"Talk on Chevalley's integral forms"의 두 판 사이의 차이

2014년 3월 27일 (목) 14:09 판

introduction

motivating questions

why do we want integral forms?
$\mathfrak{g}_{\mathbb{Z}}$ is a Lie algebra over $\mathbb{Z}$

Serre's relations

틀:수학노트 에서 가져옴
l : 리대수 $\mathfrak{g}$의 rank
$(a_{ij})$ : 카르탄 행렬
생성원 $e_i,h_i,f_i , (i=1,2,\cdots, l)$
세르 관계식
- $\left[h_i,h_j\right]=0$
- $\left[e_i,f_j\right]=\delta _{i,j}h_i$
- $\left[h_i,e_j\right]=a_{i,j}e_j$
- $\left[h_i,f_j\right]=-a_{i,j}f_j$
- $\left(\text{ad} e_i\right){}^{1-a_{i,j}}\left(e_j\right)=0$ ($i\neq j$)
- $\left(\text{ad} f_i\right){}^{1-a_{i,j}}\left(f_j\right)=0$ ($i\neq j$)
ad 는 adjoint 의 약자
- $\left(\text{ad} x\right){}^{3}\left(y\right)=[x, [x, [x, y]]]$
- $\left(\text{ad} x\right){}^{4}\left(y\right)=[x, [x, [x, [x, y]]]]$

sl(3)의 예

카르탄 행렬\[\left( \begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array} \right)\]
$i\neq j$ 일 때\[\left(\text{ad} e_i\right){}^{2}\left(e_j\right)=[e_i, [e_i,e_j]]=0\]\[\left(\text{ad} f_i\right){}^{2}\left(f_j\right)=[f_i, [f_i,f_j]]=0\]
$e_1,e_2,h_1,h_2,f_1,f_2, \left[e_1,e_2\right], \left[f_1,f_2\right]$는 리대수의 기저가 된다

UEA 에서의 관계식

카르탄행렬이 $(a_{ij})$ 로 주어지는 리대수 $\mathfrak{g}$의 UEA $U(\mathfrak{g})$ 에서 다음의 두 식

\[\left(\text{ad} e_i\right){}^{1-a_{i,j}}\left(e_j\right)=0, \quad i\neq j\] \[\left(\text{ad} f_i\right){}^{1-a_{i,j}}\left(f_j\right)=0, \quad i\neq j\]

다음과 같이 표현할 수 있다\[\sum_{k=0}^{1-a_{i,j}}(-1)^k \binom{1-a_{i,j}}{k}e_{i}^{1-a_{i,j}-k}e_{j}e_{i}^k=0\]\[\sum_{k=0}^{1-a_{i,j}}(-1)^k \binom{1-a_{i,j}}{k}f_{i}^{1-a_{i,j}-k}f_{j}f_{i}^k=0\]
풀어 쓰면 다음과 같은 형태가 된다\[x\otimes x\otimes y-2 x\otimes y\otimes x+y\otimes x\otimes x\]\[x\otimes x\otimes x\otimes y-3 x\otimes x\otimes y\otimes x+3 x\otimes y\otimes x\otimes x-y\otimes x\otimes x\otimes x\]\[x\otimes x\otimes x\otimes x\otimes y-4 x\otimes x\otimes x\otimes y\otimes x+6 x\otimes x\otimes y\otimes x\otimes x-4 x\otimes y\otimes x\otimes x\otimes x+y\otimes x\otimes x\otimes x\otimes x\]

Chevalley

리대수 $\mathfrak{sl}(2)$

$L=\langle E,F,H \rangle$
commutator

\[[E,F]=H\] \[[H,E]=2E\]$[H,F]=-2F$

general case

thm

Chevalley bases exist

Q. why is it surprising?
for example, taking $2x_{\alpha}$ instead of $x_{\alpha}$ still gives integral Lie bracket.

Kostant

universal enveloping algebra의 PBW 기저

\[\{F^kH^lE^m|k,l,m\geq 0\}\]

thm

For each choice of $r_i,s_{\alpha}\geq 0$, form the product in the given order of the elements $$ \binom{h_i}{r_i} $$ and the elements $$ \frac{x_{\alpha}^{s_{\alpha}}}{s_{\alpha}!} $$ for $i=1,\cdots, \ell$ and $\alpha\in \Phi$. Then the resulting collection if a basis for $U_{\mathbb{Z}}$ as a free $\mathbb{Z}$-module

for examplem, one can take

\[\{\frac{F^k}{k!}\binom{H}{l}\frac{E^m}{m!}|k,l,m\geq 0\}\]

$\exp(tE)$ and $\exp(tF)$ exist
$\exp(tH)$ does not exist instead $(1+t)^{H}=1+\binom{H}{1}t+\binom{H^2}{2!}t^2+\cdots$ exists

@@ 40번째 줄: / 40번째 줄: @@
 ==Chevalley==
+===리대수 <math>\mathfrak{sl}(2)</math>===
+* <math>L=\langle E,F,H \rangle</math>
+*  commutator
+:<math>[E,F]=H</math>
+:<math>[H,E]=2E</math><math>[H,F]=-2F</math>
+===general case===
 ;thm
 Chevalley bases exist
 * Q. why is it surprising?
 * for example, taking $2x_{\alpha}$ instead of $x_{\alpha}$ still gives integral Lie bracket.
+==Kostant==
+*  universal enveloping algebra의 PBW 기저
+:<math>\{F^kH^lE^m|k,l,m\geq 0\}</math>
+;thm
+For each choice of $r_i,s_{\alpha}\geq 0$, form the product in the given order of the elements
+$$
+\binom{h_i}{r_i}
+$$
+and the elements
+$$
+\frac{x_{\alpha}^{s_{\alpha}}}{s_{\alpha}!}
+$$
+for $i=1,\cdots, \ell$ and $\alpha\in \Phi$. Then the resulting collection if a basis for $U_{\mathbb{Z}}$ as a free $\mathbb{Z}$-module
+* for examplem, one can take
+:<math>\{\frac{F^k}{k!}\binom{H}{l}\frac{E^m}{m!}|k,l,m\geq 0\}</math>
+* <math>\exp(tE)</math> and <math>\exp(tF)</math> exist
+* <math>\exp(tH)</math> does not exist instead <math>(1+t)^{H}=1+\binom{H}{1}t+\binom{H^2}{2!}t^2+\cdots</math> exists
 [[분류:talks]]
 [[분류:talks and lecture notes]]