User:Jan A. Sanders/An Introduction to Leibniz Algebra Cohomology/Lecture 4

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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

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

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