Bar resolution

Definition

Suppose $G$ is a group. The bar resolution of $G$ is a long exact sequence of $\mathbb {Z} (G)$-modules:

$\dots {\mathcal {B}}_{n}(G)\to {\mathcal {B}}_{n-1}(G)\to \dots {\mathcal {B}}_{1}(G)\to {\mathcal {B}}_{0}(G)\to \mathbb {Z} \to 0$

defined by the following information.

We denote the identity element of $G$ by $1$.

The groups

The group ${\mathcal {B}}_{n}(G)$ is defined as the free abelian group on the set $G^{n+1}$, with $G$ acting on it diagonally:

$\!g\cdot (h_{0},h_{1},h_{2},\dots ,h_{n})=(gh_{0},gh_{1},gh_{2},\dots ,gh_{n})$

This group can thus be regarded as a $\mathbb {Z} (G)$-module.

As a $\mathbb {Z} G$-module, ${\mathcal {B}}_{n}(G)$ has a free generating set identified by $G^{n}$ by:

$\!(g_{1}\mid g_{2}\mid \dots \mid g_{n})\leftrightarrow (1,g_{1},g_{1}g_{2},g_{1}g_{2}g_{3},\dots ,g_{1}g_{2}g_{3}\dots g_{n})$

The notation with bars $(\ \mid \ \mid \dots \mid \ )$ is termed the bar notation.

The derivation in the original notation

The derivation $\partial _{n}$ in the original notation is given by:

$\!\partial _{n}(h_{0},h_{2},\dots ,h_{n-1},h_{n})=\sum _{i=0}^{n}(-1)^{i}(h_{0},h_{1},\dots ,{\hat {h_{i}}},\dots h_{n})$

The derivation with the bar notation

The map $\!\partial _{n}:{\mathcal {B}}_{n}(G)\to {\mathcal {B}}_{n-1}(G)$ is defined as follows:

$\!\partial _{n}(g_{1}\mid g_{2}\mid \dots \mid g_{n})=g_{1}\cdot (g_{2}\mid g_{3}\mid \dots \mid g_{n})-(g_{1}g_{2}\mid g_{3}\mid \dots \mid g_{n})+(g_{1}\mid g_{2}g_{3}\mid \dots \mid g_{n})-\dots +(-1)^{n-1}(g_{1}\mid g_{2}\mid \dots \mid g_{n-1}g_{n})+(-1)^{n}(g_{1}\mid g_{2}\mid \dots \mid g_{n-1})$

In the more precise summation notation:

$\!\partial _{n}(g_{1}\mid g_{2}\mid \dots \mid g_{n})=g_{1}\cdot (g_{2}\mid g_{3}\mid \dots \mid g_{n})+\left[\sum _{i=1}^{n-1}(-1)^{i}(g_{1}\mid g_{2}\mid \dots \mid g_{i-1}\mid g_{i}g_{i+1}\mid \dots \mid g_{n})\right]-(g_{1}\mid g_{2}\mid \dots \mid g_{n-1})$

Particular cases

$\!\partial _{1}$

In the comma notation, we have:

$\!\partial _{1}(h_{0},h_{1})=(h_{1})-(h_{0})$

The source of $\partial _{1}$ is ${\mathcal {B}}_{1}(G)$, which is a free $\mathbb {Z} (G)$ module on $(g)_{g\in G}$. The target of $\partial _{1}$ is ${\mathcal {B}}_{0}(G)$, which is simply $\mathbb {Z} G$.

$\!\partial _{1}(g)=g\cdot ()-()$

$\!\partial _{2}$

In the comma notation, we have:

$\!\partial _{2}(h_{0},h_{1},h_{2})=(h_{1},h_{2})-(h_{0},h_{2})+(h_{0},h_{1})$

The source of $\partial _{2}$ is ${\mathcal {B}}_{2}(G)$, which is a free $\mathbb {Z} (G)$-module on $G\times G$. The target of $\partial _{2}$ is ${\mathcal {B}}_{1}(G)$, which is the free $\mathbb {Z} G$-module on $G$.

$\!\partial _{2}(g_{1}\mid g_{2})=g_{1}\cdot (g_{2})-(g_{1}g_{2})+(g_{1})$

$\!\partial _{3}$

In the comma notation, we have:

$\!\partial _{3}(h_{0},h_{1},h_{2},h_{3})=(h_{1},h_{2},h_{3})-(h_{0},h_{2},h_{3})+(h_{0},h_{1},h_{3})-(h_{0},h_{1},h_{2})$

In the bar notation, we have:

$\!\partial _{3}(g_{1}\mid g_{2}\mid g_{3})=g_{1}\cdot (g_{2}\mid g_{3})-(g_{1}g_{2}\mid g_{3})+(g_{1}\mid g_{2}g_{3})-(g_{1}\mid g_{2})$