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:

$$\! (g_1 \mid g_2 \mid \dots \mid g_n) \leftrightarrow (1,g_1,g_1g_2,g_1g_2g_3,\dots,g_1g_2g_3\dots g_n)$$

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:

$$\! \partial_n(g_1 \mid g_2 \mid \dots \mid g_n) = g_1 \cdot (g_2 \mid g_3 \mid \dots \mid g_n) - (g_1g_2 \mid g_3 \mid \dots \mid g_n) + (g_1 \mid g_2g_3 \mid \dots \mid g_n) - \dots + (-1)^{n-1}(g_1 \mid g_2 \mid \dots \mid g_{n-1}g_n) + (-1)^n(g_1 \mid g_2 \mid \dots \mid g_{n-1})$$

In the more precise summation notation:

$$\! \partial_n(g_1 \mid g_2 \mid \dots \mid g_n) = g_1 \cdot (g_2 \mid g_3 \mid \dots \mid g_n) + \left[\sum_{i=1}^{n-1} (-1)^i (g_1 \mid g_2 \mid \dots \mid g_{i-1} \mid g_ig_{i+1} \mid \dots \mid g_n) \right] - (g_1 \mid g_2 \mid \dots \mid g_{n-1})$$

$$\! \partial_1$$
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 $$G$$.

$$\! \partial_2(g_1 \mid g_2) = g_1 \cdot (g_2) - (g_1g_2) + (g_1)$$

$$\! \partial_3$$
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:

$$\! \partial_3(g_1 \mid g_2 \mid g_3) = g_1 \cdot (g_2 \mid g_3) - (g_1g_2 \mid g_3) + (g_1 \mid g_2g_3) - (g_1 \mid g_2)$$