lecture 12

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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

Construction of an extension

Now one turns the question around. What is the situation if one has Lie algebras ${\mathfrak{g}}$ (projective) and ${\mathfrak{h}}$ .

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definition

If $\alpha^0\in End(\mathfrak{h})$ can be written as

$\alpha^0 h=[\alpha,h],\quad \alpha,h\in\mathfrak{h}$

one says that $\alpha^0$ is inner and we denote the space of inner endomorphisms by $\mathfrak{ad}(\mathfrak{h})$ . If

$\alpha^0[h_1,h_2]=[\alpha^0 h_1,h_2]+[h_1,\alpha^0 h_2]$

then $\alpha^0$ is called a derivation and the space of derivations is denoted by $\mathfrak{der}(\mathfrak{h})$ .

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proposition

If $\alpha^0, \beta^0\in\mathfrak{der}(\mathfrak{h})$ then $[\alpha^0,\beta^0]\in\mathfrak{der}(\mathfrak{h})$ .

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proposition

$[\mathfrak{der}(\mathfrak{h}),\mathfrak{ad}(\mathfrak{h})]\subset \mathfrak{ad}(\mathfrak{h})$

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proof

$[\mathfrak{d},\mathrm{ad}(\alpha)]h=\mathfrak{d}[\alpha,h]-[\alpha,\delta h]=[\mathfrak{d}\alpha,h]=\mathrm{ad}(\mathfrak{d}\alpha)h$

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definition

Let

$\mathfrak{out}(\mathfrak{h})=\mathfrak{der}(\mathfrak{h})/\mathfrak{ad}(\mathfrak{h})$

To see how one should define a Lie algebra structure on the sum of ${\mathfrak{g}}$ and ${\mathfrak{h}}$ , it pays to have a look at ${\mathfrak{k}}$ . Every element in $y\in{\mathfrak{k}}$ can be uniquely written as $\sigma(x)+\iota(h)$ , $x\in{\mathfrak{g}}, h\in{\mathfrak{h}}$ : take $x=p(y)$ and $h=\iota^{-1} (1-\sigma p)(y)$ . Let $\phi:{\mathfrak{k}}\rightarrow {\mathfrak{h}}\oplus_R{\mathfrak{g}}$ be defined by $\phi(\sigma(x)+\iota(h))=(h,x)$ . Then

$\phi([\sigma(x)+\iota(h_1),\sigma(y)+\iota(h_2)])=\phi([\sigma(x),\sigma(y)]+[\iota(h_1),\sigma(y)]+[\sigma(x),\iota(h_2)]+\iota([h_1,h_2]))$

$=\phi(\sigma([x,y])-\iota(h^2(x,y))+[\iota(h_1),\sigma(y)]+[\sigma(x),\iota(h_2)]+\iota([h_1,h_2]))$

$=([h_1,h_2]+{\mathbf{d}^{(0)}}(x)h_2-{\mathbf{d}^{(0)}}(y)h_1-h^2(x,y),[x,y])$

Suppose $d^{(0)}:\mathfrak{g}\rightarrow\mathfrak{out}(\mathfrak{h})$ is a Lie algebra homomorphism. Let

$d^{(0)}=[\mathbf{d}^{(0)}]$

with

(1)

$\mathbf{d}^{(0)}([x,y])=ad(h^2(x,y))+\mathbf{d}^{(0)}(x)\mathbf{d}^{(0)}(y) -\mathbf{d}^{(0)}(y)\mathbf{d}^{(0)}(x)$

(Since $d^{(0)}\in\mathfrak{out}(\mathfrak{h})$ is a representation, $\mathbf{d}^{(0)}([x,y])-\mathbf{d}^{(0)}(x)\mathbf{d}^{(0)}(y) +\mathbf{d}^{(0)}(y)\mathbf{d}^{(0)}(x)$ must be an inner derivation, represented by an element in $\mathfrak{h}$ depending linearly on $x$ and $y$ , and denoted by $h^2(x,y)$ .)

