User:Jan A. Sanders/An Introduction to Leibniz Algebra Cohomology/Lecture 4
Contents |
Construction of an extension
We now turn the question around. What is the situation if we have Leibniz algebras \({\mathfrak{g}}\) (projective) and \({\mathfrak{a}}\) (abelian), and \([a^2]\in H^2({\mathfrak{g}},{\mathfrak{a}})\) (or in \(\in H_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})\) in the Lie algebra case)? If a Leibniz section (that is, a section which is a Leibniz 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 \([a^2]\in H^2({\mathfrak{g}},{\mathfrak{a}})\) is not zero? To see how one should define a Leibniz 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 \[\phi([\sigma(x)+\iota(a_1),\sigma(y)+\iota(a_2)]=\phi([\sigma(x),\sigma(y)]+[\iota(a_1),\sigma(y)]+[\sigma(x),\iota(a_2)])\] \[=\phi(\sigma([x,y])-\iota(a^2(x,y))+[\iota(a_1),\sigma(y)]+[\sigma(x),\iota(a_2)])\] \[=({d_+^{(0)}}(x)a_2-{d_-^{(0)}}(y)a_1-a^2(x,y),[x,y])\]
definition
One now defines for an arbitrary representation \({d_\pm^{(0)}}\) and \(a^2\in C^2({\mathfrak{g}},{\mathfrak{a}})\ ,\) \[[(a_1,x),(a_2,y)]=\] \[({d_+^{(0)}}(x)a_2-{d_-^{(0)}}(y)a_1-a^2(x,y),[x,y])\]
lemma
The new bracket \([\cdot,\cdot]_{a^2} \) obeys the Jacobi identity as long as \(a^2\in Z^2({\mathfrak{g}},{\mathfrak{a}})\ .\)
proof
\[[[(a_1,x),(a_2,y)],(a_3.z)]-[(a_1,x),[(a_2,x),(a_3,z)]]+[(a_2,y),[(a_1,x),(a_3,y)]]\] \[=[({d_+^{(0)}}(x)a_2-{d_-^{(0)}}(y)a_1-a^2(x,y),[x,y]),(a_3,z)]\] \[-[(a_1,x),({d_+^{(0)}}(y)a_3-{d_-^{(0)}}(z)a_2-a^2(y,z),[y,z])]\] \[+[(a_2,y),({d_+^{(0)}}(x)a_3-{d_-^{(0)}}(z)a_1-a^2(x,z),[x,z])]\] \[=({d_+^{(0)}}([x,y])a_3-{d_-^{(0)}}(z){d_+^{(0)}}(x)a_2+{d_-^{(0)}}(z){d_-^{(0)}}(y)a_1+{d_-^{(0)}}(z)a^2(x,y)-a^2([x,y],z),[[x,y],z])\] \[-({d_+^{(0)}}(x){d_+^{(0)}}(y)a_3-{d_+^{(0)}}(x){d_-^{(0)}}(z)a_2-{d_+^{(0)}}(x)a^2(y,z)-{d_-^{(0)}}([y,z])a_1-a^2(x,[y,z],[x,[y,z]])\] \[+({d_+^{(0)}}(y){d_+^{(0)}}(x)a_3-{d_+^{(0)}}(y){d_-^{(0)}}(z)a_1-{d_+^{(0)}}(y)a^2(x,z)-{d_-^{(0)}}([x,z])a_2-a^2(y,[x,z],[y,[x,z]])\] \[=({d_-^{(0)}}([y,z])a_1-{d_+^{(0)}}(y){d_-^{(0)}}(z)a_1+{d_-^{(0)}}(z){d_-^{(0)}}(y)a_1\] \[-{d_-^{(0)}}([x,z])a_2-{d_-^{(0)}}(z){d_+^{(0)}}(x)a_2+{d_+^{(0)}}(x){d_-^{(0)}}(z)a_2\] \[+{d_+^{(0)}}([x,y])a_3-{d_+^{(0)}}(x){d_+^{(0)}}(y)a_3 +{d_+^{(0)}}(y){d_+^{(0)}}(x)a_3\] \[+{d_+^{(0)}}(x)a^2(y,z)-{d_+^{(0)}}(y)a^2(x,z)+{d_-^{(0)}}(z)a^2(x,y)\] \[-a^2([x,y],z)-a^2(y,[x,z])+a^2(x,[y,z]),[[x,y],z] -[x,[y,z]] +[y,[x,z]])\] \[=(d^{2}a^2(x,y,z),0)\] \[=(0,0).\]\(\square\)
antisymmetry
In the Lie algebra case one needs \(a^2\in Z_{\wedge}^2({\mathfrak{g}},{\mathfrak{a}})\) in order to make the bracket antisymmetric.
