## Lectures on Chevalley groups

Last update: 31 July 2013

## §3. The Chevalley groups

We wish to study automorphisms of ${V}^{k}$ of the form $\text{exp}t{X}_{\alpha }$ $\left(t\in k,\alpha \in \Sigma \right),$ where

$exptXα= Σn=0∞ tnXαn/n!$

The right side of the above expression is interpreted as follows. Since ${X}_{\alpha }^{n}/n!\in {𝒰}_{ℤ},$ we have an action of ${X}_{\alpha }^{n}/n!$ on $M\text{.}$ Thus, we get an action of ${\lambda }^{n}{X}_{\alpha }^{n}/n!$ on $M{\otimes }_{ℤ}ℤ\left[\lambda \right]\text{.}$ Since ${X}_{\alpha }^{n}$ acts as zero for $n$ sufficiently large, we see that $\underset{n=0}{\overset{\infty }{\Sigma }}{\lambda }^{n}{X}_{\alpha }^{n}/n!$ acts on $M{\otimes }_{ℤ}ℤ\left[\lambda \right]$ and hence on $M{\otimes }_{ℤ}ℤ\left[\lambda \right]{\otimes }_{ℤ}k\text{.}$ Following this last action by the homomorphism of $M{\otimes }_{ℤ}ℤ\left[\lambda \right]{\otimes }_{ℤ}k$ into ${V}^{k}=M{\otimes }_{ℤ}k$ given by $\lambda \to t,$ we get an action of $\underset{n=0}{\overset{\infty }{\Sigma }}{t}^{n}{X}_{\alpha }^{n}/n!$ on ${V}^{k}\text{.}$

We will write ${x}_{\alpha }\left(t\right)$ for $\text{exp}t{X}_{\alpha }$ and ${𝔛}_{\alpha }$ for the group $\left\{{x}_{\alpha }\left(t\right) | t\in k\right\}$ (clearly ${x}_{\alpha }\left(t\right)$ is additive in $t\text{).}$ Our main object of study is the group $G$ generated by all ${𝔛}_{\alpha }$ $\left(\alpha \in \Sigma \right)\text{.}$ We will call it a Chevalley group.

Exercise: Interpret $\underset{n=0}{\overset{\infty }{\Sigma }}{t}^{n}\left(\genfrac{}{}{0}{}{H}{n}\right)$ $\left(H\in {ℋ}_{ℤ},t\in k,t\ne -1\right)\text{.}$

Lemma 14: Let $𝒜$ be an associative algebra, $A\in 𝒜,$ and let ${d}_{A}$ be the derivation of $𝒜,$ ${d}_{A}={\ell }_{A}-{r}_{A}$ where ${\ell }_{A}B=AB,$ ${r}_{A}B=BA,$ $B\in 𝒜\text{.}$ Suppose $\text{exp} {d}_{A},$ $\text{exp} {\ell }_{A},$ $\text{exp} {r}_{A},$ and $\text{exp} A$ have meaning and that the usual rules of exponentiation apply. Then $\text{exp} {d}_{A}={\ell }_{\text{exp} A}{r}_{\text{exp}\left(-A\right)}$ $\text{(}=$ conjugation by $\text{exp} A\text{).}$ Proof. $\text{exp} {d}_{A}=\text{exp} {\ell }_{A} \text{exp}\left(-{r}_{A}\right)={\ell }_{\text{exp} A}{r}_{\text{exp}\left(-A\right)}\text{.}$ $\square$

Lemma 15: Let $\alpha ,\beta$ be roots with $\alpha +\beta \ne 0\text{.}$ Then in the ring of formal power series in two variables $t,u$ over ${𝒰}_{ℤ},{𝒰}_{ℤ}\left[\left[t,u\right]\right],$ we have the identity