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definition

One defines

$[(h_1,x_1),(h_2,x_2)]=([h_1,h_2]+{\mathbf{d}^{(0)}}(x_1)h_2-{\mathbf{d}^{(0)}}(x_2)h_1-h^2(x_1,x_2),[x_1,x_2])$

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lemma

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

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proof

$[[(h_1,x_1),(h_2,x_2)],(h_3.x_3)]-[(h_1,x_1),[(h_2,x_2),(h_3,x_3)]]+[(h_2,x_2),[(h_1,x_1),(h_3,x_3)]]$

$=[([h_1,h_2]+{\mathbf{d}^{(0)}}(x_1)h_2-{\mathbf{d}^{(0)}}(x_2)h_1-h^2(x_1,x_2),[x_1,x_2]),(h_3,x_3)]$

$-[(h_1,x_1),([h_2,h_3]+{\mathbf{d}^{(0)}}(x_2)h_3-{\mathbf{d}^{(0)}}(x_3)h_2-h^2(x_2,x_3),[x_2,x_3])]$

$+[(h_2,x_2),([h_1,h_3]+{\mathbf{d}^{(0)}}(x_1)h_3-{\mathbf{d}^{(0)}}(x_3)h_1-h^2(x_1,x_3),[x_1,x_3])]$

$=([[h_1,h_2]+{\mathbf{d}^{(0)}}(x_1)h_2-{\mathbf{d}^{(0)}}(x_2)h_1-h^2(x_1,x_2),h_3]+\mathbf{d}^{(0)}([x_1,x_2])h_3 -\mathbf{d}^{(0)}(x_3)([h_1,h_2]+{\mathbf{d}^{(0)}}(x_1)h_2-{\mathbf{d}^{(0)}}(x_2)h_1-h^2(x_1,x_2))-h^2([x_1,x_2],x_3),[[x_1,x_2],x_3])$

$-([h_1,[h_2,h_3]+{\mathbf{d}^{(0)}}(x_2)h_3-{\mathbf{d}^{(0)}}(x_3)h_2-h^2(x_2,x_3)] +\mathbf{d}^{(0)}(x_1)([h_2,h_3]+{\mathbf{d}^{(0)}}(x_2)h_3-{\mathbf{d}^{(0)}}(x_3)h_2-h^2(x_2,x_3)) -\mathbf{d}^{(0)}([x_2,x_3])h_1-h^2(x_1,[x_2,x_3]),[x_1,[x_2,x_3]])$

$+([h_2,[h_1,h_3]+{\mathbf{d}^{(0)}}(x_1)h_3-{\mathbf{d}^{(0)}}(x_3)h_1-h^2(x_1,x_3)] +\mathbf{d}^{(0)}(x_2)([h_1,h_3]+{\mathbf{d}^{(0)}}(x_1)h_3-{\mathbf{d}^{(0)}}(x_3)h_1-h^2(x_1,x_3)) -\mathbf{d}^{(0)}([x_1,x_3])h_2-h^2(x_2,[x_1,x_3]), [x_2,[x_1,x_3]])$

$=([{\mathbf{d}^{(0)}}(x_1)h_2,h_3]-[{\mathbf{d}^{(0)}}(x_2)h_1,h_3]-[h^2(x_1,x_2),h_3]+\mathbf{d}^{(0)}([x_1,x_2])h_3 -\mathbf{d}^{(0)}(x_3)[h_1,h_2]-\mathbf{d}^{(0)}(x_3){\mathbf{d}^{(0)}}(x_1)h_2+\mathbf{d}^{(0)}(x_3){\mathbf{d}^{(0)}}(x_2)h_1 +\mathbf{d}^{(0)}(x_3)h^2(x_1,x_2)-h^2([x_1,x_2],x_3)$

$-[h_1,{\mathbf{d}^{(0)}}(x_2)h_3]+[h_1,{\mathbf{d}^{(0)}}(x_3)h_2]+[h_1,h^2(x_2,x_3)] -\mathbf{d}^{(0)}(x_1)[h_2,h_3]-\mathbf{d}^{(0)}(x_1){\mathbf{d}^{(0)}}(x_2)h_3+\mathbf{d}^{(0)}(x_1){\mathbf{d}^{(0)}}(x_3)h_2 +\mathbf{d}^{(0)}(x_1)h^2(x_2,x_3) +\mathbf{d}^{(0)}([x_2,x_3])h_1+h^2(x_1,[x_2,x_3])$

$+[h_2,{\mathbf{d}^{(0)}}(x_1)h_3]-[h_2,{\mathbf{d}^{(0)}}(x_3)h_1]-[h_2,h^2(x_1,x_3)] +\mathbf{d}^{(0)}(x_2)[h_1,h_3]+\mathbf{d}^{(0)}(x_2){\mathbf{d}^{(0)}}(x_1)h_3-\mathbf{d}^{(0)}(x_2){\mathbf{d}^{(0)}}(x_3)h_1 -\mathbf{d}^{(0)}(x_2)h^2(x_1,x_3)-\mathbf{d}^{(0)}([x_1,x_3])h_2-h^2(x_2,[x_1,x_3]), 0)$

