lecture 2

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< An introduction to Leibniz algebra cohomology

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Author: Dr. Jan A. Sanders, Vrije Universiteit Amsterdam

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On to the third lecture

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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

$\iota \qquad \quad \ p$

(1)

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

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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

(2)

$k^2(x,y)=\sigma([x,y]_{\mathfrak{g}})-[\sigma(x),\sigma(y)]_{\mathfrak{k}}.$

One has

(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

(4)

$\iota(a^2(x,y))=k^2(x,y).$

Now define $d_\pm^{(0)}:\mathfrak{g}\rightarrow End(\mathfrak{a})$ by

(5)

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

(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.

$\iota d^2 a^2(x,y,z)= \iota d_+^{(0)}(x)a^2(y,z) -\iota d_+^{(0)}(y)a^2(x,z) +\iota d_-^{(0)}(z)a^2(x,y)-\iota a^2([x,y],z)-\iota a^2(y,[x,z])+\iota a^2(x,[y,z])$

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