$( exp tXα, exp uXβ ) =Πexp cij tiujXiα+jβ$

where $\left(A,B\right)=AB{A}^{-1}{B}^{-1},$ where the product on the right is taken over all roots $i\alpha +j\beta$ $\left(i,j\in {ℤ}^{+}\right)$ arranged in some fixed order, and where the ${c}_{ij}\text{'s}$ are integers depending on $\alpha ,\beta ,$ and the chosen ordering, but not on $t$ or $u\text{.}$ Furthermore ${C}_{11}={N}_{\alpha ,\beta }\text{.}$ Proof. In $𝒰\left[\left[t,u\right]\right]$ set $f\left(t,u\right)=\left(\text{exp} t{X}_{\alpha },\text{exp} u{X}_{\beta }\right)\Pi \text{exp} \left(-{c}_{ij}{t}^{i}{u}^{j}{X}_{i\alpha +j\beta }\right)$ where ${c}_{ij}\in ℂ\text{.}$ We shall show that we may choose the ${c}_{ij}\text{'s}$ in $ℤ$ such that $f\left(t,u\right)=1\text{.}$ We note that $t\frac{d}{dt}\left(\text{exp} t{X}_{\alpha }\right)=t{X}_{\alpha }\text{exp} t{X}_{\alpha }\text{.}$ Thus, using the product rule we get $tddt f(t,u) = tXαf(t,u) + exp(tXα) exp(uXβ) exp(-tXα) exp(-uXβ) · Πexp ( -cijti ujXiα+jβ ) + Σ ( exp tXα, exp uXβ ) · Πiα+jβ>kα+ℓβ exp(-cijtiujXiα+jβ) ·(ckℓktkuℓXkα+ℓβ) · Πiα+jβ≤kα+ℓβ exp(-cijtiujXiα+jβ).$ We bring the terms $-t{X}_{\alpha }$ and $\left(*\right)$ $-{c}_{k\ell }k{t}^{k}{u}^{\ell }{X}_{k\alpha +\ell \beta }$ to the front using, e.g., the relations $(exp uXβ) (-tXα)= (exp ad uXβ) (-tXα) exp uXβ$ (see Lemma 14) and $(exp ad uXβ)· (-tXα)=-tXα- Nβ,αtu Xα+β-….$ We get an expression of the form $A$ $f\left(t,u\right)$ with $A\in ℒ\left[\left[t,u\right]\right]\text{.}$ Because $f\left(t,u\right)$ is homogeneous of degree 0 relative to the grading $t\to -\alpha ,$ $u\to -\beta ,$ ${X}_{\gamma }\to \gamma ,$ $A$ is also, and from formulas such as those above we see that ${c}_{k\ell }$ is involved in the term $\left(*\right)$ above but otherwise only in terms of degree $>k+\ell$ in $t$ and $u\text{.}$ Thus $A=\underset{k,\ell \ge 1}{\Sigma }\left(-{c}_{k\ell }+{p}_{k\ell }\right){t}^{k}{u}^{\ell }{X}_{k\alpha +\ell \beta }$ with ${p}_{k\ell }$ a polynomial in ${c}_{ij}\text{'s}$ for which $i+j Now we may inductively determine values of ${c}_{k\ell }\in ℂ$ using the lexicographic ordering of the ${c}_{ij}\text{'s}$ such that $A=0\text{.}$ Then $t\frac{d}{dt}f\left(t,u\right)=0$ implies $f\left(t,u\right)=f\left(0,u\right)=1\text{.}$ To show that the ${c}_{ij}\text{'s}$ are integers, we examine the coefficient of ${t}^{i}{u}^{j}$ in the definition of $f\left(t,u\right)\text{.}$ This coefficient is $-{c}_{ij}{X}_{i\alpha +j\beta }+$ (terms coming from exponentials of multiples of ${X}_{k\alpha +\ell \beta }$ with $k+\ell Using induction, we see that ${c}_{ij}{X}_{i\alpha +j\beta }\in {𝒰}_{ℤ}\text{.}$ Hence, ${c}_{ij}\in ℤ,$ by Theorem 2. If $i=j=1,$ the coefficient is $-{c}_{11}{X}_{\alpha +\beta }+{N}_{\alpha ,\beta }{X}_{\alpha +\beta },$ so that ${c}_{11}={N}_{\alpha ,\beta }\text{.}$ $\square$

Examples: (a) If $\alpha +\beta$ is not a root, the right side of the formula in Lemma 15 is $1\text{.}$ (b) If $\alpha +\beta$ is the only root of the form $i\alpha +j\beta ,$ the right side is $\text{exp} {N}_{\alpha ,\beta }tu,$ and ${N}_{\alpha ,\beta }=±\left(r+1\right),$ with $r=r\left(\alpha ,\beta \right)$ as in Theorem 1. (c) If all the roots have one length, the right side is 1 in case (a) and $\text{exp}\left(±tu\right)$ in case (b).

