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

## 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.

### proof

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

### proposition - trace form invariant

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

### proof

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

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

### proof

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 $\flat(x)(y)=K_\mathfrak{a}(x,y)$

### proposition

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

### proof

$\flat([x,y])(z)=K_\mathfrak{a}([x,y],z)\ :$ $=-K_\mathfrak{a}(y,[x,z])\ :$ $=-\flat(y)([x,z])\ :$ $=d_1^{1}(x)\flat(y)(z)$ 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 $$x-\sharp(\flat(x))\in \ker K_\mathfrak{a}\ .$$

### proposition

$$\ker K_\mathfrak{a}$$ is an ideal.

### proof

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.

### definition

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 $$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\ .$$

### proof

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

### proposition

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

### proof

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 $$x=0\ .$$

### proposition

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 $$\mathfrak{g}\ .$$

### proof

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 $$\mathfrak{g}$$

### corollary

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

### remark

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.

### proof

Let $$\mathfrak{h}$$ be a subalgebra.

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

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

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

### proof

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.

### proposition

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

### proof

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.

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

### proof

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

### proof

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 $$\mathrm{Rad\ }\mathfrak{g}\ .$$

### proposition

$$\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}$$ is semisimple.

### proof

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 $$\mathfrak{g}/\mathrm{Rad\ }\mathfrak{g}\ .$$

### proposition

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

### proof

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 $$\mathfrak{g}\ .$$

### proof

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.

### proof

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 $0=\mathfrak{g}_0\subset\mathfrak{g}_1\subset\dots\subset\mathfrak{g}_n=\mathfrak{g},\quad \dim\mathfrak{g}_i=i$

### proof

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.

### lemma

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

### proof

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.

### remark

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