# User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 2

## Central extensions

The definition of projective is important when $$R$$ is not $$\mathbb{R}\ ,$$ $$\mathbb{C}$$ or in general a field of characteristic zero, or when the dimension of the Lie algebra is not finite, otherwise it can be skipped.

### definition - projective

A module $$\mathfrak{g}$$ is projective if for every surjective morphism of modules $$\alpha:\mathfrak{k}\rightarrow\mathfrak{g}\rightarrow 0$$ and every morphism $$\beta:\mathfrak{h}\rightarrow\mathfrak{g}\ ,$$ there exists a morphism $$\gamma:\mathfrak{h}\rightarrow\mathfrak{k}$$ such that $$\beta=\alpha\gamma\ .$$

In particular, if $$\mathfrak{g}$$ is projective and $$\alpha$$ surjective, there exists a $$\gamma:\mathfrak{g}\rightarrow\mathfrak{k}$$ such that $$id_{\mathfrak{g}}=\alpha\gamma\ ,$$ that is, $$\gamma$$ is a section of $$\alpha\ .$$

When $$\mathfrak{g}$$ is free, that is, $$\mathfrak{g}$$ has an $$R$$-basis $$\{g_\iota\}_{\iota\in I}\ ,$$ one can find for each $$g_\iota$$ a preimage $$k_\iota$$ under $$\alpha\ .$$

If $$\beta(h)=\sum_{\iota\in I} b_\iota g_\iota, b_\iota\in R\ ,$$ (finite sum, since $$\{g_\iota\}_{\iota\in I}$$ is a basis) one defines $$\gamma(h)=\sum_{\iota\in I} b_\iota k_\iota\ .$$

When $$\mathrm{rank}_R \mathfrak{g}=\#I<\infty$$ this can be done explicitly.

If $$R$$ is a field with characteristic zero, it suffices to require $$\mathrm{dim}_R \mathfrak{g}<\infty$$ in order to conclude to the existence of sections.

Consider the exact sequence of Lie algebras with Lie algebra morphisms

$\iota \qquad \quad \ p$

$\tag{1} 0 \quad \rightarrow \quad \mathfrak{a} \quad \rightarrow \quad \mathfrak{k} \quad \rightarrow \quad \mathfrak{g} \quad \rightarrow 0 \ .$

### problem statement

If $$\mathfrak{g}$$ is projective as an $$R$$-module, is there a section of $$p$$ which is also a Lie algebra morphism, that is, is it also projective as a Lie algebra?

To answer this question, first assume $$\mathfrak{a}$$ to be abelian, that is, $$[x,y]=0$$ for all $$x,y\in\mathfrak{a}\ .$$

Let $$\sigma_1$$ be a section, but not necessarily a Lie algebra morphism (Exists, since $$\mathfrak{g}$$ is projective).

Define $$\sigma_2\in C_{\wedge}^2(\mathfrak{g},\mathfrak{k})\ ,$$ (antisymmetric 2-forms) by $\tag{2} \sigma_2(x,y)=[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}}-\sigma_1([x,y]_{\mathfrak{g}}).$

One has $\tag{3} p \sigma_2(x,y)=p[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}}-p\sigma_1([x,y]_{\mathfrak{g}})=[p\sigma_1(x),p\sigma_1(y)]_{\mathfrak{g}}-[x,y]_{\mathfrak{g}}=0.$

Since $$\sigma_2(x,y)\in\ker p=\mathrm{im\ }\iota\ ,$$ one can define $$a_2\in C_{\wedge}^2(\mathfrak{g},\mathfrak{a})$$ by $\tag{4} \iota(a_2(x,y))=\sigma_2(x,y).$

Now define $$d_1:\mathfrak{g}\rightarrow End(\mathfrak{a})$$ by $\tag{5} \iota(d_1(x)a)=[\sigma_1(x),\iota(a)]_{\mathfrak{k}}$

This is well-defined since, for instance, $$p[\sigma_1(x),\iota(a)]_{\mathfrak{k}}=[p\sigma_1(x),p\iota(a)]_{\mathfrak{g}}=0\ .$$

Moreover, $$d_1$$ is a representation, since $\iota(d_1([x,y]_{\mathfrak{g}})a) = [\sigma_1([x,y]),\iota(a)]_{\mathfrak{k}}\ :$

