User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 8

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    The trace and Killing form

    Let \( R\) be \(\mathbb{C}\) and \(\dim_\mathbb{C}\mathfrak{a}<\infty\ .\) Then define \(K_\mathfrak{a}\in C^2(\mathfrak{g},\mathbb{C})\) by \[ K_\mathfrak{a}(x,y)=\mathrm{tr}(d_1(x) d_1(y))\] In the case \(\mathfrak{a}=\mathfrak{g}\) and \(d_1=\mathrm{ad}\ ,\) this is called the Killing form. In general, one calls \(K_\mathfrak{a}\) the trace form.

    example - of a trace form

    Let \(\mathfrak{g}=\mathfrak{sl}_2\) and \(\mathfrak{a}=\R^2\ ,\) with the standard representation (see Lecture 1). Then \[ K_{\R^2}(M,M)=0, \quad K_{\R^2}(M,N)=1,\quad K_{\R^2}(M,H)=0,\] \[ K_{\R^2}(N,M)=1, \quad K_{\R^2}(N,N)=0,\quad K_{\R^2}(N,H)=0,\] \[ K_{\R^2}(H,M)=0, \quad K_{\R^2}(H,N)=0,\quad K_{\R^2}(H,H)=2.\]

    proposition - trace form symmetric

    \(K_\mathfrak{a}\) is symmetric.


    This follows from \(\mathrm{tr}(AB)=\mathrm{tr}(BA)\).\(\square\)

    proposition - trace form invariant

    \( K_\mathfrak{a} \) is \(\mathfrak{g}\)-invariant, that is, \(K_\mathfrak{a}\in C^2(\mathfrak{g},\mathbb{C})^\mathfrak{g}\ .\)


    Given the trivial action of \(\mathfrak{g}\) on \(\mathbb{C}\ ,\) one has \[ d_1^{2}(x)K_\mathfrak{a}(y,z)=-K_\mathfrak{a}([x,y],z)-K_\mathfrak{a}(y,[x,z])\ :\] \[=-\mathrm{tr}(d_1([x,y]) d_1(z))-\mathrm{tr}(d_1(y) d_1([x,z]))\ :\] \[=-\mathrm{tr}(d_1(x) d_1(y) d_1(z))+\mathrm{tr}(d_1(y) d_1(x) d_1(z)) -\mathrm{tr}(d_1(y) d_1(x)d_1(z))+\mathrm{tr}(d_1(y) d_1(z)d_1(x))\ :\] \[=0\]

    proposition - \(d^2 K_\mathfrak{a} \) antisymmetric

    \[d^2 K_\mathfrak{a}\in C_{\wedge}^3(\mathfrak{g},\mathbb{C})\]


    From the \(\mathfrak{g}\)-invariance it follows that \[d^2 K_\mathfrak{a}(x,y,z)=K_\mathfrak{a}(x,[y,z])\] Furthermore, \[K_\mathfrak{a}(x,[z,y])=-K_\mathfrak{a}(x,[y,z])\] and \[K_\mathfrak{a}(z,[x,y])=-K_\mathfrak{a}(z,[y,x])=K_\mathfrak{a}([y,z],x)=K_\mathfrak{a}(x,[y,z])\]\(\square\)

    corollary - nontrivial third cohomology

    Let \(\mathfrak{g}\) be a Lie algebra. Then \[[d^2 K_\mathfrak{a}]\in H_{\wedge}^3(\mathfrak{g},\mathbb{C})\] Observe that this class is not trivial, since \(K_\mathfrak{a}\) is symmetric, not antisymmetric.

    musical maps

    Let \(\mathfrak{g}^\star=C^1(\mathfrak{g},\mathbb{C})\) and define \( \flat: \mathfrak{g}\rightarrow \mathfrak{g}^\star\) by \[ \flat(x)(y)=K_\mathfrak{a}(x,y)\]


    \[ \flat\in Hom_\mathfrak{g}(\mathfrak{g},\mathfrak{g}^\star)\]


    \[ \flat([x,y])(z)=K_\mathfrak{a}([x,y],z)\ :\] \[=-K_\mathfrak{a}(y,[x,z])\ :\] \[=-\flat(y)([x,z])\ :\] \[=d_1^{1}(x)\flat(y)(z)\] or \( \flat([x,y])=d_1^{1}(x)\flat(y)\quad \square\ .\)

    Define \[ \sharp:\mathfrak{g}^\star\rightarrow \mathfrak{g}\] by \[ K_\mathfrak{a}(\sharp(c_1),y)=c_1(y)\] Then \[ K_\mathfrak{a}(x,y)=\flat(x)(y)=K_\mathfrak{a}(\sharp(\flat(x)),y)\ ,\] or \(x-\sharp(\flat(x))\in \ker K_\mathfrak{a}\ .\)


    \(\ker K_\mathfrak{a}\) is an ideal.


