# Outline:¶

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

# Homological algebra recap¶

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

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

$$\cdots \to C_2 \xrightarrow{\partial_2} C_1 \xrightarrow{\partial_1} C_0 \xrightarrow{\varepsilon} M \to 0$$

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:

$$\text{Tor}^R_n(N,M) := H_n(N \otimes_R C_{\bullet})$$

# Homology of a group¶

Let $G$ be a (discrete) group. Want to do the above with the integral group ring

$$R = \mathbb{Z}G$$

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 $N = \mathbb{Z}$?

### Free actions and free resolutions¶

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

$$\cdots \to C_2 E \to C_1 E \to C_0 E \xrightarrow{\varepsilon} \mathbb{Z} \to 0$$

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

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

### Co-invariants and tensoring with $\mathbb{Z}$¶

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

$$M_G := M/\langle gm - m\rangle.$$

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

$$\left(\mathbb{Z}S\right)_G \cong \mathbb{Z}(S/G).$$

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

$$(S_n E)/G \cong S_n (E/G),$$

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, $$H_\bullet X \cong H_\bullet G.$$

# $K(G,1)$-spaces¶

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

1. $X$ is a path-connected space (CW-complex)
2. $\pi_1(X) = G$
3. 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 $x_0$).

### Examples of $K(G,1)$-spaces¶

• 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

### $K(G,1)$-spaces exist!¶

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

#### Two constructions¶

• 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)

## References¶

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