$= -[\sigma_2(x,y),\iota(a)]_{\mathfrak{k}} +[[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}},\iota(a)]_{\mathfrak{k}}\ :$
$= -[\iota a_2(x,y),\iota(a)]_{\mathfrak{k}} +[\sigma_1(x),[\sigma_1(y),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}-[\sigma_1(y),[\sigma_1(x),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}\ :$
$=- \iota[a_2(x,y),a]_{\mathfrak{a}}+[\sigma_1(x),\iota(d_1(y)a)]_{\mathfrak{k}}-[\sigma_1(y),\iota(d_1(x)a)]_{\mathfrak{k}}\ :$
$= \iota(d_1(x)d^{(0)}(y)a) -\iota(d_1(y)d_1(x)a)$

or $$d_1([x,y]_{\mathfrak{g}})=[d_1(x),d_1(y)]_{\mathfrak{k}}\ .$$

In the sequel we drop the subscript of the bracket.

Suppose now there exists a $$\delta\sigma_1\in C^1(\mathfrak{g},\mathfrak{k})$$ such that $$\sigma_1+\delta\sigma_1$$ is a Lie algebra homomorphism and a section of $$p\ .$$ Then $x=p(\sigma_1+\delta\sigma_1)(x)=p\sigma_1(x)+p \delta\sigma_1(x)=x+p \delta\sigma_1(x),$ implying that $$\delta\sigma_1 (x)\in\ker(p)=\mathrm{im\ }\iota\ .$$ Define $$a_1\in C^1(\mathfrak{g},\mathfrak{a})$$ by $$\iota a_1(x)=\delta\sigma_1 (x)\ .$$ Then (by assumption!) $0 =(\sigma_1+\delta\sigma_1)([x,y])-[(\sigma_1+\delta\sigma_1)(x),(\sigma_1+\delta\sigma_1)(y)]\ :$ $=\sigma_1([x,y])+\delta\sigma_1([x,y])-[\sigma_1(x),\sigma_1(y)]-[\delta\sigma_1(x),\sigma_1(y)]-[\sigma_1(x),\delta\sigma_1(y)]-[\delta\sigma_1(x),\delta\sigma_1(y)]\ :$

$= -\sigma_2(x,y) +\delta\sigma_1([x,y])-[\iota a_1(x),\sigma_1(y)]-[\sigma_1(x),\iota a_1(y)]-[\iota a_1(x),\iota a_1(y)]\ :$ $= -\iota(a_2(x,y)+(d_1(x)a_1(y)-d_1(y)a_1(x)-a_1([x,y])-[a_1(x),a_1(y)])\ :$ $= -\iota((a_2+d^1 a_1)(x,y)).$ This implies that the existence of such a $$\delta\sigma_1$$ is equivalent to $$a_2\in\mathrm{im\ }d^1\ .$$

On the other hand, if it would turn out that $$d^2 a_2\neq 0\ ,$$ this would be a definite obstruction of the existence of such a $$\delta\sigma_1\ .$$

We can, however, rule out the last possibility. $\iota d^2 a_2(x,y,z)= \iota d_1(x)a_2(y,z) -\iota d_1(y)a_2(x,z) +\iota d_1(z)a_2(x,y)-\iota a_2([x,y],z)-\iota a_2(y,[x,z])+\iota a_2(x,[y,z])\ :$

$=-[\sigma_1(x),\sigma_2(y,z)]-[\sigma_1(y),\sigma_2(x,z)]-[\sigma_2(x,y),\sigma_1(z)]-\sigma_2([x,y],z)-\sigma_2(y,[x,z])+\sigma_2(x,[y,z])\ :$
$=[\sigma(x),\sigma([y,z])-[\sigma(y),\sigma(z)]]-[\sigma(y),\sigma([x,z])-[\sigma(x),\sigma(z)]]-[\sigma([x,y])-[\sigma(x),\sigma(y)],\sigma(z)]\ :$
$-\sigma_1([[x,y],z])+[\sigma_1([x,y]),\sigma_1(z)]-\sigma_1([y,[x,z]])+[\sigma_1(y),\sigma_1([x,z]]+\sigma_1([x,[y,z]])-[\sigma_1(x),\sigma_1([y,z])]\ :$
$= -[\sigma(x),[\sigma(y),\sigma(z)]]+[\sigma(y),[\sigma(x),\sigma(z)]] +[[\sigma(x),\sigma(y)],\sigma(z)] -\sigma_1([[x,y],z]+[y,[x,z]]-[x,[y,z]]) \ :$
$=0.$

It follows that $$d^2 a_2=0\ .$$