$=([-\mathbf{d}^{(0)}(x_1)[h_2,h_3]+[h_2,{\mathbf{d}^{(0)}}(x_1)h_3]+[{\mathbf{d}^{(0)}}(x_1)h_2,h_3]$

$+\mathbf{d}^{(0)}(x_2)[h_1,h_3]-[h_1,{\mathbf{d}^{(0)}}(x_2)h_3] -[{\mathbf{d}^{(0)}}(x_2)h_1,h_3]$

$-[h_2,{\mathbf{d}^{(0)}}(x_3)h_1]+[h_1,{\mathbf{d}^{(0)}}(x_3)h_2]-\mathbf{d}^{(0)}(x_3)[h_1,h_2]$

$-[h^2(x_1,x_2),h_3]+\mathbf{d}^{(0)}([x_1,x_2])h_3 -\mathbf{d}^{(0)}(x_1){\mathbf{d}^{(0)}}(x_2)h_3+\mathbf{d}^{(0)}(x_2){\mathbf{d}^{(0)}}(x_1)h_3$

$-\mathbf{d}^{(0)}(x_3){\mathbf{d}^{(0)}}(x_1)h_2 +\mathbf{d}^{(0)}(x_1){\mathbf{d}^{(0)}}(x_3)h_2-\mathbf{d}^{(0)}([x_1,x_3])h_2-[h_2,h^2(x_1,x_3)]$

$+[h_1,h^2(x_2,x_3)]+\mathbf{d}^{(0)}(x_3){\mathbf{d}^{(0)}}(x_2)h_1 +\mathbf{d}([x_2,x_3])h_1-\mathbf{d}^{(0)}(x_2){\mathbf{d}^{(0)}}(x_3)h_1$

$+\mathbf{d}^{(0)}(x_3)h^2(x_1,x_2)-h^2([x_1,x_2],x_3)$

$+\mathbf{d}^{(0)}(x_1)h^2(x_2,x_3) +h^2(x_1,[x_2,x_3])$

$-\mathbf{d}^{(0)}(x_2)h^2(x_1,x_3)-h^2(x_2,[x_1,x_3]), 0)$

$=(\mathbf{d}^2h^2(x_1,x_2,x_3),0)$

$=(0,0)$

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proposition

$\mathbf{d}^2h^2(x_1,x_2,x_3)\in\mathcal{C}(\mathfrak{h})$

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proof

$\mathrm{ad}(\mathbf{d}^2h^2(x_1,x_2,x_3))$

$=\mathrm{ad}(\mathbf{d}^{(0)}(x_1)h^2(x_2,x_3))-\mathbf{d}^{(0)}(x_2)h^2(x_1,x_3)+\mathbf{d}^{(0)}(x_3)h^2(x_1,x_2) -h^2([x_1,x_2],x_3)-h^2(x_2,[x_1,x_3])+h^2(x_1,[x_2,x_3]))$

$=[\mathbf{d}^{(0)}(x_1),\mathrm{ad}(h^2(x_2,x_3))]-[\mathbf{d}^{(0)}(x_2),\mathrm{ad}(h^2(x_1,x_3))] +[\mathbf{d}^{(0)}(x_3),\mathrm{ad}(h^2(x_1,x_2))] -\mathrm{ad}(h^2([x_1,x_2],x_3))-\mathrm{ad}(h^2(x_2,[x_1,x_3]))+\mathrm{ad}(h^2(x_1,[x_2,x_3]))$

$=[\mathbf{d}^{(0)}(x_1),\mathbf{d}^{(0)}([x_2,x_3])]-[\mathbf{d}^{(0)}(x_2),\mathbf{d}^{(0)}([x_1,x_3])] +[\mathbf{d}^{(0)}(x_3),\mathbf{d}^{(0)}([x_1,x_2])] -\mathrm{ad}(h^2([x_1,x_2],x_3))-\mathrm{ad}(h^2(x_2,[x_1,x_3]))+\mathrm{ad}(h^2(x_1,[x_2,x_3]))$

$=\mathbf{d}^{(0)}([x_1,[x_2,x_3]]) -\mathbf{d}^{(0)}([[x_1,x_2],x_3])-\mathbf{d}^{(0)}([x_2,[x_1,x_3]])$

$=0$

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corollary

If $\mathfrak{h}$ has no center, one has automatically $\mathbf{d}^2h^2=0$ , which implies that the Jacobi identity is always satisfied. Another way of saying this, is that the choice $d^{(0)}\in\mathfrak{out}(\mathfrak{h})$ determines an extension

$\iota_0 \qquad \quad \ p_0$

(2)

$0 \quad \rightarrow \quad \mathfrak{ad}(\mathfrak{h}) \quad \rightarrow \quad \mathfrak{k}_0 \quad \rightarrow \quad \mathfrak{g} \quad \rightarrow 0$ .

