$\DeclareMathOperator{\ZZ}{\mathbb{Z}}$

$\DeclareMathOperator{\SS}{\mathbb{S}}$

$\DeclareMathOperator{\RP}{\mathbb{RP}}$

- Homological algebra recap
- Homology of a group
- $K(G,1)$-spaces

Let $R$ be a ring, $M$ an $R$-module,

A **projective resolution** of $M$ is an exact sequence of $R$-modules

such that each $C_i$ is a projective $R$-module.

Let $N$ be another $R$-module.

Tensoring $N$ with $C_\bullet$, we get another chain complex

$$ \cdots \to N \otimes C_2 \to N \otimes C_1\to N \otimes C_0 \to 0.$$Taking the homology of this complex yields:

$$ \text{Tor}^R_n(N,M) := H_n(N \otimes_R C_{\bullet})$$Let $G$ be a (discrete) group. Want to do the above with the **integral group ring**

and its modules.

The **group homology** of $G$ is
$$ H_{\bullet}G := \text{Tor}^{\mathbb{Z}G}_{\bullet}(\ZZ,\ZZ)$$

which is the homology of the chain complex

$$ \cdots \to \ZZ \otimes_{\mathbb{Z}G} C_2 \to \ZZ \otimes_{\mathbb{Z}G} C_1 \to \ZZ \otimes_{\mathbb{Z}G} C_0 \to 0$$where

$$\cdots \to C_2 \to C_1 \to C_0 \to \mathbb{Z} \to 0$$is a projective resolution of $\mathbb{Z}$.

How do we get projective resolutions of $M = \mathbb{Z}$?

Why do we tensor with $N = \mathbb{Z}$?

**Topology has the answers!**

** Proposition.** Let $G$ act freely on a space $E$.

Then $S_n E$ is a free $G$-set and $C_n E = \mathbb{Z}S_n E$ is a free $\mathbb{Z}G$-module.

** Lemma.** If $S$ is a free $G$-set, then $\mathbb{Z}S$ is a free $\mathbb{Z}G$-module.

So we get a chain complex of *free* $G$-modules

and if $E$ is *contractible*, $\tilde{H}_\bullet E = 0$, so this is exact.

We get a *free* resolution of $\mathbb{Z}$!

Let $M$ be a $\mathbb{Z}G$-module. The **co-invariants** of $M$ are

** Lemma.** For $S$ an arbitrary $G$-set,

** Proposition.** Let $X$ be a space with $\pi_1(X) = G$ and contractible universal cover $E$. Then

and thus,

$$(C_\bullet E)_G \cong C_\bullet X.$$** Proposition.** $$M_G \cong \mathbb{Z} \otimes_{\mathbb{Z}G} M.$$

**Putting it together...**

Let $X$ be a *path-connected* space with $\pi_1(X) = G$ and *contractible* universal cover $E$.

Then $G$ acts freely on $E$, so we get a free resolution of $\ZZ$:

$$ \cdots \to C_1 E \to C_0 E \xrightarrow{\varepsilon} \ZZ \to 0,$$Tensoring $C_\bullet E$ with $\ZZ$ yields

$$\cdots \to \ZZ \otimes C_1 E \to \ZZ \otimes C_0 E \to 0,$$which is isomorphic to

$$\cdots \to C_1 X \to C_0 X \to 0.$$Taking homology, $$H_\bullet X \cong H_\bullet G.$$

$X$ is a $K(G,1)$-**space (complex)** if:

- $X$ is a path-connected space (CW-complex)
- $\pi_1(X) = G$
- Its universal cover $E$ is contractible

** Theorem.** The homotopy type of a $K(G,1)$-complex is uniquely determined by $G$.

We could have thus defined:

$$H_\bullet G :=\, H_\bullet\, K(G,1)$$** Proposition.** [Hatcher, 1B.9]

Let $X$ be a connected CW-complex, $Y$ a $K(G,1)$-complex. Then every

$$\varphi: \pi_1(X,x_0) \to \pi_1(Y,y_0)$$is induced by

$$f: (X,x_0) \to (Y,y_0),$$and $f$ is unique up to homotopy (relative to $x_0$).

- Common spaces with
*contractible*universal covering space:- $\SS^1$ is a $K(\ZZ,1)$
- Compact surfaces with
*infinite*$\pi_1$ (i.e. all except $\SS^2$ and $\RP^2$) - Can "fill-in" non-contractible covering space with cells

- $K(A,1)$ for abelian $A$:
- Lens spaces: $\SS^\infty/\ZZ_m$ is a $K(\ZZ_m,1)$
- $K(G,1) \times K(H,1)$ is a $K(G\times H,1)$
- If $K(G,1)$ is finite-dimensional, then $G$ is torsion-free (hence infinite!)

- Amalgamated free products:
- Any graph is a $K(\ZZ * \cdots * \ZZ,1)$
- $X \cup_f Y$, with $f: A \to X$ injective, is a $K(G *_K H,1)$
- Graphs of groups

In fact, $K(G,1)$-*complexes* exist!

- From group presentation:
- Construct a surface $X_G$ (generators are 1-cells, relations are 2-cells)
- Make $E$ contractible by attaching $n$-cells (infinitely many!)
- Not functorial
- Different presentations can give different (non-homotopic) $X_G$

- Classifying space construction:
- $EG = $ complex whose $n$-cells are
*all*$n$-tuples $(g_0,g_1,\dots,g_n), g_i \in G$ - $g\cdot (g_0,\dots,g_n) = (gg_0,\dots,gg_n)$
- $BG = EG/G$ is the
**classifying space**of $G$ - Functorial (group homomorphisms give cellular maps)

- $EG = $ complex whose $n$-cells are

- [Kenneth Brown] Cohomology of Groups
*(first two chapters)* - [Allen Hatcher] Algebraic Topology
*(various places. Just search for "*`K(G,1)`

")