# User:Jan A. Sanders/An Introduction to Leibniz Algebra Cohomology/Lecture 2

## Central extensions

A module $$\mathfrak{g}$$ is projective if for every surjective morphism of modules $$\alpha:\mathfrak{k}\rightarrow\mathfrak{g}\rightarrow 0$$ and every morphism $$\beta:\mathfrak{h}\rightarrow\mathfrak{g}\ ,$$ there exists a morphism $$\gamma:\mathfrak{h}\rightarrow\mathfrak{k}$$ such that $$\beta=\alpha\gamma\ .$$ In particular, if $$\mathfrak{g}$$ is projective and $$\alpha$$ surjective, there exists a $$\gamma:\mathfrak{g}\rightarrow\mathfrak{k}$$ such that $$id_{\mathfrak{g}}=\alpha\gamma\ ,$$ that is, $$\gamma$$ is a section of $$\alpha\ .$$

Consider the exact sequence of Leibniz algebras with Leibniz algebra morphisms

$\iota \qquad \quad \ p$

$\tag{1} 0 \quad \rightarrow \quad \mathfrak{a} \quad \rightarrow \quad \mathfrak{k} \quad \rightarrow \quad \mathfrak{g} \quad \rightarrow 0 \ .$

### question

If $$\mathfrak{g}$$ is projective as an $$R$$-module, is there a section of $$p$$ which is also a Leibniz algebra morphism, that is, is it also projective as a Leibniz algebra?

To answer this question, first assume $$\mathfrak{a}$$ to be abelian, that is, $$[x,y]=0$$ for all $$x,y\in\mathfrak{a}\ .$$ Let $$\sigma$$ be a section, but not necessarily a Leibniz algebra morphism (Exists, since $$\mathfrak{g}$$ is projective). Define $$k^2\in C^2(\mathfrak{g},\mathfrak{k})$$ by $\tag{2} k^2(x,y)=\sigma([x,y]_{\mathfrak{g}})-[\sigma(x),\sigma(y)]_{\mathfrak{k}}.$

One has $\tag{3} p k^2(x,y)=p\sigma([x,y]_{\mathfrak{g}})-p[\sigma(x),\sigma(y)]_{\mathfrak{k}}=[x,y]_{\mathfrak{g}}-[p\sigma(x),p\sigma(y)]_{\mathfrak{g}}=0.$

Since $$k^2(x,y)\in\ker p=\mathrm{im\ }\iota\ ,$$ one can define $$a^2\in C^2(\mathfrak{g},\mathfrak{a})$$ (or $$a^2\in C_{\wedge}^2(\mathfrak{g},\mathfrak{a})$$ in the Lie algebra case, where we denote by $$C_{\wedge}^n(\mathfrak{g},\mathfrak{a})$$ the antisymmetric forms in $$C^n(\mathfrak{g},\mathfrak{a})$$) by $\tag{4} \iota(a^2(x,y))=k^2(x,y).$

Now define $$d_\pm^{(0)}:\mathfrak{g}\rightarrow End(\mathfrak{a})$$ by $\tag{5} \iota(d_+^{(0)}(x)a)=[\sigma(x),\iota(a)]_{\mathfrak{k}}$

$\tag{6} \iota(d_-^{(0)}(x)a)=-[\iota(a),\sigma(x)]_{\mathfrak{k}}$

This is well-defined since, for instance, $$p[\sigma(x),\iota(a)]_{\mathfrak{k}}=[p\sigma(x),p\iota(a)]_{\mathfrak{g}}=0\ .$$ Observe that $$[\iota(d_{+}^{(0)}(x)a),y]=[\iota(d_{-}^{(0)}(x)a),y]\ .$$

Moreover, $$d_\pm^{(0)}$$ is a representation, since $\iota(d_+^{(0)}([x,y])a) = [\sigma([x,y]),\iota(a)]_{\mathfrak{k}}\ :$

$= [k^2(x,y),\iota(a)]_{\mathfrak{k}} +[[\sigma(x),\sigma(y)],\iota(a)]_{\mathfrak{k}}\ :$
$= [\iota a^2(x,y),\iota(a)]_{\mathfrak{k}} +[\sigma(x),[\sigma(y),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}-[\sigma(y),[\sigma(x),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}\ :$
$= \iota[a^2(x,y),a]_{\mathfrak{a}}+[\sigma(x),\iota(d_+^{(0)}(y)a)]_{\mathfrak{k}}-[\sigma(y),\iota(d_{+}^{(0)}(x)a)]_{\mathfrak{k}}\ :$
$= \iota(d_+^{(0)}(x)d_+^{(0)}(y)a) -\iota(d_+^{(0)}(y)d_{+}^{(0)}(x)a)$

or $$d_+^{(0)}([x,y])=d_+^{(0)}(x)d_+^{(0)}(y) -d_+^{(0)}(y)d_{+}^{(0)}(x)\ ,$$ and (the following checks can be skipped in the Lie algebra case) $\iota(d_-^{(0)}([x,y])a) = -[\iota(a),\sigma([x,y])]_{\mathfrak{k}}\ :$