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proposition

$\bar{\mathbf{d}}^2 \bar{h}^2-\mathbf{d}^2h^2\in B_\wedge^3(\mathfrak{g},\mathcal{Z}(\mathfrak{h}))$ .

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proof

Let $\bar{\mathbf{d}}^{(0)}(x)=\mathbf{d}^{(0)}+\mathrm{ad}(h^1(x))$ . Then

$\mathrm{ad}(\bar{h}^2(x_1,x_2))=\bar{\mathbf{d}}^{(0)}([x_1,x_2])-[\bar{\mathbf{d}}^{(0)}(x_1),\bar{\mathbf{d}}^{(0)}(x_2)]$

$=\mathbf{d}^{(0)}([x_1,x_2])+\mathrm{ad}(h^1([x_1,x_2]))-[\mathbf{d}^{(0)}(x_1),\mathbf{d}^{(0)}(x_2)] -[\mathbf{d}^{(0)}(x_1),\mathrm{ad}(h^1(x_2))]-[\mathrm{ad}(h^1(x_1)),\mathbf{d}^{(0)}(x_2)]- [\mathrm{ad}(h^1(x_1)),\mathrm{ad}(h^1(x_2))]$

$=\mathrm{ad}(h^2(x_1,x_2))+\mathrm{ad}(h^1([x_1,x_2]))-\mathrm{ad}(\mathbf{d}^{(0)}(x_1)h^1(x_2)]+ \mathrm{ad}(\mathbf{d}^{(0)}(x_2)h^1(x_1))-\mathrm{ad}([h^1(x_1),h^1(x_2)])$

$=\mathrm{ad}(h^2(x_1,x_2)-\mathbf{d}^1h^1(x_1,x_2)-[h^1(x_1),h^1(x_2)])$

This implies

$\bar{h}^2=h^2-\mathbf{d}^1h^1-[h^1,h^1]+\eta^2.\quad \eta^2\in C_\wedge^2(\mathfrak{g},\mathcal{Z}(\mathfrak{h}))$

Then

$\bar{\mathbf{d}}^2 \bar{h}^2(x_1,x_2,x_3)=\bar{\mathbf{d}}^{(0)}(x_1)\bar{h}^2(x_2,x_3)-\bar{\mathbf{d}}^{(0)}(x_2)\bar{h}^2(x_1,x_3) +\bar{\mathbf{d}}^{(0)}(x_3)\bar{h}^2(x_1,x_2)- \bar{h}^2([x_1,x_2],x_3)- \bar{h}^2(x_2,[x_1,x_3])+ \bar{h}^2(x_1,[x_2,x_3])$

$=(\mathbf{d}^{(0)}(x_1)+\mathrm{ad}(h^1(x_1)))(h^2(x_2,x_3)-\mathbf{d}^1h^1(x_2,x_3)-[h^1,h^1](x_2,x_3)+ \eta^2(x_2,x_3))$

$-(\mathbf{d}^{(0)}(x_2)+\mathrm{ad}(h^1(x_2)))(h^2(x_1,x_3)-\mathbf{d}^1h^1(x_1,x_3)-[h^1,h^1](x_1,x_3)+ \eta^2(x_1,x_3))$

$+(\mathbf{d}^{(0)}(x_3)+\mathrm{ad}(h^1(x_3)))(h^2(x_1,x_2)-\mathbf{d}^1h^1(x_1,x_2)-[h^1,h^1](x_1,x_2)+ \eta^2(x_1,x_2))$

$+h^2([x_1,x_2],x_3)-\mathbf{d}^1h^1([x_1,x_2],x_3)-[h^1([x_1,x_2]),h^1(x_3)]+\eta^2([x_1,x_2],x_3)$

