# Notes for Lie algebra and Lie group

Yaoxiaothu/ 3月 3, 2012/ 数学笔记/ 0 comments

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

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

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

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

$[I,J]=span\{[x,y]:x\in I,\quad y\in J\}$
and $[L,L]$ is called the derivative algebra of L.

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

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

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

• 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.

For solvable Lie algebra we have

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

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, $\lambda([x,y])=0$.

• 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.

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 $W_{\lambda}\neq 0$.

$xyv=[x,y]v+yxv=\lambda(x)yv$
$\Rightarrow yv\in W_{\lambda}$

0.3. Jordan Chevalley decomposition

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

• 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.

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}