2 Lie algebra cohomology in degree two

Let \(\mathbb {K}\) be a field and let \(\mathfrak {g}\) be a Lie algebra over \(\mathbb {K}\). Fix also a vector space \(\mathfrak {a}\) over \(\mathbb {K}\), (interpreted, when necessary, as an abelian Lie algebra, i.e., all Lie brackets in \(\mathfrak {a}\) are taken to be zero).

Definition 1 Lie algebra 1-cocycle

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A 1-cocycle of the Lie algebra \(\mathfrak {g}\) with coefficients in the vector space \(\mathfrak {a}\) is a linear map

\begin{align*} \beta \colon \mathfrak {g}\to \mathfrak {a}. \end{align*}

The set of all such 1-cocycles is denoted by \(C^1(\mathfrak {g}, \mathfrak {a})\).

Definition 2 Lie algebra 2-cocycle

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A 2-cocycle of the Lie algebra \(\mathfrak {g}\) with coefficients in the vector space \(\mathfrak {a}\) is a bilinear map

\begin{align*} \gamma \colon \mathfrak {g}\times \mathfrak {g}\to \mathfrak {a}\end{align*}

such that for all \(X \in \mathfrak {g}\) we have the antisymmetry condition

\begin{align} \label{eq:LieTwoCocycle.self} \gamma (X,X) = \; & 0 \end{align}

and for all \(X,Y,Z \in \mathfrak {g}\) we have the Leibnitz rule

\begin{align} \label{eq:LieTwoCocycle.leibniz} \gamma (X,[Y,Z]) = \; & \gamma ([X,Y],Z) + \gamma (Y,[X,Z]) . \end{align}

The set of all such 2-cocycles is denoted by \(C^2(\mathfrak {g}, \mathfrak {a})\).

Lemma 3 Skew-symmetry of 2-cocycles

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For any \(\gamma \in C^2(\mathfrak {g},\mathfrak {a})\) and any \(X,Y \in \mathfrak {g}\), we have the skew-symmetry property

\begin{align*} \gamma (X,Y) = - \gamma (Y,X) . \end{align*}

Proof ▶

The Leibnitz rule 2 applied to \(X+Y\) gives

\begin{align*} 0 = \; & \gamma (X+Y,X+Y) \\ = \; & \gamma (X,X) + \gamma (X,Y) + \gamma (Y,X) + \gamma (Y,Y) \end{align*}

by bilinearity of \(\gamma \). The first and the last terms in the last expression vanish by antisymmetry 1, and the asserted skew-symmetry equation follows.

Lemma 4 Lie algebra 1-cocycles form a vector space

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The set \(C^1(\mathfrak {g}, \mathfrak {a})\) of 1-cocycles of \(\mathfrak {g}\) with coefficients in \(\mathfrak {a}\) forms a vector space over \(\mathbb {K}\).

Proof ▶

By definition, \(C^1(\mathfrak {g}, \mathfrak {a})\) is the space of linear maps \(\mathfrak {g}\to \mathfrak {a}\), and such linear maps form a vector space.

Lemma 5 Lie algebra 2-cocycles form a vector space

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The set \(C^2(\mathfrak {g}, \mathfrak {a})\) of 2-cocycles of \(\mathfrak {g}\) with coefficients in \(\mathfrak {a}\) forms a vector space over \(\mathbb {K}\).

Proof ▶

The conditions defining \(C^2(\mathfrak {g}, \mathfrak {a})\) are linear, so this is staightforward.

Definition 6 Lie algebra 2-coboundary

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Given a 1-cocycle \(\beta \in C^1(\mathfrak {g}, \mathfrak {a})\), we define the coboundary \(\partial \beta \) of \(\beta \) to be the bilinear map

\begin{align*} \partial \beta \colon \; & \mathfrak {g}\times \mathfrak {g}\to \mathfrak {a}\\ \end{align*}

given by

\begin{align*} \partial \beta (X, Y) = \; & \beta ([X,Y]) . \end{align*}

We then have \(\partial \beta \in C^2(\mathfrak {g},\mathfrak {a})\). The mapping \(\partial \colon C^1(\mathfrak {g},\mathfrak {a}) \to C^2(\mathfrak {g},\mathfrak {a})\) is linear. Its range is denoted \(B^2(\mathfrak {g},\mathfrak {a}) \subset C^2(\mathfrak {g},\mathfrak {a})\) and called the set of 2-coboundaries of the Lie algebra \(\mathfrak {g}\) with coefficients in \(\mathfrak {a}\).

Definition 7 Lie algebra 2-cohomology

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The vector space

\begin{align*} H^2(\mathfrak {g},\mathfrak {a}) := C^2(\mathfrak {g},\mathfrak {a}) \, / \, B^2(\mathfrak {g},\mathfrak {a}) \end{align*}

is called the Lie algebra cohomology in degree 2 of \(\mathfrak {g}\) with coefficients in \(\mathfrak {a}\).

Lemma 8 Cohomology of abelian Lie algebras

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If \(\mathfrak {g}\) is abelian, i.e., \([\mathfrak {g},\mathfrak {g}] = 0\), then the canonical projection

\begin{align*} C^2(\mathfrak {g},\mathfrak {a}) \to H^2(\mathfrak {g},\mathfrak {a}) \end{align*}

is a linear isomorphism.

Proof ▶

The projection is surjective by construction, so it suffices to show that it is also injective. The kernel of the projection is \(B^2(\mathfrak {g},\mathfrak {a}) = \mathrm{Im}\; \partial \). In view of Definition 6, abelianity of \(\mathfrak {g}\) implies \(\partial \beta = 0\) for any \(\beta \in C^1(\mathfrak {g},\mathfrak {a})\). Therefore \(B^2(\mathfrak {g},\mathfrak {a}) = 0\), and the kernel of the projection is trivial, so the projection is indeed injective.