    Let \(y\in\ker K_\mathfrak{a}\ ,\) that is \(K_\mathfrak{a}(x,y)=0\) for all \(x\in\mathfrak{g}\ .\)

    Then it follows from the invariance of \(K_\mathfrak{a}\) that \[ K_\mathfrak{a}([y,x],z)+K_\mathfrak{a}(y,[x,z])=0\] and therefore \(K_\mathfrak{a}([y,x],z)=0\) for all \(z\in\mathfrak{g}\ .\)

    This shows that \([\mathfrak{g},\ker K_\mathfrak{a}]\subset \ker K_\mathfrak{a}\ .\)

    The statement that \([\ker K_\mathfrak{a},\mathfrak{g}]\subset \ker K_\mathfrak{a}\) follows by a symmetry argument.


    A Lie algebra \(\mathfrak{g}\) is called simple if \([\mathfrak{g},\mathfrak{g}]\neq 0\) and \(\mathfrak{g}\) contains no ideals besides \(0\) and itself.

    proposition - simple Lie algebra

    If \(\mathfrak{g}\) is simple, then \(\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\ .\)


    \([\mathfrak{g},\mathfrak{g}]\neq 0\) is an ideal, so it must equal \([\mathfrak{g},\mathfrak{g}]=\mathfrak{g}\ .\)


    If \( K_\mathfrak{a} \) is nonzero, and \(\mathfrak{g}\) is simple, then \(\flat\) is injective.


    Let \( x\in\ker\flat\ .\)

    Then \( 0=\flat(x)(y)=K_\mathfrak{a}(x,y)\) for all \(y\in\mathfrak{g}\ ,\) that is, \(x\in \ker K_\mathfrak{a}\ .\)

    But \( \ker K_\mathfrak{a}\) must be zero, so \(x=0\ .\)


    Let \(\mathfrak{h}\) be an ideal in \(\mathfrak{g}\ .\) Define \[\mathfrak{h}^\perp=\{x\in\mathfrak{g}|K_\mathfrak{g}(x,y)=0 \quad \forall y\in \mathfrak{h}\}\] Then \(\mathfrak{h}^\perp\) is an ideal in \(\mathfrak{g}\ .\)


    Let \(g\in\mathfrak{g}\ ,\) \( h\in\mathfrak{h}\) and \(k\in\mathfrak{h}^\perp\ .\) Then \[K_\mathfrak{g}([g,k],h)=-K_\mathfrak{g}(k,[g,h])=0\] This shows that \([\mathfrak{g},\mathfrak{h}^\perp]\subset \mathfrak{h}^\perp\) and similarly \([\mathfrak{h}^\perp,\mathfrak{g}]\subset \mathfrak{h}^\perp\ .\)

    definition - derived series, solvable

    One defines a series of ideals of \(\mathfrak{g}\ ,\) the derived series, as follows. \[\mathfrak{g}^{(0)}=\mathfrak{g}\] \[\mathfrak{g}^{(i+1)}=[\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}]\] If, for some \(n\in\mathbb{N}\ ,\) \(\mathfrak{g}^{(n)}=0\) then \(\mathfrak{g}\) is called solvable.

    well defined

    \( \mathfrak{g}^{(0)}\) is an ideal in \(\mathfrak{g}\ .\) Suppose that \(\mathfrak{g}^{(i)}\) is an ideal for \( i=0,\dots,n\ .\) Then \[ [\mathfrak{g},\mathfrak{g}^{(n+1)}]=[\mathfrak{g},[\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]]\ :\] \[ \subset [[\mathfrak{g},\mathfrak{g}^{(n)}],\mathfrak{g}^{(n)}]+[\mathfrak{g}^{(n)},[\mathfrak{g},\mathfrak{g}^{(n)}]]\ :\] \[\subset [\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]=\mathfrak{g}^{(n+1)}\] The inclusion \( [\mathfrak{g}^{(n+1)},\mathfrak{g}]\subset \mathfrak{g}^{(n+1)}\) follows in a similar way. By induction it follows that all the \( g^{(i)}\)'s are ideals in \(\mathfrak{g}\)


