User:Jan A. Sanders/An Introduction to Leibniz Algebra Cohomology/Lecture 9
the Casimir operator
Let \(\tilde{\mathfrak{g}}=\mathfrak{g}/\ker d_+^{(0)}\) (This makes sense, since \(\ker d_+^{(0)}\) is an ideal). Since \([x,y]+[y,x]\in\ker d_+^{(0)}\ ,\) \(\tilde{\mathfrak{g}}\) is a Lie algebra. A trace form \( K_\mathfrak{a}\) on \(\mathfrak{g}\) induces a trace form \(\tilde{K}_\mathfrak{a}\) on \(\tilde{\mathfrak{g}}\) by \[\tilde{K}_\mathfrak{a}([x],[y])=K_\mathfrak{a}(x,y)\] Suppose \(\dim_\mathbb{C}\tilde{\mathfrak{g}}=n<\infty\ .\) Let \( e_1,\cdots,e_n\) be a basis of \(\tilde{\mathfrak{g}}\ .\) If \(\tilde{K}_\mathfrak{a}\) is nondegenerate, then define \(e^1,\cdots,e^n\) to be the dual basis with respect to \(\tilde{K}_\mathfrak{a}\ ,\) that is, \(\tilde{K}_\mathfrak{a}(e_i,e^j)=\delta_i^j\ .\)
example
Let, for \(\mathfrak{g}=\tilde{\mathfrak{g}}=\mathfrak{sl}_2\ ,\) the basis be given by \[e_1=M,\quad e_2=N,\quad e_3=H\] Then a dual basis is given by \[ e^1=N,\quad e^2=M,\quad e^3=\frac{1}{2} H\]
proposition
Suppose \( [e_i,e_j]=\sum_{k=1}^n c_{ij}^k e_k\ .\) Then \([e^i,e_j]=\sum_{k=1}^n c_{jk}^i e^k\ .\)
proof
The structure constants \(c_{ij}^k\) can be expressed in terms of the trace form as follows. \[ \tilde{K}_\mathfrak{a}([e_i,e_j],e^k)=\sum_{s=1}^n c_{ij}^s \tilde{K}_\mathfrak{a}(e_s,e^k)=\sum_{s=1}^n c_{ij}^s \delta_s^k=c_{ij}^k\] Let \([e^i,e_j]=\sum_{k=1}^n d_{jk}^i e^k\ .\) Then \[\tilde{K}_\mathfrak{a}(e_k,[e^i,e_j])=\sum_{s=1}^n d_{js}^i \tilde{K}_\mathfrak{a}(e_k,e^s)=\sum_{s=1}^n d_{js}^i \delta_k^s=d_{jk}^i\] The result follows from the \(\mathfrak{g}\)-invariance of \(\tilde{K}_\mathfrak{a}\ :\) \[\tilde{K}_\mathfrak{a}(e_k,[e^i,e_j])=-\tilde{K}_\mathfrak{a}(e_k,[e_j,e^i])=\tilde{K}_\mathfrak{a}([e_j,e_k],e^i)=c_{jk}^i\]
corollary
\(c_{ij}^k\) is fully antisymmetric in its indices.
corollary
\[ [x,e^i]=-\sum_{k=1}^n \tilde{K}_\mathfrak{a}(x,[e_k,e^i])e^k\] and \( [x,e_i]=\sum_{k=1}^n \tilde{K}_\mathfrak{a}(x,[e_i,e^k])e_k\)
definition
Define the Casimir operator \(\gamma^{0}\) by \[\gamma^{0}=\sum_{i=1}^n d_+^{(0)}(e^i)d_+^{(0)}(e_i) \in End(\mathfrak{a})\]
remark
For the Casimir to exist one only needs finite-dimensionality of \(\mathfrak{g}\ ,\) \(\mathfrak{a}\) can be infinite dimensional. Only when \(K_{\mathfrak{a}}\) plays a role, one assumes \(\mathfrak{a}\) to be finite-dimensional, just so that one does not have to worry about traces in infinite-dimensional spaces.
