lecture 8b

From Scholarpedia

< An introduction to Lie algebra cohomology
This article has not been peer-reviewed or accepted for publication yet; It may be unfinished, contain inaccuracies, or unapproved changes.

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

On to the ninth lecture

Back to the eighth lecture

Contents

[edit]

the Casimir operator

Let \tilde{\mathfrak{g}}=\mathfrak{g}/\ker d^{(0)} (This makes sense, since \ker d^{(0)} 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.

[edit]

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
[edit]

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.

[edit]

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
[edit]

corollary

c_{ij}^k is fully antisymmetric in its indices.

[edit]

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
[edit]

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})
[edit]

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.

[edit]

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)
[edit]

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
[edit]

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.

[edit]

lemma

Suppose \dim\mathfrak{a}<\infty. Then

\gamma^0 d^{(0)}(x)=d^{(0)}(x)\gamma^0
[edit]

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
[edit]

corollary

The map \gamma^0 is a \mathfrak{g}-endomorphism.

[edit]

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.

[edit]

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 mth 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.

[edit]

theorem

Let d^{(0)} be a representation. 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).

[edit]

remark

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

[edit]

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

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

[edit]

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}).

[edit]

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}).

[edit]

corollary

Let d^{(0)} 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.

[edit]

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.

[edit]

lemma

If [\tilde{\mathfrak{g}},\tilde{\mathfrak{g}}]=\tilde{\mathfrak{g}} then H^1(\tilde{\mathfrak{g}},\mathbb{C})=0.

[edit]

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).

[edit]

corollary

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.

[edit]

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].

[edit]

proposition

The \mathfrak{g}^i are ideals of \mathfrak{g}.

[edit]

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.

[edit]

definition

\mathfrak{g} is called nilpotent if there is an n\in\mathbb{N} such that \mathfrak{g}^n=0.

[edit]

proposition

A nilpotent Lie algebra is solvable, but a solvable Lie algebra need not be nilpotent.

[edit]

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.

[edit]

proposition

If \mathfrak{g} is nilpotent, then so are all subalgebras and homomorphic images.

[edit]

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}
[edit]

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.

[edit]

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.

[edit]

proposition

If \mathfrak{g} is nilpotent and nonzero, then so is \mathcal{Z}(\mathfrak{g})\neq 0.

[edit]

proof

Let n be the minimal order such that \mathfrak{g}^n=0, then \mathfrak{g}^{n-1}\subset \mathcal{Z}(\mathfrak{g}).

[edit]

lemma

If x\in \mathfrak{gl}(V) is nilpotent, then ad(x) is nilpotent.

[edit]

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.

[edit]

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.

[edit]

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.

[edit]

theorem (Engel)

If all all elements of \mathfrak{g} are ad-nilpotent, then \mathfrak{g} is nilpotent.

[edit]

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.

[edit]

corollary

If \mathfrak{g} is solvable, then [\mathfrak{g},\mathfrak{g}] is nilpotent.

[edit]

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).

[edit]

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.

[edit]

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.

[edit]

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.

[edit]

proof




In progress...





[edit]

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])
[edit]

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)
[edit]

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.

[edit]

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.

[edit]

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.

[edit]

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}).

[edit]

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}.

[edit]

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.

[edit]

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.

[edit]

proof

This follows from the fact that H^2(\tilde{\mathfrak{g}},\mathfrak{a})=0.

[edit]

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.


On to the ninth lecture

Back to the seventh lecture

Retrieved from "http://www.scholarpedia.org/article/An_introduction_to_Lie_algebra_cohomology/lecture_8b"
For authors