Talk on Chevalley's integral forms

introduction

motivating questions

• why do we want integral forms?
• what are good bases?
• $\mathfrak{g}_{\mathbb{Z}}$ 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 or non-trivial?
• tentative answer : can we check the Jacobi identity?
• 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

• See [H] chapter 26?
• 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

refs

• [H] J. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer, (1972).