Notes for Lie algebra and Lie group

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Notes for Lie algebra and Lie group 0.1. basic fact of Lie algebra

Definition 1 (Normalizer). if K is a subspace of L, the normalizer of K to be the normalizer of K to be

$N_{L}(K)=\{x\in L|\quad [x,K]\subset K\}$

Exercise 1. Prove

$N_{L}(K)$ is a Lie sub-algebra.
K is a Lie sub-algebra $\Leftrightarrow$ $K\subset N_{L}(K)$

Definition 2 (centeralizer). Let Z be a subset of L, the centeralizer of Z is defined to be

$C_{L}(Z)=\{x\in L|\quad[x,Z]=0\}$

Exercise 2. Prove that

$C_{L}(Z)$ is a Lie sub-alg of L

Definition 3 (center).

$Z(L)=C_{L}(L)={x\in L|\quad [x,z]=0,\forall z\in L}$

Exercise 3. Z(L)is and ideal of L.

Definition 4 (derivative algebra). Let I, J be ideals of L

$[I,J]=span\{[x,y]:x\in I,\quad y\in J\}$
and

$[L,L]$ is called the derivative algebra of L.

Definition 5 (simple Lie algebra). A Lie algebra L is simple if L has nontrival ideals and it is not abelian.

Remark 1. If L has no ideals except itself and 0. then

$Z(L)=0$ or L.

0.2. Solvable Lie algebra and nilpotent Lie algebra All lie algebra in the course is f.d. unless otherwise specified.

Definition 6 (Solvable). Define the derived of a Lie alg L to be a sequence of ideals.

$L^{(0)}=L,\quad L^{(1)}=[L,L],\quad \dots L^{(n+1)}=[L^{(n)},L^{(n)}]$
L is {\bf solvable} if

$L^{(n)}$ =0, for some n.

Definition 7 (Nilpotent). Define the tower central series of L to be

$L^{0}=L,\quad L^{1}=[L,L],\quad L^{2}=[L^{1},L]\quad \dots L^{n+1}=[L^{n},L]$
L is {\bf nilpotent} if

$L^{n}$ =0,for some n.

Remark 2.

$L^{(i)}\subset L^{i}$

$\rightarrow$ nilpotent lie algebra is solvable.

Lemma 8. if

$[L,L]$ is nilpotent, then L is solvable.

Proof . the proof is obvious, for

$[L,L]$ is nilpotent, hence solvable, then L is solvable.

Remark 3.

A nilpotent sub lie alg of $gl(n,F)$ does not mean that every element is nilpotent matrice. for example $FI_{n}\subset gl(n,F)$ ,this is a abel lie algebra.
Abel Lie algebra is nilpotent.

Theorem 9 (Engel’s theorem). Let V be a f.d. vec space, if

$L \subset gl(V)$ is a sub lie algebra. such that each x

$\in L$ is a nipotent endmorphism. Then

$\exists v\in V$ , and

$v\neq 0$ , such that

$xv=0$ , for all

$x\in L$ .

For solvable Lie algebra we have

Theorem 10 (Lie’s theorem). Let

$L\subset gl(V)$ be a solvable Lie algebra over F (char F=0, and

$\bar{F}=F$ ), then V contains common eigenvector for all

$x \in L$ .

Remark 4. If a sub algebra L of

$gl(n,F)$ consists of nilpotent matrices, then L is nilpotent.

Lemma 11. Char F=0, Let L be a Lie algebra, K be an ideal, and let

$\rho: L\rightarrow gl(V)$ are repres of L. Suppose v is a none zero element of V, such that

$xv=\lambda(x)v$ , for all

$x\in K$ , and some linear functional

$\lambda$ of K, then

$\lambda|_{[L,K]}=0$

Proof . Fix

$y\in L$ , let n be the largest number such that

$v, yv,\dots y^{n}v$ are linearly independent, denote the subspace then spanned by W. Let

$x\in K$ , then

$xv=\lambda(x)v$
$xyv=[x,y]v+yxv=\lambda([x,y])v+\lambda(x)yv$
$xy^{2}v=[x,y]yv+yxyv=[[x,y],y]v+2y[x,y]v+y^{2}xv$

i.e. relative to this basis of W, x is an upper triangular matrix whose diagonal values equal

$\lambda(x)$ . hence

$tr_{W}(x)=n\lambda(x)$ .

In particular $tr_{W}([x,y])=n\lambda([x,y])$ . however both x, y preserve W, hence their commutator has trace 0 on W, In particle $n\lambda([x,y])=0$ , for Char F=0, $\lambda([x,y])=0$ .

Proposition 12. Let L be a Lie algebra:

If L is solvable (nilpotent), then all its sub algebras and homomorphic images are solvable (nilpotent).
if I is solvable ideals such that $L/I$ is solvable, then L is solvable.
$L/Z(L)$ is nilpotent, so is L.
if I, J are solvable ideals, so is I+J.

