}\) This definition is dual to the definition of an injective \(G\)-module, but this symmetry is a bit misleading: it is much easier to find projectives than injectives. For example, any \(G\)-module which is a free module over the ring \(\ZZ[G]\) is projective such as \(\ZZ[G]\) itself!

Definition 3.3.3.

A projective resolution of \(M\) is an exact sequence \(\cdots \to P_1 \to P_0 \to M \to 0\) of \(G\)-modules in which the \(P_i\) are projective. Given such a resolution, take coinvariants to get a complex

\begin{equation*} \cdots \stackrel{d_2}{\to} P_2 \stackrel{d_1}{\to} P_1 \stackrel{d_0}{\to} P_0 \to 0, \end{equation*}

then put \(H_i(G, M) = \ker(d_{i-1})/\im(d_i)\text{.}\) Again, this is canonically independent of the resolution and functorial, and there is a long exact sequence which starts out

\begin{equation*} \cdots \to H_1(G, M'') \stackrel{\delta}{\to} H_0(G, M') \to H_0(G, M) \to H_0(G, M'') \to 0. \end{equation*}

Definition 3.3.4.

We say that \(M\) is acyclic (for homology) if \(H_i(G,M) =0\) for \(i>0\text{.}\) As with group cohomology, we can replace a projective resolution with an acyclic resolution and get the same homology groups. For example, induced modules are again acyclic and the analogue of Shapiro’s lemma holds (key point: any free \(\ZZ[H]\)-module induces to a free \(\ZZ[G]\)-module).

Remark 3.3.5.

One can give a concrete description of homology as well, but we won’t need it for our purposes. Even without one, though, we can calculate \(H_1(G, \ZZ)\text{,}\) using the exact sequence

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

By the long exact sequence in homology,

\begin{equation*} 0 = H_1(G, \ZZ[G]) \to H_1(G, \ZZ) \to H_0(G, I_G) \to H_0(G, \ZZ[G]) \end{equation*}

is exact, i.e. \(0 \to H_1(G, \ZZ) \to I_G/I_G^2 \to \ZZ[G]/I_G\) is exact. The last map is induced by \(I_G \hookrightarrow \ZZ[G]\) and so is the zero map. Thus \(H_1(G, \ZZ) \cong I_G/I_G^2\text{;}\) recall that in Exercise 3, it was shown that the map \(g \mapsto [g] - 1\) defines an isomorphism \(G^{\ab} \to I_G/I_G^2\text{.}\) This can be thought of as an algebraic analogue of the fact that the first homology group of a (reasonable) topological space equals the abelianization of the fundamental group.

Subsection The Tate groups

We now “fit together” the long exact sequences of cohomology and homology to get a doubly infinite exact sequence.

Definition 3.3.6.

Let \(M\) be a \(G\)-module. Define the map \(\Norm_G: M \to M\) by

\begin{equation*} \Norm_G(m) = \sum_{g \in G} m^g. \end{equation*}

Then \(\Norm_G\) induces a homomorphism

\begin{equation*} \Norm_G: H_0(G,M) = M_G \to M^G = H^0(G,M). \end{equation*}

Remark 3.3.7.

You might be wondering why \(\Norm_G\) is called a “norm” rather than a “trace”. The reason is that in practice, our modules \(M\) will most often be groups which are most naturally written multiplicatively, e.g., the nonzero elements of a field.

Definition 3.3.8.

We now define the Tate cohomology groups (or Tate homology groups if you prefer) as follows:

\begin{equation*} H_T^i = \begin{cases}H^i(G, M) \amp i > 0 \\ M^G/\Norm_G M \amp i=0 \\ \ker(\Norm_G)/MI_G \amp i=-1 \\ H_{-i-1}(G,M) \amp i\lt -1. \end{cases} \end{equation*}

Lemma 3.3.9.

For any short exact sequence \(0 \to M' \to M \to M'' \to 0\) of \(G\)-modules, we have a canonical exact sequence

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

which extends infinitely in both directions.

Proof.

Since we already have long exact sequences for homology and cohomology, the only remaining issue is exactness between \(H^{-2}_T(G, M'')\) and \(H^1_T(G, M')\) inclusive. This follows by diagram-chasing, as in the proof of the snake lemma (Lemma 3.1.16) on the commutative diagram Figure 3.3.10 with exact rows, noting that the diagram remains commutative with the dashed arrows added.

Remark 3.3.11.

If \(M\) is an induced \(G\)-module, then \(H^i_T(G,M) = 0\) for all \(i\) (see Exercise 1. That is, induced modules are acyclic for all of cohomology, homology, and Tate (co)homology.

Subsection Extended functoriality revisited

The extended functoriality for cohomology groups (Definition 3.2.21) has analogues for homology groups and Tate cohomology groups, but under more restrictive conditions.

Definition 3.3.12.

Again, let \(M\) be a \(G\)-module and \(M'\) a \(G'\)-module, and consider a homomorphism \(\alpha: G' \to G\) of groups and a homomorphism \(\beta: M \to M'\) of abelian which are compatible in the sense of Definition 3.2.21. We would like to obtain canonical homomorphisms \(H_i(G, M) \to H_i(G', M')\text{,}\) but for this we need to add an additional condition to ensure that \(M \to M'\) induces a well-defined map \(M_G \to M'_{G'}\text{.}\) For instance, this holds if \(\alpha\) is surjective.

For Tate cohomology groups, there is a further complication that the map \(M^G \to (M')^{G'}\) does not necessarily induce a map \(\Norm(M) \to \Norm(M')\text{.}\) However, this does occur if \(\alpha\) is injective, so for instance we have well-defined restriction maps \(\Res: H^0_T(G, M) \to H^0_T(H, M)\) whenever \(H\) is a subgroup of \(G\text{.}\)

Remark 3.3.13.

In Example 3.2.22, we used the restriction and corestriction maps to show that for \(G\) a finite group and \(M\) a G-module, the groups \(H^i(G, M)\) are torsion groups killed by \(\#G\) for all \(i > 0\text{.}\) While we cannot extend the corestriction map to Tate cohomology, we may still argue directly that \(H^0_T(G, M)\) is killed by \(\#G\text{.}\)

Exercises Exercises

1.

Prove that if \(M\) is an induced \(G\)-module, then \(H^i_T(G,M) = 0\) for all \(i \in \ZZ\text{.}\)

Hint.

Use the fact that induced \(G\)-modules are acyclic for both cohomology and homology to reduce to checking the cases \(i = -1, 0\text{.}\) Another option is to extend Shapiro’s lemma to Tate cohomology groups.

2.

Let \(G \subseteq H\) be an inclusion of finite groups. Show that via the identification from Remark 3.3.5, the map \(\Res: H^{-2}_T(G, \ZZ) \to H^{-2}_T(H,\ZZ)\) corresponds to the transfer (Verlagerung) map \(V: G^{\ab} \to H^{\ab}\text{.}\) This provides another way to derive the existence of the latter.

Prev Top Next