# An introduction to Lie algebra cohomology/lecture 8b

## Contents

### reference

We follow Humphreys, 1972 closely for those parts that are not explicitly concerned with cohomology.

### the field

Although we use $$\mathbb{C}$$ as our field, one can also do the initial part of this section with an arbitrary field.

At some point we will need the field to have characteristic zero, and a bit later we want it to be closed.

The latter condition can be slightly relaxed, but we need to find the roots of the characteristic equation of the ad-action of elements in the Lie algebra.

## the Casimir operator

### definition - dual basis

Let $$\tilde{\mathfrak{g}}=\mathfrak{g}/\ker d_1$$ (This makes sense, since $$\ker d_1$$ is an ideal).

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

$[x,e^i]=-\sum_{k=1}^n \tilde{K}_\mathfrak{a}(x,[e_k,e^i])e^k$ and $$[x,e_i]=\sum_{k=1}^n \tilde{K}_\mathfrak{a}(x,[e_i,e^k])e_k$$

### definition - Casimir

Define the Casimir operator $$\gamma$$ by $\gamma=\sum_{i=1}^n d_1(e^i)d_1(e_i) \in End(\mathfrak{a})$

### 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 $$\gamma\ .$$

The definition of $$\gamma$$ 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 $\gamma=\sum_{i=1}^n d_1(f^i)d_1(f_i)$

### corollary

If the dual basis is chosen with respect to $$K_\mathfrak{a}\ ,$$ then $\mathrm{tr }(\gamma)=\sum_{i=1}^n \mathrm{tr }(d_1(e^i)d_1(e_i))=\sum_{i=0}^n K_\mathfrak{a}(e^i,e_i)=n$

### example

In the case $$\mathfrak{g}=\mathfrak{sl}_2$$ and $$\mathfrak{a}=\R^2\ ,$$ with the standard representation, one has $\gamma=d_1(e^1)d_1(e_1)+d_1(e^2)d_1(e_3)+d_1(e^3)d_1(e_3)\ :$ $=d_1(N)d^{(0)}(M)+d_1(M)d^{(0)}(N)+\frac{1}{2}d_1(H)d_1(H)\ :$ $=\begin{bmatrix} 0&0\\1&0\end{bmatrix}\begin{bmatrix} 0&1\\0&0\end{bmatrix}+ \begin{bmatrix} 0&1\\0&0\end{bmatrix}\begin{bmatrix} 0&0\\1&0\end{bmatrix}+\frac{1}{2} \begin{bmatrix} 1&0\\0&-1\end{bmatrix}\begin{bmatrix} 1&0\\0&-1\end{bmatrix}\ :$ $=\frac{3}{2}\begin{bmatrix} 1&0\\0&1\end{bmatrix}$ One checks that indeed $$\mathrm{tr\ }\gamma=3\ .$$

### lemma

Suppose $$\dim\mathfrak{a}<\infty\ .$$ Then $\gamma d_1(x)=d_1(x)\gamma$

### proof

$\gamma d_1(x)-d_1(x)\gamma\ :$ $=\sum_{i=1}^n d_1(e^i)d_1(e_i)d_1(x)-d_1(x)\sum_{i=1}^n d_1(e^i)d_1(e_i)\ :$ $=\sum_{i=1}^n d_1(e^i)d_1([e_i,x])+\sum_{i=1}^n d_1(e^i)d_1(x)d_1(e_i)-\sum_{i=1}^n d_1(e^i)d_1(x)d(e_i)+\sum_{i=1}^n d_1([e^i,x])d_1(e_i)\ :$ $=\sum_{i=1}^n d_1(e^i)d_1([e_i,x])+\sum_{i=1}^n d_1([e^i,x])d_1(e_i)\ :$ $=-\sum_{i,j=1}^n \tilde{K}_{\mathfrak{a}}(x,[e_i,e^j])d_1(e^i)d_1(e_j)+\sum_{i,j=1}^n \tilde{K}_{\mathfrak{a}}(x,[e_j,e^i]) d_1(e^j) d_1(e_i)\ :$ $=0$

### corollary

The map $$\gamma$$ is a $$\mathfrak{g}$$-endomorphism.