    For \(i\leq j \ ,\) \(\mathfrak{g}^{(j)}\) is an ideal in \(\mathfrak{g}^{(i)}\ .\)


    If \(\mathfrak{g}\) is solvable (that is, \(\mathfrak{g}^{(n)}=0\) for some \(n\)), then it contains an abelian ideal (namely \(\mathfrak{g}^{(n-1)}\)).

    proposition - solvable

    If \(\mathfrak{g}\) is solvable, then all its subalgebras and homomorphic images are.


    Let \( \mathfrak{h}\) be a subalgebra.

    Then \( \mathfrak{h}^{(0)}\subset\mathfrak{g}^{(0)}\ .\)

    Assume \(\mathfrak{h}^{(i)}\subset\mathfrak{g}^{(i)}\ .\)

    Then \[ \mathfrak{h}^{(i+1)}=[\mathfrak{h}^{(i)},\mathfrak{h}^{(i)}]\subset [\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}]=\mathfrak{g}^{(i+1)}\] and the statement is proved by induction.

    Similarly, let \(\phi:\mathfrak{g}\rightarrow \mathfrak{h}\) be surjective, and assume \(\phi:\mathfrak{g}^{(i)}\rightarrow \mathfrak{h}^{(i)}\) to be surjective.

    Then \[\phi(\mathfrak{g}^{(i+1)})=\phi([\mathfrak{g}^{(i)},\mathfrak{g}^{(i)}])=[\phi(\mathfrak{g}^{(i)}),\phi(\mathfrak{g}^{(i)})]= [\mathfrak{h}^{(i)},\mathfrak{h}^{(i)}]=\mathfrak{h}^{(i+1)}\]

    proposition - solvable quotient

    If \(\mathfrak{h}\) is a solvable ideal such that \(\mathfrak{g}/\mathfrak{h}\) is solvable, then \(\mathfrak{g}\) is solvable.


    Say \((\mathfrak{g}/\mathfrak{h})^{(n)}=0\ .\)

    Let \(\pi:\mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{h}\) be the canonical projection.

    Then \(\pi(\mathfrak{g}^{(n)})=(\mathfrak{g}/\mathfrak{h})^{(n)}=0\) or \(\mathfrak{g}^{(n)}\subset \mathfrak{h}\ .\)

    Since \(\mathfrak{h}^{(m)}=0\ ,\) \(\mathfrak{g}^{(n+m)}=(\mathfrak{g}^{(n)})^{(m)}\subset \mathfrak{h}^{(m)}=0\ ,\) implying the statement.


    If \(\mathfrak{h}, \mathfrak{k}\) are solvable ideals of \(\mathfrak{g}\ ,\) then so is \(\mathfrak{h}+\mathfrak{k}\ .\)


    One has \[ (\mathfrak{h}+ \mathfrak{k})/\mathfrak{k}\equiv \mathfrak{h}/(\mathfrak{h}\cap \mathfrak{k}) \] Since \(\mathfrak{h}\) is solvable, so is \(\mathfrak{h}/(\mathfrak{h}\cap \mathfrak{k}) \ .\) But this implies that \(\mathfrak{h}+\mathfrak{k}\ ,\) since \(\mathfrak{k}\) is solvable.

    proposition - radical

    If \(\dim \mathfrak{g}<\infty\ ,\) there exists a unique maximal solvable ideal in \(\mathfrak{g}\ ,\) the radical of \(\mathfrak{g}\ ,\) denoted by \( \mathrm{Rad\ }\mathfrak{g}\ .\)


    Let \(\mathfrak{s}\) be a maximal solvable ideal in \(\mathfrak{g}\ .\)

    Suppose \(\mathfrak{h}\) is another solvable ideal.

