lecture 4

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< An introduction to Lie algebra cohomology

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Author: Dr. Jan A. Sanders, Vrije Universiteit Amsterdam
Author: Dr. Sara Lombardo, Vrije Universiteit Amsterdam

Back to the third lecture

On to the fifth lecture

1 Construction of an extension

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Construction of an extension

We now turn the question around. What is the situation if we have Lie algebras ${\mathfrak{g}}$ (projective) and ${\mathfrak{a}}$ (abelian), and $[a^2]\in H_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$ ? If a Lie section (that is, a section which is a Lie algebra morphism) is possible, then one can view ${\mathfrak{k}}$ as the direct sum of ${\mathfrak{g}}$ and ${\mathfrak{a}}$ . What is the situation if $[a^2]\in H_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$ is not zero? To see how one should define a Lie algebra structure on the sum of ${\mathfrak{g}}$ and ${\mathfrak{a}}$ , it pays to have a look at ${\mathfrak{k}}$ . Every element in $y\in{\mathfrak{k}}$ can be uniquely written as $\sigma(x)+\iota(a)$ , $x\in{\mathfrak{g}}, a\in{\mathfrak{a}}$ : take $x=p(y)$ and $a=\iota^{-1} (1-\sigma p)(y)$ . Let $\phi:{\mathfrak{k}}\rightarrow {\mathfrak{a}}\oplus_R{\mathfrak{g}}$ be defined by $\phi(\sigma(x))+\iota(a))=(a,x)$ . Then

$\phi([\sigma(x)+\iota(a_1),\sigma(y)+\iota(a_2)]=\phi([\sigma(x),\sigma(y)]+[\iota(a_1),\sigma(y)]+[\sigma(x),\iota(a_2)])$

$=\phi(\sigma([x,y])-\iota(a^2(x,y))+[\iota(a_1),\sigma(y)]+[\sigma(x),\iota(a_2)])$

$=({d^{(0)}}(x)a_2-{d^{(0)}}(y)a_1-a^2(x,y),[x,y])$

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definition

One now defines for an arbitrary representation ${d^{(0)}}$ and $a^2\in C_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$ ,

$[(a_1,x),(a_2,y)]=({d^{(0)}}(x)a_2-{d^{(0)}}(y)a_1-a^2(x,y),[x,y])$

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lemma

The new bracket $[\cdot,\cdot]_{a^2}$ obeys the Jacobi identity as long as $a^2\in Z_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$ .

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proof

$[[(a_1,x),(a_2,y)],(a_3.z)]-[(a_1,x),[(a_2,x),(a_3,z)]]+[(a_2,y),[(a_1,x),(a_3,y)]]$

$=[({d^{(0)}}(x)a_2-{d^{(0)}}(y)a_1-a^2(x,y),[x,y]),(a_3,z)]$

$-[(a_1,x),({d^{(0)}}(y)a_3-{d^{(0)}}(z)a_2-a^2(y,z),[y,z])]$

$+[(a_2,y),({d^{(0)}}(x)a_3-{d^{(0)}}(z)a_1-a^2(x,z),[x,z])]$

$=({d^{(0)}}([x,y])a_3-{d^{(0)}}(z){d^{(0)}}(x)a_2+{d^{(0)}}(z){d^{(0)}}(y)a_1+{d^{(0)}}(z)a^2(x,y)-a^2([x,y],z),[[x,y],z])$

$-({d^{(0)}}(x){d^{(0)}}(y)a_3-{d^{(0)}}(x){d^{(0)}}(z)a_2-{d^{(0)}}(x)a^2(y,z)-{d^{(0)}}([y,z])a_1-a^2(x,[y,z],[x,[y,z]])$

$+({d^{(0)}}(y){d^{(0)}}(x)a_3-{d^{(0)}}(y){d^{(0)}}(z)a_1-{d^{(0)}}(y)a^2(x,z)-{d^{(0)}}([x,z])a_2-a^2(y,[x,z],[y,[x,z]])$

$=({d^{(0)}}([y,z])a_1-{d^{(0)}}(y){d^{(0)}}(z)a_1+{d^{(0)}}(z){d^{(0)}}(y)a_1$

$-{d^{(0)}}([x,z])a_2-{d^{(0)}}(z){d^{(0)}}(x)a_2+{d^{(0)}}(x){d^{(0)}}(z)a_2$

$+{d^{(0)}}([x,y])a_3-{d^{(0)}}(x){d^{(0)}}(y)a_3 +{d^{(0)}}(y){d^{(0)}}(x)a_3$

$+{d^{(0)}}(x)a^2(y,z)-{d^{(0)}}(y)a^2(x,z)+{d^{(0)}}(z)a^2(x,y)$

$-a^2([x,y],z)-a^2(y,[x,z])+a^2(x,[y,z]),[[x,y],z] -[x,[y,z]] +[y,[x,z]])$

$=(d^{2}a^2(x,y,z),0)$

$=(0,0).$ $\square$

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antisymmetry

One needs $a^2\in Z_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$ in order to make the bracket antisymmetric.

