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

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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 $d^2 a_2=0\ .$

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