well defined
The \(e_i, e^i\) stand for equivalence classes, but taking different representatives does not change the value of \(\gamma^0\ .\)
The definition of \(\gamma^0\) is also independent of the choice of basis. Let \( f_i=\sum_{k=1}^n A_i^k e_k\ ,\) with \(A\) an invertible matrix, be another basis, with dual basis \(f^i\ .\) Let \(f^i=\sum_{k=1}^n B_k^i e^k\ .\) Then \[\delta_j^i=K_\mathfrak{b}(e^i,e_j)=\sum_{k,l=1^n}B_k^i A_j^l K_\mathfrak{b}(f^k,f_l)=\sum_{k,l=1^n}B_k^i A_j^l\delta_l^k=\sum_{k=1}^n B_k^i A_j^k\] This shows that \[\gamma^{0}=\sum_{i=1}^n d_+^{(0)}(f^i)d_+^{(0)}(f_i)\]
corollary
If the dual basis is chosen with respect to \(K_\mathfrak{a}\ ,\) then \[ \mathrm{tr }(\gamma^0)=\sum_{i=1}^n \mathrm{tr }(d_+^{(0)}(e^i)d_+^{(0)}(e_i))=\sum_{i=0}^n K_\mathfrak{a}(e^i,e_i)=n\]
example
In the case \(\mathfrak{g}=\mathfrak{sl}_2\) and \(\mathfrak{a}=\R^2\ ,\) with the standard representation, one has \[\gamma^{0}=d^{(0)}(e^1)d^{(0)}(e_1)+d^{(0)}(e^2)d^{(0)}(e_3)+d^{(0)}(e^3)d^{(0)}(e_3)\ :\] \[=d^{(0)}(N)d^{(0)}(M)+d^{(0)}(M)d^{(0)}(N)+\frac{1}{2}d^{(0)}(H)d^{(0)}(H)\ :\] \[=\begin{bmatrix} 0&0\\1&0\end{bmatrix}\begin{bmatrix} 0&1\\0&0\end{bmatrix}+ \begin{bmatrix} 0&1\\0&0\end{bmatrix}\begin{bmatrix} 0&0\\1&0\end{bmatrix}+\frac{1}{2} \begin{bmatrix} 1&0\\0&-1\end{bmatrix}\begin{bmatrix} 1&0\\0&-1\end{bmatrix}\ :\] \[=\frac{3}{2}\begin{bmatrix} 1&0\\0&1\end{bmatrix}\] One checks that indeed \( \mathrm{tr\ }\gamma^0=3\ .\)
lemma
Suppose \(\dim\mathfrak{a}<\infty\ .\) Then \[\gamma^0 d_+^{(0)}(x)=d_+^{(0)}(x)\gamma^0\]
proof
\[\gamma^0 d_+^{(0)}(x)-d_+^{(0)}(x)\gamma^0\ :\] \[=\sum_{i=1}^n d_+^{(0)}(e^i)d_+^{(0)}(e_i)d_+^{(0)}(x)-d_+^{(0)}(x)\sum_{i=1}^n d_+^{(0)}(e^i)d_+^{(0)}(e_i)\ :\] \[=\sum_{i=1}^n d_+^{(0)}(e^i)d_+^{(0)}([e_i,x])+\sum_{i=1}^n d_+^{(0)}(e^i)d_+^{(0)}(x)d_+^{(0)}(e_i)-\sum_{i=1}^n d_+^{(0)}(e^i)d_+^{(0)}(x)d_+(e_i)+\sum_{i=1}^n d_+([e^i,x])d_+(e_i)\ :\] \[=\sum_{i=1}^n d_+^{(0)}(e^i)d_+^{(0)}([e_i,x])+\sum_{i=1}^n d_+([e^i,x])d_+(e_i)\ :\] \[=-\sum_{i,j=1}^n \tilde{K}_{\mathfrak{a}}(x,[e_i,e^j])d_+^{(0)}(e^i)d_+^{(0)}(e_j)+\sum_{i,j=1}^n \tilde{K}_{\mathfrak{a}}(x,[e_j,e^i]) d_+^{(0)}(e^j) d_+^{(0)}(e_i)\ :\] \[=0\]
corollary
If the representation is even, then \(\gamma^0\) is a \(\mathfrak{g}\)-endomorphism.
lemma
Let \(\alpha\in \mathrm{End}_\mathfrak{g}(\mathfrak{a})\ ,\) with \(\dim\mathfrak{a}<\infty\ .\) Then \(\mathfrak{a}=\mathfrak{a}_0\oplus\mathfrak{a}_1\ ,\) where \(\mathfrak{a}_i\) is invariant under \(\alpha\) and \(\mathfrak{g}\ .\) Moreover, of one denotes the restriction of \(\alpha\) to \(\mathfrak{a}_i\) by \(\alpha_i\ ,\) one has that \(\alpha_0\) is nilpotent and \(\alpha_1\) is invertible.
