lecture 8

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< An introduction to Leibniz algebra cohomology

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Author: Dr. Jan A. Sanders, Vrije Universiteit Amsterdam

Back to the seventh lecture

On to lecture 8b

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The trace and Killing form

Let $R$ be $\mathbb{C}$ and $\dim_\mathbb{C}\mathfrak{a}<\infty$ . Then define $K_\mathfrak{a}\in C^2(\mathfrak{g},\mathbb{C})$ by

$K_\mathfrak{a}(x,y)=\mathrm{tr}(d_+^{(0)}(x) d_+^{(0)}(y))$

In the case $\mathfrak{a}=\mathfrak{g}$ and $d_+^{(0)}=\mathrm{ad}_+$ , this is called the Killing form. In general, one calls $K_\mathfrak{a}$ the trace form.

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example

Let $\mathfrak{g}=\mathfrak{sl}_2$ and $\mathfrak{a}=\R^2$ , with the standard representation (see Lecture 1). Then

$K_{\R^2}(M,M)=0, \quad K_{\R^2}(M,N)=1,\quad K_{\R^2}(M,H)=0,$

$K_{\R^2}(N,M)=1, \quad K_{\R^2}(N,N)=0,\quad K_{\R^2}(N,H)=0,$

$K_{\R^2}(H,M)=0, \quad K_{\R^2}(H,N)=0,\quad K_{\R^2}(H,H)=2.$

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proposition

$K_\mathfrak{a}$ is symmetric.

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proof

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

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proposition

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

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proof

Given the trivial action of $\mathfrak{g}$ on $\mathbb{C}$ , one has

$d^{(2)}(x)K_\mathfrak{a}(y,z)=-K_\mathfrak{a}([x,y],z)-K_\mathfrak{a}(y,[x,z])$

$=-\mathrm{tr}(d_+^{(0)}([x,y]) d_+^{(0)}(z))-\mathrm{tr}(d_+^{(0)}(y) d_+^{(0)}([x,z]))$

$=-\mathrm{tr}(d_+^{(0)}(x) d_+^{(0)}(y) d_+^{(0)}(z))+\mathrm{tr}(d_+^{(0)}(y) d_+^{(0)}(x) d_+^{(0)}(z)) -\mathrm{tr}(d_+^{(0)}(y) d_+^{(0)}(x)d_+^{(0)}(z))+\mathrm{tr}(d_+^{(0)}(y) d_+^{(0)}(z)d_+^{(0)}(x))$

$=0$

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proposition

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

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proof

From the $\mathfrak{g}$ -invariance it follows that

$d^2 K_\mathfrak{a}(x,y,z)=K_\mathfrak{a}(x,[y,z])$

Furthermore,

$K_\mathfrak{a}(x,[z,y])=\mathrm{tr}(d_+^{(0)}(x) d_+^{(0)}([z,y]))$

$=-\mathrm{tr}(d_+^{(0)}(x) d_+^{(0)}([y,z]))$

$=-K_\mathfrak{a}(x,[y,z])$

and

$K_\mathfrak{a}(z,[x,y])=-K_\mathfrak{a}(z,[y,x])=K_\mathfrak{a}([y,z],x)=K_\mathfrak{a}(x,[y,z])$ $\square$

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corollary

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.

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musical maps

Let $\mathfrak{g}^\star=C^1(\mathfrak{g},\mathbb{C})$ and define $\flat: \mathfrak{g}\rightarrow \mathfrak{g}^\star$ by

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

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proposition

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

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proof

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

$=-K_\mathfrak{a}(y,[x,z])$

$=-\flat(y)([x,z])$

$=d^{(1)}(x)\flat(y)(z)$

or $\flat([x,y])=d^{(1)}(x)\flat(y)\quad \square$ .

Define

$\sharp:\mathfrak{g}^\star\rightarrow \mathfrak{g}$

$K_\mathfrak{a}(\sharp(c^1),y)=c^1(y)$

Then

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

or $x-\sharp(\flat(x))\in \ker K_\mathfrak{a}$ .

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proposition

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

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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.

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definition

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

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proposition

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

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proof

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

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proposition

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

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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 $x=0$ .

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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 $\mathfrak{g}$ .

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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 $[\mathfrak{h}^\perp,\mathfrak{g}]\subset \mathfrak{h}^\perp$ .

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definition

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.

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well defined

$\mathfrak{g}^{(0)}$ is an ideal in $\mathfrak{g}$ . Suppose that $\mathfrak{(i)}$ 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 $\mathfrak{g}$

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corollary

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

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remark

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

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proposition

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

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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

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

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proposition

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

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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.

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proposition

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

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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.

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propostion

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

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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, $\mathfrak{h}\subset\mathfrak{s}$ .

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definition

A Leibniz algebra $\mathfrak{g}$ is called semisimple if $\mathrm{Rad\ }\mathfrak{g}=0$ .

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proposition

If $\mathfrak{g}$ is simple, it is semisimple

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

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proposition

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

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

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proposition

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

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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.

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theorem

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 $\mathfrak{g}$ .

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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 $\mathfrak{g}$ .

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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 $i=1,\dots,n$ .

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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.

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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.

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lemma

Let $\mathfrak{g}$ be solvable. Then there exists a flag of ideals

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

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proof

Let $d_+^{(0)}:\mathfrak{g}\rightarrow \mathfrak{gl}(\mathfrak{a})$ be a finite dimensional representation of $\mathfrak{g}$ . Then $d_+^{(0)}(\mathfrak{g})$ is solvable, and stabilizes a flag in $\mathfrak{a}$ . Take $\mathfrak{a}=\mathfrak{g}$ and $d_+^{(0)}=\mathrm{ad}_+$ , then the $\mathfrak{g}_i$ are ideals (since they are $\mathfrak{g}$ -invariant) and they obey the flag condition.

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lemma

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

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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.

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remark

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

On to lecture 8b

Back to the seventh lecture

Invited by:

Dr. Eugene M. Izhikevich, Editor-in-Chief of Scholarpedia, the peer-reviewed open-access encyclopedia

Retrieved from "http://www.scholarpedia.org/article/An_introduction_to_Leibniz_algebra_cohomology/lecture_8"

lecture 8

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Contents

The trace and Killing form

example

proposition

proof

proposition

proof

proposition

proof

corollary

musical maps

proposition

proof

proposition

proof

definition

proposition

proof

proposition

proof

proposition

proof

definition

well defined

corollary

remark

proposition

proof

proposition

proof

proposition

proof

propostion

proof

definition

proposition

proof

proposition

proof

proposition

proof

theorem

proof

definition (flag)

theorem (Lie)

proof

lemma

proof

lemma

proof

remark

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