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

## 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

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