User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 2

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    Central extensions

    The definition of projective is important when \(R\) is not \(\mathbb{R}\ ,\) \(\mathbb{C}\) or in general a field of characteristic zero, or when the dimension of the Lie algebra is not finite, otherwise it can be skipped.

    definition - projective

    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 \(\alpha\ .\)

    When \(\mathfrak{g}\) is free, that is, \(\mathfrak{g}\) has an \(R\)-basis \(\{g_\iota\}_{\iota\in I}\ ,\) one can find for each \(g_\iota\) a preimage \(k_\iota\) under \(\alpha\ .\)

    If \(\beta(h)=\sum_{\iota\in I} b_\iota g_\iota, b_\iota\in R\ ,\) (finite sum, since \(\{g_\iota\}_{\iota\in I}\) is a basis) one defines \(\gamma(h)=\sum_{\iota\in I} b_\iota k_\iota\ .\)

    When \(\mathrm{rank}_R \mathfrak{g}=\#I<\infty\) this can be done explicitly.

    If \(R\) is a field with characteristic zero, it suffices to require \(\mathrm{dim}_R \mathfrak{g}<\infty\) in order to conclude to the existence of sections.

    Consider the exact sequence of Lie algebras with Lie algebra morphisms

    \[ \iota \qquad \quad \ p\]

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

    problem statement

    If \(\mathfrak{g}\) is projective as an \(R\)-module, is there a section of \(p\) which is also a Lie algebra morphism, that is, is it also projective as a Lie 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 \(\sigma_1\) be a section, but not necessarily a Lie algebra morphism (Exists, since \(\mathfrak{g}\) is projective).

    Define \(\sigma_2\in C_{\wedge}^2(\mathfrak{g},\mathfrak{k})\ ,\) (antisymmetric 2-forms) by \[\tag{2} \sigma_2(x,y)=[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}}-\sigma_1([x,y]_{\mathfrak{g}}).\]

    One has \[\tag{3} p \sigma_2(x,y)=p[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}}-p\sigma_1([x,y]_{\mathfrak{g}})=[p\sigma_1(x),p\sigma_1(y)]_{\mathfrak{g}}-[x,y]_{\mathfrak{g}}=0. \]

    Since \(\sigma_2(x,y)\in\ker p=\mathrm{im\ }\iota\ ,\) one can define \(a_2\in C_{\wedge}^2(\mathfrak{g},\mathfrak{a})\) by \[\tag{4} \iota(a_2(x,y))=\sigma_2(x,y). \]

    Now define \(d_1:\mathfrak{g}\rightarrow End(\mathfrak{a})\) by \[\tag{5} \iota(d_1(x)a)=[\sigma_1(x),\iota(a)]_{\mathfrak{k}}\]

    This is well-defined since, for instance, \(p[\sigma_1(x),\iota(a)]_{\mathfrak{k}}=[p\sigma_1(x),p\iota(a)]_{\mathfrak{g}}=0\ .\)

    Moreover, \(d_1\) is a representation, since \[\iota(d_1([x,y]_{\mathfrak{g}})a) = [\sigma_1([x,y]),\iota(a)]_{\mathfrak{k}}\ :\]

    \[ = -[\sigma_2(x,y),\iota(a)]_{\mathfrak{k}} +[[\sigma_1(x),\sigma_1(y)]_{\mathfrak{k}},\iota(a)]_{\mathfrak{k}}\ :\]
    \[ = -[\iota a_2(x,y),\iota(a)]_{\mathfrak{k}} +[\sigma_1(x),[\sigma_1(y),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}-[\sigma_1(y),[\sigma_1(x),\iota(a)]_{\mathfrak{k}}]_{\mathfrak{k}}\ :\]
    \[ =- \iota[a_2(x,y),a]_{\mathfrak{a}}+[\sigma_1(x),\iota(d_1(y)a)]_{\mathfrak{k}}-[\sigma_1(y),\iota(d_1(x)a)]_{\mathfrak{k}}\ :\]
    \[ = \iota(d_1(x)d^{(0)}(y)a) -\iota(d_1(y)d_1(x)a)\]

    or \(d_1([x,y]_{\mathfrak{g}})=[d_1(x),d_1(y)]_{\mathfrak{k}}\ .\)

    In the sequel we drop the subscript of the bracket.

