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

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==introduction==
 
==introduction==
 
===motivating questions===
 
===motivating questions===
−
* why do we want integral forms?
+
* why do we want integral forms of an algebra?
 
* what are good bases?
 
* 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?
 
* how can we check the consistency of Chevalley basis?
  
  
−
==highest weight representations of $\mathfrak{sl}_2$==
+
==integral forms==
 +
* $A$ algebra over $\mathbb{C}$ (for any field $F$ of characteristic 0)
 +
;def
 +
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 $A_\mathbb{Z}$.
 +
* Chevalley 1955, integral forms for finite-dimensional simple Lie algebras
 +
** His work led to the construction of Chevalley groups
 +
* Kostant 1966, integral forms for the UEAs of simple Lie algebras
 +
** 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 (according to [[The fake monster formal group by Borcherds]])
 +
 
 +
 
 +
==review of basics on $\mathfrak{sl}_2$==
 +
===Lie algebra <math>\mathfrak{sl}(2)</math>===
 +
* <math>\mathfrak{g}=\mathbb{C}\langle E,F,H \rangle</math>
 +
* commutator
 +
:<math>
 +
[E,F]=H \\
 +
[H,E]=2E \\
 +
[H,F]=-2F
 +
</math>
 +
* <math>\mathfrak{g}_{\mathbb{Z}}=\mathbb{\mathbb{Z}}\langle E,F,H \rangle</math> is an integral form (so $\mathfrak{g}_{\mathbb{Z}}$ is a Lie algebra over $\mathbb{Z}$)
 +
 
 +
===UEA===
 +
* universal enveloping algebra의 PBW 기저 <math>\{F^kH^lE^m|k,l,m\geq 0\}</math>
 +
* what's $U(\mathfrak{g})_{\mathbb{Z}}$?
 +
 
 +
 
 +
===finite dimensional representations===
 
* <math>V</math> :유한차원인 기약표현
 
* <math>V</math> :유한차원인 기약표현
 
* <math>V=\oplus_{\lambda\in\mathbb{C}}V_{\lambda}</math>, <math>V_{\lambda}=\{v\in V|Hv=\lambda v\}</math>
 
* <math>V=\oplus_{\lambda\in\mathbb{C}}V_{\lambda}</math>, <math>V_{\lambda}=\{v\in V|Hv=\lambda v\}</math>
19번째 줄: 44번째 줄:
 
:<math>F v_j=(j+1)v_{j+1}</math>
 
:<math>F v_j=(j+1)v_{j+1}</math>
 
:<math>E v_j=(\lambda -j+1)v_{j-1}</math>
 
:<math>E v_j=(\lambda -j+1)v_{j-1}</math>
−
* <math>\{v_j|j\geq 0\}</math> 가 생성하는 벡터공간이 유한차원인 $\mathfrak{g}$-모듈이 되려면, <math>\lambda\in\mathbb{Z}, \lambda\geq 0</math> 이 만족되어야 한다
+
* to get a finite dimensional $\mathfrak{g}$-module $V$ spanned by <math>\{v_j|j\geq 0\}</math>, we need <math>\lambda\in\mathbb{Z}, \lambda\geq 0</math>
 +
;Question.
 +
where do $\frac{F^j}{j!}$ come from?
  