definition
We denote the new Leibniz algebra by \({\mathfrak{g}}{\oplus}_{a^2} {\mathfrak{a}}\ ,\) the semidirect product of \({\mathfrak{g}}\) and \({\mathfrak{a}}\) induced by \(a^2 \in Z^2({\mathfrak{g}},{\mathfrak{a}})\ .\)
theorem
Let \(a^2\) be as defined in lecture two. Then
- \[ \mathfrak{k}\simeq \mathfrak{a}\oplus_{a^2} \mathfrak{g}\ .\]
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. \[\tilde{\sigma}(x)=(a^1(x),x),\quad a^1\in C^1(\mathfrak{g},\mathfrak{a})\] Then \[\iota(\tilde{d}_+^{(0)}(x)a=[\tilde{\sigma}(x),\iota(a)]\ :\] \[=[(a^1(x),x),(a,0)]\ :\] \[=(d_+^{(0)}(x)a,0)\] and \[\iota(\tilde{d}_-^{(0)}(x)a=-[\iota(a),\tilde{\sigma}(x)]\ :\] \[=-[((a,0),a^1(x),x)]\ :\] \[=(d_-^{(0)}(x)a,0)\] It follows that \(\tilde{d}_\pm^{(0)}=d_\pm^{(0)}\ ,\) which implies that the induced coboundary operators will be the same. Now \[\iota(\tilde{a}^2(x_1,x_2))=\tilde{\sigma}([x_1,x_2])-[\tilde{\sigma}(x_1),\tilde{\sigma}(x_2)]\ :\] \[=(a^1([x_1,x_2]),[x_1,x_2])-[(a^1(x_1),x_1),(a^1(x_2),x_2)]\ :\] \[=(a^1([x_1,x_2]),[x_1,x_2])-(d_+^{(0)}(x_1)a^1(x_2)-d_-^{(0)}(x_2)a^1(x_1)-a^2(x_1,x_2),[x_1,x_2])\ :\] \[=(a^2(x_1,x_2)-d^1 a^1(x_1,x_2),0)\] and this implies \[\tilde{a}^2=a^2-d^1 a^1\ .\] Furthermore
- \[\mathfrak{k}\simeq \mathfrak{a}\oplus_{a^2} \mathfrak{g}\simeq \mathfrak{a}\oplus_{\tilde{a}^2} \mathfrak{g}\]
and if \(a^1\in Z^1(\mathfrak{g},\mathfrak{a})\) one has
- \[\mathfrak{a}\oplus_{a^2} \mathfrak{g}= \mathfrak{a}\oplus_{\tilde{a}^2} \mathfrak{g}\]
theorem
To the short exact sequence of Leibniz algebras \[0\rightarrow\mathfrak{a}\rightarrow\mathfrak{k}\rightarrow\mathfrak{g}\rightarrow 0\] is associated an \[[a^2]\in H^2(\mathfrak{g},\mathfrak{a})\] such that for any \(a^2\in[a^2]\) one has
- \[\mathfrak{k}\simeq \mathfrak{a}\oplus_{a^2} \mathfrak{g}\]