Corollary: If $\text{exp} t{X}_{\alpha },$ etc. in the formula in Lemma 15 are replaced by ${x}_{\alpha }\left(t\right),$ etc., then the resulting equation holds for all $t,u\in k\text{.}$

We call a set $S$ of roots closed if $\alpha ,\beta \in S,$ $\alpha +\beta \in \Sigma$ implies $\alpha +\beta \in S\text{.}$ The following are examples of closed sets of roots: (a) $P=$ set of all positive roots. (b) $P-\left\{\alpha \right\},$ $\alpha$ a simple root. (c) ${P}_{r}=\left\{\alpha | \text{ht} \alpha \ge r,r\ge 1\right\}\text{.}$

We shall call a subset $I$ of a closed set $S$ an ideal if $\alpha \in I,$ $\beta \in S,$ $\alpha +\beta \in S$ implies $\alpha +\beta \in I\text{.}$ We see that (a), (b) and (c) above are ideals in $P\text{.}$

Lemma 16: Let $I$ be an ideal in the closed set $S\text{.}$ Let ${𝔛}_{S}$ and ${𝔛}_{I}$ denote the groups generated by all ${𝔛}_{\alpha }$ $\text{(}\alpha \in S$ and $\alpha \in I,$ respectively). If $\alpha \in S$ implies $-\alpha \notin S,$ then ${𝔛}_{I}$ is a normal subgroup of ${𝔛}_{S}\text{.}$ Proof. This follows immediately from Lemma 15. $\square$

Lemma 17: Let $S$ be a closed set of roots such that $\alpha \in S$ implies $-\alpha \notin S,$ then every element of ${𝔛}_{S}$ can be written uniquely as $\underset{\alpha \in S}{\Pi }{x}_{\alpha }\left({t}_{\alpha }\right)$ where ${t}_{\alpha }\in k$ and the product is taken in any fixed order. Proof. We shall first prove the lemma in the case in which the ordering is consistent with heights; i.e., $\text{ht} \alpha <\text{ht} \beta$ implies $\alpha <\beta \text{.}$ If ${\alpha }_{1}$ is the first element of $S,$ then $S-\left\{{\alpha }_{1}\right\}$ is an ideal in $S\text{.}$ Hence ${𝔛}_{S}={𝔛}_{{\alpha }_{1}}{𝔛}_{S-\left\{{\alpha }_{1}\right\}}\text{.}$ Using induction on the size of $S,$ we see ${𝔛}_{S}=\Pi {𝔛}_{\alpha }\text{.}$ Now suppose $y\in {𝔛}_{S},$ $y=\Pi {x}_{\alpha }\left({t}_{\alpha }\right)\text{.}$ Since ${X}_{{\alpha }_{1}}^{k}\ne 0,$ there is a weight vector $v\in M$ corresponding to a weight $\lambda$ such that ${X}_{{\alpha }_{1}}v\ne 0\text{.}$ Now $yv=v+{t}_{{\alpha }_{1}}{X}_{{\alpha }_{1}}v+z$ where $v\in {V}_{\lambda },$ ${t}_{{\alpha }_{1}}{X}_{{\alpha }_{1}}v\in {V}_{\lambda +{\alpha }_{1}},$ and $z$ is a sum of terms from other weight spaces. Hence ${t}_{{\alpha }_{1}}\in k$ is uniquely determined by $y\text{.}$ Since ${x}_{{\alpha }_{1}}{\left({t}_{{\alpha }_{1}}\right)}^{-1}y\in {𝔛}_{S-\left\{{\alpha }_{1}\right\}},$ we may complete the proof of this case by induction. $\square$

The proof Lemma 17 for an arbitrary ordering follows immediately from:

Lemma 18: Let $𝔛$ be a group with subgroups ${𝔛}_{1},{𝔛}_{2},\dots ,{𝔛}_{r}$ such that:

 (a) $𝔛={𝔛}_{1}{𝔛}_{2}\dots {𝔛}_{r},$ with uniqueness of expression. (b) ${𝔛}_{i}{𝔛}_{i+1}\dots {𝔛}_{r}$ is a normal subgroup of $𝔛$ for $i=1,2,\dots ,r\text{.}$
If $p$ is any permutation of $1,2,\dots ,r$ then $𝔛={𝔛}_{p1}{𝔛}_{p2}\dots {𝔛}_{pr}$ with uniqueness of expression. Proof. (Exercise) Consider $𝔛/{𝔛}_{r}$ and use induction. $\square$

Corollary 1: The map $t\to {x}_{\alpha }\left(t\right)$ is an isomorphism of the additive group of $k$ onto ${𝔛}_{\alpha }\text{.}$

Corollary 2: Let $P$ be the set of all positive roots and let $U={𝔛}_{P}\text{.}$ Then $U=\Pi {𝔛}_{\alpha }$ with uniqueness of expression, where the product is taken over all $\alpha \in P$ arranged in any fixed order.

