Talk on Chevalley's integral forms
introduction
motivating questions
- why do we want integral forms?
- what are good bases?
- Kostant found that the good integral forms are the ones with a structural base and showed that the universal enveloping algebras of finite dimensional semisimple Lie algebras have a structural base
- The fake monster formal group by Borcherds
- $\mathfrak{g}_{\mathbb{Z}}$ is a Lie algebra over $\mathbb{Z}$
- how can we check the consistency of Chevalley basis?
highest weight representations of $\mathfrak{sl}_2$
- \(V\) :유한차원인 기약표현
- \(V=\oplus_{\lambda\in\mathbb{C}}V_{\lambda}\), \(V_{\lambda}=\{v\in V|Hv=\lambda v\}\)
- \(\lambda\in \mathbb{C}\) 에 대하여, 다음의 조건을 만족하는 highest weight vector \(v_0\) 를 정의
\[Ev_0=0\] \[Hv_0=\lambda v_0\]
- \(v_j:=\frac{F^j}{j!}v_0\) 로 정의하면, 다음 관계가 만족된다
\[H v_j=(\lambda -2j)v_j\] \[F v_j=(j+1)v_{j+1}\] \[E v_j=(\lambda -j+1)v_{j-1}\]
- \(\{v_j|j\geq 0\}\) 가 생성하는 벡터공간이 유한차원인 $\mathfrak{g}$-모듈이 되려면, \(\lambda\in\mathbb{Z}, \lambda\geq 0\) 이 만족되어야 한다
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\[x\otimes x\otimes y-2 x\otimes y\otimes x+y\otimes x\otimes x\]\[x\otimes x\otimes x\otimes y-3 x\otimes x\otimes y\otimes x+3 x\otimes y\otimes x\otimes x-y\otimes x\otimes x\otimes x\]\[x\otimes x\otimes x\otimes x\otimes y-4 x\otimes x\otimes x\otimes y\otimes x+6 x\otimes x\otimes y\otimes x\otimes x-4 x\otimes y\otimes x\otimes x\otimes x+y\otimes x\otimes x\otimes x\otimes x\]
integral forms
- Let $\mathbb{Z}$ denote the integers. If $A$ is an algebra, over a field $\mathbb{F}$ of characteristic 0, define an integral form $A_\mathbb{Z}$ of $A$ to be a $\mathbb{Z}$-algebra such that $A_\mathbb{Z}\otimes_\mathbb{Z}\mathbb{F}=A$.
- An integral basis for $A$ is a $\mathbb{Z}$-basis for $A_\mathbb{Z}$.
- the theory integral forms for finite-dimensional simple Lie algebras was first studied by Chevalley in 1955. His work led to the construction of Chevalley groups
- Kostant found an integral form for the UEA of simple Lie algebra $\mathfrak{g}$
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.
procedure
- Lemma 7.3
The structure constants $N_{\alpha,\beta}$ for extraspecial pairs $(\alpha,\beta)$ can be chosen as arbitrary non-zero elements of $\mathbb{C}$ , by appropriate choice of the elements $e_{\alpha}$.
- Proposition 7.4
All the structure constants $N_{\alpha,\beta}$ are determined by the structure constants for extraspecial pairs.
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?
- a nice property of this integral form is
$$ \Delta(Z_{\alpha}) = \sum_{0\leq\beta\leq\alpha}Z_{\beta} \otimes Z_{\alpha−\beta}. $$
- for example, 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
example
- let us compute $e^2f^2$
$$ e^2f^2=2 h^2-8 fe-2 h+f^2e^2+4 fhe $$
- thus
$$ e^{(2)}f^{(2)}=\frac{h^2}{2}-2fe-\frac{h}{2}+f^{(2)}e^{(2)}+fhe $$
- so we cannot use $\frac{h^k}{k!}$ as elements of integral basis
- that's where $\binom{h}{2}=\frac{h^2}{2}-\frac{h}{2}$ comes from
remarks on Chevalley groups
- see theorem (6.11) of Curtis, 'Chevalley groups and related topics'
- For arbitrary field $k$ and a faithful representation $V$ of $\mathfrak{g}$, we can define the Chevalley group $G_{V,k}$.
- it actually depends on $k$ and the lattice of weights $\Gamma_{V}$ of $\mathfrak{g}$-module $V$
refs
- [H] J. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer, (1972).