  
−
==Serre's relations==
+
==base of $\mathfrak{g}$ and structure constants==
−
* {{수학노트|url=세르_관계식_(Serre_relations)}} 에서 가져옴
+
===basis===
−
* l : 리대수 <math>\mathfrak{g}</math>의 rank
+
* on $\mathfrak{g}$, we have a non-deg bilinear form $(\cdot,\cdot)$.
−
* <math>(a_{ij})</math> : 카르탄 행렬
+
* fix $\mathfrak{h}$
−
* 생성원 <math>e_i,h_i,f_i , (i=1,2,\cdots, l)</math>
+
* $\Delta$ : root system
−
* 세르 관계식
+
* $\Pi$ : simple system (base of $\Delta$)
−
** <math>\left[h_i,h_j\right]=0</math>
+
* Cartan decomposition
−
** <math>\left[e_i,f_j\right]=\delta _{i,j}h_i</math>
+
$$
−
** <math>\left[h_i,e_j\right]=a_{i,j}e_j</math>
+
\mathfrak{g}=\mathfrak{h}\oplus \left(\oplus_{\alpha\in \Delta} \mathfrak{g}_{\alpha}\right)
−
** <math>\left[h_i,f_j\right]=-a_{i,j}f_j</math>
+
$$
−
** <math>\left(\text{ad} e_i\right){}^{1-a_{i,j}}\left(e_j\right)=0</math> (<math>i\neq j</math>)
+
* fix $H_{\alpha}$ uniquely for each $\alpha\in \Delta$ by
−
** <math>\left(\text{ad} f_i\right){}^{1-a_{i,j}}\left(f_j\right)=0</math> (<math>i\neq j</math>)
+
$$
−
*  ad 는 adjoint 의 약자
+
\beta(H_{\alpha})=2\frac{(\alpha,\beta)}{(\alpha,\alpha)}\,\quad \beta\in \mathfrak{h}^{*}
−
** <math>\left(\text{ad} x\right){}^{3}\left(y\right)=[x, [x, [x, y]]]</math>
+
$$
−
** <math>\left(\text{ad} x\right){}^{4}\left(y\right)=[x, [x, [x, [x, y]]]]</math>
+
* we can choose $x_{\alpha}\in \mathfrak{g}_{\alpha}$ so that
 +
$$[x_{\alpha},x_{-\alpha}]=h_{\alpha}$$
 +
* structure constants $n_{\alpha,\beta}$
 +
$$[x_{\alpha},x_{\beta}]=n_{\alpha,\beta}x_{\alpha+\beta}$$
 +
* $n_{\alpha,\beta}\neq 0$ only if $\alpha+\beta\in \Delta$
 +
* $n_{\alpha,\beta}$ is not fixed by the above condition
  
−
 
+
===structure constants===
−
===sl(3)의 예===
+
* taken from [[Lie Algebras of Finite and Affine Type by Carter]]
−
 
+
;Lemma 7.3
−
*  카르탄 행렬:<math>\left( \begin{array}{cc}  2 & -1 \\  -1 & 2 \end{array} \right)</math>
+
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 $e_{\alpha}$.
−
* <math>i\neq j</math> 일 때:<math>\left(\text{ad} e_i\right){}^{2}\left(e_j\right)=[e_i, [e_i,e_j]]=0</math>:<math>\left(\text{ad} f_i\right){}^{2}\left(f_j\right)=[f_i, [f_i,f_j]]=0</math>
 
−
* $e_1,e_2,h_1,h_2,f_1,f_2, \left[e_1,e_2\right], \left[f_1,f_2\right]$는 리대수의 기저가 된다
 
−
 
 
−
 
 
−
===UEA 에서의 관계식===
 
−
 
 
−
*  카르탄행렬이 <math>(a_{ij})</math> 로 주어지는 리대수 <math>\mathfrak{g}</math>의 UEA <math>U(\mathfrak{g})</math> 에서 다음의 두 식
 
−
:<math>\left(\text{ad} e_i\right){}^{1-a_{i,j}}\left(e_j\right)=0, \quad i\neq j</math>
 
−
:<math>\left(\text{ad} f_i\right){}^{1-a_{i,j}}\left(f_j\right)=0, \quad i\neq j</math>
 
−
*  다음과 같이 표현할 수 있다:<math>\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</math>:<math>\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</math>
 
−
*  풀어 쓰면 다음과 같은 형태가 된다:<math>x\otimes x\otimes y-2 x\otimes y\otimes x+y\otimes x\otimes x</math>:<math>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</math>:<math>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</math>
 
−
 
 
−
 
 
−
==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 $\mathfrak{g}$
 
  
  
 +
;Proposition 7.4
 +
All the structure constants $n_{\alpha,\beta}$ are determined by the structure constants for extraspecial pairs.
  
