\documentclass[12pt]{article}
\usepackage{amsfonts, amsthm, amsmath}
\usepackage[all]{xy}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}
\setlength{\topmargin}{0in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\parskip}{0pt}
\setlength{\parindent}{20pt}

\def\AA{\mathbb{A}}
\def\CC{\mathbb{C}}
\def\FF{\mathbb{F}}
\def\PP{\mathbb{P}}
\def\QQ{\mathbb{Q}}
\def\RR{\mathbb{R}}
\def\ZZ{\mathbb{Z}}
\def\gotha{\mathfrak{a}}
\def\gothb{\mathfrak{b}}
\def\gothm{\mathfrak{m}}
\def\gotho{\mathfrak{o}}
\def\gothp{\mathfrak{p}}
\def\gothq{\mathfrak{q}}
\def\gothr{\mathfrak{r}}
\DeclareMathOperator{\ab}{ab}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\coker}{coker}
\DeclareMathOperator{\Cor}{Cor}
\DeclareMathOperator{\disc}{Disc}
\DeclareMathOperator{\Frob}{Frob}
\DeclareMathOperator{\Gal}{Gal}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\Ind}{Ind}
\DeclareMathOperator{\Inf}{Inf}
\DeclareMathOperator{\Norm}{Norm}
\DeclareMathOperator{\Res}{Res}
\DeclareMathOperator{\Trace}{Trace}
\DeclareMathOperator{\Cl}{Cl}

\def\head#1{\medskip \noindent \textbf{#1}.}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{cor}[theorem]{Corollary}

\begin{document}

\begin{center}
\bf
Math 254B, UC Berkeley, Spring 2002 (Kedlaya) \\
Homology of Finite Groups
\end{center}

\head{Reference} Milne, II.2; for cyclic groups, also
Neukirch, IV.7 and Lang, IX.1.

\head{Caveat} The Galois cohomology groups used in Neukirch are
not the ones we defined earlier. They are the Tate cohomology groups we
are going to define below.

\head{Leftover reinterpretation from last time (or, fun with $H^1$)}

I should have mentioned when we were discussing $H^1$ that it can be
interpreted in terms of ``principal homogeneous spaces'', an interpretation
that naturally comes up in practice fairly often. Namely, let $G$ be a
finite group and $M$ a $G$-module. Then the elements of $H^1(G, M)$ correspond
to sets $A$ with both a $G$-action and an $M$-action, subject to the
following restrictions:
\begin{enumerate}
\item[(a)] for any $a$, the map $M \to M$ given by $m \mapsto m(a)$ is a bijection;
\item[(b)] for $a \in A$, $g \in G$ and $m \in M$, $m(a)^g = m^g(a)$ 
(i.e., the $G$-action and $M$-action commute).
\end{enumerate}
(The trivial example is $A=M$ with the same $G$-action, and $M$ acting by
translation: $m(a) = m+a$.)
In particular, there is a well-defined ``subtraction'' map $A \times A \to M$
sending $(a, b)$ to the element $m$ of $M$ such that $m(a) = b$. The
correspondence is as follows: given such a set $A$, pick $a \in A$,
take the map $\rho: G \to M$ given by $\rho(g) = a^g - a$, and
let $\phi$ be the 1-cocycle with $\phi(e, g) = \rho(g)$. Details left
to the reader.

This comes up, for example, in the theory of elliptic curves.
If $L$ is a finite extension of $K$ and $E$ is an elliptic curve over
$E$, then $H^1(\Gal(L/K), E(\overline{K}))$ is the set of $K$-isomorphism
classes of curves whose Jacobians are $K$-isomorphic to $E$
(but which might not themselves be isomorphic to $E$ by virtue of not
having a $K$-rational point). 

Another example from elliptic curves: $H^1(\Gal(L/K), \Aut(E))$ parametrizes
twists of $E$, elliptic curves defined over $K$ which are $L$-isomorphic
to $E$. (E.g., $y^2 = x^3+x+1$ versus $2y^2 = x^3 + x +1$, with $L = \QQ(\sqrt{2})$.) See Silverman,
\textit{The Arithmetic of Elliptic Curves}, especially Chapter X,
for all this and more fun with $H^1$, including the infamous ``Selmer group''
and ``Tate-Shafarevich group''.

