Group Homology

group actions + homological algebra

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

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

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

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.

• Free modules are projective, and we can always get a free, hence projective, resolution
• $\varepsilon$ is an augmentation of $F_\bullet$

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:

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

• Tor is independent of the resolution of $M$

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 $N = \mathbb{Z}$?

Topology has the answers!

• $G$ acts trivially on $\mathbb{Z}$
• In fact, we get free resolutions

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.

• $S_n E$ is the set of singular $n$-simplices $\{\sigma: \Delta^n \to E\}$. The action of $G$ on $E$ induces an action on $S_n E$ by composition: $g\cdot \sigma := g \circ \sigma$
• The induced action is free because $g \circ\sigma = \sigma \iff g = e$
• $C_n E$ is the $n^{th}$ singular chain of $E$
• Although "free" appears many times, it means different things. In particular, the last statement requires proving.

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

Proof. For $s \in S$, we have $$Gx \cong G/G_s.$$

Since $S$ is the disjoint union of its orbits and $$ \mathbb{Z}\left(\coprod S_i\right) = \oplus \mathbb{Z}S_i$$ we get $$ \mathbb{Z}S \cong \oplus\mathbb{Z}\left(G/G_s\right).$$

Since $G$ acts freely, $G_s = \{e\}$ so $G/G_s = G$ and $$\mathbb{Z}S \cong \oplus \mathbb{Z}G.$$

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.

• This is the augmented chain complex
• Of course, have to check that the boundary maps are $\mathbb{Z}G$-module homomorphisms:
• If $\sigma: \Delta^n \to E$, then $\partial \sigma = \sum (-1)^i \sigma \circ \epsilon_i$ where $\epsilon_i: \Delta^{n-1} \to \Delta^n$ is the standard inclusion into the $i^{th}$ face.
• Since $g$ is post-composition while $\epsilon_i$ are pre-composition, it's easy to see that $g \cdot \partial \sigma = \partial (g \cdot \sigma)$.

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

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

• $M_G$ is the largest quotient on which $G$ acts trivially (compare to $M^G$, the largest submodule on which $G$ acts trivially)

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

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

Proof. $(\mathbb{Z}S)_G \cong (\oplus \mathbb{Z}(G/G_s))_G \cong \oplus (\mathbb{Z}(G/G_s))_G \cong \oplus \mathbb{Z}(Gs) \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,

$$(C_\bullet E)_G \cong C_\bullet X.$$

• $G$ acts freely on $E$, and is in fact a covering space action.

Proposition. $$M_G \cong \mathbb{Z} \otimes_{\mathbb{Z}G} M.$$

Proof. Note that $$ 1 \otimes g\cdot m = 1\cdot g \otimes m = 1 \otimes m $$ so the map $M \to \mathbb{Z}\otimes M, m \mapsto 1 \otimes m$ factors through $M_G$.

We also have a bilinear map $\mathbb{Z} \times M \to M_G, (a,m) \mapsto am$, yielding a map $\mathbb{Z} \otimes M \to M_G, a\otimes m \mapsto am$.

These maps are mutual inverses.

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

• In fact, the map on chains induced by $q:E\to X$ is isomorphic to $\ZZ \otimes -$

$K(G,1)$-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

• $K(G,n)$-spaces are called Eilenberg-MacLane spaces, and require $\pi_m(X) = 0$ for $m \neq n$.

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 $x_0$).

• To get the theorem, let $X$ and $Y$ both be $K(G,1)$ complexes, then use the isomorphism $\pi_1 X \cong \pi_1 Y$ to get maps $f:X \to Y, g: Y\to X$ whose compositions are homotopic to identity.
• Proof of proposition: See Hatcher Prop 1B.9:
• Map on fundamental group gives map on 1-skeleton
• Extend this to 2- and higher cells

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)")