Corollary 3: $U$ is unipotent and is superdiagonal relative to an appropriate choice of a basis for ${V}^{k}\text{.}$ Similarly, ${U}^{-}={𝔛}_{-P}$ is unipotent and is subdiagonal relative to the same choice of basis. Proof. Choose a basis of weight vectors and order them in a manner consistent with the following partial ordering of the weights: $\mu$ precedes $\nu$ if $\mu -\nu$ is a sum of positive roots. $\square$

Corollary 4: If $i\ge 1$ let ${U}_{i}$ be the group generated by all ${𝔛}_{\alpha }$ with $\text{ht} \alpha \ge 1\text{.}$ We have then:

 (a) ${U}_{i}$ is normal in $U\text{.}$ (b) $\left(U,{U}_{i}\right)\subseteq {U}_{i+1},$ in particular, $\left(U,U\right)\subseteq {U}_{2}\text{.}$ (c) $U$ is nilpotent.

Corollary 5: If $P=Q\cup R$ with $Q$ and $R$ closed sets such that $Q\cap R=\varnothing ,$ then $U={𝔛}_{Q}{𝔛}_{R}$ and ${𝔛}_{Q}\cap {𝔛}_{R}=1\text{.}$ (e.g., if $\alpha$ is a simple root, one can take $Q=\left\{\alpha \right\}$ and $R=P-\left\{\alpha \right\}\text{.}$

Example: If $ℒ={s\ell }_{\ell +1},$ we have seen that the roots correspond to pairs $\left(i,j\right)$ $i\ne j,$ the positive roots to pairs $\left(i,j\right)$ $i and that we may take ${X}_{ij}={E}_{ij},$ the usual matrix unit. Thus, ${x}_{ij}\left(t\right)=1+t{E}_{ij}\text{.}$ We see that $U=$ {all unipotent, superdiagonal matrices}, ${U}^{-}=$ {all unipotent, subdiagonal matrices}, and that $G$ is ${SL}_{\ell +1},$ the group of $\ell +1$ square matrices of determinant 1. The nontrivial commutator relations are: $\left({x}_{ij}\left(t\right),{x}_{jk}\left(u\right)\right)={x}_{ik}\left(tu\right)$ if $i,j,k$ are distinct.

Lemma 19: For any root $\alpha$ and any $t\in {k}^{*}$ define ${w}_{\alpha }\left(t\right)={x}_{\alpha }\left(t\right){x}_{-\alpha }\left(-{t}^{-1}\right){x}_{\alpha }\left(t\right)$ and ${h}_{\alpha }\left(t\right)={w}_{\alpha }\left(t\right){w}_{\alpha }{\left(1\right)}^{-1}\text{.}$ Then:

 (a) ${w}_{\alpha }\left(t\right){X}_{\beta }{w}_{\alpha }{\left(t\right)}^{-1}=c{t}^{-⟨\beta ,\alpha ⟩}{X}_{{w}_{\alpha }\beta }$ where $c=c\left(\alpha ,\beta \right)=±1$ is independent of $t,k$ and the representation chosen, and $c\left(\alpha ,\beta \right)=c\left(\alpha ,-\beta \right)\text{.}$ (b) If $v\in {V}_{\mu }^{k}$ there exists $v\prime \in {V}_{{w}_{\alpha }\mu }^{k}$ independent of $t$ such that ${w}_{\alpha }\left(t\right)v={t}^{-⟨\mu ,\alpha ⟩}v\prime \text{.}$ (c) ${h}_{\alpha }\left(t\right)$ acts "diagonally" on ${V}_{\mu }^{k}$ as multiplication by ${t}^{⟨\mu ,\alpha ⟩}\text{.}$
(Note that ${w}_{\alpha }$ is being used to denote both the defined automorphism and the reflection in the hyperplane orthogonal to $\alpha \text{).}$ Proof. We prove this assuming $k=ℂ\text{.}$ The transfer of coefficients to an arbitrary field is almost immediate. We show first that ${w}_{\alpha }\left(t\right)H{w}_{\alpha }{\left(t\right)}^{-1}={w}_{\alpha }H$ for all $H\in ℋ\text{.}$ By linearity it suffices to prove this for ${H}_{\alpha },$ for if $\alpha \left(H\right)=0$ then ${X}_{\alpha }$ commutes with $H$ so that both sides equal $H\text{.}$ If $H={H}_{\alpha },$ the left side, because of Lemma 2 and the definitions of ${x}_{\alpha }\left(t\right)$ and ${w}_{\alpha }\left(t\right),$ is an element of the three dimensional algebra $⟨{X}_{\alpha },{Y}_{\alpha },{H}_{\alpha }⟩$ whose value depends on calculations within this algebra, not on the representation chosen. Taking the usual representation in ${s\ell }_{2},$ we get $Hα= [100-1] andwα(t)= [1t01] [10-t-11] [1t01]= [0t-t-10]$ so that the desired equation follows. We next prove (b). From the definitions of ${x}_{\alpha }\left(t\right)$ and ${w}_{\alpha }\left(t\right)$ it follows that if $v′′=wα (t)v,then v′′= Σi=-∞∞ tiviwhere vi∈Vμ+iα$ (the sum is actually finite since there are only finitely many weights). Then for $H\in ℋ,$ $H{v}^{\prime \prime }=H{w}_{\alpha }\left(t\right)v={w}_{\alpha }\left(t\right){w}_{\alpha }{\left(t\right)}^{-1}H{w}_{\alpha }\left(t\right)v={w}_{\alpha }\left(t\right)\left({w}_{\alpha }H\right)v=\mu \left({w}_{\alpha }H\right){v}^{\prime \prime }=\left({w}_{\alpha }\mu \right)\left(H\right){v}^{\prime \prime }\text{.}$ Hence ${v}^{\prime \prime }$ corresponds to the weight ${w}_{\alpha }\mu =\mu -⟨\mu ,\alpha ⟩\alpha \text{.}$ Thus the only nonzero term in the sum occurs for $i=-⟨\mu ,\alpha ⟩\text{.}$ By (b) applied to the adjoint representation ${w}_{\alpha }\left(t\right){X}_{\beta }{w}_{\alpha }{\left(t\right)}^{-1}=c{t}^{-⟨\beta ,\alpha ⟩}{X}_{{w}_{\alpha }\beta }$ where $c\in ℂ$ and is independent of $t$ and of the representation chosen. Now ${w}_{\alpha }\left(1\right)$ is an automorphism of ${ℒ}_{ℤ}$ and ${X}_{\gamma }$ is a primitive element of ${ℒ}_{ℤ}$ for all $\gamma$ so $c=±1\text{.}$ Finally ${H}_{{w}_{\alpha }\beta }={w}_{\alpha }\left(1\right){H}_{\beta }{w}_{\alpha }{\left(1\right)}^{-1}=\left[{w}_{\alpha }\left(1\right){X}_{\beta }{w}_{\alpha }{\left(1\right)}^{-1},{w}_{\alpha }\left(1\right){X}_{-\beta }{w}_{\alpha }{\left(1\right)}^{-1}\right]=c\left(\alpha ,\beta \right)c\left(\alpha ,-\beta \right){H}_{{w}_{\alpha }\beta }$ so $c\left(\alpha ,\beta \right)=c\left(\alpha ,-\beta \right),$ which proves (a). Note that ${w}_{\alpha }{\left(t\right)}^{-1}={w}_{\alpha }\left(-t\right)$ so that ${h}_{\alpha }\left(t\right)={w}_{\alpha }{\left(-t\right)}^{-1}{w}_{\alpha }\left(-1\right)\text{.}$ By (b) ${w}_{\alpha }\left(-t\right)v={\left(-t\right)}^{-⟨\mu ,\alpha ⟩}v\prime$ and ${w}_{\alpha }\left(-1\right)v={\left(-1\right)}^{-⟨\mu ,\alpha ⟩}v\prime \text{.}$ Hence ${w}_{\alpha }{\left(-t\right)}^{-1}{w}_{\alpha }\left(-1\right)v={t}^{⟨\mu ,\alpha ⟩}v,$ proving (c). $\square$

