## Lectures on Chevalley groups

Last update: 18 July 2013

## §1. A basis for $ℒ$

We start with some basic properties of semisimple Lie algebras over $ℂ,$ and establish some notation to be used throughout. The assertions not proved here are proved in the standard books on Lie algebras, e.g., those of Dynkin, Jacobson or Sophus Lie (Séminaire).

Let $ℒ$ be a semisimple Lie algebra over $ℂ,$ and $ℋ$ a Cartan subalgebra of $ℒ\text{.}$ Then $ℋ$ is necessarily Abelian and $ℒ=ℋ\oplus \underset{\alpha \ne 0}{\Sigma }{ℒ}_{\alpha }$ where $\alpha \in {ℋ}^{*}$ and ${ℒ}_{\alpha }=\left\{X\in ℒ | \left[H,X\right]=\alpha \left(H\right)X$ for all $H\in ℋ\right\}\text{.}$ Note that $ℋ={ℒ}_{0}\text{.}$ The $\alpha \text{'s}$ are linear functions on $ℋ,$ called roots. We adopt the convention that ${ℒ}_{\gamma }=0$ if $\gamma$ is not a root. Then $\left[{ℒ}_{\alpha },{ℒ}_{\beta }\right]\subseteq {ℒ}_{\alpha +\beta }\text{.}$ The rank of $ℒ={\text{dim}}_{ℂ}ℋ=\ell ,$ say. The roots generate ${ℋ}^{*}$ as a vector space over $ℂ\text{.}$

Write $V$ for ${ℋ}_{Q}^{*},$ the vector space over $Q$ generated by the roots. Then ${\text{dim}}_{Q} V=\ell \text{.}$ Let $\gamma \in V\text{.}$ Since the Killing form is nondegenerate there exists an ${H}_{\gamma }^{\prime }\in ℋ$ such that $\left(H,{H}_{\gamma }^{\prime }\right)=\gamma \left(H\right)$ for all $H\in ℋ\text{.}$ Define $\left(\gamma ,\delta \right)=\left({H}_{\gamma }^{\prime },{H}_{\delta }^{\prime }\right)$ for all $\gamma ,\delta \in V\text{.}$ This is a symmetric, nondegenerate, positive definite bilinear form on $V\text{.}$

Denote the collection of all roots by $\Sigma \text{.}$ Then $\Sigma$ is a subset of the nonzero elements of $V$ satisfying:

 (0) $\Sigma$ generates $V$ as a vector space over $Q\text{.}$ (1) $\alpha \in \Sigma ⇒-\alpha \in \Sigma$ and $k\alpha \notin \Sigma$ for $k$ an integer $\ne ±1\text{.}$ (2) $2\left(\alpha ,\beta \right)/\left(\beta ,\beta \right)\in ℤ$ for all $\alpha ,\beta \in \Sigma \text{.}$ (Write $⟨\alpha ,\beta ⟩=2\left(\alpha ,\beta \right)/\left(\beta ,\beta \right)\text{.}$ These are called Cartan integers). (3) $\Sigma$ is invariant under all reflections ${w}_{\alpha }$ $\left(\alpha \in \Sigma \right)$ (where ${w}_{\alpha }$ is the reflection in the hyperplane orthogonal to $\alpha ,$ i.e., ${w}_{\alpha }v=v-2\left(v,\alpha \right)/\left(\alpha ,\alpha \right)$ $\alpha \text{).}$

Thus $\Sigma$ is a root system in the sense of Appendix I. Conversely, if $\Sigma$ is any root system satisfying condition (2), then $\Sigma$ is the root system of some Lie algebra.

The group $W$ generated by all ${w}_{\alpha }$ is a finite group (Appendix 1.6) called the Weyl group. If $\left\{{\alpha }_{1},\dots ,{\alpha }_{\ell }\right\}$ is a simple system of roots (Appendix I.8), then $W$ is generated by the ${w}_{{\alpha }_{i}}$ $\left(i=1,\dots ,n\right)$ (Appendix I.16) and every root is congruent under $W$ to a simple root (Appendix I.15).