proof
One has s decreasing sequence of subspaces \[ \mathfrak{a}\supset \alpha\mathfrak{a}\supset \alpha^2 \mathfrak{a}\supset\cdots\] where \(\alpha^m\) denotes the \(m\)th power of \(\alpha\ .\) Since \(\mathfrak{a}\) is finite-dimensional, this stabilizes, say at \(k\ .\) Define \(\mathfrak{a}_1=\alpha^k \mathfrak{a}\ .\) This is \(\alpha\)-invariant by construction, and \(\mathfrak{g}\)-invariant since \(\alpha\) commutes with the \(\mathfrak{g}\)-action on \(\mathfrak{a}\ .\) Let \(\mathfrak{\beta}_i=\ker \alpha^i\ .\) Then \[\mathfrak{b}_0\subset\mathfrak{b}_1\subset\cdots\subset\mathfrak{a}\] Again,this stabilizes, say at \(l\ .\) Let \(\mathfrak{a}_0=\mathfrak{b}_l\) and observe that \(\mathfrak{a}_0\) is \(\alpha\)-invariant and \(\mathfrak{g}\)-invariant. Let \(m=\max(k,l)\ .\) Then \[ \mathfrak{a}_0=\ker \alpha^m,\quad \mathfrak{a}_1=\mathrm{im}\alpha^m\] Take \(x\in\mathfrak{a}\ .\) Then \(\alpha^m x=\alpha^{2m} y\) for some \(y\in\mathfrak{a}\ ,\) since \(\alpha^m\mathfrak{a}=\alpha^{2m}\mathfrak{a}\ .\) Write \(x=(x-\alpha^my)+\alpha^m y\in\ker \alpha^m+\mathrm{im}\alpha^m\ .\) This implies \[\mathfrak{a}=\mathfrak{a}_0+\mathfrak{a}_1\] Let \(z\in \mathfrak{a}_0\cap\mathfrak{a}_1\ .\) This implies that \(z=\alpha^m w\) and \(\alpha^m z=0\ .\) It follows that, since \(\alpha^{2m}w=0\ ,\) \(w\in\mathfrak{a}_0\ .\) Therefore \(\alpha^mw=0\ ,\) or, in other words, \( z=0\ .\) This shows that \(\mathfrak{a}_0\cap\mathfrak{a}_1=0\) and \[\mathfrak{a}=\mathfrak{a}_0\oplus\mathfrak{a}_1\] Since \(\mathfrak{a}_1=\alpha^m\mathfrak{a}=\alpha^{m+1}\mathfrak{a}=\alpha\mathfrak{a}_1\ ,\) it follows that \(\alpha_1\) is surjective, and therefore an isomorphism. Denote the projections of \(\mathfrak{a}\rightarrow\mathfrak{a}_i\) by \(\pi_i^0\) and observe they commute with the \(\mathfrak{g}\)-action. The decomposition \(\mathfrak{a}=\mathfrak{a}_0\oplus\mathfrak{a}_1\) is called the Fitting decomposition of \(\mathfrak{a}\) with respect to \(\alpha\ .\)
theorem
Let \(d^{(0)}\) be an even representation, denoted by \(d^{(0)}\ .\) Suppose there exists a nondegenerate trace form \(\tilde{K}_\mathfrak{a}\ .\) Then \(H^m(\tilde{\mathfrak{g}},\mathfrak{a})=H^m(\tilde{\mathfrak{g}},\mathfrak{a}_0)\ .\)
remark
The assumpion that \(d^{(0)}\) be even puts us almost in the Lie algebra situation. However the forms do not need to be antisymmetric.
proof
Consider the Fitting decomposition of \(\mathfrak{a}\) with respect to \(\gamma^0\ .\) Take \([\zeta^m]\in H^m(\tilde{\mathfrak{g}},\mathfrak{a})\) and let \[\pi_1^m\zeta^m=(-1)^{m-1}\gamma^m\omega^m\] Then, since \( \gamma^{m+1}d^m=d^m\gamma^m\) and \( \pi_1^{m+1}d^m=d^m\pi_1^m\ ,\) one has \[0=\pi_1^{m+1}d^m\zeta^m=d^m\pi_1^m\zeta^m=(-1)^{m-1}d^m \gamma^m\omega^m=(-1)^{m-1}\gamma^{m+1}d^m \omega^m\] Since \(\gamma^0\) is an isomorphism on \(\mathfrak{a}_1\ ,\) this shows that \(d^m\omega^m=0\ .\)
