Talk on Chevalley's integral forms
introduction
- linear algebra : all bases are equal (half true. diagonalization)
- actually 'All bases are equal, but some bases are more equal than others'
- usually good bases : rich source of mathematics
- what are good bases?
integral forms
- $A$ : algebra (or vector space) over $\mathbb{C}$ (for any field $\mathbb{F}$ of characteristic 0)
- def
An integral form (or a $\mathbb{Z}$-form) $A_\mathbb{Z}$ of $A$ to be a $\mathbb{Z}$-algebra ($\mathbb{Z}$-module) such that $\mathbb{C}\otimes_\mathbb{Z}A_\mathbb{Z}=A$.
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 (see The fake monster formal group by Borcherds for more)
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 ($\mathfrak{g}_{\mathbb{Z}}$ is a Lie algebra over $\mathbb{Z}$)
UEA
- universal enveloping algebra $U(\mathfrak{g})$ PBW basis
\[\{F^kH^lE^m|k,l,m\geq 0\}\]
- Hopf algebra with coproduct $\Delta : U(\mathfrak{g})\to U(\mathfrak{g})$ defined by $\Delta(x)=x\otimes 1+1\otimes x$ for $x\in \mathfrak{g}$
- integral form and integral basis ? answer later
finite dimensional representations
- \(V\) : irreducible finite dimensional module
- \(V=\oplus_{\lambda\in\mathbb{C}}V_{\mu}\), \(V_{\mu}=\{v\in V|Hv=\mu v\}\)
- there exists \(v_0\neq 0\) such that
\[Ev_0=0\] \[Hv_0=\lambda v_0\]
- let $F^{(j)}:=\frac{F^j}{j!}$, $E^{(j)}:=\frac{E^j}{j!}$
- define \(v_j:=F^{(j)}v_0, j \in\mathbb{Z}_{\geq 0}\), we have
\[H v_j=(\lambda -2j)v_j\] \[F v_j=(j+1)v_{j+1}\] \[E v_j=(\lambda -j+1)v_{j-1}\]
- as $V$ is finite dimensional, there exists $l\in \in\mathbb{Z}_{\geq 0}$ such that $v_m\neq 0$ and $v_{m+1}=0$
- then $Ev_{m+1}=(\lambda-m)v_{m}=0$ and so \(\lambda-m=0\). So $\lambda \in\mathbb{Z}_{\geq 0}$
- Let $V_{\mathbb{Z}}$ be the $\mathbb{Z}$-span of \(\{v_j|j\geq 0\}\)
- as $V$ is irreducible, $V=\mathbb{C}\otimes_{\mathbb{Z}}V_{\mathbb{Z}}$
- so $V_{\mathbb{Z}}$ is an integral form for $V$ with integral basis $v_j$
- Question.
where do $F^{(j)}$ come from?
- prop
$V_{\mathbb{Z}}$ is stable under the action of $F^{(j)}$ and $E^{(j)}$ and thus stable also under the action of $\exp (tE)$ and $\exp (tF)$ (matrices with integral coefficients, key fact to define the Chevalley groups)
basis of $\mathfrak{g}$ and structure constants
basis
- simple Lie algebra $\mathfrak{g}$, we have a non-deg invariant bilinear form $(\cdot,\cdot)$.
- fix a Cartan subalgebra $\mathfrak{h}$
- $\Delta$ : root system
- $\Pi$ : fundamental system
- Cartan decomposition
$$ \mathfrak{g}=\mathfrak{h}\oplus \left(\oplus_{\alpha\in \Delta} \mathfrak{g}_{\alpha}\right) $$
- fix $H_{\alpha}\in \mathfrak{h}$ uniquely for each $\alpha\in \Delta$ by
$$ \beta(H_{\alpha})=2\frac{(\alpha,\beta)}{(\alpha,\alpha)}\,\quad \beta\in \mathfrak{h}^{*} $$
- exercise
$H_{\alpha}$ can be written as a $\mathbb{Z}$-linear combination of $H_{\alpha_i}, \alpha_i\in \Pi$.
- we can choose $x_{\alpha}\in \mathfrak{g}_{\alpha}$ so that
$$[x_{\alpha},x_{-\alpha}]=H_{\alpha}$$
- The elements $\{H_{\alpha_i} : \alpha_i\in \Pi\}$ together with elements $x_{\alpha}\in \mathfrak{g}_{\alpha}$ ($\alpha\in \Delta$) form a basis of $\mathfrak{g}$
structure constants
- multiplication in $\mathfrak{g}$
$$ [h,x_{\alpha}]=\alpha(h)x_{\alpha}\\ [x_{\alpha},x_{-\alpha}]=H_{\alpha}\\ [x_{\alpha},x_{\beta}]=n_{\alpha,\beta}x_{\alpha+\beta} $$
- structure constants $n_{\alpha,\beta}$
- $n_{\alpha,\beta}\neq 0$ only if $\alpha+\beta\in \Delta$
- $n_{\alpha,\beta}$ is not fixed by the above condition. how much freedom do we have?
- 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 $x_{\alpha}$.
- All the structure constants $n_{\alpha,\beta}$ are determined by the structure constants for extraspecial pairs.