Lemma 1: For each root $\alpha ,$ let ${H}_{\alpha }^{\prime }\in ℋ$ be such that $\left(H,{H}_{\alpha }^{\prime }\right)=\alpha \left(H\right)$ for all $H\in ℋ\text{.}$ Define ${H}_{\alpha }=2/\left(\alpha ,\alpha \right){H}_{\alpha }^{\prime }$ and ${H}_{i}={H}_{{\alpha }_{i}}$ $\left(i=1,\dots ,\ell \right)\text{.}$ Then each ${H}_{\alpha }$ is an integral linear combination of the ${H}_{i}\text{.}$ Proof. Write ${w}_{i}$ for ${w}_{{\alpha }_{i}}\in W\text{.}$ Define an action of $W$ on $ℋ$ by ${w}_{i}{H}_{j}^{\prime }={H}_{j}^{\prime }-⟨{\alpha }_{j},{\alpha }_{i}⟩{H}_{i}^{\prime }\text{.}$ Then $wiHj = 2(αj,αj) wiHj′ = 2(αj,αj) Hj′- 2(αj,αj) · 2(αi,αj) (αi,αi) Hi′ = Hj- 2(αi,αi)· 2(αi,αj) (αj,αj) Hi′ = Hj-⟨αi,αj⟩ Hi = Hwiαj$ Then since the ${w}_{i}$ generate $W,$ $w{H}_{j}$ is an integral linear combination of the ${H}_{i}$ for all $w\in W\text{.}$ Now if $\alpha$ is an arbitrary root then $\alpha =w{\alpha }_{j}$ for some $w\in W$ and some $j\text{.}$ Then ${H}_{\alpha }={H}_{w{\alpha }_{j}}=w{H}_{{\alpha }_{j}}=w{H}_{j}=$ an integral linear combination of the ${H}_{i}\text{.}$ $\square$

For every root $\alpha$ choose ${X}_{\alpha }\in {ℒ}_{\alpha },$ ${X}_{\alpha }\ne 0\text{.}$ If $\alpha +\beta \ne 0$ define ${N}_{\alpha ,\beta }$ by $\left[{X}_{\alpha },{X}_{\beta }\right]={N}_{\alpha ,\beta }{X}_{\alpha +\beta }\text{.}$ Set ${N}_{\alpha ,\beta }=0$ if $\alpha +\beta$ is not a root.

If $\alpha$ and $\beta$ are roots the $\alpha \text{-string}$ of roots through $\beta$ is the sequence $\beta -r\alpha ,\dots ,\beta ,\dots ,\beta +q\alpha$ where $\beta +i\alpha$ is a root for $-r\le i\le q$ but $\beta -\left(r+1\right)\alpha$ and $\beta +\left(q+1\right)\alpha$ are not roots.

Lemma 2: The ${X}_{\alpha }$ can be chosen so that:

 (a) $\left[{X}_{\alpha },{X}_{-\alpha }\right]={H}_{\alpha }\text{.}$ (b) If $\alpha$ and $\beta$ are roots, $\beta \ne ±\alpha ,$ and $\beta -r\alpha ,\dots ,\beta ,\dots \beta +q\alpha$ is the $\alpha \text{-string}$ of roots through $\beta$ then ${N}_{\alpha ,\beta }^{2}=q\left(r+1\right){|\alpha +\beta |}^{2}/{|\beta |}^{2}\text{.}$ Proof. See the first part of the proof of Theorem 10, p. 147 in Jacobson, Lie Algebras. $\square$

Lemma 3: If $\alpha ,\beta$ and $\alpha +\beta$ are roots, then $q\left(r+1\right){|\alpha +\beta |}^{2}/{|\beta |}^{2}={\left(r+1\right)}^{2}\text{.}$ Proof.

