# 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)}