$+h^2(x_2,[x_1,x_3])-\mathbf{d}^1h^1(x_2,[x_1,x_3])-[h^1(x_2),h^1([x_1,x_3])]+\eta^2(x_2,[x_1,x_3])$

$+h^2(x_1,[x_2,x_3])-\mathbf{d}^1h^1(x_1,[x_2,x_3])-[h^1(x_1),h^1([x_2,x_3])]+\eta^2(x_1,[x_2,x_3])$

$=\mathbf{d}^{(0)}(x_1)(h^2(x_2,x_3)-\mathbf{d}^1h^1(x_2,x_3)-[h^1()x_2),h^1(x_3)]+ \eta^2(x_2,x_3))$

$-\mathbf{d}^{(0)}(x_2)(h^2(x_1,x_3)-\mathbf{d}^1h^1(x_1,x_3)-[h^1(x_1),h^1(x_3)]+ \eta^2(x_1,x_3))$

$+\mathbf{d}^{(0)}(x_3)(h^2(x_1,x_2)-\mathbf{d}^1h^1(x_1,x_2)-[h^1(x_1),h^1(x_2)]+ \eta^2(x_1,x_2))$

$+\mathrm{ad}(h^1(x_1))(h^2(x_2,x_3)-\mathbf{d}^1h^1(x_2,x_3)-[h^1(x_2),h^1(x_3)]+ \eta^2(x_2,x_3))$

$-\mathrm{ad}(h^1(x_2))(h^2(x_1,x_3)-\mathbf{d}^1h^1(x_1,x_3)-[h^1(x_1),h^1(x_3)]+ \eta^2(x_1,x_3))$

$+\mathrm{ad}(h^1(x_3))(h^2(x_1,x_2)-\mathbf{d}^1h^1(x_1,x_2)-[h^1(x_1),h^1(x_2)]+ \eta^2(x_1,x_2))$

$-h^2([x_1,x_2],x_3)+\mathbf{d}^1h^1([x_1,x_2],x_3)+[h^1([x_1,x_2]),h^1(x_3)]+\eta^2([x_1,x_2],x_3)$

$-h^2(x_2,[x_1,x_3])+\mathbf{d}^1h^1(x_2,[x_1,x_3])+[h^1(x_2),h^1([x_1,x_3])]+\eta^2(x_2,[x_1,x_3])$

$+h^2(x_1,[x_2,x_3])-\mathbf{d}^1h^1(x_1,[x_2,x_3])-[h^1(x_1),h^1([x_2,x_3])]+\eta^2(x_1,[x_2,x_3])$

$=\mathbf{d}^2 h^2(x_1,x_2,x_3)-\mathbf{d}^{(0)}(x_1)(\mathbf{d}^{(0)}(x_2)h^1(x_3) -\mathbf{d}^{(0)}(x_3)h^1(x_2)-h^1([x_2,x_3]))-[\mathbf{d}^{(0)}(x_1)h^1(x_2),h^1(x_3)] -[h^1(x_2),\mathbf{d}^{(0)}(x_1)h^1(x_3)]+ {d}^{(0)}(x_1)\eta^2(x_2,x_3))$

$+\mathbf{d}^{(0)}(x_2)(\mathbf{d}^{(0)}(x_1)h^1(x_3)-\mathbf{d}^{(0)}(x_3)h^1(x_1)-h^1([x_1,x_3]))+[\mathbf{d}^{(0)}(x_2)h^1(x_1),h^1(x_3)] +[h^1(x_1),\mathbf{d}^{(0)}(x_2)h^1(x_3)] -{d}^{(0)}(x_2) \eta^2(x_1,x_3))$

$-\mathbf{d}^{(0)}(x_3)(\mathbf{d}^{(0)}(x_1)h^1(x_2)-\mathbf{d}^{(0)}(x_2)h^1(x_1)-h^1([x_1,x_2]))-[\mathbf{d}^{(0)}(x_3)h^1(x_1),h^1(x_2)] -[h^1(x_1),\mathbf{d}^{(0)}(x_3)h^1(x_2)]+ {d}^{(0)}(x_3)\eta^2(x_1,x_2))$

$+\mathrm{ad}(h^1(x_1))(h^2(x_2,x_3)-(\mathbf{d}^{(0)}(x_2)h^1(x_3)-\mathbf{d}^{(0)}(x_3)h^1(x_2)-h^1([x_2,x_3])))$

$-\mathrm{ad}(h^1(x_2))(h^2(x_1,x_3)-(\mathbf{d}^{(0)}(x_1)h^1(x_3)-\mathbf{d}^{(0)}(x_3)h^1(x_1)-h^1([x_1,x_3])))$