$= -[\iota(a),k^2(x,y)]_{\mathfrak{k}} -[\iota(a),[\sigma(x),\sigma(y)]_{\mathfrak{k}}]_{\mathfrak{k}}\ :$
$= -[\iota(a),\iota(a^2(x,y))]_{\mathfrak{k}}-[\sigma(x),[\iota(a),\sigma(y)]_{\mathfrak{k}}]_{\mathfrak{k}} +[[\sigma(x),\iota(a)]_{\mathfrak{k}},\sigma(y)]_{\mathfrak{k}}\ :$
$= -\iota([a,a^2(x,y)]_{\mathfrak{a}})+[\sigma(x),\iota(d_-^{(0)}(y)a)]_{\mathfrak{k}} +[\iota(d_{\pm}^{(0)}(x)a),\sigma(y)]_{\mathfrak{k}} \ :$
$= \iota(d_+^{(0)}(x)d_-^{(0)}(y)a-d_-^{(0)}(y)d_{\pm}^{(0)}(x)a)\ ,$

or $$d_-^{(0)}([x,y])=d_+^{(0)}(x)d_-^{(0)}(y)-d_-^{(0)}(y)d_{\pm}^{(0)}(x)\ .$$

Suppose now there exists a $$k^1\in C^1(\mathfrak{g},\mathfrak{k})$$ such that $$\sigma+k^1$$ is a Leibniz algebra homomorphism and a section of $$p\ .$$ Then $x=p(\sigma+k^1)(x)=p\sigma(x)+p k^1(x)=x+p k^1(x),$ implying that $$k^1 (x)\in\ker(p)=\mathrm{im\ }\iota\ .$$ Define $$a^1\in C^1(\mathfrak{g},\mathfrak{a})$$ by $$\iota a^1(x)=k^1 (x)\ .$$ Then (by assumption!) $0 =(\sigma+k^1)([x,y])-[(\sigma+k^1)(x),(\sigma+k^1)(y)]\ :$ $=\sigma([x,y])+k^1([x,y])-[\sigma(x),\sigma(y)]-[k^1(x),\sigma(y)]-[\sigma(x),k^1(y)]-[k^1(x),k^1(y)]\ :$

$= k^2(x,y) +k^1([x,y])-[\iota a^1(x),\sigma(y)]-[\sigma(x),\iota a^1(y)]-[\iota a^1(x),\iota a^1(y)]\ :$ $= \iota(a^2(x,y)-(d_+^{(0)}(x)a^1(y)-d_-^{(0)}(y)a^1(x)-a^1([x,y])-[a^1(x),a^1(y)])\ :$ $= \iota((a^2-d^1 a^1)(x,y)).$ This implies that the existence of such a $$k^1$$ is equivalent to $$a^2\in\mathrm{im\ }d^1\ .$$ On the other hand, if it would turn out that $$d^2 a^2\neq 0\ ,$$ this would be a definite obstruction of the existence of such a $$k^1\ .$$ We can, however, rule out the last possibility. $\iota d^2 a^2(x,y,z)= \iota d_+^{(0)}(x)a^2(y,z) -\iota d_+^{(0)}(y)a^2(x,z) +\iota d_-^{(0)}(z)a^2(x,y)-\iota a^2([x,y],z)-\iota a^2(y,[x,z])+\iota a^2(x,[y,z])\ :$

$=[\sigma(x),k^2(y,z)]-[\sigma(y),k^2(x,z)]-[k^2(x,y),\sigma(z)]-k^2([x,y],z)-k^2(y,[x,z])+k^2(x,[y,z])\ :$
$=[\sigma(x),\sigma([y,z])-[\sigma(y),\sigma(z)]]-[\sigma(y),\sigma([x,z])-[\sigma(x),\sigma(z)]]-[\sigma([x,y])-[\sigma(x),\sigma(y)],\sigma(z)]\ :$
$-\sigma([[x,y],z])+[\sigma([x,y]),\sigma(z)]-\sigma([y,[x,z]])+[\sigma(y),\sigma([x,z]]+\sigma([x,[y,z]])-[\sigma(x),\sigma([y,z])]\ :$
$= -[\sigma(x),[\sigma(y),\sigma(z)]]+[\sigma(y),[\sigma(x),\sigma(z)]] +[[\sigma(x),\sigma(y)],\sigma(z)]\ :$
$-\sigma([[x,y],z]+[y,[x,z]]-[x,[y,z]]) \ :$
$=0.$

It follows that $$d^2 a^2=0\ .$$