  
 
==Chevalley==
 
==Chevalley==
 
* a synthesis between the theory of Lie groups and the theory of finite groups  
 
* a synthesis between the theory of Lie groups and the theory of finite groups  
−
===리대수 <math>\mathfrak{sl}(2)</math>===
+
 
−
* <math>L=\langle E,F,H \rangle</math>
 
−
*  commutator
 
−
:<math>
 
−
[E,F]=H \\
 
−
[H,E]=2E \\
 
−
[H,F]=-2F
 
−
</math>
 
  
 
===observation===
 
===observation===
−
* from the root system, we can fix $h_{\alpha}$ uniquely for each $\alpha\in \Delta$
+
* if we make another choice $x_{\alpha}'=u_{\alpha}x_{\alpha}$ with $u_{\alpha}u_{-\alpha}=1$, then structure constants satisfy the following property
−
* 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}
 
n_{\alpha,\beta}'n_{-\alpha,-\beta}'=n_{\alpha,\beta}n_{-\alpha,-\beta}
 
$$
 
$$
 
;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 $\beta-p\alpha\in \Delta$. ($\alpha$ string through $\beta$)
  
 +
;lemma
 +
It is possible to choose basis elements $x_{\alpha}\in \mathfrak{g}_{\alpha}$ such that $[x_{\alpha},x_{-\alpha}]=H_{\alpha}$, and $n_{-\alpha,-\beta}=-n_{\alpha,\beta}$ for all $\alpha$ and $\beta$. For this choice of $x_{\alpha}$, we have $n_{\alpha,\beta}=\pm (p+1)$
  
 +
Hint : Use the Chevalley involution $\sigma :\mathfrak{g}\to \mathfrak{g}$. It is an involution with $\sigma(h)=-h$ for any $h\in \mathfrak{h}$ and $\sigma(\mathfrak{g}_{\alpha})=\mathfrak{g}_{-\alpha}$.
  
−
===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===
 
−
* taken from [[Lie Algebras of Finite and Affine Type by Carter]]
 
−
;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.
 
  
 +
===Chevalley basis===
 +
;thm (Chevalley 1955)
 +
The elements $\{H_{\alpha_i} : \alpha_i\in \Pi\}$ together with elements $X_{\alpha}\in \mathfrak{g}_{\alpha}$ ($\alpha\in \Delta$) chosen to satisfy $[X_{\alpha},X_{-\alpha}]=H_{\alpha}$ and $[X_{\alpha},X_{\beta}]=\pm (p+1) X_{\alpha+\beta}$ (if $\alpha+\beta\in \Delta)$ form a basis for a $\mathbb{Z}$-form $\mathfrak{g}_{\mathbb{Z}}$ of $\mathfrak{g}$.
  
  
 +
* 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==
 
==Kostant==
−
* universal enveloping algebra의 PBW 기저
+
* Let $\{X_{\alpha}\}$ and $\{H_{\alpha_i}\}$ be a Chevalley basis for $\mathfrak{g}$
−
:<math>\{F^kH^lE^m|k,l,m\geq 0\}</math>
+
* let $\Delta^{+}=\{\alpha_1,\cdots, \alpha_N\}$
−
;thm
+
* for $Q=(q_1,\cdots, q_N)$ with $q_i$ non-negative integers, put
−
For each choice of $r_i,s_{\alpha}\geq 0$, form the product in the given order of the elements
+
$$
 +
e_{\pm Q}=\prod_{i=1}^N (X_{\pm \alpha_i}^{q_i}{(q_i)!}
 
$$
 
$$
−
\binom{h_i}{r_i}
+
* for $x\in \mathfrak{g}$ and $s\in \mathbb{Z}_{\geq 0}$, put
 
$$
 
$$
−
and the elements
+
\binom{x}{s}=\frac{x(x-1)\cdots (x-s+1)}{s!}\in U(\mathfrak{g})
 
$$
 
$$
−
\frac{x_{\alpha}^{s_{\alpha}}}{s_{\alpha}!}
+
* let $n$ be the rank of $\mathfrak{g}$ for each $n$-tuple $P=(p_i)_{1\leq i \leq n}$, define
 
$$
 
$$
−
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
+
h_{P}=\prod_{i=1}^{n}\binom{H_{\alpha_i}}{p_i}
−
* 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}.
+
;thm (Kostant 1966)
 +
The elements
 
$$
 
$$
−
* for example, one can take
+
\{e_{-Q}h_Pe_{S}\}
 +
$$ for all $Q,P,S$ form an integral basis for $U_{\mathbb{Z}}$.
 +
 