$+\mathrm{ad}(h^1(x_3))(h^2(x_1,x_2)-(\mathbf{d}^{(0)}(x_1)h^1(x_2)-\mathbf{d}^{(0)}(x_2)h^1(x_1)-h^1([x_1,x_2])))$

$+\mathbf{d}^1h^1([x_1,x_2],x_3)+[h^1([x_1,x_2]),h^1(x_3)]+\eta^2([x_1,x_2],x_3)$

$+\mathbf{d}^1h^1(x_2,[x_1,x_3])+[h^1(x_2),h^1([x_1,x_3])]+\eta^2(x_2,[x_1,x_3])$

$-\mathbf{d}^1h^1(x_1,[x_2,x_3])-[h^1(x_1),h^1([x_2,x_3])]+\eta^2(x_1,[x_2,x_3])$

$=\mathbf{d}^2 h^2(x_1,x_2,x_3)$

$-\mathbf{d}^{(0)}(x_1)\mathbf{d}^{(0)}(x_2)h^1(x_3) +\mathbf{d}^{(0)}(x_2)\mathbf{d}^{(0)}(x_1)h^1(x_3)+[h^1(x_3),h^2(x_1,x_2)] + {d}^{(0)}(x_1)\eta^2(x_2,x_3))$

$+\mathbf{d}^{(0)}(x_1)\mathbf{d}^{(0)}(x_3)h^1(x_2)-\mathbf{d}^{(0)}(x_3)\mathbf{d}^{(0)}(x_1)h^1(x_2)-[h^1(x_2),h^2(x_1,x_3)] -{d}^{(0)}(x_2) \eta^2(x_1,x_3))$

$-\mathbf{d}^{(0)}(x_2)\mathbf{d}^{(0)}(x_3)h^1(x_1)+\mathbf{d}^{(0)}(x_3)\mathbf{d}^{(0)}(x_2)h^1(x_1) +[h^1(x_1),h^2(x_2,x_3)] +{d}^{(0)}(x_3)\eta^2(x_1,x_2))$

$+\mathbf{d}^1h^1([x_1,x_2],x_3)+\mathbf{d}^{(0)}(x_3)h^1([x_1,x_2])+\eta^2([x_1,x_2],x_3)$

$+\mathbf{d}^1h^1(x_2,[x_1,x_3])-\mathbf{d}^{(0)}(x_2)h^1([x_1,x_3])+\eta^2(x_2,[x_1,x_3])$

$-\mathbf{d}^1h^1(x_1,[x_2,x_3])+\mathbf{d}^{(0)}(x_1)h^1([x_2,x_3])+\eta^2(x_1,[x_2,x_3])$

$=\mathbf{d}^2 h^2(x_1,x_2,x_3) + {d}^{(0)}(x_1)\eta^2(x_2,x_3))-{d}^{(0)}(x_2) \eta^2(x_1,x_3))+{d}^{(0)}(x_3)\eta^2(x_1,x_2))$

$+\mathbf{d}^1h^1([x_1,x_2],x_3)+\mathbf{d}^{(0)}(x_3)h^1([x_1,x_2])-\mathbf{d}^{(0)}([x_1,x_2])h^1(x_3)$

$+\mathbf{d}^1h^1(x_2,[x_1,x_3])-\mathbf{d}^{(0)}(x_2)h^1([x_1,x_3])+\mathbf{d}^{(0)}([x_1,x_3])h^1(x_2)$

$-\mathbf{d}^1h^1(x_1,[x_2,x_3])+\mathbf{d}^{(0)}(x_1)h^1([x_2,x_3])-\mathbf{d}^{(0)}([x_2,x_3])h^1(x_1)$

$=\mathbf{d}^2 h^2(x_1,x_2,x_3) + {d}^2\eta^2(x_1,x_2,x_3)$

$-h^1([[x_1,x_2],x_3])-h^1([x_2,[x_1,x_3]])+h^1([x_1,[x_2,x_3]])$

$=\mathbf{d}^2 h^2(x_1,x_2,x_3)+ {d}^2\eta^2(x_1,x_2,x_3)$

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proposition

$\mathbf{d}^2 h^2\in Z^3(\mathfrak{g},\mathcal{Z}(\mathfrak{h}))$

[edit]

proof

$d^3 \mathbf{d}^2 h^2(x_1,x_2,x_3,x_4)=d^{(0)}(x_1)\mathbf{d}^2 h^2(x_2,x_3,x_4) -d^{(0)}(x_2)\mathbf{d}^2 h^2(x_1,x_3,x_4)+d^{(0)}(x_3)\mathbf{d}^2 h^2(x_1,x_2,x_4)-d^{(0)}(x_4)\mathbf{d}^2 h^2(x_1,x_2,x_3)$

