# Ch. 6: Group Homology and Cohomology (Wednesday, March 10)

:::{.lemma title="?"}
Last time: started setting up group homology.
For $G$ a group and $A\in \modsleft{G}$, we think of $\ZZ$ as a trivial \(G\dash\)module and

1. $A_G \cong \ZZ \tensor_{\ZZ G} A$, the $G\dash$coinvariants.

2. $A^G \cong \Hom_{\ZZ G}( \ZZ, A)$.
  the $G\dash$invariants, this is the largest $G\dash$trivial submodule of $A$

:::
 
:::{.definition title="?"}
For $A\in \modsleft{G}$,

1. $H_*(G; A) \da L_*(\wait))G (A)$ are the **homology groups of $G$ with coefficients in $A$**.
  It is isomorphic to $\Tor_*^{\ZZ G}(\ZZ, A)$ by (1) in the lemma above.
  In particular, $H_0(G; A) \cong A_G$.

2. $H^*(G; A) \da R^*(\wait)^G(A)$ is the **cohomology of $G$ with coefficients in $A$**.
  It is isomorphic to $\Ext^*_{\ZZ G}(\ZZ, A)$ by (2) in the lemma.
  In particular, $H^0(G; A) \cong A^G$.
:::

\todo[inline]{Ask about contructing resolutions: take any "augmentation" map and iterate kernels? Different resolution lengths?}

:::{.example title="?"}
For $G = \ts{ 1 }$, for any $A\in \modsleft{G}$ we have $A^G = A = A_G$.
Forgetful functors are usually exact, and in this case $(\wait)^G, (\wait)_G: \modsleft{G} \to \Ab$ is really a forgetful functor and thus exact.
Here $H_n(G; A) = 0 = H^n(G; A)$ for $n>0$.
:::

:::{.example title="?"}
Let $G$ be infinite cyclic, which we'll write multiplicatively to prevent the notation from conflicting with the addition on $\ZZ G$, so $G\da T = \gens{ t }= \ts{ t^n \st n\in \ZZ }$. 
Then $\ZZ G = \ZZ[t, t ^{-1} ]$ are integral Laurent polynomials, since we're taking integer linear combinations of various $t^n$.
Computing $H_*(T, A) \cong \Tor_*^{\ZZ T}(\ZZ, A)$ and $H^*(T; A) \cong \Ext^*_{\ZZ T}(\ZZ, A)$ using a projective resolution of $\ZZ$ as a $\ZZ T\dash$module, since the first slot Ext requires an injective resolution in the opposite category.
It suffices to take a free resolution:
\[
\cdots \to P_2 \to P_1 \to P_0 \to \ZZ \to 0 \da
\cdots \to 0\to
\ZZ T \mapsvia{\times (t-1)} 
\ZZ T \mapsvia{\ev_1} \ZZ \to 0
.\]
Note that the resolution ends here because the multiplication $\times(t-1)$ is injective on polynomials rings.
Thus $H_{>\geq 2}(T; A) = H^{\geq 2}(T; A) = 0$. 
The zeroth terms are invariants/coinvariants.
For $\Tor$, we apply $\wait \tensor_{\ZZ T} A$ to this resolution to obtain
\[
0\to FP_1 \to FP_0 \to 0 
&\da
0 \to \ZZ T \tensor_{\ZZ T} A \mapsvia{(t-1) \tensor \one} \ZZ T \tensor_{\ZZ T} A \to 0\\
&=
0 \to A \mapsvia{(t-1) \tensor \one} A \to 0
.\]

One can check that 

- $\ker (t-1) \tensor \one = A^T = H_1(T; A)$ is equal to the invariants and 
- $\coker (t-1) \tensor \one = A_T = H_0(T; A)$ is equal to the coinvariants.

The second fact had to be true, but the first is surprising!