 +
;proof
 +
See '''[H]''' chapter 26.
 +
 
 +
 
 +
===example===
 +
* for $\mathfrak{g}=\mathfrak{sl}_2$,
 
:<math>\{\frac{F^k}{k!}\binom{H}{l}\frac{E^m}{m!}|k,l,m\geq 0\}</math>
 
:<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
 
−
===example===
 
 
* let us compute $e^2f^2$
 
* let us compute $e^2f^2$
 
$$
 
$$
143번째 줄: 145번째 줄:
 
* so we cannot use $\frac{h^k}{k!}$ as elements of integral basis
 
* 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
 
* that's where $\binom{h}{2}=\frac{h^2}{2}-\frac{h}{2}$ comes from
 +
 +
 +
===properties===
 +
* <math>\exp(tE)</math> and <math>\exp(tF)</math> exist in $U_{\mathbb{Z}}[[t]]$
 +
* <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 in $U_{\mathbb{Z}}[[t]]$
 +
* a nice property of this integral form is
 +
$$
 +
\Delta(Z_{\alpha}) = \sum_{0\leq\beta\leq\alpha}Z_{\beta} \otimes Z_{\alpha−\beta}.
 +
$$
 +
where $\Delta : U(\mathfrak{g})\to U(\mathfrak{g})$ is the coproduct defined by
 +
$$
 +
\Delta(x)=x\otimes 1+1\otimes x
 +
$$
 +
for $x\in \mathfrak{g}$
  
  
149번째 줄: 165번째 줄:
 
* For arbitrary field $k$ and a faithful representation $V$ of $\mathfrak{g}$, we can define the Chevalley group $G_{V,k}$.
 
* 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 $V$
 
* it actually depends on $k$ and the lattice of weights $\Gamma_{V}$ of $\mathfrak{g}$-module $V$
 +
  
 
==refs==
 
==refs==

2014년 4월 1일 (화) 00:14 판

introduction

motivating questions

  • why do we want integral forms of an algebra?
  • what are good bases?
  • how can we check the consistency of Chevalley basis?


integral forms

  • $A$ algebra over $\mathbb{C}$ (for any field $F$ of characteristic 0)
def

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 $A_\mathbb{Z}$.

  • Chevalley 1955, integral forms for finite-dimensional simple Lie algebras
    • His work led to the construction of Chevalley groups
  • Kostant 1966, integral forms for the UEAs of simple Lie algebras
    • 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 (according to The fake monster formal group by Borcherds)


review of basics on $\mathfrak{sl}_2$

Lie algebra \(\mathfrak{sl}(2)\)

  • \(\mathfrak{g}=\mathbb{C}\langle E,F,H \rangle\)
  • commutator

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

  • \(\mathfrak{g}_{\mathbb{Z}}=\mathbb{\mathbb{Z}}\langle E,F,H \rangle\) is an integral form (so $\mathfrak{g}_{\mathbb{Z}}$ is a Lie algebra over $\mathbb{Z}$)

UEA

  • universal enveloping algebra의 PBW 기저 \(\{F^kH^lE^m|k,l,m\geq 0\}\)
  • what's $U(\mathfrak{g})_{\mathbb{Z}}$?


finite dimensional representations

  • \(V\) :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

\[H v_j=(\lambda -2j)v_j\] \[F v_j=(j+1)v_{j+1}\] \[E v_j=(\lambda -j+1)v_{j-1}\]

  • to get a finite dimensional $\mathfrak{g}$-module $V$ spanned by \(\{v_j|j\geq 0\}\), we need \(\lambda\in\mathbb{Z}, \lambda\geq 0\)
Question.

where do $\frac{F^j}{j!}$ come from?


base of $\mathfrak{g}$ and structure constants

basis

  • on $\mathfrak{g}$, we have a non-deg bilinear form $(\cdot,\cdot)$.
  • fix $\mathfrak{h}$
  • $\Delta$ : root system
  • $\Pi$ : simple system (base of $\Delta$)
  • Cartan decomposition

$$ \mathfrak{g}=\mathfrak{h}\oplus \left(\oplus_{\alpha\in \Delta} \mathfrak{g}_{\alpha}\right) $$

  • fix $H_{\alpha}$ uniquely for each $\alpha\in \Delta$ by

$$ \beta(H_{\alpha})=2\frac{(\alpha,\beta)}{(\alpha,\alpha)}\,\quad \beta\in \mathfrak{h}^{*} $$

  • we can choose $x_{\alpha}\in \mathfrak{g}_{\alpha}$ so that

$$[x_{\alpha},x_{-\alpha}]=h_{\alpha}$$

  • structure constants $n_{\alpha,\beta}$

$$[x_{\alpha},x_{\beta}]=n_{\alpha,\beta}x_{\alpha+\beta}$$

  • $n_{\alpha,\beta}\neq 0$ only if $\alpha+\beta\in \Delta$
  • $n_{\alpha,\beta}$ is not fixed by the above condition

structure constants

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.