Then define \[ \mu^{m-1}(x_1,\cdots,x_{m-1})=\sum_{i=1}^n d^{(0)}(e^i)\omega^m(x_1,\cdots,x_{m-1},e_i)\] (Here one needs the trace form to be nondegenerate, in order to define the dual basis). Then (and here we use the assumption that \(d^{(0)}\) is even) \[0=\sum_{i=1}^n d^{(0)}(e^i)d^m\omega^m(x_1,\dots,x_m,e_i)\ :\] \[=\sum_{i=1}^n (-1)^{m} d^{(0)}(e^i)d^{(0)}(e_i) \omega^m(x_1,\dots,x_{m})\ :\] \[+\sum_{k=1}^{m}\sum_{i=1}^n(-1)^{k-1} d^{(0)}(e^i) d^{(0)}(x_k) \omega^m(x_1,\dots,\hat{x}_k,\dots,x_m,e_i)\ :\] \[-\sum_{k=1}^{m}\sum_{i=1}^n (-1)^{k-1} d^{(0)}(e^i)\omega^m(x_1,\dots,\hat{x}_k,\dots,[x_k,e_i])\ :\] \[-\sum_{k=1}^{m}\sum_{l=1}^{k-1}\sum_{i=1}^n(-1)^{l-1} d^{(0)}(e^i) \omega^m(x_1,\dots,\hat{x}_l,\dots,[x_l,x_k],\dots,e_i)\ :\] \[=(-1)^m\gamma^0\omega^m(x_1,\dots,x_{m})\ :\] \[+\sum_{k=1}^{m}\sum_{i=1}^n(-1)^{k-1} d^{(0)}(x_k) d^{(0)}(e^i)\omega^m(x_1,\dots,\hat{x}_k,\dots,x_m,e_i)\ :\] \[-\sum_{k=1}^{m}\sum_{l=1}^{k-1}\sum_{i=1}^n(-1)^{l-1} d^{(0)}(e^i) \omega^m(x_1,\dots,\hat{x}_l.\dots,[x_l,x_k],\dots,e_i)\ :\] \[-\sum_{k=1}^{m}\sum_{i=1}^n(-1)^{k-1} d^{(0)}([x_k,e^i])\omega^m(x_1,\dots,\hat{x}_k,\dots,e_i)\ :\] \[-\sum_{k=1}^{m}\sum_{i=1}^n (-1)^{k-1} d^{(0)}(e^i)\omega^m(x_1,\dots,\hat{x}_k,\dots,[x_k,e_i])\ :\] \[=-\pi_1^m\zeta^m(x_1^1,\dots,x_{m}^m)+d^{m-1}\mu^{m-1}(x_1,\dots,x_m)\ :\] \[+\sum_{k=1}^{m}\sum_{i=1}^n \sum_{p=1}^n(-1)^{k-1} K_\mathfrak{a}(x_k,[e_p,e^i])d^{(0)}(e^p)\omega^m(x_1,\dots,\hat{x}_k,\dots,x_m,e_i)\ :\] \[-\sum_{k=1}^{m}\sum_{i=1}^n \sum_{p=1}^n (-1)^{k-1}K_\mathfrak{a}(x_k,[e_i,e^p]) d^{(0)}(e^i)\omega^m(x_1,\dots,\hat{x}_k.\dots,x_m,e_p)\ :\] \[=-\pi_1^m\zeta^m(x_1^1,\dots,x_{m}^m)+d^{m-1}\mu^{m-1}(x_1,\dots,x_m)\] and the theorem is proved.\(\square\)
theorem
Let \(M=\dim(\mathfrak{a}_0)\ .\) Then \(H^m(\tilde{\mathfrak{g}},\mathfrak{a}_0)=\bigoplus_{\iota=1}^M H^m(\tilde{\mathfrak{g}},\mathbb{C})\ .\)
proof
Since \(\gamma^0\) is nilpotent, its trace on \(\mathfrak{a}_0\) is zero. But this implies that the representation vanishes on \(\mathfrak{a}_0\ ,\) since \(\mathrm{tr\ }\gamma_0^0=n\ ,\) where \( n\) is the number of basis vectors \(e_\iota\) of \(\mathfrak{a}_0\) such that \(d^{(0)}(e_\iota)\neq 0\ .\) Therefore \(H^m(\tilde{\mathfrak{g}},\mathfrak{a}_0)=\bigoplus_{\iota=1}^M H^m(\tilde{\mathfrak{g}},\mathbb{C})\ .\)
corollary
Let \(d^{(0)}\) be a nontrivial even representation, such that \(\mathfrak{a}\) is irreducible, that is, it contains no \(\mathfrak{g}\)-invariant subspaces. Suppose there exists a nondegenerate trace form \(\tilde{K}_\mathfrak{a}\ .\) Then \(H^m(\tilde{\mathfrak{g}},\mathfrak{a})=0\ .\)
proof
Since the representation is irreducible, one has either \(\mathfrak{a}=\mathfrak{a}_0\) or \(\mathfrak{a}=\mathfrak{a}_1\ .\) But in the first case the representation would be trivial, which is excluded. Therefore one is in the second case and the statement follows.
lemma
If \([\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}]=\tilde{\mathfrak{g}}\) then \(H^1(\tilde{\mathfrak{g}},\mathbb{C})=0\ .\)
proof
Since the representation is trivial, \( d^1\omega^1=0\) implies \(\omega^1([x,y])=0\) for all \(x,y\in\tilde{\mathfrak{g}}\ .\) But this implies that \( \omega^1(z)=0\) for all \( z\in\tilde{\mathfrak{g}}\ ,\) since every \(z\) can be written as a finite linear combination of commutators. It follows that \(\omega^1=0\) and therefore trivial (with the representation zero, the only way a one form can be trivial is by being zero).