    Suppose now there exists a \(\delta\sigma_1\in C^1(\mathfrak{g},\mathfrak{k})\) such that \(\sigma_1+\delta\sigma_1\) is a Lie algebra homomorphism and a section of \(p\ .\) Then \[ x=p(\sigma_1+\delta\sigma_1)(x)=p\sigma_1(x)+p \delta\sigma_1(x)=x+p \delta\sigma_1(x), \] implying that \(\delta\sigma_1 (x)\in\ker(p)=\mathrm{im\ }\iota\ .\) Define \(a_1\in C^1(\mathfrak{g},\mathfrak{a})\) by \(\iota a_1(x)=\delta\sigma_1 (x)\ .\) Then (by assumption!) \[ 0 =(\sigma_1+\delta\sigma_1)([x,y])-[(\sigma_1+\delta\sigma_1)(x),(\sigma_1+\delta\sigma_1)(y)]\ :\] \[ =\sigma_1([x,y])+\delta\sigma_1([x,y])-[\sigma_1(x),\sigma_1(y)]-[\delta\sigma_1(x),\sigma_1(y)]-[\sigma_1(x),\delta\sigma_1(y)]-[\delta\sigma_1(x),\delta\sigma_1(y)]\ :\]

    \[ = -\sigma_2(x,y) +\delta\sigma_1([x,y])-[\iota a_1(x),\sigma_1(y)]-[\sigma_1(x),\iota a_1(y)]-[\iota a_1(x),\iota a_1(y)]\ :\] \[ = -\iota(a_2(x,y)+(d_1(x)a_1(y)-d_1(y)a_1(x)-a_1([x,y])-[a_1(x),a_1(y)])\ :\] \[= -\iota((a_2+d^1 a_1)(x,y)).\] This implies that the existence of such a \(\delta\sigma_1\) 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 \(\delta\sigma_1\ .\)

    We can, however, rule out the last possibility. \[ \iota d^2 a_2(x,y,z)= \iota d_1(x)a_2(y,z) -\iota d_1(y)a_2(x,z) +\iota d_1(z)a_2(x,y)-\iota a_2([x,y],z)-\iota a_2(y,[x,z])+\iota a_2(x,[y,z])\ :\]

    \[=-[\sigma_1(x),\sigma_2(y,z)]-[\sigma_1(y),\sigma_2(x,z)]-[\sigma_2(x,y),\sigma_1(z)]-\sigma_2([x,y],z)-\sigma_2(y,[x,z])+\sigma_2(x,[y,z])\ :\]
    \[=[\sigma(x),\sigma([y,z])-[\sigma(y),\sigma(z)]]-[\sigma(y),\sigma([x,z])-[\sigma(x),\sigma(z)]]-[\sigma([x,y])-[\sigma(x),\sigma(y)],\sigma(z)]\ :\]
    \[ -\sigma_1([[x,y],z])+[\sigma_1([x,y]),\sigma_1(z)]-\sigma_1([y,[x,z]])+[\sigma_1(y),\sigma_1([x,z]]+\sigma_1([x,[y,z]])-[\sigma_1(x),\sigma_1([y,z])]\ :\]
    \[= -[\sigma(x),[\sigma(y),\sigma(z)]]+[\sigma(y),[\sigma(x),\sigma(z)]] +[[\sigma(x),\sigma(y)],\sigma(z)] -\sigma_1([[x,y],z]+[y,[x,z]]-[x,[y,z]]) \ :\]

    It follows that \(d^2 a_2=0\ .\)

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