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definition

We denote the new Lie algebra by ${\mathfrak{g}}{\oplus}_{a^2} {\mathfrak{a}}$ , the semidirect product of ${\mathfrak{g}}$ and ${\mathfrak{a}}$ induced by $a^2 \in Z_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})$ .

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theorem

Let $a^2$ be as defined in lecture two. Then

$\mathfrak{k}\simeq \mathfrak{a}\oplus_{a^2} \mathfrak{g}$ .

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proof

Consider $\mathfrak{a}\oplus_{a^2}\mathfrak{g}$ , where the representation $d^{(0)}$ and the form $a^2\in Z_{\wedge}^2(\mathfrak{g},\mathfrak{a})$ are constructed as in the second lecture. Let $\phi: \mathfrak{a}\oplus_{a^2}\mathfrak{g}\rightarrow \mathfrak{k}$ be defined by $\phi((a,x))=\iota(a)+\sigma(x)$ . Then

$\phi([(a_1,x_1),(a_2,x_2)])=\phi((d^{(0)}(x_1)a_2-d^{(0)}(x_2)a_1-a^2(x_1,x_2),[x_1,x_2]))$

$=\iota(d^{(0)}(x_1)a_2-d^{(0)}(x_2)a_1-a^2(x_1,x_2))+\sigma([x_1,x_2])$

$=[\sigma(x_1),\iota(a_2)]+[\iota(a_1),\sigma(x_2)]+[\sigma(x_1),\sigma(x_2)]$

$=[\iota(a_1)+\sigma(x_1),\iota(a_2)+\sigma(x_2)]$

$=[\phi((a_1,x_1)),\phi((a_2,x_2))].$

Let now $\psi:\mathfrak{k}\rightarrow \mathfrak{a}\oplus_{a^2}\mathfrak{g}$ be defined by

$\psi(x)=(\iota^{-1}(x-\sigma(p(x))),p(x))$ .

Then

$\psi([x,y])=(\iota^{-1}([x,y]-\sigma(p([x,y]))),p([x,y]))$

$=(\iota^{-1}([x,y]-\sigma([p(x),p(y)])),[p(x),p(y)])$

$=(\iota^{-1}([x,y]-[\sigma(p(x)),\sigma(p(y))]-\iota a^2(p(x),p(y))),[p(x),p(y))])$

$=(\iota^{-1}([x-\sigma(p(x)),\sigma(p(y))]+[\sigma(p(x)),y-\sigma(p(y))]-\iota a^2(p(x),p(y))),[p(x),p(y))])$

$=(\iota^{-1}([\sigma(p(x)),y-\sigma(p(y))])+\iota^{-1}([x-\sigma(p(x),\sigma(p(y))])-a^2(p(x),p(y)),[p(x),p(y)])$

$=(d^{(0)}(p(x))\iota^{-1}(y-\sigma(p(y)))-d^{(0)}(p(y))\iota^{-1}(x-\sigma(p(x))-a^2(p(x),p(y)),[p(x),p(y)])$

$=[(\iota^{-1}(x-\sigma(p(x))),p(x)),(\iota^{-1}(y-\sigma(p(y))),p(y))]$

$=[\psi(x),\psi(y)]$

This implies that $\phi$ and $\psi$ are Leibniz algebra homomorphisms. Furthermore,

$\phi(\psi(x))=\phi((\iota^{-1}(x-\sigma(p(x))),p(x)))$

$= x-\sigma(p(x))+\sigma(p(x)$

$=x$

and

$\psi(\phi(a,x))=\psi(\iota(a)+\sigma(x))$

$=(\iota^{-1}(\iota(a)+\sigma(x)-\sigma(p(\iota(a)+\sigma(x))),\sigma(p(\iota(a)+\sigma(x)))$

$=(\iota^{-1}(\iota(a)+\sigma(x)-\sigma(x)),\sigma(x))$

$=(a,\sigma(x))$

This $\phi$ and $\psi$ are Lie algebra isomorphisms. One has

$\mathfrak{k}\simeq \mathfrak{a}\oplus_{a^2} \mathfrak{g}\quad\square$

What if one now applies the construction in the second lecture to $\mathfrak{a}\oplus_{a^2} \mathfrak{g}$ ? The maps $\iota$ and $p$ are given by

$\iota(a)=(a,0)$

$p((a,x))=x$

From the definition of the bracket it follows that $p$ is a Lie algebra homomorphism, for $\iota$ this is trivial since $\mathfrak{a}$ is abelian. One chooses a section $\tilde{\sigma}$ as follows.