corollary
Let \(d^{(0)}\) be an even representation, denoted by \(d^{(0)}\ .\) Suppose there exists a nondegenerate trace form \(\tilde{K}_\mathfrak{a}\) and \([\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}]=\tilde{\mathfrak{g}}\ .\) Then \(H^1(\tilde{\mathfrak{g}},\mathfrak{a})=H^1(\tilde{\mathfrak{g}},\mathfrak{a}_0)=\bigoplus_{\iota=1}^{\dim(\mathfrak{a}_0)} H^1(\tilde{\mathfrak{g}},\mathbb{C})=0\ .\)
definition
Define the lower central series of a Lie algebra by \(\mathfrak{g}^0=\mathfrak{g}\) and \(\mathfrak{g}^{i+1}=[\mathfrak{g},\mathfrak{g}^i]\ .\)
proposition
The \(\mathfrak{g}^i\) are ideals of \(\mathfrak{g}\ .\)
proof
For \( i=0\) this is trivial. Suppose \(\mathfrak{g}^i\) is an ideal. Then \[[\mathfrak{g},\mathfrak{g}^{i+1}]=[\mathfrak{g},[\mathfrak{g},\mathfrak{g}^i]]\subset [\mathfrak{g},\mathfrak{g}^i]=\mathfrak{g}^{i+1}\] The proposition follows by induction.
definition
\(\mathfrak{g}\) is called nilpotent if there is an \(n\in\mathbb{N}\) such that \(\mathfrak{g}^n=0\ .\)
proposition
A nilpotent Lie algebra is solvable, but a solvable Lie algebra need not be nilpotent.
proof
The first part follows from \(\mathfrak{g}^{(i)}\subset \mathfrak{g}^i\ .\) An algebra that is solvable bit not nilpotent is the Lie algebra of upper triangular matrices in \(\mathfrak{gl}_n\ .\)
proposition
If \(\mathfrak{g}\) is nilpotent, then so are all subalgebras and homomorphic images.
proof
Let \( \mathfrak{h}\) be a subalgebra. Then \( \mathfrak{h}^{0}\subset\mathfrak{g}^{0}\ .\) Assume \(\mathfrak{h}^{i}\subset\mathfrak{g}^{i}\ .\) Then \[ \mathfrak{h}^{i+1}=[\mathfrak{g},\mathfrak{h}^{i}]\subset [\mathfrak{g},\mathfrak{g}^{i}]=\mathfrak{g}^{i+1}\] 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 \[\phi(\mathfrak{g}^{i+1})=\phi([\mathfrak{g},\mathfrak{g}^{i}])=[\phi(\mathfrak{g}),\phi(\mathfrak{g}^{i})]= [\mathfrak{h},\mathfrak{h}^{i}]=\mathfrak{h}^{i+1}\]
proposition
Let \(\mathcal{Z}(\mathfrak{g})\) denote the center of \(\mathfrak{g}\ ,\) that is, \[\mathcal{Z}(\mathfrak{g})=\{x\in \mathfrak{g}|[x,y]=0 \quad \forall y\in\mathfrak{g}\}\] If \(\mathfrak{g}/\mathcal{Z}(\mathfrak{g})\) is nilpotent, then \(\mathfrak{g}\) is nilpotent.
proof
Say \(\mathfrak{g}^n\subset \mathcal{Z}(\mathfrak{g})\ ,\) then \(\mathfrak{g}^{n+1}=[\mathfrak{g},\mathfrak{g}^{n}]\subset [\mathfrak{g},\mathcal{Z}(\mathfrak{g})]=0\ .\)
proposition
If \(\mathfrak{g}\) is nilpotent and nonzero, then so is \(\mathcal{Z}(\mathfrak{g})\neq 0\ .\)
proof
Let \( n \) be the minimal order such that \(\mathfrak{g}^n=0\ ,\) then \(\mathfrak{g}^{n-1}\subset \mathcal{Z}(\mathfrak{g})\ .\)
lemma
If \(x\in \mathfrak{gl}(V)\) is nilpotent, then \( ad(x) \) is nilpotent.