### lemma - Fitting decomposition

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 $$d_1$$ be a representation.

Suppose there exists a nondegenerate trace form $$\tilde{K}_\mathfrak{a}\ .$$

Let $$\mathfrak{a}_0\oplus\mathfrak{a}_1$$ be the Fitting decomposition with respect to $$\gamma\ .$$

Then $$H^m(\tilde{\mathfrak{g}},\mathfrak{a})=H^m(\tilde{\mathfrak{g}},\mathfrak{a}_0)\ .$$

### remark

Contrary to the usual statement of this theorem, the forms do not need to be antisymmetric.

### proof

Consider the Fitting decomposition of $$\mathfrak{a}$$ with respect to $$\gamma\ .$$ Take $$[\zeta_m]\in H^m(\tilde{\mathfrak{g}},\mathfrak{a})$$ and let $\pi^m\zeta_m=(-1)^{m-1}\gamma^m\omega_m$ Then, since $$\gamma^{m+1}d^m=d^m\gamma^m$$ and $$\pi^{m+1}d^m=d^m\pi^m\ ,$$ one has $0=\pi^{m+1}d^m\zeta_m=d^m\pi^m\zeta_m=(-1)^{m-1}d^m \gamma^m\omega_m=(-1)^{m-1}\gamma^{m+1}d^m \omega_m$ Since $$\gamma$$ is an isomorphism on $$\mathfrak{a}_1\ ,$$ this shows that $$d^m\omega_m=0\ .$$

Then define $\mu_{m-1}(x_1,\cdots,x_{m-1})=\sum_{i=1}^n d_1(e^i)\omega_m(x_1,\cdots,x_{m-1},e_i)$ (Here one needs the trace form to be nondegenerate, in order to define the dual basis). Then $0=\sum_{i=1}^n d_1(e^i)d^m\omega_m(x_1,\dots,x_m,e_i)\ :$ $=\sum_{i=1}^n (-1)^{m} d_1(e^i)d_1(e_i) \omega_m(x_1,\dots,x_{m})\ :$ $+\sum_{k=1}^{m}\sum_{i=1}^n(-1)^{k-1} d_1(e^i) d_1(x_k) \omega_m(x_1,\dots,\hat{x}_k,\dots,x_m,e_i)\ :$ $-\sum_{k=1}^{m}\sum_{i=1}^n (-1)^{k-1} d_1(e^i)\omega_m(x_1,\dots,\hat{x}_k,\dots,[x_k,e_i])\ :$ $-\sum_{k=1}^{m}\sum_{l=1}^{k-1}\sum_{i=1}^n(-1)^{l-1} d_1(e^i) \omega_m(x_1,\dots,\hat{x}_l,\dots,[x_l,x_k],\dots,e_i)\ :$ $=(-1)^m\gamma\omega_m(x_1,\dots,x_{m})\ :$ $+\sum_{k=1}^{m}\sum_{i=1}^n(-1)^{k-1} d_1(x_k) d_1(e^i)\omega_m(x_1,\dots,\hat{x}_k,\dots,x_m,e_i)\ :$ $-\sum_{k=1}^{m}\sum_{l=1}^{k-1}\sum_{i=1}^n(-1)^{l-1} d_1(e^i) \omega_m(x_1,\dots,\hat{x}_l.\dots,[x_l,x_k],\dots,e_i)\ :$ $-\sum_{k=1}^{m}\sum_{i=1}^n(-1)^{k-1} d_1([x_k,e^i])\omega_m(x_1,\dots,\hat{x}_k,\dots,e_i)\ :$ $-\sum_{k=1}^{m}\sum_{i=1}^n (-1)^{k-1} d_1(e^i)\omega_m(x_1,\dots,\hat{x}_k,\dots,[x_k,e_i])\ :$ $=-\pi^m\zeta_m(x_1^1,\dots,x_{m}^m)+d^{m-1}\mu_{m-1}(x_1,\dots,x_m)\ :$ $+\sum_{k=1}^{m}\sum_{i=1}^n \sum_{p=1}^n(-1)^{k-1} K_\mathfrak{a}(x_k,[e_p,e^i])d_1(e^p)\omega_m(x_1,\dots,\hat{x}_k,\dots,x_m,e_i)\ :$ $-\sum_{k=1}^{m}\sum_{i=1}^n \sum_{p=1}^n (-1)^{k-1}K_\mathfrak{a}(x_k,[e_i,e^p]) d_1(e^i)\omega_m(x_1,\dots,\hat{x}_k.\dots,x_m,e_p)\ :$ $=-\pi^m\zeta_m(x_1^1,\dots,x_{m}^m)+d^{m-1}\mu_{m-1}(x_1,\dots,x_m)$ 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 $$\gamma_0=\gamma|\mathfrak{a}_0$$ is nilpotent, its trace on $$\mathfrak{a}_0$$ is zero.