$-\mathbf{d}^2 h^2([x_1,x_2],x_3,x_4)-\mathbf{d}^2 h^2(x_2,[x_1,x_3],x_4)-\mathbf{d}^2 h^2(x_2,x_3,[x_1,x_4])+\mathbf{d}^2 h^2(x_1,[x_2,x_3],x_4)+\mathbf{d}^2 h^2(x_1,x_3,[x_2,x_4])-\mathbf{d}^2 h^2(x_1,x_2,[x_3,x_4])$

$=d^{(0)}(x_1)(\mathbf{d}^{(0)}(x_2)h^2(x_3,x_4)-\mathbf{d}^{(0)}(x_3)h^2(x_2,x_4)+\mathbf{d}^{(0)}(x_4)h^2(x_2,x_3) -h^2([x_2,x_3],x_4)-h^2(x_3,[x_2,x_4])+h^2(x_2,[x_3,x_4]))$

$-d^{(0)}(x_2)(\mathbf{d}^{(0)}(x_1)h^2(x_3,x_4)-\mathbf{d}^{(0)}(x_3)h^2(x_1,x_4)+\mathbf{d}^{(0)}(x_4)h^2(x_1,x_3) -h^2([x_1,x_3],x_4)-h^2(x_3,[x_1,x_4])+h^2(x_1,[x_3,x_4]))$

$+d^{(0)}(x_3)(\mathbf{d}^{(0)}(x_1)h^2(x_2,x_4)-\mathbf{d}^{(0)}(x_2)h^2(x_1,x_4)+\mathbf{d}^{(0)}(x_4)h^2(x_1,x_2) -h^2([x_1,x_2],x_4)-h^2(x_2,[x_1,x_4])+h^2(x_1,[x_2,x_4]))$

$-d^{(0)}(x_4)(\mathbf{d}^{(0)}(x_1)h^2(x_2,x_3)-\mathbf{d}^{(0)}(x_2)h^2(x_1,x_3)+\mathbf{d}^{(0)}(x_3)h^2(x_1,x_2) -h^2([x_1,x_2],x_3)-h^2(x_2,[x_1,x_3])+h^2(x_1,[x_2,x_3]))$

$-\mathbf{d}^{(0)}([x_1,x_2])h^2(x_3,x_4)+\mathbf{d}^{(0)}(x_3)h^2([x_1,x_2],x_4) -\mathbf{d}^{(0)}(x_4)h^2([x_1,x_2],x_3)$

$+h^2([[x_1,x_2],x_3],x_4)+h^2(x_3,[[x_1,x_2],x_4])-h^2([x_1,x_2],[x_3,x_4])$

$-\mathbf{d}^{(0)}(x_2)h^2([x_1,x_3],x_4)+\mathbf{d}^{(0)}([x_1,x_3])h^2(x_2,x_4) -\mathbf{d}^{(0)}(x_4)h^2(x_2,[x_1,x_3])$

$+h^2([x_2,[x_1,x_3]],x_4)+h^2([x_1,x_3],[x_2,x_4])-h^2(x_2,[[x_1,x_3],x_4])$

$-\mathbf{d}^{(0)}(x_2)h^2(x_3,[x_1,x_4])+\mathbf{d}^{(0)}(x_3)h^2(x_2,[x_1,x_4]) -\mathbf{d}^{(0)}([x_1,x_4])h^2(x_2,x_3)$

$+h^2([x_2,x_3],[x_1,x_4])+h^2(x_3,[x_2,[x_1,x_4]])-h^2(x_2,[x_3,[x_1,x_4]])$

$+\mathbf{d}^{(0)}(x_1)h^2([x_2,x_3],x_4)-\mathbf{d}^{(0)}([x_2,x_3])h^2(x_1,x_4) +\mathbf{d}^{(0)}(x_4)h^2(x_1,[x_2,x_3])$