    Then \(\mathfrak{s}+\mathfrak{h}\supset \mathfrak{s}\) is solvable, and by the maximality, \(\mathfrak{s}+\mathfrak{h}= \mathfrak{s}\ ,\) that is, \(\mathfrak{h}\subset\mathfrak{s}\ .\)

    definition - semisimple

    A Lie algebra \(\mathfrak{g}\) is called semisimple if \(\mathrm{Rad\ }\mathfrak{g}=0\ .\)

    proposition - simple implies semisimple

    If \(\mathfrak{g}\) is simple, it is semisimple


    For a simple Leibniz algebra the derived series is stationary, that is, \(\mathfrak{g}^{(i)}=\mathfrak{g}\) for all \(i\in\mathbb{N}\ .\)

    The only other possible ideal is \(0\ ,\) so this must be \(\mathrm{Rad\ }\mathfrak{g}\ .\)


    \(\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\) is semisimple.


    Let \([\mathfrak{h}]\) be a nonzero solvable ideal in \(\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\ .\)

    Then \(\mathfrak{h}+\mathrm{Rad\ }\mathfrak{g}\) strictly contains \(\mathrm{Rad\ }\mathfrak{g}\ ,\) which is in contradiction with its maximality.

    Thus \(\mathfrak{h}\subset \mathrm{Rad\ }\mathfrak{g}\ ,\) that is, \([\mathfrak{h}]\) is the zero ideal in \(\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\ .\)


    If \(\ker K_\mathfrak{g}=0\ ,\) then \(\mathfrak{g}\) is semisimple.


    Let \(\mathfrak{h}\) be an abelian ideal of \(\mathfrak{g}\ .\) Take \(h\in\mathfrak{h}, g\in\mathfrak{g}\ .\) Then \( ad(h)ad(g)\) maps \( \mathfrak{g}\) to \(\mathfrak{h}\ .\) Thus \( (ad(h)ad(g))^2=0\ .\) This implies that \[ K_\mathfrak{g}(h,g)=\mathrm{tr}(ad(h)ad(g))=0\] In other words, \(\mathfrak{h}\subset\ker K_\mathfrak{g}=0\ .\) If there are no abelian ideals, then there are no solvable ideals besides \(0\ ,\) that is, \(\mathfrak{g}\) is semisimple.

    theorem - common eigenvector

    Let \(\mathfrak{g}\) be a solvable subalgebra of \(\mathfrak{gl}(\mathfrak{a})\ ,\) \(\dim\mathfrak{a}<\infty\ .\) If \(\mathfrak{a}\neq 0\ ,\) then \(\mathfrak{a}\) contains a common eigenvector for all endomorphisms in \(\mathfrak{g}\ .\)


    Induction on \(\dim\mathfrak{g}\ .\) Since \(\mathfrak{g}\) is solvable, it properly contains \(\mathfrak{g}^{(1)}=[\mathfrak{g},\mathfrak{g}]\ ,\) otherwise \(\mathfrak{g}^{(i)}=\mathfrak{g}\) for \( i\in\mathbb{N}\ .\)

    Since \(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]\) is abelian, subspaces are ideals.

    Take a subspace of codimension one.

    Then the inverse image \(\mathfrak{h}\) in \(\mathfrak{g}\) is an ideal of codimension one which includes \([\mathfrak{g},\mathfrak{g}]\ .\)

    \( \mathfrak{h}\) is solvable, and by the induction assumption there exists a vector \(a\in\mathfrak{a}\) such that \(a \) is an eigenvector for each \(h\in\mathfrak{h}\ ,\) that is, \[ h a=\lambda(h)a,\quad\lambda\in C^1(\mathfrak{h},\mathbb{C})\] (the exceptional case here is when \(\dim \mathfrak{h}=0\ .\)

    In that case, \(\mathfrak{g}\) onedimensional and abelian, so one takes an eigenvector of a generator of \(\mathfrak{g}\)). Let \[\mathcal{W}=\{a\in\mathfrak{a}|x a=\lambda(x)a \quad \forall x\in \mathfrak{h}\}\] Now for \(x\in\mathfrak{g}\) and \(y\in\mathfrak{h}\) one finds \[ y x w=x y w-[x,y] w=\lambda(y) x w-\lambda([x,y])w\ .\] If one can prove that \(\lambda([x,y])=0\) then \(\mathcal{W}\) is invariant under the action of \(\mathfrak{g}\ .\)

    Fix \(x\in \mathfrak{g}\ ,\) \(w\in\mathcal{W}\ .\)

    Let \( n>0 \) be the smallest integer such that \(w, xw, \dots, x^n w\) are linearly dependent.