But this implies that the representation vanishes on $$\mathfrak{a}_0\ ,$$ since $$\mathrm{tr\ }\gamma_0=n\ ,$$ where $$n$$ is the number of basis vectors $$e_\iota$$ of $$\mathfrak{a}_0$$ such that $$d_1(e_\iota)\neq 0\ .$$

Therefore $$H^m(\tilde{\mathfrak{g}},\mathfrak{a}_0)=\bigoplus_{\iota=1}^M H^m(\tilde{\mathfrak{g}},\mathbb{C})\ ,$$ where the action of $$\tilde{\mathfrak{g}}$$ on $$\mathbb{C}$$ is supposed to be trivial, as usual.

### corollary

Let $$d_1$$ be a nontrivial 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

Suppose there exists a nondegenerate trace form $$\tilde{K}_\mathfrak{a}$$ and $$[\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}]=\tilde{\mathfrak{g}}\ .$$

Let $$M=\dim(\mathfrak{a}_0)\ .$$

Then $$H^1(\tilde{\mathfrak{g}},\mathfrak{a})=H^1(\tilde{\mathfrak{g}},\mathfrak{a}_0)=\bigoplus_{\iota=1}^{M} H^1(\tilde{\mathfrak{g}},\mathbb{C})=0\ .$$

### definition -lower central series

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 $$\mathrm{ad}(x)$$ is nilpotent. In particular, if $$x^n=0$$ then $$\mathrm{ad}^{2n}(x)=0\ .$$

### 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 $$x^n=0\ ,$$ then $$(\lambda_x-\rho_x)^{2n}=0$$ (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 $$0<\dim V<\infty\ .$$ If $$\mathfrak{g}$$ consists of nilpotent endomorphisms, then there exists $$0\neq v\in V$$ such that $$d^{(0)}(\mathfrak{g})v=0\ .$$

### proof

The proof is by induction on $$\dim\mathfrak{g}\ .$$ The statement is obvious if the dimension is zero, since any $$v\in V$$ will do.

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 $d_1(y)d_1(x)w=d_1(x)d_1(y)w-d_1([x,y])w=0\ .$ This implies that $$d_1(x)w\in \mathcal{W}\ ,$$ 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 $$d_1(x)v=0\ .$$ This implies that $$d_1(\mathfrak{g})v=0\ ,$$ 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 $$h\in\mathfrak{h}\cap\mathcal{Z}(\mathfrak{g})\neq 0\ .$$

### the field

As remarked in the beginning of this lecture, at this point we need our field to have characteristic zero and we also assume it to be closed.

### 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, since one can take its eigenvectors as a basis of $$\mathfrak{a}\ .$$

(In we work in a general field, one requires here that the roots of the minimal polynomial are contained in the field; such a field is called a splitting field relative to $$x\ .$$)

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

Let $$\lambda_1,\cdots,\lambda_k$$ be the distinct eigenvalues of $$x$$ with multiplicities $$m_1,\cdots,m_k\ .$$ Its characteristic polynomial is then $$\chi(\lambda)=\prod_{i=1}^k (\lambda-\lambda_i)^{m_i}\ .$$ Let $$V_i=\mathrm{ker} (x-\lambda_i)^{m_i}\ ,$$ then $$V=\bigoplus_{i=1}^k V_i\ .$$

Using the Chinese Remainder Theorem we find a polynomial $$p(\lambda)$$ such that $$p(\lambda)=\lambda_i \mathrm{\ mod\ } (\lambda-\lambda_i)^{m_i}$$ and $$p(\lambda)=0 \mathrm{\ mod\ } \lambda\ .$$

Let $$q(\lambda)=\lambda-p(\lambda)\ .$$

Then put $$x_s=p(x)$$ and $$x_n=q(x)\ .$$ Since both are polynomial in $$x\ ,$$ they commute.