For $\Ext^*$, we apply the contravariant $\Hom_{\ZZ T}(\wait, A)$ to obtain
\[
0
\to \Hom_{\ZZ T}(\ZZ T, A)
\mapsvia{\wait \circ (t-1)} 
\Hom_{\ZZ T}(\ZZ T, A)
\to 0
.\]

One checks

- $\coker( \wait \circ (t-1)) = A_T = H^1(T; A)$ (surprising!) and
- $\ker( \wait \circ (t-1)) = A^T = H^0(T; A)$
:::

:::{.remark}
See exercise 6.1.2 for $kG\dash$modules for $k\in \Ring$ arbitrary.
:::

:::{.question}
What can we say about $H_0$ and $H^0$ for more general groups?
:::

## $H_0$ for Groups

:::{.definition title="Augmentation Maps"}
Define the **augmentation map**
\[
\eps: \ZZ G\to \ZZ \\
\sum n_i g_i &\mapsto \sum n_i
,\]
which is a ring morphism.
Define $\ideal{I} \da \ker \eps$ to be the **augmentation ideal**.
:::

:::{.observation}
There is a basis of $\ZZ G$ as a \(\ZZ\dash\)module given by 
\[
\mathcal{B}\da B_1 \union B_2 \da \ts{ 1 } \union \ts{ g-1 \st 1\neq g\in G } 
.\]
Note that $\eps(g-1) = 0$, so $\ideal{I}$ is a free \(\ZZ\dash\)module with basis $B_2$.
Here the kernel should be expected to have codimension 1!
We also have $\ZZ G/ \ideal{I} \cong \ZZ$ as rings, where the left-hand side is a \(G\dash\)module.
Letting $\bar{\wait}$ denote coset/equivalence class representatives, we have
\[
g \bar{1} = \bar{g1} = \bar{g} = \bar{1}
,\]
and so the action $G \actson \ZZ G/ \ideal{I}$ is trivial.

:::{.fact}
For $R$ a ring and $\ideal{I} \normal R$ a (left? right?) ideal and $M\in \rmod$,
\[
R/I \tensor_R M \cong M/IM
.\]
:::

So for any \( A\in \modsleft{G} \) we have 
\[
H_0(G; A) 
&= A_G \\
&\cong \ZZ \tensor_{\ZZ G} A \\
&= \Tor_0^{\ZZ G}(\ZZ; A) \\
&= \ZZ G/\ideal{I} \tensor_{\ZZ G} A \\
&\cong A/ \ideal{I} A
.\]

:::

:::{.example title="?"}
\envlist

- $H_0(G; \ZZ) \cong \ZZ/ \ideal{I} \ZZ \cong \ZZ$, where $\ideal{I} \ZZ = 0$ since $\ZZ$ is the trivial \(G\dash\)module and $(g-1)a = ga-1a=a-a=0$.

- $H_0(G; \ZZ G) \cong \ZZ G/ \ideal{I} \cong \ZZ$.

- $H_0(G; \ideal{I} ) \cong \ideal{I} / \ideal{I}^2$.

:::

:::{.example title="?"}
Noting that $A = \ZZ G$ is projective in \( \modsleft{\ZZ G} \), so $H_n(G; \ZZ G) = 0$ for $n>0$, using that this was a version of $\Tor$ and projective implies flat.
:::

## $H^0$ for Groups

:::{.definition title="Norm Element"}
Let $G$ be a finite group, then the **norm element** is defined by 
\[
N = \sum_{g\in G} g\in \ZZ G
.\]
:::

:::{.remark}
For $h\in G$,
\[
hN = \sum_g hg = \sum_{g'\in G} g' = N
,\]
and so $N \in (\ZZ G)^g$.
Similarly $Nh = N$ and so $Z(\ZZ G)$ is in the center.

> Note the two different $Z$s here!

:::

:::{.lemma title="?"}
Let $G$ be finite, then 
\[
H^0(G; \ZZ G) = (\ZZ G)^G = \ZZ N
,\]
which is a two-sided ideal of $\ZZ G$ that is isomorphic to $\ZZ$.
:::

:::{.proof title="?"}
The inclusion $\ZZ N \subseteq (\ZZ G)^G$ is clear from the previous remark, so it remains to show the other inclusion.
Suppose 
\[
a\in \sum_{g\in G} n_g g \in (\ZZ G)^G
.\]
Then for all $h\in G$, we have
\[
a = ha = \sum n_g h_g
.\]
Now note that the $g$ are a free $\ZZ\dash$basis for $\ZZ G$, so we can equate coefficients of $h$ to find that $n_h = n_1$.
Since $h$ was arbitrary, we have $a = n_1 N \in \ZZ N$.
:::

:::{.remark}
Exercise 6.1.3 shows that $H^0(G; \ZZ G) = 0$ when $G$ is infinite, in which case $\ideal{I} = \ts{ a \in \ZZ G \st N a = 0 }$ is the annihilator of the norm element.
Next class we'll start on spectral sequences.
:::