\head{Homology}

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 details.)

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 $g \in G$. In yet other words,
$M_G = M/M I_G$, where $I_G$ is the \emph{augmentation ideal} of the
group algebra $\ZZ[G]$:
\[
I_G = \{\sum_{g \in G} z_g[g]: \sum_g z_g = 0\}.
\]
Or if you like, $M_G = M \otimes_{\ZZ[G]} \ZZ$. 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 
$0 \to M' \to M \to M'' \to 0$, 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.

A $G$-module $M$ is \emph{projective} if for any surjection $N \to N'$
of $G$-modules and any map $\phi: M \to N'$, there exists a map
$\psi: M \to N$ lifting $\phi$. This is the reverse notion to injective;
but it's much easier to find projectives than injectives. For example,
any $G$-module which is a free module over the ring $\ZZ[G]$ is projective,
e.g., $\ZZ[G]$ itself!

Given a projective resolution $\cdots \to P_1 \to P_0 \to M \to 0$ of
a $G$-module $M$ (an exact sequence in which the $P_i$ are injective), 
take coinvariants to get a no longer exact complex
\[
\cdots \stackrel{d_2}{\to} P_2 \stackrel{d_1}{\to} P_1 \stackrel{d_0}{\to}
P_0 \to 0,
\]
then put $H_i(G, M) = \ker(d_{i-1})/\im(d_i)$. Again, this is canonically
independent of the resolution and functorial, and there is a long
exact sequence which starts out
\[
\cdots \to H_1(G, M'') \to \stackrel{\delta}{\to} H_0(G, M') \to H_0(G, M) \to H_0(G, M'') \to 0.
\]
Also, you can replace the projective resolution by an acyclic resolution
(where here $M$ being acyclic means $H_i(G,M) =0$ for $i>0$) and get
the same homology groups. But since free modules are projective, this is
rarely useful.

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 $H_1(G, \ZZ)$, using the exact
sequence
\[
0 \to I_G \to \ZZ[G] \to \ZZ \to 0.
\]
By the long exact sequence in homology,
\[
0 = H_1(G, \ZZ[G]) \to H_1(G, \ZZ) \to H_0(G, I_G) \to H_0(G, \ZZ[G])
\]
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 $H_1(G, \ZZ) \cong I_G/I_G^2$, which on a previous
homework we showed was isomorphic to $G^{\ab}$.

\head{The Tate groups}

We now ``fit together'' the long exact sequences of cohomology and homology
to get a doubly infinite exact sequence. Define the map $\Norm_G: M \to M$
by
\[
\Norm_G(m) = \sum_{g \in G} m^g.
\]
(It looks like it should be called
``trace'', but in practice our modules $M$ will be groups which are most
naturally written multiplicatively, i.e., the nonzero elements of a field.)
Then $\Norm_G$ induces a homomorphism
\[
\Norm_G: H_0(M,G) = M_G \to M^G = H^0(M,G).
\]
Now define
\[
H_T^i = \begin{cases}
H^i(G, M) & i > 0 \\
M^G/\Norm_G(M) & i=0 \\
\ker(\Norm_G)/MI_G & i=-1 \\
H_{-i-1}(G,M) & i<-1;
\end{cases}
\]
then for any short exact sequence $0 \to M' \to M \to M'' \to 0$,
we get an exact sequence
\[
\cdots
\to H^{i-1}_T(G, M'') \to H^i_T(G, M') \to H^i_T(G, M) \to H^i_T(G, M'')
\to H^{i+1}_T(G, M') \to \cdots
\]
which extends infinitely in both directions.
(The $T$ stands for Tate, who had a major part in introducing
Galois cohomology into algebraic number theory.)

