Notes for Lie algebra and Lie group
Notes for Lie algebra and Lie group 0.1. basic fact of Lie algebra
定义 1 (Normalizer). if K is a subspace of L, the normalizer of K to be the normalizer of K to be
\[N_{L}(K)=\{x\in L|\quad [x,K]\subset K\}\]
\[N_{L}(K)=\{x\in L|\quad [x,K]\subset K\}\]
习题 1. Prove
- $N_{L}(K)$ is a Lie sub-algebra.
- K is a Lie sub-algebra $\Leftrightarrow$ $K\subset N_{L}(K)$
定义 2 (centeralizer). Let Z be a subset of L, the centeralizer of Z is defined to be
\[C_{L}(Z)=\{x\in L|\quad[x,Z]=0\}\]
\[C_{L}(Z)=\{x\in L|\quad[x,Z]=0\}\]
习题 2. Prove that $C_{L}(Z)$ is a Lie sub-alg of L
定义 3 (center). \[Z(L)=C_{L}(L)={x\in L|\quad [x,z]=0,\forall z\in L}\]
习题 3. Z(L)is and ideal of L.
定义 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.
\[[I,J]=span\{[x,y]:x\in I,\quad y\in J\}\]
and $[L,L]$ is called the derivative algebra of L.
定义 5 (simple Lie algebra). A Lie algebra L is simple if L has nontrival ideals and it is not abelian.
注记 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.
定义 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.
\[L^{(0)}=L,\quad L^{(1)}=[L,L],\quad \dots L^{(n+1)}=[L^{(n)},L^{(n)}]\]
L is {\bf solvable} if $L^{(n)}$=0, for some n.
定义 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.
\[L^{0}=L,\quad L^{1}=[L,L],\quad L^{2}=[L^{1},L]\quad \dots L^{n+1}=[L^{n},L]\]
L is {\bf nilpotent} if $L^{n}$=0,for some n.
注记 2. $L^{(i)}\subset L^{i}$ $\rightarrow$ nilpotent lie algebra is solvable.
引理 8. if $[L,L]$is nilpotent, then L is solvable.
证明 . the proof is obvious, for $[L,L]$ is nilpotent, hence solvable, then L is solvable.
注记 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.
定理 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
定理 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 $x \in L$.
注记 4. If a sub algebra L of $gl(n,F)$ consists of nilpotent matrices, then L is nilpotent.
引理 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 $\lambda|_{[L,K]}=0$
证明 . 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, $\lambda([x,y])=0$.
命题 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.
定理 13. let L be a nilpotent Lie alg, and I is a non-zero ideal then $I\cap Z(L)\neq \emptyset$.
证明 . $ad:L\rightarrow gl(L)$, 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 $W_{\lambda}\neq 0$.
推论 14. $W_{\lambda}$ is surepres of $V$.
证明 . $\forall y\in L, v \in W_{L}, x\in K, [y,x]\in K$.
\[ xyv=[x,y]v+yxv=\lambda(x)yv\]
$\Rightarrow yv\in W_{\lambda}$
\[ xyv=[x,y]v+yxv=\lambda(x)yv\]
$\Rightarrow yv\in W_{\lambda}$
证明 (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 $W_{\lambda}$
定义 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 $
推论 16. Irreducible representation of solvable algebra are one dimensional.
注记 5. Lie’s theorem is true for any repres.
0.3. Jordan Chevalley decomposition
定理 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.
引理 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.
定理 19 (Jordan-Chevally decomp). ($\bar{F}=F$, dim V<$\infty$)
Let x $\in End(V)$
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 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 $x_{s}(B)\subset A$, $x_m(B)\subset A$.
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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