proof
Define \(\lambda_x, \rho_x\in\mathrm{End}(\mathrm{End}(V))\) by \[ \lambda_x y=xy,\quad \rho_x y=yx\] These are nilpotent, since for instance, \(\lambda_x^n=\lambda_{x^n}\ .\) If \(x^n=0\ ,\) than \((\lambda_x-\rho_x)^{2n}=0\) (since \(\lambda_x \rho_x=\rho_x\lambda_x\)). This proves the statement, since \(\mathrm{ad}(x)=\lambda_x-\rho_x\ .\)
theorem
Let \(\mathfrak{g}\) be a subalgebra of \(\mathfrak{gl}(V)\ ,\) with \(\dim V<\infty\ .\) If \(\mathfrak{g}\) consists of nilpotent endomorphisms and \(V\neq 0\ ,\) then there exists \(v\in V\) such that \(v\neq 0\) and \(d^{(0)}(\mathfrak{g})v=0\ .\)
proof
The proof is by induction on \(\dim\mathfrak{g}\ .\) The statement is obvious if the dimension is zero. Suppose \(\mathfrak{h}\) is a subalgebra of \(\mathfrak{g}\ .\) Then \(\mathfrak{h}\) acts via \(\mathrm{ad}\) as a Lie algebra of nilpotent linear transformations on \(\mathfrak{g}\ ,\) and therefore on \(\mathfrak{g}/\mathfrak{h}\ .\) Since \(\dim\mathfrak{h}<\dim\mathfrak{g}\) one can use the induction hypothesis to conclude that there exists a vextor \(x+\mathfrak{h}\ ,\) \( x\notin \mathfrak{h}\ ,\) such that \([y,x]=0\) for any \(y\in \mathfrak{h}\ .\) Thus \(\mathfrak{h}\) is properly contained in its normalizer \[ N_\mathfrak{g}(\mathfrak{h})=\{x\in\mathfrak{g}|[x,\mathfrak{h}]\subset\mathfrak{h}\}\] The normalizer is a subalgebra, so if one takes \(\mathfrak{h}\) to be a maximal proper subalgebra, then its normalizer must be the whole \(\mathfrak{g}\ ,\) that is to say, \(\mathfrak{h}\) is an ideal in \(\mathfrak{g}\ .\) Take \(0\neq x\in\mathfrak{g}/\mathfrak{h}\) and let \(\mathfrak{x}\) be the subalgebra generated by \(x\ .\) Then the inverse image of \(\mathfrak{x}\) in \(\mathfrak{g}\) is a subalgebra properly containing \(\mathfrak{h}\ ,\) that is, it is \(\mathfrak{g}\ .\) This only makes sense if there is basically one such \(x\ ,\) and it follows that \(\dim\mathfrak{g}/\mathfrak{h}=1\ .\) One writes \[ \mathfrak{g}=\mathfrak{h}+ \mathbb{C} x\ .\] By induction, \(\mathcal{W}=\{v\in V|d^{(0)}(\mathfrak{h})v=0\}\) is nonzero. One has for \(x\in\mathfrak{g}\ ,\) \(y\in\mathfrak{h}\) and \( w\in\mathcal{W}\) that \[d^{(0)}(y)d^{(0)}(x)w=d^{(0)}(x)d^{(0)}(y)w-d^{(0)}([x,y])w=0\ .\] This implies that \(d^{(0)}(x)w\in \mathcal{W}\ ,\) that is, \(\mathcal{W}\) is invariant under \(\mathfrak{g}\ .\) Take \(x\in\mathfrak{x}\) as before. Then (since \(\dim\mathfrak{x}=1\)) there exists a nonzero \( v\in\mathcal{W}\) such that \(d^{(0)}(x)v=0\ .\) This implies that \(d^{(0)}(\mathfrak{g})v=0\ ,\) as desired.
theorem (Engel)
If all all elements of \(\mathfrak{g}\) are ad-nilpotent, then \(\mathfrak{g}\) is nilpotent.
proof
Identifying \(\mathrm{ad\ }(x)\) with a nilpotent element in \(\mathrm{End}(\mathfrak{g})\ ,\) one conludes to the existence of an \(x\in\mathfrak{g}\) such that \(\mathrm{ad\ }(\mathfrak{g})x=0\ ,\) that, \(x\in \mathcal{Z}(\mathfrak{g})\neq 0\ .\) Then \(\mathfrak{g}/\mathcal{Z}(\mathfrak{g})\) again consists of ad-nilpotent elements and \(\dim \mathfrak{g}/\mathcal{Z}(\mathfrak{g})< \dim \mathfrak{g}\ .\) Using induction on the dimension, one concludes that \(\mathfrak{g}/\mathcal{Z}(\mathfrak{g})\) is nilpotent. It follows that \(\mathfrak{g}\) is nilpotent.
corollary
If \(\mathfrak{g}\) is solvable, then \([\mathfrak{g},\mathfrak{g}]\) is nilpotent.
lemma
Let \(\mathfrak{g}\) be nilpotent and \(\mathfrak{h}\) a nonzero ideal of \(\mathfrak{g}\ .\) Then \(\mathfrak{h}\cap\mathcal{Z}(\mathfrak{g})\neq 0\) (and in particular, \(\mathcal{Z}(\mathfrak{g})\neq 0\)).
proof
If \(\mathfrak{g}^n=0\) then \( (\mathrm{ad\ }(x))^n=0\ .\) Consider \( \mathfrak{h}\) as the representation space (with \(d^{(0)}=\mathrm{ad}\)). Then there exist an element \(h\in\mathfrak{h}\) such that \[ ad(\mathfrak{g})h=0\ .\] This is equivalent with saying that \(h\in\mathfrak{h}\cap\mathcal{Z}(\mathfrak{g})\neq 0\ .\)
definition + remarks
One calls \(x\in\mathrm{End}(\mathfrak{a})\) semisimple if the roots of its minimal polynomial over \(\mathbb{C}\) are all distinct. This is equivalent to saying that \( x\) is diagonizable. If two endomorphisms commute, they can be simultaneously diagonalized. A semisimple endomorphism remains semisimple when restricted to an invariant subspace.
proposition
Let \(\mathfrak{a}\) be a finite dimensional vectorspace over \(\mathbb{C}\ ,\) \(x\in\mathrm{End}(\mathfrak{a})\ .\) There exist unique \(x_s, x_n\in\mathrm{End}(\mathfrak{a})\) such that \(x=x_s+x_n\ ,\) \(x_s\) is semisimple, \(x_n\) is nilpotent and \(x_s\) and \(x_n\) commute.