We use two facts:

 $\left(*\right)$ $r-q=⟨\beta ,\alpha ⟩\text{.}$
(For ${w}_{\alpha }$ maps $\beta -r\alpha$ to $\beta +q\alpha$ so $\beta +q\alpha ={w}_{\alpha }\left(\beta -r\alpha \right)=\beta -r\alpha -2\left(\beta -r\alpha ,\alpha \right)/\left(\alpha ,\alpha \right)\alpha =\beta -⟨\beta ,\alpha ⟩\alpha +r\alpha \text{).}$
 $\left(**\right)$ In the $\alpha \text{-string}$ of roots through $\beta$ at most two root lengths occur.
(For if $V\prime$ is the vector space over $Q$ generated by $\alpha$ and $\beta$ and $\Sigma \prime =\Sigma \cap V,$ then $\Sigma \prime$ is a root system and every root in the $\alpha \text{-string}$ of roots through $\beta$ belongs to $\Sigma \prime \text{.}$ Now $V\prime$ is two dimensional; so a system of simple roots for $\Sigma \prime$ has at most two elements. Since every root in $\Sigma \prime$ is conjugate under the Weyl group of $\Sigma \prime$ to a simple root, $\Sigma \prime$ and hence the $\alpha \text{-string}$ of roots through $\beta$ has at most two root lengths). We must show that $q{|\alpha +\beta |}^{2}/{|\beta |}^{2}=r+1\text{.}$ Now by $\left(*\right)\text{:}$

$r+1-q |α+β|2 /|β|2 = q+⟨β,α⟩+ 1-q (α+β,α+β) /(β,β) = ⟨β,α⟩+1- q|α|2/ |β|2-q ⟨α,β⟩ = (⟨β,α⟩+1) ( 1-q|α|2 /|β|2 ) .$

Set $A=⟨\beta ,\alpha ⟩+1$ and $B=1-q{|\alpha |}^{2}/{|\beta |}^{2}\text{.}$ We must show $A=0$ or $B=0\text{.}$

If $|\alpha |\ge |\beta |$ then $|⟨\beta ,\alpha ⟩|=2|\left(\beta ,\alpha \right)|/{|\alpha |}^{2}\le 2|\left(\beta ,\alpha \right)|/{|\beta |}^{2}=|⟨\alpha ,\beta ⟩|\text{.}$ By Schwarz's inequality $⟨\beta ,\alpha ⟩⟨\alpha ,\beta ⟩=4{\left(\alpha ,\beta \right)}^{2}/|\alpha ||\beta |\le 4$ with equality if and only if $\mathrm{\alpha \mathrm{=k\beta \text{.}}}$ Since $\alpha$ and $\beta$ are roots and $\alpha \ne ±\beta$ we have $\alpha \ne k\beta$ so $⟨\beta ,\alpha ⟩⟨\alpha ,\beta ⟩<4\text{.}$ Then since $|⟨\beta ,\alpha ⟩|\le |⟨\alpha ,\beta ⟩|$ we have $⟨\beta ,\alpha ⟩=-1,$ $0,$ or 1. If $⟨\beta ,\alpha ⟩=-1$ then $A=0\text{.}$ If $⟨\beta ,\alpha ⟩\ge 0$ then $|\beta +2\alpha |>|\beta +\alpha |>|\beta |\text{.}$ Since there are only two root lengths $\beta +2\alpha$ is not a root and hence $q=1\text{.}$ Since $|\beta +\alpha |>|\alpha |$ and $|\beta +\alpha |>|\beta |$ and at most two root lengths occur $|\alpha |=|\beta |\text{.}$ Hence $B=0\text{.}$