Lemma 20: Write ${\omega }_{\alpha }$ for ${w}_{\alpha }\left(1\right)\text{.}$ Then:

 (a) ${\omega }_{\alpha }{h}_{\beta }\left(t\right){\omega }_{\alpha }^{-1}={h}_{{w}_{\alpha }\beta }\left(t\right)=$ an expression as a product of $h\text{'s,}$ independent of the representation space. (b) ${\omega }_{\alpha }{x}_{\beta }\left(t\right){\omega }_{\alpha }^{-1}={x}_{{w}_{\alpha }\beta }\left(ct\right)$ with $c$ as in Lemma 19(a). (c) ${h}_{\alpha }\left(t\right){x}_{\beta }\left(u\right){h}_{\alpha }{\left(t\right)}^{-1}={x}_{\beta }\left({t}^{⟨\beta ,\alpha ⟩}u\right)\text{.}$ Proof. To prove (a) we apply both sides to $v\in {V}_{\mu }^{k}\text{.}$ ${\omega }_{\alpha }{h}_{\beta }{\omega }_{\alpha }^{-1}v={\omega }_{\alpha }{t}^{⟨{w}_{\alpha }\mu ,\beta ⟩}{\omega }_{\alpha }^{-1}v$ (by Lemma 19 (c) applied to ${\omega }_{\alpha }^{-1}v\in {V}_{{w}_{\alpha }\mu }^{k}\text{)}={t}^{⟨{w}_{\alpha }\mu ,\beta ⟩}v={t}^{⟨\mu ,{w}_{\alpha }\beta ⟩}v={h}_{{w}_{\alpha }\beta }\left(t\right)v\text{.}$ By Lemma 19(a) ${\omega }_{\alpha }{X}_{\beta }{\omega }_{\alpha }^{-1}=c{X}_{{w}_{\alpha }\beta }\text{.}$ Exponentiating this gives (b). By Lemma 19(c) applied to the adjoint representation ${h}_{\alpha }\left(t\right){X}_{\beta }{h}_{\alpha }{\left(t\right)}^{-1}={t}^{⟨\beta ,\alpha ⟩}{X}_{\beta }\text{.}$ Exponentiating this gives (c). $\square$

Denote by (R) the following set of relations:

 (R1) ${x}_{\alpha }\left(t\right){x}_{\alpha }\left(u\right)={x}_{\alpha }\left(t+u\right)\text{.}$ (R2) $\left({x}_{\alpha }\left(t\right){x}_{\beta }\left(u\right)\right)=\Pi {X}_{i\alpha +j\beta }\left({c}_{ij}{t}^{i}{u}^{j}\right)$ $\left(\alpha +\beta \ne 0\right)$ with the ${c}_{ij}$ as in Lemma 15. (R3) ${w}_{\alpha }\left(t\right)={x}_{\alpha }\left(t\right){x}_{-\alpha }\left(-{t}^{-1}\right){x}_{\alpha }\left(t\right)\text{.}$ (R4) ${h}_{\alpha }\left(t\right)={w}_{\alpha }\left(t\right){w}_{\alpha }{\left(1\right)}^{-1}\text{.}$ (R5) ${\omega }_{\alpha }={w}_{\alpha }\left(1\right)\text{.}$ (R6) ${\omega }_{\alpha }{h}_{\beta }\left(t\right){\omega }_{\alpha }^{-1}=$ some expression as a product of $h\text{'s}$ (independent of the representation space). (R7) ${\omega }_{\alpha }{x}_{\beta }\left(t\right){\omega }_{\alpha }^{-1}={x}_{{w}_{\alpha }\beta }\left(ct\right)$ $c$ as in Lemma 19(a). (R8) ${h}_{\alpha }\left(t\right){x}_{\beta }\left(u\right){h}_{\alpha }{\left(t\right)}^{-1}={x}_{\beta }\left({t}^{⟨\beta ,\alpha ⟩}u\right)\text{.}$

Since all the relations in (R) are independent of the representation space chosen, results proved using only the relations (R) will be independent of the representation space chosen. Such results will be labeled (E) (usually for existence). Results proved using other information will be labeled (U) (usually for uniqueness).