- see Lie Algebras of Finite and Affine Type by Carter for more
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 $\beta$)
- remark
it is the minimum $p\in \mathbb{Z}_{\geq 0}$ such that $$ \left(\text{ad} x_{-\alpha}\right)^{p}\left(x_{\beta}\right)=0 $$
- 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 $\sigma(x_{\alpha})=x_{-\alpha}$.
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 $\mathfrak{g}$.
Kostant
integral form
- Let $\{X_{\alpha}\}$ and $\{H_{\alpha_i}\}$ be a Chevalley basis for $\mathfrak{g}$
- Let $U(\mathfrak{g})_{\mathbb{Z}}$ be the $\mathbb{Z}$-subalgebra of $U(\mathfrak{g})$ generated by $X_{\alpha}^{(n)}=X_{\alpha}^{n}/n!$ for all $\alpha\in \Delta$ and $n\in \mathbb{Z}_{\geq 0}$.
- it is an integral form for $U(\mathfrak{g})$
- can we describe its integral basis?
basis
- let $\Delta^{+}=\{\alpha_1,\cdots, \alpha_N\}$
- for $Q=(q_1,\cdots, q_N)$ with $q_i\in \mathbb{Z}_{\geq 0}$, put
$$ e_{Q}=\prod_{i=1}^N X_{\alpha_i}^{(q_i)} $$
- for $S=(s_1,\cdots, s_N)$ with $q_i\in \mathbb{Z}_{\geq 0}$, put
$$ f_{S}=\prod_{i=1}^N X_{-\alpha_i}^{(s_i)} $$
- 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 $l$ be the rank of $\mathfrak{g}$ for each $l$-tuple $P=(p_1,\cdots, p_l)$, define
$$ h_{P}=\prod_{i=1}^{l}\binom{H_{\alpha_i}}{p_i} $$
- thm (Kostant 1966)
The elements $$ \{f_{Q}h_Pe_{S}|Q\in\mathbb{Z}_{\geq 0}^{N},P\in\mathbb{Z}_{\geq 0}^{l},S\in\mathbb{Z}_{\geq 0}^{N}\} $$ form an integral basis for $U(\mathfrak{g})_{\mathbb{Z}}$.
- 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. In general, we have
$$ E^{(m)}F^{(n)}=\sum_{j=0}^k F^{(n-j)}\binom{H-m-n+2j}{j}E^{(m-j)} $$ where $k=\min(m,n)$
- exercise
Let $j,k\in\mathbb{Z}_{\geq 0}$. The polynomial $\binom{x-j}{k}$ can be written as a $\mathbb{Z}$-linear combination of $\binom{x}{i}$'s.
properties
- \(\exp(tE)\) and \(\exp(tF)\) exist in $U(\mathfrak{g})_{\mathbb{Z}}t$
- a nice property of this integral form is
$$ \Delta(Z_{\alpha}) = \sum_{0\leq\beta\leq\alpha}Z_{\beta} \otimes Z_{\alpha−\beta}. $$ where $Z_{\alpha}=f_{Q}h_Pe_{S}$, $\alpha=(Q,P,S)\in\mathbb{Z}_{\geq 0}^{N+l+N}$ and $\Delta : U(\mathfrak{g})\to U(\mathfrak{g})$ is the coproduct defined by $$ \Delta(x)=x\otimes 1+1\otimes x $$ for $x\in \mathfrak{g}$
- partial ordering on $\mathbb{Z}_{\geq 0}^{N+l+N}$
remarks on Chevalley groups
- def
An admissible integral form of a $\mathfrak{g}$-module $V$ is an integral form $M$ such that $U(\mathfrak{g})_{\mathbb{Z}}\cdot M\subseteq M$
- prop
Let $V$ be a finite dimensional $\mathfrak{g}$-module. Then $V$ has an admissible integral form. If $V$ is irreducible and $v_0$ is a highest weight vector, then $U(\mathfrak{g})_{\mathbb{Z}}.v_0$ is an admissible integral form of $V$.
- Let $V$ a faithful representation of $\mathfrak{g}$ and $M$ an admissible integral form
- Choose an integral basis $\{m_1,\cdots, m_d\}$of $M$. Then $e_{\alpha}(t):=\exp \left(t\rho(X_{\alpha})\right)\in GL_{d}(\mathbb{Z}[t])$
- now let $k$ arbitrary field and $M^k=k\otimes_{\mathbb{Z}}M$ which is a vector space over $k$
- def
The Chevalley group $G_{V,k}$ is the subgroup of $GL(M^k)$ generated by all $e_{\alpha}(u),\, \alpha\in \Delta, u\in k$ regarded as a $k$-linear transformation of $M^k$
- it actually depends on $k$ and the lattice of weights $\Gamma_{V}$ of $\mathfrak{g}$-module $V$
- When $\Gamma_V=Q$, Q the root lattice, we call it an adjoint Chevalley group
- thm (Chevalley-Dickson theorem)
Let $G$ be an adjoint Chevalley group. If $|k|=2$, suppose $\Delta$ is not of type $A_1,B_2$ or $G_2$. If $|k|=3$, suppose that $\Delta$ is not of type $A_1$. Then $G$ is a simple group.
See (Curtis, 'Chevalley groups and related topics' thm 6.11) for a proof.
refs
- [H] J. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer, (1972).