"Talk on Chevalley's integral forms"의 두 판 사이의 차이
imported>Pythagoras0 |
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* for $Q=(q_1,\cdots, q_N)$ with $q_i$ non-negative integers, put | * for $Q=(q_1,\cdots, q_N)$ with $q_i$ non-negative integers, put | ||
$$ | $$ | ||
| − | e_{\ | + | e_{Q}=\prod_{i=1}^N (X_{\alpha_i}^{q_i}{(q_i)!} |
| + | $$ | ||
| + | and | ||
| + | $$ | ||
| + | f_{Q}=\prod_{i=1}^N (X_{-\alpha_i}^{q_i}{(q_i)!} | ||
$$ | $$ | ||
* for $x\in \mathfrak{g}$ and $s\in \mathbb{Z}_{\geq 0}$, put | * for $x\in \mathfrak{g}$ and $s\in \mathbb{Z}_{\geq 0}$, put | ||
| 125번째 줄: | 129번째 줄: | ||
The elements | The elements | ||
$$ | $$ | ||
| − | \{ | + | \{f_{Q}h_Pe_{S}\} |
$$ for all $Q,P,S$ form an integral basis for $U_{\mathbb{Z}}$. | $$ for all $Q,P,S$ form an integral basis for $U_{\mathbb{Z}}$. | ||
2014년 4월 1일 (화) 00:23 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
$$ e_{Q}=\prod_{i=1}^N (X_{\alpha_i}^{q_i}{(q_i)!} $$ and $$ f_{Q}=\prod_{i=1}^N (X_{-\alpha_i}^{q_i}{(q_i)!} $$
- 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 $$ \{f_{Q}h_Pe_{S}\} $$ 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).