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 \(d^2 a^2=0\ .\)