Lemma 21: Let $U$ be the group generated by all ${𝔛}_{\alpha }$ $\left(\alpha >0\right),$ $H$ the group generated by all ${h}_{\alpha }\left(t\right)$ and $B$ the group generated by $U$ and $H\text{.}$ Then:

 (a) $U$ is normal in $B$ and $B=UH\text{.}$ (E) (b) $U\cap H=1\text{.}$ (U) Proof. Since conjugation by ${h}_{\alpha }\left(t\right)$ preserves ${𝔛}_{\beta }$ (by (R8)) $U$ is normal in $B$ and (a) holds. Relative to an appropriate basis of $V$ any element of $U\cap H$ is both diagonal and unipotent, hence $=1\text{.}$ $\square$

Example: In ${SL}_{n}$ $H=$ {diagonal matrices}, $U=$ {unipotent superdiagonal matrices}, $B=$ {superdiagonal matrices}.

Lemma 22: Let $N$ be the group generated by all ${w}_{\alpha }\left(t\right),$ $H$ be the subgroup generated by all ${h}_{\alpha }\left(t\right),$ and $W$ the Weyl group. Then:

 (a) $H$ is normal in $N\text{.}$ (E) (b) There exists a homomorphism $\phi$ of $W$ onto $N/H$ such that $\phi \left({w}_{\alpha }\right)=H{w}_{\alpha }\left(t\right)$ for all roots $\alpha \text{.}$ (E) (c) $\phi$ is an isomorphism. (U) Proof. Since by (R6) conjugation by ${\omega }_{\alpha }$ preserves $H$ and by (R4) and (R5) ${w}_{\alpha }\left(t\right)={h}_{\alpha }\left(t\right){\omega }_{\alpha },$ (a) holds. Since $H{w}_{\alpha }\left(t\right)=H{w}_{\alpha }\left(t\right){w}_{\alpha }{\left(1\right)}^{-1}{w}_{\alpha }\left(1\right)=H{w}_{\alpha }\left(1\right),$ $H{w}_{\alpha }\left(t\right)$ is independent of $t\text{.}$ Write ${\stackrel{ˆ}{w}}_{\alpha }=H{w}_{\alpha }\left(t\right)\text{.}$ Then since ${w}_{\alpha }\left(1\right)\in {\stackrel{ˆ}{w}}_{\alpha }$ and ${w}_{\alpha }\left(-1\right)\in {\stackrel{ˆ}{w}}_{\alpha },$ $1={w}_{\alpha }\left(1\right){w}_{\alpha }\left(-1\right)\in {\stackrel{ˆ}{w}}_{\alpha }^{2}\text{.}$ Hence $\left(*\right)$ ${\stackrel{ˆ}{w}}_{\alpha }^{2}=1\text{.}$ Also ${\omega }_{\beta }={w}_{\beta }\left(1\right)\in {\stackrel{ˆ}{w}}_{\beta }$ so ${\omega }_{\alpha }{\omega }_{\beta }{\omega }_{\alpha }^{-1}\in {\stackrel{ˆ}{w}}_{\alpha }{\stackrel{ˆ}{w}}_{\beta }{\stackrel{ˆ}{w}}_{\alpha }^{-1}\text{.}$ But ${\omega }_{\alpha }{\omega }_{\beta }{\omega }_{\alpha }^{-1}={\omega }_{\alpha }{x}_{\beta }\left(1\right){x}_{-\beta }{x}_{-\beta }\left(-1\right){x}_{\beta }\left(1\right){\omega }_{\alpha }^{-1}$ (by (R3)) $={x}_{{w}_{\alpha }\beta }\left(c\right){x}_{-{w}_{\alpha }\beta }\left(-c\right){x}_{{w}_{\alpha }\beta }\left(c\right)$ (by (R7)) $={\omega }_{{w}_{\alpha }\beta }^{c}\in {\stackrel{ˆ}{w}}_{{w}_{\alpha }\beta }\text{.}$ Thus $\left(*\right)$ ${\stackrel{ˆ}{w}}_{\alpha }{\stackrel{ˆ}{w}}_{\beta }{\stackrel{ˆ}{w}}_{\alpha }^{-1}={\stackrel{ˆ}{w}}_{{w}_{\alpha }\beta }\text{.}$ By Appendix IV. 40 the relations $\left(*\right)$ form a defining set for $W\text{.}$ Thus there exists a homomorphism $\phi :W\to N/H$ such that $\phi {w}_{\alpha }={\stackrel{ˆ}{w}}_{\alpha }=H{w}_{\alpha }\left(t\right)\text{.}$ $\phi$ is clearly onto. Suppose $w\in \text{ker} \phi \text{.}$ If $w={w}_{{\alpha }_{1}}{w}_{{\alpha }_{2}}\dots ,$ a product of reflections, then ${w}_{{\alpha }_{1}}\left(1\right){w}_{{\alpha }_{2}}\left(1\right)\dots =h\in H\text{.}$ Conjugating ${𝔛}_{\alpha }$ by ${w}_{{\alpha }_{1}}\left(1\right){w}_{{\alpha }_{2}}\left(1\right)\dots$ we get ${𝔛}_{w\alpha }$ and conjugating by $h$ we get ${𝔛}_{\alpha }\text{.}$ Hence ${𝔛}_{w\alpha }={𝔛}_{\alpha }$ for all roots $\alpha \text{.}$ Since $w\alpha =\alpha$ for all $\alpha$ implies $w=1$ the proof is completed by: $\square$