$-h^2([x_1,[x_2,x_3]],x_4)-h^2([x_2,x_3],[x_1,x_4])+h^2(x_1,[[x_2,x_3],x_4])$

$+\mathbf{d}^{(0)}(x_1)h^2(x_3,[x_2,x_4])-\mathbf{d}^{(0)}(x_3) h^2(x_1,[x_2,x_4]) +\mathbf{d}^{(0)}([x_2,x_4])h^2(x_1,x_3)$

$-h^2([x_1,x_3],[x_2,x_4])-h^2(x_3,[x_1,[x_2,x_4]])+h^2(x_1,[x_3,[x_2,x_4]])$

$-\mathbf{d}^{(0)}(x_1)h^2(x_2,[x_3,x_4])+\mathbf{d}^{(0)}(x_2)h^2(x_1,[x_3,x_4]) -\mathbf{d}^{(0)}([x_3,x_4])h^2(x_1,x_2)$

$+h^2([x_1,x_2],[x_3,x_4])+h^2(x_2,[x_1,[x_3,x_4]])-h^2(x_1,[x_2,[x_3,x_4]])$

$=-\mathbf{d}^{(0)}(x_4)\mathbf{d}^{(0)}(x_3)h^2(x_1,x_2)+\mathbf{d}^{(0)}(x_3) \mathbf{d}^{(0)}(x_4)h^2(x_1,x_2)-\mathbf{d}^{(0)}([x_3,x_4])h^2(x_1,x_2)$

$-\mathbf{d}^{(0)}(x_2)\mathbf{d}^{(0)}(x_4)h^2(x_1,x_3)+\mathbf{d}^{(0)}(x_4)\mathbf{d}^{(0)}(x_2)h^2(x_1,x_3) +\mathbf{d}^{(0)}([x_2,x_4])h^2(x_1,x_3)$

$-\mathbf{d}^{(0)}(x_3) \mathbf{d}^{(0)}(x_2)h^2(x_1,x_4)+\mathbf{d}^{(0)}(x_2)\mathbf{d}^{(0)}(x_3)h^2(x_1,x_4) -\mathbf{d}^{(0)}([x_2,x_3])h^2(x_1,x_4)$

$+\mathbf{d}^{(0)}(x_1)\mathbf{d}^{(0)}(x_4)h^2(x_2,x_3)-\mathbf{d}^{(0)}(x_4)\mathbf{d}^{(0)}(x_1)h^2(x_2,x_3) -\mathbf{d}^{(0)}([x_1,x_4])h^2(x_2,x_3)$

$-\mathbf{d}^{(0)}(x_1)\mathbf{d}^{(0)}(x_3)h^2(x_2,x_4)\mathbf{d}^{(0)}(x_3) \mathbf{d}^{(0)}(x_1)h^2(x_2,x_4)+\mathbf{d}^{(0)}([x_1,x_3])h^2(x_2,x_4)$

$+\mathbf{d}^{(0)}(x_1)\mathbf{d}^{(0)}(x_2)h^2(x_3,x_4) -\mathbf{d}^{(0)}(x_2)\mathbf{d}^{(0)}(x_1)h^2(x_3,x_4) -\mathbf{d}^{(0)}([x_1,x_2])h^2(x_3,x_4)$

$=-\mathrm{ad}(h^2(x_3,x_4))h^2(x_1,x_2)+\mathrm{ad}(h^2(x_2,x_4))h^2(x_1,x_3)-\mathrm{ad}(h^2(x_2,x_3))h^2(x_1,x_4)$

$-\mathrm{ad}(h^2(x_1,x_4))h^2(x_2,x_3)+\mathrm{ad}(h^2(x_1,x_3))h^2(x_2,x_4)-\mathrm{ad}(h^2(x_1,x_2))h^2(x_3,x_4)$

$=0$

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corollary

The choice of $d^{(0)}\in\mathfrak{out}(\mathfrak{h})$ determines a cohomology class $[\mathbf{d}^2 h^2]\in H^3(\mathfrak{g},\mathcal{Z}(\mathfrak{h}))$ .

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proposition

One can see $\mathfrak{h}$ as a semidirect sum $\mathcal{Z}(\mathfrak{h})\oplus_{[\nu]}\mathfrak{ad}(\mathfrak{h})$ , where $[\nu]\in H^2(\mathfrak{ad}(\mathfrak{h}),\mathcal{Z}(\mathfrak{h}))^\mathfrak{g}$ .

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question

Why $\mathfrak{g}$ invariant?

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definition

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

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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 $a^2\in Z^2(\mathfrak{g},\mathfrak{a})$ 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 Leibniz 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.