Central group extension

A group extension of $𝐴$ by $𝐵$

1 \to 𝐵 e x p ↪ 𝐺 𝜋 ↠ 𝐴 \to 1

is called central iff $𝐵 ↪ 𝐺$ is contained within the centre $𝑍 (𝐺)$ , group whence $𝐵$ is abelian. In what follows we write $𝐵$ additively and $𝐺$ and $𝐴$ multiplicatively, and write $e 𝑏 = e (𝑏)$ for any $𝑏 \in 𝐵$ ,

Second cohomology

Identifying $𝐵$ with the [[Abelian groups as Z-modules|corresponding $ℤ$ -module]] equipped with the trivial representation of $𝐺$ (thus a $ℤ [𝐺]$ -module) may consider the Group cohomology, where the 2-cochains $𝐶 2 (𝐺, 𝐵)$ are maps¹

𝜀 0 : 𝐴 \times 𝐴 \to 𝐵

and the 2-cocycles $𝑍 2 (𝐺, 𝐵)$ are 2-cochains such that

𝜀 0 (𝑎, 𝑏) + 𝜀 0 (𝑎 𝑏, 𝑐) = 𝜀 0 (𝑏, 𝑐) + 𝜀 0 (𝑎, 𝑏 𝑐) \forall 𝑎, 𝑏, 𝑐 \in 𝐺

and the 2-coboundaries $𝐵 2 (𝐺, 𝐵)$ are 2-cochains such that

𝜀 0 (𝑎 𝑏) = 𝜂 (𝑎 𝑏) - 𝜂 (𝑎) - 𝜂 (𝑏) \forall 𝑎, 𝑏 \in 𝐺

for some 1-cochain $𝜂 : 𝐴 \to 𝐵$ . Thus, in particular, $ℤ$ -bilinear maps $𝐴 \times 𝐴 \to 𝐵$ are 2-cocycles. The second cohomology group is then

𝐻 2 (𝐴, 𝐵) = 𝑍 2 (𝐴, 𝐵) / 𝐵 2 (𝐴, 𝐵)

Correspondence between 2-cocycles and central extensions

Given any \Set-section $𝑠 (-) : 𝐴 \to 𝐺$ of $𝜋$ we have $𝐺 = {𝑠 𝑎 e 𝑏 : 𝑎 \in 𝐴; 𝑏 \in 𝐵}$ ; and $𝑠 𝑎 𝑠 𝑏 = 𝑠 𝑎 𝑏 e 𝜀 0 (𝑎, 𝑏)$ defines a 2-cycle. Conversely let $𝜀 0 : 𝐴 \times 𝐴 \to 𝐵$ be a 2-cocycle. Then the set $𝐵 \times 𝐴$ is a group under the following multiplication

(𝑝, 𝑎) \cdot (𝑞, 𝑏) = (𝑝 + 𝑞 + 𝜀 0 (𝑎, 𝑏), 𝑎 𝑏)

with identity $(- 𝜀 0 (1, 1), 1)$ , and we have the above central extension where

𝜋 : (𝑝, 𝑎) \mapsto 𝑎 e x p : 𝑝 \mapsto (𝑝 - 𝜀 0 (1, 1), 1)

and for the associated section $𝑠 (-) : 𝑎 \mapsto (0, 𝑎)$ we have $𝑠 𝑎 𝑠 𝑏 = 𝑠 𝑎 𝑏 e 𝜀 0 (𝑎, 𝑏)$ . Note $𝑠 1 = 1$ iff $𝜀 0 (𝑎, 1) = 𝜀 0 (1, 𝑎) = 0$ for all $𝑎 \in 𝐴$ .