    Let \(\mathcal{W}_0=0\) and \(\mathcal{W}_i\) be the subspace of \(\mathfrak{a}\) spanned by \(w, xw,\dots, x^{i-1} w\ .\)

    It follows that \(\dim\mathcal{W}_n=n\) and \(W_{n+i}=W_n, i\geq 0\ .\)

    Each \(\mathcal{W}_i\) is invariant under \(y\in\mathfrak{h}\ .\)

    The matrix of \(y\) is upper triangular with eigenvalue \(\lambda(y)\) on the diagonal.

    This implies \(\mathrm{tr}_{\mathcal{W}_i}(y)=i\lambda(y)\ .\)

    Since \([x,y]\in\mathfrak{h}\ ,\) one also has \[\mathrm{tr}_{\mathcal{W}_n}([x,y])=i\lambda([x,y])\] Both \(x\) and \(y\) leave \(\mathcal{W}_n\) invariant, so the trace of \([x,y]\) must be zero.

    Thus \(n\lambda([x,y])=0\ .\)

    This shows that \(\mathcal{W}\) is invariant under the action of \(\mathfrak{g}\ .\)

    Write \(\mathfrak{g}=\mathfrak{h}+\mathbb{C} z\ .\) Let \(w_0 \in\mathcal{W}\) be an eigenvector of \(z\) (acting on \(\mathcal{W}\)).

    Then \(w_0\) is a common eigenvector of \(\mathfrak{g}\ .\)

    definition - flag

    Let \(\mathfrak{a}\) be a finite dimensional vectorspace (\(\dim\mathfrak{a}=n\)). A flag is a chain of subspaces \[0=\mathfrak{a}_0\subset\mathfrak{a}_1\subset\dots\subset\mathfrak{a}_n=\mathfrak{a},\quad \dim\mathfrak{a}_i=i\] If \(x\in\mathrm{End}(\mathfrak{a})\ ,\) one says that \( x\) leaves the flag invariant if \(x \mathfrak{a}_i\subset \mathfrak{a}_i\) for \(i=1,\dots,n\ .\)

    theorem (Lie)

    Let \(\mathfrak{g}\) be a solvable subalgebra of \(\mathfrak{gl}(\mathfrak{a}), \dim\mathfrak{a}=n<\infty\ .\) Then \( \mathfrak{g}\) leaves a flag in \(\mathfrak{a}\) invariant.


    It follows from the proof above that there exists a codimension one \( \mathfrak{g}\)-invariant subspace.

    Let that be \(\mathfrak{a}_{n-1}\ .\)

    Repeat the argument starting with \(\mathfrak{a}_{n-1}\) instead of \(\mathfrak{a}_{n}\) and use induction.

    lemma - flag of ideals

    Let \( \mathfrak{g}\) be solvable. Then there exists a flag of ideals \[ 0=\mathfrak{g}_0\subset\mathfrak{g}_1\subset\dots\subset\mathfrak{g}_n=\mathfrak{g},\quad \dim\mathfrak{g}_i=i\]


    Let \(d_1:\mathfrak{g}\rightarrow \mathfrak{gl}(\mathfrak{a})\) be a finite dimensional representation of \(\mathfrak{g}\ .\)

    Then \(d_1(\mathfrak{g})\) is solvable, and stabilizes a flag in \(\mathfrak{a}\ .\)

    Take \(\mathfrak{a}=\mathfrak{g}\) and \(d_1=\mathrm{ad}\ ,\) then the \(\mathfrak{g}_i\) are ideals (since they are \(\mathfrak{g}\)-invariant) and they obey the flag condition.


    Let \(\mathfrak{g}\) be solvable. Then \(x\in\mathfrak{g}^{(1)}\) implies that \( \mathrm{ad}_\mathfrak{g}(x)\) is nilpotent.


    From the flag of ideals construct a basis. relative to this basis the matrix of \(\mathrm{ad}_\mathfrak{g}(y), y\in \mathfrak{g}\ ,\) is upper triangular.

    Thus the matrix of \(\mathrm{ad}_\mathfrak{g}(x), x\in \mathfrak{g}^{(1)}\) is strictly upper triangular, and therefore nilpotent.


    In the next lecture it is shown that this implies that \(\mathfrak{g}^{(1)}\) is nilpotent (to be defined).

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