proof
In progress...
definition
Define a representation of \(\tilde{\mathfrak{g}}\) on \(\tilde{\mathfrak{g}}'=C^1(\tilde{\mathfrak{g}},\mathbb{C})\) as follows: \[(b^{(0)}(x)c^1)(y)=-c^1([x,y])\]
well defined
\[ (b^{(0)}([x,y])c^1)(z)=-c^1([[x,y],z])\ :\] \[=-c^1([x,[y,z]])+c^1([y,[x,z]])\ :\] \[=(b^{(0)}(x)c^1)([y,z])-(b^{(0)}(y)c^1)([x,z])\ :\] \[= -(b^{(0)}(y)b^{(0)}(x)c^1)(z)+(b^{(0)}(x)b^{(0)}(y)c^1)(z)\ :\] \[=([b^{(0)}(x),b^{(0)}(y)]c^1)(z)\]
lemma
Let \(\tilde{\mathfrak{g}}\) be a Lie algebra. Suppose there exists a nondegenerate trace form \(K_{\tilde{\mathfrak{g}}'}\ .\) If \([\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}]=\tilde{\mathfrak{g}}\) then \(H^2(\tilde{\mathfrak{g}},\mathbb{C})=0\ .\)
remark
The following proofs rely on the fact that \(\tilde{\mathfrak{g}}\) is semisimple. This is proved in the literature, but not yet in these notes. Alternatively, one could require that \(H^1(\tilde{\mathfrak{g}},\cdot)=0\ .\)
proof
Let for \(m\geq 1\) a map \(\phi:C^m(\tilde{\mathfrak{g}},\mathbb{C})\rightarrow C^{m-1}(\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}')\) be given by \[ (\phi u^m)(x_1,\dots,x_{m-1})(x)=u^m(x_1,\dots,x_{m-1},x)\] Since \[ [(b^{m-1}\phi u^m)(x_1,\dots,x_m)](x)=\sum_{i=1}^m (-1)^{i-1} b^{(0)}(x_i) \phi u^m (x_1,\dots,\hat{x}_i,\dots,x_m)(x)-\sum_{i<j}(-1)^{i-1} \phi u^m (x_1,\dots,\hat{x}_i,\dots,[x_i,x_j]\dots,x_m)(x)\ :\] \[ =-\sum_{i=1}^m (-1)^{i-1} u^m (x_1,\dots,\hat{x}_i,\dots,x_m,[x_i,x])-\sum_{i<j}(-1)^{i-1} u^m (x_1,\dots,\hat{x}_i,\dots,[x_i,x_j]\dots,x_m,x)\ :\] \[ = - d^m u^m (x_1,\dots,x_m,x)\ :\] \[ =- [\phi d^m u^m (x_1,\dots,x_m)](x)\] This implies that \(b^{m-1}\phi=-\phi d^m\) and in particular that \(\phi:Z^m(\tilde{\mathfrak{g}},\mathbb{C})\rightarrow Z^{m-1}(\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}')\ .\) Take \(\omega^2\in Z^2(\tilde{\mathfrak{g}},\mathbb{C})\ .\) Then \(b^1 \phi\omega^2 =0\ .\) this implies that there exists a \( \beta^1\in C^{0}(\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}')=\tilde{\mathfrak{g}}'\) such that \[\omega^2(x,y)= \phi\omega^2(x)(y)=b^0\beta^1(x)(y)=b^{(0)}(x)\beta^1(y)=-\beta^1([x,y])=d^1\beta^1(x,y)\] This proves that \(\omega^2=d^1\beta^1\ .\)
remark
In the Lie algebra case with antisymmetric forms, these cohomology results were obtained by Whitehead. There is not an analogous result for \(H_{\wedge}^3(\tilde{\mathfrak{g}},\mathbb{C})\ .\) This is related to the fact that \([d^2 K_{\mathfrak{g}'}]\in H_{\wedge}^3(\tilde{\mathfrak{g}},\mathbb{C})\ .\)
theorem (Weyl)
Suppose \(\tilde{\mathfrak{g}}\) and \(\mathfrak{a}\) are finite dimensional. If \([\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}]=\tilde{\mathfrak{g}}\) then \(\mathfrak{a}\) is completely reducible, that is, if \(\mathfrak{b}\) is a \(\tilde{\mathfrak{g}}\)-invariant subspace of \(\mathfrak{a}\ ,\) then there exists a \(\tilde{\mathfrak{g}}\)-invariant direct summand to \(\mathfrak{b}\ .\)
proof