If $|\alpha |<|\beta |,$ then $|\alpha +\beta |\le |\beta |$ (since otherwise three root lengths would occur). Hence $\left(\alpha ,\beta \right)<0$ so $⟨\alpha ,\beta ⟩<0\text{.}$ Then $|\beta -\alpha |>|\beta |>|\alpha |$ so $\beta -\alpha$ is not a root and hence $r=0\text{.}$ As above $⟨\alpha ,\beta ⟩⟨\beta ,\alpha ⟩<4$ and $|⟨\alpha ,\beta ⟩|<|⟨\beta ,\alpha ⟩|$ so $⟨\alpha ,\beta ⟩=-1,0,$ or 1. Hence $⟨\alpha ,\beta ⟩=-1\text{.}$ Then by $\left(*\right)$ $q=-⟨\beta ,\alpha ⟩=⟨\beta ,\alpha ⟩/⟨\alpha ,\beta ⟩={|\beta |}^{2}/{|\alpha |}^{2}\text{.}$ Hence $B=0\text{.}$

$\square$

We collect these results in:

Theorem 1: The ${H}_{i}$ $\left(i=1,2,\dots ,\ell \right)$ chosen as in Lemma 1 together with the ${X}_{\alpha }$ chosen as in Lemma 2 form a basis for $ℒ$ relative to which the equations of structure are as follows (and, in particular, fare integral):

 (a) $\left[{H}_{i},{H}_{j}\right]=0$ (b) $\left[{H}_{i},{X}_{\alpha }\right]=⟨\alpha ,{\alpha }_{i}⟩{X}_{\alpha }$ (c) $\left[{X}_{\alpha },{X}_{-\alpha }\right]={H}_{\alpha }=$ an integral linear combination of the ${H}_{i}\text{.}$ (d) $\left[{X}_{\alpha },{X}_{\beta }\right]=±\left(r+1\right){X}_{\alpha +\beta }$ if $\alpha +\beta$ is a root. (e) $\left[{X}_{\alpha },{X}_{\beta }\right]=0$ if $\alpha +\beta \ne 0$ and $\alpha +\beta$ is not a root. Proof. (a) holds since $ℋ$ is abelian. (b) holds since $\left[{H}_{\beta },{X}_{\alpha }\right]=a\left({H}_{\beta }\right){X}_{\alpha }=⟨\alpha ,\beta ⟩{X}_{\alpha }\text{.}$ (c) follows from the choice of the ${X}_{\alpha }$ and the ${H}_{i}$ and from Lemma 1. (d) follows from Lemma 2(b) and Lemma 3. (e) holds since $\left[{ℒ}_{\alpha },{ℒ}_{\beta }\right]=0$ if $\alpha +\beta$ is not a root. $\square$

Remarks:

 (a) Such a basis is called a Chevalley basis. It is unique up to sign changes and automorphisms of $ℒ\text{.}$ (b) ${X}_{\alpha },{X}_{-\alpha }$ and ${H}_{\alpha }$ span a 3-dimensional subalgebra isomorphic to ${s𝓁}_{2}$ $\text{(}2×2$ matrices of trace 0). $Xα⟷ X-α⟷  Hα⟷ [100-1]$ (c) As an example let $ℒ={s𝓁}_{\ell +1}\text{.}$
Then $ℋ=$ {diagonal matrices} is a Cartan subalgebra. For $i,j=1,\dots ,\ell +1,i\ne j,$ def±ne $\alpha =\alpha \left(i,j\right)$ by $\alpha \left(i,j\right):\text{diag}\left({a}_{1},\dots ,{a}_{\ell +1}\right)\to {a}_{i}-{a}_{j}\text{.}$ Then the $\alpha \left(i,j\right)$ are the roots. Let ${E}_{i,j}$ be the matrix unit with $1$ in the $\left(i,j\right)$ position and $0$ elsewhere. Then ${X}_{\alpha }={E}_{i,j},$ ${X}_{-\alpha }={E}_{j,i},$ ${H}_{\alpha }={E}_{i,i}-{E}_{j,j}$ and ${H}_{i}={E}_{i,i}-{E}_{i+1,i+1}\text{.}$

Exercise: If only one root length occurs then all coefficients in (d) of Theorem 1 are $±1\text{.}$ Otherwise $±2$ and $±3$ can occur.

## Notes and References

This is a typed excerpt of Lectures on Chevalley groups by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson. This work was partially supported by Contract ARO-D-336-8230-31-43033.