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:

$g\cdot (h_0,h_1,h_2,\dots ,h_n)=(g{h}_0,g{h}_1,g{h}_2,\dots ,g{h}_n)$

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:

${\partial }_n(h_0,h_2,\dots ,h_{n-1},h_n)=\sum _{i=0}^n(-1{)}^i(h_0,h_1,\dots ,\widehat{h_i},\dots h_n)$

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:

${\partial }_n(g_1\mid g_2\mid \dots \mid g_n)=g_1\cdot (g_2\mid g_3\mid \dots \mid g_n)+[\sum _{i=1}^{n-1}(-1{)}^i(g_1\mid g_2\mid \dots \mid g_{i-1}\mid g_i{g}_{i+1}\mid \dots \mid g_n)]-(g_1\mid g_2\mid \dots \mid g_{n-1})$

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:

${\partial }_3(g_1\mid g_2\mid g_3)=g_1\cdot (g_2\mid g_3)-(g_1{g}_2\mid g_3)+(g_1\mid g_2{g}_3)-(g_1\mid g_2)$