Theorem 13. let L be a nilpotent Lie alg, and I is a non-zero ideal then

$I\cap Z(L)\neq \emptyset$ .

Proof .

$ad:L\rightarrow gl(L)$ , and I is ideal of L, hence I is a subrepres of the adjoint repres, since L is nilpotent for each

$x\in L$ ,

$ad(x)|_{L}$ is nilpotent, and

$ad(x)|_{I}$ is nilpotent.

Using Engel’s theorem, applied to $L\rightarrow gl(I)$ , there exists $y\in I$ , $y\neq 0$ , such that $ad(x)y=[x,y]=0$ , for all $x\in L$ , then $y\in Z(L)$ .

$\Rightarrow I\cap Z(L)\neq 0$

Continuing the lemma, define:

$W=W_{\lambda}=\{v\in V: xv=\lambda(x)v, \forall x\in K\}$
this lemma assumes

$W_{\lambda}\neq 0$ .

Corollary 14.

$W_{\lambda}$ is surepres of

$V$ .

Proof .

$\forall y\in L, v \in W_{L}, x\in K, [y,x]\in K$ .

$xyv=[x,y]v+yxv=\lambda(x)yv$

$\Rightarrow yv\in W_{\lambda}$

Proof (Lie’s proof). use induction on dim L. if dim L is solvable,

$[L,L]\neq L$ , let K be any codim 1 subspace of L that contains [L,L], then K is lie algebra (ideal), hence K is solvable ideal. Write

$L=K\oplus Fy$ , for some

$y\in L$ By induction, there exist a functional

$\lambda$ on K such that

$W_{\lambda}\neq 0$ , by lemma, y preserves W. Finally choose v to be any eigenvector of y on

$W_{\lambda}$

Definition 15 (irreducible). A repres V of L is called irreducible if V has no other subpres other than 0 and V. and a repres

$\rho L\rightarrow gl(V)$ is called faithful if

$ker\rho=0$

Corollary 16. Irreducible representation of solvable algebra are one dimensional.

Remark 5. Lie’s theorem is true for any repres.

0.3. Jordan Chevalley decomposition

Theorem 17. Let V be a f.d vec space over F, An endomorphism s

$\in End(V)$ is {\bf semi-simple} is one of the following conditions hold:\\

the minimal polynomial has no repeated roots.
the matrix is diagonalizable.
V admit a basis consisting of eigenvalues of s.

Lemma 18.

The sum of two commutable semi-simple endmorphism is again semi-simple.
if $s\in End V$ is semi-simple and s preserves s subspace W of V, then the restriction of s to W is semi-simple.
if s is semi-simple and nilpotent, then s=0.

Theorem 19 (Jordan-Chevally decomp). (

$\bar{F}=F$ , dim V<

$\infty$ )
Let x

$\in End(V)$

there exists unique $x_{s},x_{n}\in End(V)$ , such that $x=x_s+x_n$ , $x_s$ is semi-simple, $x_n$ is nilpotent, and $x_sx_n=x_nx_s$ .
there exist polynomials $P(T),Q(T)\in F(T)$ with no constants such that $x_s=p(x),x_{n}=q(x)$ .
if $A\subset B\subset V$ ,and $x(B)\subset A$ , then $x_{s}(B)\subset A$ , $x_m(B)\subset A$ .

\begin{proof}
suppose the characterization of x is

$(T-\lambda_{1})^{m_1}\dots(T-\lambda_{n})^{m_n}$ , where all

$\lambda_i$ are distinct. Then

$V=\oplus_{i=1}^{k}V_{\lambda_i}$ ,

$V\lambda_{i}=ker(x-\lambda_{i})$ .

let p(T) be any polynomials such that $P(T)=\lambda_{i} (mod (T-\lambda_i)^{m_{i}})$ , $\quad P(T)=0(mod T)$ if 0 is not an eigenvalue.

the set $x_{s}=p(x), x_{n}=x-p(x)=q(x)$ . (The existence of p(T) is just because Chinese Remainder theorem.) On each $V_{\lambda_i}$ , $x_{s}=\lambda_i$ , and $x_{n}$ is nilpotent on $V_{\lambda_i}$ .

hence $x_{s}$ is semi-simple $x_{n}$ is nilpotent. since $x_s$ and $x_n$ are polynomials in x, $x_{s}$ and $x_{n}$ commutes with each other. what is left is to prove uniquess.

$x=x_s+x_n=x’_s+x’_n$ , it is obvious that all four $x_s, x’_{s},x_n,x’_n$ commutes with each other. hence $x_{s}-x’_{s}$ is semi-simple, nilpotent. $x_s=x’_s$ .
\end{proof}

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