One has $$x_s-\lambda_i |V_i=0\ ,$$ that is, $$x_s$$acts diagonally on $$V\ ,$$ since on each $$V_i$$ the characteristic polynomial is $$(\lambda-\lambda_i)^{m_i}\ .$$

Furthermore $$x_n=x-x_s$$ is nilpotent, since on each $$V_i$$ it obeys its own characteristic equation $$x_n^{m_i}=0\ ,$$ so with $$m=\mathrm{max}_{i=1,\ldots,k}m_i$$ one has $$x_n^m=0\ .$$

Any other such decomposition $$x=s+n$$ would lead to $$x_s-s=n-x_n\ .$$ Since $$s$$ and $$n$$ commute, they also commute with $$x$$ and therefore with $$x_s$$ and $$x_n\ .$$

Since the sum of commuting semisimple operators is semisimple and the sum of nilpotent operators nilpotent, and the only operator that is both semisimple and nilpotent is $$0\ ,$$ one must conclude that $$s=x_s$$ and $$n=x_n\ .$$

### lemma

$$\mathrm{Der}(\mathfrak{g})$$ contains the semisimple and nilpotent parts in $$\mathrm{End}(\mathfrak{g})$$ of its elements.

### proof

If $$\delta\in \mathrm{Der}(\mathfrak{g})\ ,$$ let $$\delta_s, \delta_n\in \mathrm{End}(\mathfrak{g})$$ be its semisimple and nilpotent part, respectively. We show that $$\delta_s\in \mathrm{Der}(\mathfrak{g})\ .$$

For $$\lambda\in\mathbb{C}\ ,$$ let $$\mathfrak{g}_\lambda=\left\{x\in\mathfrak{g}|(\delta-\lambda)^k x=0\mathrm{\ for\ some\ } k \right\}\ .$$ Then $$\delta_s$$ acts on $$\mathfrak{g}_\lambda$$ by multiplication by $$\lambda\ .$$

One verifies that $$[\mathfrak{g}_\lambda,\mathfrak{g}_\mu]\subset\mathfrak{g}_{\lambda+\mu}\ :$$

One has $$(\delta-(\lambda+\mu))^n[x,y]=\sum_{i=0}^n\binom{n}{i}[(\delta-\lambda)^{n-i} x,(\delta-\mu)^i y]\ .$$

Indeed, for $$n=1$$ this reads $$(\delta-(\lambda+\mu))[x,y]=[\delta x,y]+[x,\delta y]-(\lambda+\mu)[x,y]=[(\delta-\lambda) x,y]+[x,(\delta-\mu) y]$$ and the general inductive step is now standard.

Thus one has $$\delta_s[x,y]=[\delta_s x,y]+[x,\delta_s y]$$ for $$x\in\mathfrak{g}_\lambda, y\in \mathfrak{g}_\mu\ .$$

Since $$\mathfrak{g}=\bigoplus_\lambda \mathfrak{g}_\lambda\ ,$$ it follows that $$\delta_s$$ is a derivation.

### definition

Define a representation of $$\tilde{\mathfrak{g}}$$ on $$\tilde{\mathfrak{g}}'=C^1(\tilde{\mathfrak{g}},\mathbb{C})$$ as follows: $(b_1(x)c_1)(y)=-c_1([x,y])$

### well defined

$(b_1([x,y])c_1)(z)=-c_1([[x,y],z])\ :$ $=-c_1([x,[y,z]])+c_1([y,[x,z]])\ :$ $=(b_1(x)c_1)([y,z])-(b_1(y)c_1)([x,z])\ :$ $= -(b_1(y)b_1(x)c_1)(z)+(b_1(x)b_1(y)c_1)(z)\ :$ $=([b_1(x),b_1(y)]c_1)(z)$

