"Talk on Chevalley's integral forms"의 두 판 사이의 차이
imported>Pythagoras0 |
imported>Pythagoras0 |
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57번째 줄: | 57번째 줄: | ||
$$[x_{\alpha},x_{\beta}]=n_{\alpha,\beta}x_{\alpha+\beta}$$ | $$[x_{\alpha},x_{\beta}]=n_{\alpha,\beta}x_{\alpha+\beta}$$ | ||
is not fixed by the above condition | is not fixed by the above condition | ||
− | * but if we make another choice $x_{\alpha}'=u_{\alpha}x_{\alpha}$ with $u_{\alpha}u_{\alpha} | + | * but if we make another choice $x_{\alpha}'=u_{\alpha}x_{\alpha}$ with $u_{\alpha}u_{-\alpha}=1$, then structure constants satisfy the following property |
$$ | $$ | ||
n_{\alpha,\beta}'n_{-\alpha,-\beta}'=n_{\alpha,\beta}n_{-\alpha,-\beta} | n_{\alpha,\beta}'n_{-\alpha,-\beta}'=n_{\alpha,\beta}n_{-\alpha,-\beta} | ||
63번째 줄: | 63번째 줄: | ||
;lemma | ;lemma | ||
The number $n_{\alpha,\beta}n_{-\alpha,-\beta}$ is given by $-(p+1)^2$ where $p$ is the largest integer $\geq 0$ such that $\beta-p\alpha\in \Delta$ | The number $n_{\alpha,\beta}n_{-\alpha,-\beta}$ is given by $-(p+1)^2$ where $p$ is the largest integer $\geq 0$ such that $\beta-p\alpha\in \Delta$ | ||
− | |||
===general case=== | ===general case=== |
2014년 3월 29일 (토) 00:47 판
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
- a synthesis between the theory of Lie groups and the theory of finite groups
리대수 \(\mathfrak{sl}(2)\)
- \(L=\langle E,F,H \rangle\)
- commutator
\[ [E,F]=H \\ [H,E]=2E \\ [H,F]=-2F \]
observation
- from the root system, we can fix $h_{\alpha}$ uniquely for each $\alpha\in \Delta$
- we can choose $x_{\alpha}$ so that $[x_{\alpha},x_{-\alpha}]=h_{\alpha}$
- the structure constants $n_{\alpha,\beta}$ where
$$[x_{\alpha},x_{\beta}]=n_{\alpha,\beta}x_{\alpha+\beta}$$ is not fixed by the above condition
- but if we make another choice $x_{\alpha}'=u_{\alpha}x_{\alpha}$ with $u_{\alpha}u_{-\alpha}=1$, then structure constants satisfy the following property
$$ n_{\alpha,\beta}'n_{-\alpha,-\beta}'=n_{\alpha,\beta}n_{-\alpha,-\beta} $$
- lemma
The number $n_{\alpha,\beta}n_{-\alpha,-\beta}$ is given by $-(p+1)^2$ where $p$ is the largest integer $\geq 0$ such that $\beta-p\alpha\in \Delta$
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).