\head{Finite cyclic groups}

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

\begin{theorem}
Let $G$ be a finite cyclic group and $M$ a $G$-module. Then there is
a canonical, functorial isomorphism $H^i_T(G,M) \to H^{i+2}_T(G,M)$ for all
$i \in \ZZ$.
\end{theorem}
\begin{proof}
Choose a generator $g$ of $G$.
We start with the four-term exact sequence of $G$-modules
\[
0 \to \ZZ \to \ZZ[G] \to \ZZ[G] \to \ZZ \to 0
\]
in which the first map is $1 \mapsto [1]$, the second map is
$[h] \mapsto [hg] - [h]$, and the third map is $[h] \mapsto 1$.
Since everything in sight is a free abelian group, we can tensor over $\ZZ$
with $M$ and get another exact sequence:
\[
0 \to M \to M \otimes_\ZZ \ZZ[G] \to M \otimes_\ZZ \ZZ[G] \to M \to 0.
\]
The terms in the middle are just $\Ind^G_{\{e\}}(M)$, 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
\[
0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \stackrel{h}{\to} D \to 0
\]
is exact and $B$ and $C$ have all Tate groups zero, then
there is a canonical isomorphism $H^{i+2}_T(G, A) \to H^i_T(G, D)$.
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
\[
H^{i+2}(G, A) \cong H^{i+1}(G, B/\im(f)) = H^{i+1}(G, B/\ker(g))
\cong H^i(G,D).
\]
\end{proof}
In particular, 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:
\[
\xymatrix{
 & H^{-1}_T(G, M) \ar[r] & H^{-1}_T(G, M'') \ar[dr] & \\
H^{-1}_T(G, M') \ar[ru] & & & H^0_T(G, M') \ar[dl] \\
 & H^{0}_T(G, M'') \ar[lu] & H^{0}_T(G, M) \ar[l] &
}
\]
If the groups $H^i_T(G, M)$ are finite, we define the
\emph{Herbrand quotient} 
\[
h(M) = \#H^0_T(G,M) / \#H^{-1}_T(G, M).
\]
Then from the exactness of the hexagon, if $M', M, M''$ all have
Herbrand quotients, then
\[
h(M) = h(M') h(M'').
\]
Moreover, if two of $M', M, M''$ have Herbrand quotients, 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
\[
H^{-1}_T(G, M') \to H^{-1}_T(G, M)
\to H^{-1}(G, M'')
\]
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.

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 the work for free''; once one group
is computed directly, at least the order of the other will be automatically
known.

One special case is easy to work out: if $M$ is finite, then $h(M) = 1$,
because 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 $m \mapsto m^g - m$, so
$M_G$ and $M^G$ have the same order, as do $H^{-1}$ and $H^0$.

\head{Extended functoriality}

We already saw that if we have a homomorphism of $G$-modules, we get
induced homomorphisms on homology and cohomology, and thus also on the Tate
groups. But what if we want to relate $G$-modules for different groups $G$,
as will happen in our study of class field theory? It turns out that in
a suitable sense, the Tate groups are also functorial with respect to changing
groups.

