Bar resolution

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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})

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_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)