\\title{Notes for Lie algebra and Lie group}
\n\\subsection{basic fact of Lie algebra}
\n\\begin{defn}[Normalizer]
\nif K is a subspace of L, the normalizer of K to be the normalizer of K to be
\n\\[N_{L}(K)=\\{x\\in L|\\quad [x,K]\\subset K\\}\\]
\n\\end{defn}
\n\\begin{excs}
\nProve
\n\\begin{itemize}
\n\\item $N_{L}(K)$ is a Lie sub-algebra.
\n\\item K is a Lie sub-algebra $\\Leftrightarrow$ $K\\subset N_{L}(K)$
\n\\end{itemize}
\n\\end{excs}
\n\\begin{defn}[centeralizer]
\nLet Z be a subset of L, the centeralizer of Z is defined to be
\n\\[C_{L}(Z)=\\{x\\in L|\\quad[x,Z]=0\\}\\]
\n\\end{defn}
\n\\begin{excs}
\nProve that $C_{L}(Z)$ is a Lie sub-alg of L
\n\\end{excs}
\n\\begin{defn}[center]
\n\\[Z(L)=C_{L}(L)={x\\in L|\\quad [x,z]=0,\\forall z\\in L}\\]
\n\\end{defn}
\n\\begin{excs}
\nZ(L)is and ideal of L.
\n\\end{excs}
\n\\begin{defn}[derivative algebra]
\nLet I, J be ideals of L
\n\\[[I,J]=span\\{[x,y]:x\\in I,\\quad y\\in J\\}\\]
\nand $[L,L]$ is called the derivative algebra of L.
\n\\end{defn}
\n\\begin{defn}[simple Lie algebra]
\nA Lie algebra L is simple if L has nontrival ideals and it is not abelian.
\n\\end{defn}
\n\\begin{rem}
\nIf L has no ideals except itself and 0. then $Z(L)=0$ or L.
\n\\end{rem}
\n\\subsection{Solvable Lie algebra and nilpotent Lie algebra}
\nAll lie algebra in the course is f.d. unless otherwise specified.
\n\\begin{defn}[Solvable]
\nDefine the derived of a Lie alg L to be a sequence of ideals.
\n\\[L^{(0)}=L,\\quad L^{(1)}=[L,L],\\quad \\dots L^{(n+1)}=[L^{(n)},L^{(n)}]\\]
\nL is {\\bf solvable} if $L^{(n)}$=0, for some n.
\n\\end{defn}
\n\\begin{defn}[Nilpotent]
\nDefine the tower central series of L to be
\n\\[L^{0}=L,\\quad L^{1}=[L,L],\\quad L^{2}=[L^{1},L]\\quad \\dots L^{n+1}=[L^{n},L]\\]
\nL is {\\bf nilpotent} if $L^{n}$=0,for some n.
\n\\end{defn}
\n\\begin{rem}
\n$L^{(i)}\\subset L^{i}$ $\\rightarrow$ nilpotent lie algebra is solvable.
\n\\end{rem}
\n\\begin{lem}
\nif $[L,L]$is nilpotent, then L is solvable.
\n\\end{lem}
\n\\begin{proof}
\nthe proof is obvious, for $[L,L]$ is nilpotent, hence solvable, then L is solvable.
\n\\end{proof}
\n\\begin{rem}
\n\\begin{itemize}
\n\\item 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.
\n\\item Abel Lie algebra is nilpotent.
\n\\end{itemize}
\n\\end{rem}
\n\\begin{thm}[Engel’s theorem]
\nLet 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 $x\\in L$.
\n\\end{thm}
\nFor solvable Lie algebra we have
\n\\begin{thm}[Lie’s theorem]
\nLet $L\\subset gl(V)$ be a solvable Lie algebra over F (char F=0, and $\\bar{F}=F$), then V contains common eigenvector for all $x \\in L$.
\n\\end{thm}
\n\\begin{rem}
\nIf a sub algebra L of $gl(n,F)$ consists of nilpotent matrices, then L is nilpotent.
\n\\end{rem}
\n\\begin{lem}
\nChar 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 $\\lambda|_{[L,K]}=0$
\n\\end{lem}
\n\\begin{proof}
\nFix $y\\in L$, 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
\n\\begin{itemize}
\n\\item $xv=\\lambda(x)v$
\n\\item $xyv=[x,y]v+yxv=\\lambda([x,y])v+\\lambda(x)yv$
\n\\item $xy^{2}v=[x,y]yv+yxyv=[[x,y],y]v+2y[x,y]v+y^{2}xv$
\n\\end{itemize}
\n\\end{proof}
\ni.e. relative to this basis of W, x is an upper triangular matrix whose diagonal values equal $\\lambda(x)$. hence $tr_{W}(x)=n\\lambda(x)$.<\/p>\n
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, $\\lambda([x,y])=0$.