Proof

That $𝐺 = {𝑠 𝑎 e 𝑏 : 𝑎 \in 𝐴; 𝑏 \in 𝐵}$ follows from the the fact cosets of $𝐵$ partition $𝐺$ . Next we claim
$e 𝜀 0 (𝑎, 𝑏) = 𝑠 - 1 𝑎 𝑏 𝑠 𝑎 𝑠 𝑏$
defines a 2-cocycle. Note that $𝜋 (𝑠 - 1 𝑎 𝑏 𝑠 𝑎 𝑠 𝑏) = 1$ , hence the formula is well-defined. Letting $l n$ denote the inverse of $e x p$ , we have
$= 𝜀 0 (𝑎, 𝑏) - 𝜀 0 (𝑎, 𝑏 𝑐) + 𝜀 0 (𝑎 𝑏, 𝑐) - 𝜀 0 (𝑏, 𝑐) = l n (𝑠 - 1 𝑎 𝑏 𝑠 𝑎 𝑠 𝑏) - l n (𝑠 - 1 𝑎 𝑏 𝑐 𝑠 𝑎 𝑠 𝑏 𝑐) + l n (𝑠 - 1 𝑎 𝑏 𝑐 𝑠 𝑎 𝑏 𝑠 𝑐) - l n (𝑠 - 1 𝑏 𝑐 𝑠 𝑏 𝑠 𝑐) = l n (𝑠 - 1 𝑎 𝑏 𝑠 𝑎 𝑠 𝑏) + l n (𝑠 - 1 𝑏 𝑐 𝑠 - 1 𝑎 𝑠 𝑎 𝑏 𝑠 𝑐) - l n (𝑠 - 1 𝑏 𝑐 𝑠 𝑏 𝑠 𝑐) = l n (𝑠 - 1 𝑎 𝑏 𝑠 𝑎 𝑠 𝑏) + l n (𝑠 - 1 𝑐 𝑠 - 1 𝑏 𝑠 - 1 𝑎 𝑠 𝑎 𝑏 𝑠 𝑐) = l n (𝑠 - 1 𝑎 𝑏 𝑠 𝑎 𝑠 𝑏) + l n (𝑠 - 1 𝑏 𝑠 - 1 𝑎 𝑠 𝑎 𝑏) = 0$
as required, where we have used centrality of $𝑠 - 1 𝑏 𝑠 - 1 𝑎 𝑠 𝑎 𝑏$ .

Now given a 2-cocycle $𝜀 0 \in 𝑍 2 (𝐴, 𝐵)$ we define the following multiplication on the set $𝐵 \times 𝐴$
$(𝑝, 𝑎) \cdot (𝑝, 𝑏) = (𝑝 + 𝑞 + 𝜀 0 (𝑎, 𝑏), 𝑎 𝑏)$
which clearly constitutes a monoid since
$(𝑝, 𝑎) \cdot (- 𝜀 0 (1, 1), 1) = (𝑝 - 𝜀 0 (1, 1) + 𝜀 0 (𝑎, 1), 𝑎) = (𝑝 + 𝜀 0 (𝑎, 1 \cdot 1) - 𝜀 0 (𝑎 \cdot 1, 1), 𝑎) = (𝑝, 𝑎)$
and likewise on the right. The inverse is easily seen to be given by
$(𝑝, 𝑎) - 1 = (- 𝑝 - 𝜀 0 (𝑎, 𝑎 - 1) - 𝜀 0 (1, 1), 𝑎 - 1)$
Thus the given multiplication makes the set $𝐵 \times 𝐴$ a group which we denote $𝐺$ . Clearly we have the central extension
$1 \to 𝐵 e x p ↪ 𝐺 𝜋 ↠ 𝐴 \to 1$
where $e x p$ and $𝜋$ are given above. Letting $𝑠 𝑎 = (0, 𝑎)$ , we find Noting that
$𝜀 0 (𝑎, 1) = 𝜀 0 (1, 1) + 𝜀 0 (𝑎, 1 \cdot 1) - 𝜀 0 (𝑎 \cdot 1, 1) = 𝜀 0 (1, 1)$
now
$(0, 𝑎 𝑏) e 𝜀 0 (𝑎, 𝑏) = (0, 𝑎 𝑏) (𝜀 0 (𝑎, 𝑏) - 𝜀 0 (1, 1), 1) = (𝜀 0 (𝑎, 𝑏) - 𝜀 0 (1, 1) + 𝜀 0 (𝑎 𝑏, 1)) = (𝜀 0 (𝑎, 𝑏), 1) = (0, 𝑎) (0, 𝑏)$
as claimed.

This correspondence has the property

Central extensions are equivalent iff their 2-cocycles for some sections are cohomologous. Thus there is a bijection between $𝐻 2 (𝐴, 𝐵)$ and equivalence classes of extensions.