Let $M$ be a $G$-module and $M'$ a $G'$-module. Suppose we are given
a homomorphism $\alpha: G' \to G$ of groups and a homomorphism
$\beta: M \to M'$ of abelian groups (note that they go in opposite
directions!). We say these are \emph{compatible} if
$\beta(m^{\alpha(g)}) = \beta(m)^g$ for all $g \in G$ and $m \in M$.
In this case, one gets canonical homomorphisms $H^i(G, M) \to H^i(G', M')$,
$H_i(G, M) \to H_i(G', M')$ (construct them on pairs of injective/projective
resolutions, then show that any two choices are homotopic),
and $H^i_T(G, M) \to H^i_T(G,M)$. The principal examples are as follows.
(For the rest of this section I write $H(G,M)$ to mean either a cohomology
group, a homology group, or a Tate group of some index. If there are more
than one such group in a display, they all have the same type and index.)
\begin{enumerate}
\item[(a)] If $H$ is a subgroup of $G$, $M$ is an $H$-module, and
$M' = \Ind_H^G(M) = M \otimes_{\ZZ[H]} \ZZ[G]$, then we have a map $M'
\to M$ in which $m \otimes g \mapsto m^g$. The resulting homomorphisms
\[
H(G, \Ind_H^G(M)) \to H(H, M)
\]
are all isomorphisms; this was Shapiro's
Lemma for cohomology, and it also holds for homology and Tate groups.
\item[(b)] If $H$ is a subgroup of $G$, $M$ is a $G$-module, and
$M'$ is just $M$ with all but the $H$-action forgotten, we get
the \emph{restriction homomorphisms}
\[
\Res: H(G, M) \to H(H, M).
\]
Another way to get the same map: use the homomorphism of $G$-modules
$M \to \Ind^G_H(M)$ sending $m$ to $\sum_i m \otimes g_i$, where
$g_i$ runs over a set of right coset representatives of $H$ in $G$,
then apply Shapiro's Lemma to get
\[
H(G, M) \to H(G, \Ind^G_H(M)) \stackrel{\sim}{\to} H(H, M).
\]
\item[(c)]
If $H$ is a \emph{normal} subgroup of $G$, let $\alpha$ be the surjection
$G \to G/H$, and let $\beta$ be the injection $M^H \hookrightarrow M$.
Note that $G/H$ acts on $M^H$; in this case, we get the \emph{inflation
homomorphisms}
\[
\Inf: H(G/H, M^H) \to H(G, M).
\]
\item[(d)]
I pointed out in class earlier that homology and cohomology groups don't
seem to carry a nontrivial $G$-action, because you compute them by taking
invariants and coinvariants. This can be reinterpreted in terms of
extended functoriality: let $\alpha: G \to G$ be the conjugation by some
fixed $h$: $g \mapsto h^{-1}gh$, and let $\beta: M \to M$ be the map
$m \mapsto m^h$. Then the induced homomorphisms $H(G,M) \to H(G,M)$ are
all the identity map.
\item[(e)]
If $M$ is a $G$-module and $H$ is a subgroup of $G$,
then there is a canonical $G$-module homomorphism $M \otimes_{\ZZ[H]}
\ZZ[G] = \Ind^G_H(M) \to M$
sending $m \otimes g$ to $m^g$. (Note: we first restrict $M$ to be an
$H$-module, then induce it back up.) This gives maps
$H(G, \Ind^G_H(M)) \to H(G,M)$, which together with the isomorphisms
of Shapiro's lemma gives what are called the \emph{corestriction
homomorphisms}:
\[
\Cor: H(H, M) \stackrel{\sim}{\to} H(G, \Ind^G_H(M)) \to H(G, M).
\]
\item[(f)]
The composition $\Cor \circ \Res$ is induced by the homomorphism of
$G$-modules $M \to \Ind^G_H(M) \to M$ given by
\[
m \mapsto \sum_i m \otimes g_i \to \sum_i m = [G:H]m.
\]
Thus $\Cor \circ \Res$ acts as multiplication by $[G:H]$ on each 
(co)homology group. Bonus consequence: taking $H$ to be the trivial group,
then the group in the middle is isomorphic to $H(H, M) = 0$. So
every (co)homology group for $G$ is killed by $\#G$, and in particular
is a torsion group. In fact, if $M$ is finitely generated as an abelian
group, this means
$H(G, M)$ is always finite, because each of these will be finitely generated
and torsion. (Of course, this won't happen in many of
our favorite examples, e.g., $H(\Gal(L/K), L^*)$ for $L$ and $K$ fields.)
\end{enumerate}

\head{Exercises}

\begin{enumerate}
\item
The periodicity of the Tate groups for $G$ cyclic means that there is
a canonical isomorphism between $H^{-1}_T(G, M)$ and $H^1_T(G,M)$,
i.e., between $\ker(\Norm_G)/MI_G$ and the set of equivalence classes
of 1-cocycles. What is this isomorphism explicitly? I.e, given an
element of $\ker(\Norm_G)/MI_G$, what is the corresponding 1-cocycle?
\item
Put $K = \QQ_p(\sqrt{p})$. Compute the Herbrand quotient of
$K^*$ as a $G$-module for $G = \Gal(\QQ_p(\sqrt{p})/\QQ_p)$. (Hint:
use the exact sequence $1 \to \gotho_K^* \to K^* \to \ZZ \to 1$.)
\item
Show that $\Res: H^{-2}_T(G, \ZZ) \to H^{-2}_T(H,\ZZ)$ corresponds to
the transfer (Verlagerung) map $G^{\ab} \to H^{\ab}$.
\end{enumerate}

\end{document}