Let \(\mathfrak{b}\) be a \(\tilde{\mathfrak{g}}\)-invariant subspace of \(\mathfrak{a}\ .\) The idea of the proof is as follows. Let \(P_\mathfrak{b}\) be the projector on \(\mathfrak{b}\ .\) If \(P_\mathfrak{b}\) commutes with the \(\mathfrak{g}\)-action, we are done, since then we find a direct summand by letting \(1-P_\mathfrak{b}\) act on \(\mathfrak{a}\ .\) To make \(P_\mathfrak{b}\) commute with the action, one perturbs it with another map \(c^0\ .\) In order for \(P_\mathfrak{b}+c^0\) to be a projection on \(\mathfrak{b}\) one needs that \(\mathrm{im\ }c^0 \subset \mathfrak{b}\) and \(\mathfrak{b}\subset \ker c^0\) (since \(P_\mathfrak{b}\) is the identity on \(\mathfrak{b}\)). These considerations lead to the following definition. Define \(\mathcal{W}\) to be the space of all \(A\in\mathrm{End}(\mathfrak{a})\) such that \[ \mathrm{im\ }A\subset \mathfrak{b}\subset \ker A\ .\] Then \(\mathcal{W}\) is a subspace: Let \(a\in \mathfrak{a}, b\in\mathfrak{b}\) and \(A,B \in \mathcal{W}\ .\) Then \((A+B)b = Ab +Bb=0\) and \( (A+B)a=Aa+Ab \in \mathfrak{b}\ .\) Define a representation \(\delta^{(0)}\) of \(\tilde{\mathfrak{g}}\) on \(\mathcal{W}\) by \[ \delta^{(0)}(x)A=[d^{(0)}(x),A]_{\mathrm{End}(\mathfrak{a})}\] Let \(P_\mathfrak{b}\) be a projector on \(\mathfrak{b}\) as a vectorspace. Then \([d^{(0)}(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a}}\in \mathcal{W}\ .\) Therefore \( c^1\ ,\) defined by \[ c^1(x)=[d^{(0)}(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}\] is a linear map from \(\tilde{\mathfrak{g}}\) to \(\mathcal{W}\ ,\) that is, \(c^1\in C^1(\tilde{\mathfrak{g}},\mathcal{W})\ .\) Observe that one cannot say\[c^1=\delta^0 P_\mathfrak{b}\] for the simple reason that \(P_\mathfrak{b}\notin\mathcal{W}\ .\) Then \[ \delta^1 c^1(x,y)=\delta^{(0)}(x)c^1(y)-\delta^{(0)}(y)c^1(x)-c^1([x,y])\ :\] \[=\delta^{(0)}(x)[d^{(0)}(y),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}-\delta^{(0)}(y)[d^{(0)}(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})} -[d^{(0)}([x,y]),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}\ :\] \[=[d^{(0)}(x),[d^{(0)}(y),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}]_{\mathrm{End}(\mathfrak{a})} -[d^{(0)}(y),[d^{(0)}(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}]_{\mathrm{End}(\mathfrak{a})} -[d^{(0)}([x,y]),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}\ :\] \[=[[d^{(0)}(x),d^{(0)}(y)]_{\mathrm{End}(\mathfrak{a})},P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})} -[d^{(0)}([x,y]),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}\ :\] \[=0\] Since \(H^1(\tilde{\mathfrak{g}},\mathcal{W})=0\ ,\) one has \(c^1=\delta^0c^0\ .\) Then, with \(\mathcal{P}_\mathfrak{b}=P_\mathfrak{b}-c^0\ ,\) \[[d^{(0)}(x),\mathcal{P}_\mathfrak{b}]=c^1(x)-\delta^{(0)}(x)c^0=c^1(x)-\delta^0c^0(x)=0\] One has \( \mathcal{P}_\mathfrak{b}a\in \mathfrak{b}\) for \(a\in\mathfrak{a}\) and \(\mathcal{P}_\mathfrak{b}b=P_\mathfrak{b}b=b\) for \(b\in\mathfrak{b}\ .\) The conclusion is that \( \mathcal{P}_\mathfrak{b}\) is a projector on \(\mathfrak{b}\) as a \(\tilde{\mathfrak{g}}\)-module (and therefore \((1-\mathcal{P}_\mathfrak{b}) \) is a projector on the complementary subspace). Since \(\mathfrak{a}\) is finite-dimensional, the result can be proved using induction.
theorem
Suppose \(\tilde{\mathfrak{g}}\) and \(\mathfrak{a}\) are finite dimensional. Suppose there exists a nondegenerate trace form \(K_{\tilde{\mathfrak{g}}'}\ .\) If \([\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}]=\tilde{\mathfrak{g}}\) then any extension of \(\tilde{\mathfrak{g}}\) by \(\mathfrak{a}\) is trivial.
proof
This follows from the fact that \(H^2(\tilde{\mathfrak{g}},\mathfrak{a})=0\ .\) \mathfrak{g}^{(1)}
references
- Humphreys, James E. Introduction to Lie algebras and representation theory. Graduate Texts in Mathematics, Vol. 9. Springer-Verlag, New York-Berlin, 1972. xii+169 pp.