Proof

Consider the central extension
$1 \to 𝐵 e x p ↪ 𝐺 𝜋 ↠ 𝐴 \to 1$
and let $𝑠 (-), 𝑡 (-) : 𝐴 ↪ 𝐺$ be \Set-sections of $𝜋$ , and consider the corresponding 2-cycles $𝜀 0, 𝜂 0 \in 𝑍 2 (𝐴, 𝐵)$ defined by
$𝑠 𝑎 𝑠 𝑏 = 𝑠 𝑎 𝑏 e 𝜀 0 (𝑎, 𝑏) 𝑡 𝑎 𝑡 𝑏 = 𝑡 𝑎 𝑏 e 𝜂 0 (𝑎, 𝑏)$
Then, taking into account the fact $𝜋 (𝑥) = 1$ implies $𝑥 \in 𝑍 (𝐺)$ ,
$e 𝜀 0 (𝑎, 𝑏) - 𝜂 0 (𝑎, 𝑏) = 𝑠 𝑎 𝑠 𝑏 𝑡 - 1 𝑏 𝑡 - 1 𝑎 𝑡 𝑎 𝑏 𝑠 - 1 𝑎 𝑏 = 𝑠 𝑏 𝑡 - 1 𝑏 𝑠 𝑎 𝑡 - 1 𝑎 𝑡 𝑎 𝑏 𝑠 - 1 𝑎 𝑏$
so
$𝜀 0 (𝑎, 𝑏) - 𝜂 0 (𝑎, 𝑏) = l n (𝑡 𝑎 𝑏 𝑠 - 1 𝑎 𝑏) - l n (𝑡 𝑎 𝑠 - 1 𝑎) - l n (𝑡 𝑏 𝑠 - 1 𝑏) \in 𝐵 2 (𝐴, 𝐵)$
thus different sections of $𝜋$ give cohomologous 2-cocycles. It immediately follows that equivalent central extensions will give cohomologous 2-cocycles.

For the converse, it is sufficient to show that given a central extension with a section $𝑠 (-)$ such that $𝑠 1 = 1$ and a corresponding 2-cycle $𝜀 0 : 𝐴 \times 𝐴 \to 𝐵$ , the induced extension on G' =_{\Set} B \times A is equivalent. We show that the following commutes

where
$𝜋' : (𝑝, 𝑎) \mapsto 𝑎 e x p' : 𝑝 \mapsto (𝑝 - 𝜀 0 (1, 1), 1) = (𝑝, 1) 𝜑 : 𝑠 𝑎 e 𝑝 \mapsto (𝑝, 𝑎)$
and $𝜑 : 𝐺 \to 𝐺'$ is an isomorphism. Note that for every $𝑔 \in 𝐺$ , $𝑔 = 𝑠 𝑎 e 𝑝$ for unique $𝑎 \in 𝐴$ and $𝑝 \in 𝐵$ , so $𝜑$ is a well-defined bijection. Further, for any $𝑎, 𝑏 \in 𝐴$ and $𝑝, 𝑞 \in 𝐵$
$𝜑 (𝑠 𝑎 e 𝑝 𝑠 𝑏 e 𝑞) = 𝜑 (𝑠 𝑎 𝑠 𝑏 e 𝑝 + 𝑞) = 𝜑 (𝑠 𝑎 𝑏 e 𝑝 + 𝑞 + 𝜀 0 (𝑎, 𝑏)) = (𝑝 + 𝑞 + 𝜀 0 (𝑎, 𝑏), 𝑎 𝑏) = (𝑝, 𝑎) (𝑞, 𝑏) = 𝜑 (𝑠 𝑎 e 𝑝) 𝜑 (𝑠 𝑏 e 𝑞)$
so $𝜑$ is a group isomorphism, and
$𝜑 (e x p 𝑝) = 𝜑 (𝑠 1 e 𝑝) = (𝑝, 1) = e x p' 𝑝 𝜋' 𝜑 (𝑠 𝑎 e 𝑝) = 𝜋' (𝑝, 𝑎) = 𝑎 = 𝜋 (𝑠 𝑎 e 𝑝)$
so the diagram commutes, as required.