Chevalley

  • a synthesis between the theory of Lie groups and the theory of finite groups


observation

  • 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 $\beta-p\alpha\in \Delta$. ($\alpha$ string through $\beta$)

lemma

It is possible to choose basis elements $x_{\alpha}\in \mathfrak{g}_{\alpha}$ such that $[x_{\alpha},x_{-\alpha}]=H_{\alpha}$, and $n_{-\alpha,-\beta}=-n_{\alpha,\beta}$ for all $\alpha$ and $\beta$. For this choice of $x_{\alpha}$, we have $n_{\alpha,\beta}=\pm (p+1)$

Hint : Use the Chevalley involution $\sigma :\mathfrak{g}\to \mathfrak{g}$. It is an involution with $\sigma(h)=-h$ for any $h\in \mathfrak{h}$ and $\sigma(\mathfrak{g}_{\alpha})=\mathfrak{g}_{-\alpha}$.


Chevalley basis

thm (Chevalley 1955)

The elements $\{H_{\alpha_i} : \alpha_i\in \Pi\}$ together with elements $X_{\alpha}\in \mathfrak{g}_{\alpha}$ ($\alpha\in \Delta$) chosen to satisfy $[X_{\alpha},X_{-\alpha}]=H_{\alpha}$ and $[X_{\alpha},X_{\beta}]=\pm (p+1) X_{\alpha+\beta}$ (if $\alpha+\beta\in \Delta)$ form a basis for a $\mathbb{Z}$-form $\mathfrak{g}_{\mathbb{Z}}$ of $\mathfrak{g}$.


  • 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

  • Let $\{X_{\alpha}\}$ and $\{H_{\alpha_i}\}$ be a Chevalley basis for $\mathfrak{g}$
  • let $\Delta^{+}=\{\alpha_1,\cdots, \alpha_N\}$
  • for $Q=(q_1,\cdots, q_N)$ with $q_i$ non-negative integers, put

$$ e_{\pm Q}=\prod_{i=1}^N (X_{\pm \alpha_i}^{q_i}{(q_i)!} $$

  • for $x\in \mathfrak{g}$ and $s\in \mathbb{Z}_{\geq 0}$, put

$$ \binom{x}{s}=\frac{x(x-1)\cdots (x-s+1)}{s!}\in U(\mathfrak{g}) $$

  • let $n$ be the rank of $\mathfrak{g}$ for each $n$-tuple $P=(p_i)_{1\leq i \leq n}$, define

$$ h_{P}=\prod_{i=1}^{n}\binom{H_{\alpha_i}}{p_i} $$

thm (Kostant 1966)

The elements $$ \{e_{-Q}h_Pe_{S}\} $$ for all $Q,P,S$ form an integral basis for $U_{\mathbb{Z}}$.

proof

See [H] chapter 26.


example

  • for $\mathfrak{g}=\mathfrak{sl}_2$,

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

  • 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


properties

  • \(\exp(tE)\) and \(\exp(tF)\) exist in $U_{\mathbb{Z}}t$
  • \(\exp(tH)\) does not exist instead \((1+t)^{H}=1+\binom{H}{1}t+\binom{H^2}{2!}t^2+\cdots\) exists in $U_{\mathbb{Z}}t$
  • a nice property of this integral form is

$$ \Delta(Z_{\alpha}) = \sum_{0\leq\beta\leq\alpha}Z_{\beta} \otimes Z_{\alpha−\beta}. $$ where $\Delta : U(\mathfrak{g})\to U(\mathfrak{g})$ is the coproduct defined by $$ \Delta(x)=x\otimes 1+1\otimes x $$ for $x\in \mathfrak{g}$


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).


related items

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