### 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 $$\phi^m:C^m(\tilde{\mathfrak{g}},\mathbb{C})\rightarrow C^{m-1}(\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}')$$ be given by $(\phi^m u_m)(x_1,\dots,x_{m-1})(x)=u_m(x_1,\dots,x_{m-1},x)$ Since $[(b^{m-1}\phi^m u_m)(x_1,\dots,x_m)](x)=\ :$ $=\sum_{i=1}^m (-1)^{i-1} b_1(x_i) \phi^m u_m (x_1,\dots,\hat{x}_i,\dots,x_m)(x)-\sum_{i<j}(-1)^{i-1} \phi^m u_m (x_1,\dots,\hat{x}_i,\dots,[x_i,x_j]\dots,x_m)(x)\ :$ $=-\sum_{i=1}^m (-1)^{i-1} u_m (x_1,\dots,\hat{x}_i,\dots,x_m,[x_i,x])-\sum_{i<j}(-1)^{i-1} u_m (x_1,\dots,\hat{x}_i,\dots,[x_i,x_j]\dots,x_m,x)\ :$ $= - d^m u_m (x_1,\dots,x_m,x)\ :$ $=- [\phi^{m+1} d^m u_m (x_1,\dots,x_m)](x)$ This implies that $$b^{m-1}\phi^m=-\phi^{m+1} d^m$$ and in particular that $$\phi^m:Z^m(\tilde{\mathfrak{g}},\mathbb{C})\rightarrow Z^{m-1}(\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}')\ .$$

Take $$\omega_2\in Z^2(\tilde{\mathfrak{g}},\mathbb{C})\ .$$ Then $$b^1 \phi^2\omega_2 =0\ .$$

It follows from the assumptions that $$H^1( \tilde{\mathfrak{g}},\tilde{\mathfrak{g}}')=0\ .$$

This implies that there exists a $$\beta_1\in C^{0}(\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}')=\tilde{\mathfrak{g}}'$$ such that $$\phi\omega_2=b\beta_1$$ and $\omega_2(x,y)= \phi^2\omega_2(x)(y)=b\beta_1(x)(y)=b_1(x)\beta_1(y)=-\beta_1([x,y])=d^1\beta_1(x,y)$ This proves that $$\omega_2=d^1\beta_1\ .$$

### remark

These cohomology results were obtained by Whitehead in the antisymmetric case.

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 $$\delta_1$$ of $$\tilde{\mathfrak{g}}$$ on $$\mathcal{W}$$ by $\delta_1(x)A=[d_1,A]_{\mathrm{End}(\mathfrak{a})}$ Let $$P_\mathfrak{b}$$ be a projector on $$\mathfrak{b}$$ as a vectorspace. Then $$[d_1(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a}}\in \mathcal{W}\ .$$ Therefore $$c^1\ ,$$ defined by $c^1(x)=[d_1(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}$ 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$c^1=\delta P_\mathfrak{b}$ for the simple reason that $$P_\mathfrak{b}\notin\mathcal{W}\ .$$ Then $\delta^1 c^1(x,y)=\delta_1(x)c^1(y)-\delta_1(y)c^1(x)-c^1([x,y])\ :$ $=\delta_1(x)[d^{(0)}(y),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}-\delta^{(0)}(y)[d_1(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})} -[d_1([x,y]),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}\ :$ $=[d_1(x),[d_1(y),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}]_{\mathrm{End}(\mathfrak{a})} -[d_1(y),[d_1(x),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}]_{\mathrm{End}(\mathfrak{a})} -[d_1([x,y]),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}\ :$ $=[[d_1(x),d_1(y)]_{\mathrm{End}(\mathfrak{a})},P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})} -[d_1([x,y]),P_\mathfrak{b}]_{\mathrm{End}(\mathfrak{a})}\ :$ $=0$ Since $$H^1(\tilde{\mathfrak{g}},\mathcal{W})=0\ ,$$ one has $$c^1=\delta c^0\ .$$ Then, with $$\mathcal{P}_\mathfrak{b}=P_\mathfrak{b}-c^0\ ,$$ $[d_1(x),\mathcal{P}_\mathfrak{b}]=c^1(x)-\delta_1(x)c^0=c^1(x)-\delta c^0(x)=0$ 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 $$H^2(\tilde{\mathfrak{g}},\mathfrak{a})=0\ .$$