\n\\begin{prop}
\nLet L be a Lie algebra:
\n\\begin{itemize}
\n\\item If L is solvable (nilpotent), then all its sub algebras and homomorphic images are solvable (nilpotent).
\n\\item if I is solvable ideals such that $L\/I$ is solvable, then L is solvable.
\n\\item $L\/Z(L)$ is nilpotent, so is L.
\n\\item if I, J are solvable ideals, so is I+J.
\n\\end{itemize}
\n\\end{prop}
\n\\begin{thm}
\nlet L be a nilpotent Lie alg, and I is a non-zero ideal then $I\\cap Z(L)\\neq \\emptyset$.
\n\\end{thm}
\n\\begin{proof}
\n$ad:L\\rightarrow gl(L)$, 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.<\/p>\n
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 $y\\in Z(L)$.<\/p>\n
$\\Rightarrow I\\cap Z(L)\\neq 0$
\n\\end{proof}
\nContinuing the lemma, define:
\n\\[W=W_{\\lambda}=\\{v\\in V: xv=\\lambda(x)v, \\forall x\\in K\\}\\]
\nthis lemma assumes $W_{\\lambda}\\neq 0$.
\n\\begin{cor}
\n$W_{\\lambda}$ is surepres of $V$.
\n\\end{cor}
\n\\begin{proof}
\n$\\forall y\\in L, v \\in W_{L}, x\\in K, [y,x]\\in K$.
\n\\[ xyv=[x,y]v+yxv=\\lambda(x)yv\\]
\n$\\Rightarrow yv\\in W_{\\lambda}$
\n\\end{proof}
\n\\begin{proof}[Lie’s proof]
\nuse 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 $W_{\\lambda}$
\n\\end{proof}
\n\\begin{defn}[irreducible]
\nA 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 $
\n\\end{defn}
\n\\begin{cor}
\nIrreducible representation of solvable algebra are one dimensional.
\n\\end{cor}
\n\\begin{rem}
\nLie’s theorem is true for any repres.
\n\\end{rem}
\n\\subsection{Jordan Chevalley decomposition}
\n\\begin{thm}
\nLet 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:\\\\
\n\\begin{itemize}
\n\\item the minimal polynomial has no repeated roots.
\n\\item the matrix is diagonalizable.
\n\\item V admit a basis consisting of eigenvalues of s.
\n\\end{itemize}
\n\\end{thm}
\n\\begin{lem}
\n\\begin{itemize}
\n\\item The sum of two commutable semi-simple endmorphism is again semi-simple.
\n\\item 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.
\n\\item if s is semi-simple and nilpotent, then s=0.
\n\\end{itemize}
\n\\end{lem}
\n\\begin{thm}[Jordan-Chevally decomp]
\n($\\bar{F}=F$, dim V<$\\infty$)
\nLet x $\\in End(V)$
\n\\begin{itemize}
\n\\item 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 $x_sx_n=x_nx_s$.
\n\\item there exist polynomials $P(T),Q(T)\\in F(T)$ with no constants such that $x_s=p(x),x_{n}=q(x)$.
\n\\item if $A\\subset B\\subset V$,and $x(B)\\subset A$, then $x_{s}(B)\\subset A$, $x_m(B)\\subset A$.
\n\\end{itemize}
\n\\begin{proof}
\nsuppose the characterization of x is $(T-\\lambda_{1})^{m_1}\\dots(T-\\lambda_{n})^{m_n}$, where all $\\lambda_i$ are distinct. Then $V=\\oplus_{i=1}^{k}V_{\\lambda_i}$, $V\\lambda_{i}=ker(x-\\lambda_{i})$.<\/p>\n
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.<\/p>\n
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 $V_{\\lambda_i}$.<\/p>\n
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.<\/p>\n
$x=x_s+x_n=x’_s+x’_n$, 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. $x_s=x’_s$.
\n\\end{proof}
\n\\end{thm}<\/p>\n","protected":false},"excerpt":{"rendered":"
\\title{Notes for Lie algebra and Lie gro …<\/p>\n