Lemma 23: If $\alpha$ and $\beta$ are distinct roots then ${𝔛}_{\alpha }\ne {𝔛}_{\beta }\text{.}$ Proof. We know that ${𝔛}_{\alpha }$ is nontrivial. If $\alpha$ and $\beta$ have the same sign, the result follows from Lemma 17. If they have opposite signs, then one is superdiagonal unipotent, the other subdiagonal (relative to an appropriate basis), and the result again follows. $\square$

Convention: If $n\in N$ represents $w\in W$ (under $\phi :W\to N/H\text{)}$ we will write $wB$ $\left(Bw\right)$ in place of $nB$ $\left(Bn\right)\text{.}$

Lemma 24: If $\alpha$ is a simple root then

 $B\cup B{w}_{\alpha }B$ is a group. (E) Proof. Let $S=B\cup B{w}_{\alpha }B\text{.}$ Since $B$ is a group and $\phi \left({w}_{\alpha }\right)=\phi {\left({w}_{\alpha }\right)}^{-1},$ $S$ is closed under inversion, and since ${S}^{2}\subseteq BB\cup BB{w}_{\alpha }B\cup B{w}_{\alpha }BB\cup B{w}_{\alpha }B{w}_{\alpha }B\subseteq S\cup B{w}_{\alpha }B{w}_{\alpha }B$ it suffices to show ${w}_{\alpha }B{w}_{\alpha }\subseteq S\text{.}$ We first show that ${𝔛}_{-\alpha }\subseteq S\text{.}$ If $1\ne y\in {𝔛}_{-\alpha }$ then there exists $t\in {k}^{*}$ such that $y={x}_{-\alpha }\left(t\right)={x}_{\alpha }\left({t}^{-1}\right){w}_{\alpha }\left(-{t}^{-1}\right){x}_{\alpha }\left({t}^{-1}\right)\in B{w}_{\alpha }B\text{.}$ Hence ${𝔛}_{-\alpha }\subseteq S\text{.}$ Now let $P$ be the collection of all positive roots. Then ${w}_{\alpha }B{w}_{\alpha }={w}_{\alpha }B{w}_{\alpha }^{-1}={w}_{\alpha }{𝔛}_{\alpha }{𝔛}_{P-\left\{\alpha \right\}}H{w}_{\alpha }^{-1}={w}_{\alpha }{𝔛}_{\alpha }{w}_{\alpha }^{-1}{w}_{\alpha }{𝔛}_{P-\left\{\alpha \right\}}{w}_{\alpha }^{-1}{w}_{\alpha }H{w}_{\alpha }^{-1}={𝔛}_{-\alpha }{𝔛}_{P-\left\{\alpha \right\}}H$ (since ${w}_{\alpha }$ preserves $P-\left\{\alpha \right\}$ by Appendix I.11) $\subseteq SB=S\text{.}$ $\square$

Lemma 25: If $w\in W$ and $\alpha$ is a simple root, then:

 (a) If $w\alpha >0$ (i.e. if $N\left(w{w}_{\alpha }\right)=N\left(w\right)+1$ (see Appendix II.17)) then $BwB·B{w}_{\alpha }B\subseteq Bw{w}_{\alpha }B\text{.}$ (E) (b) In any case $BwB·B{w}_{\alpha }B\subseteq Bw{w}_{\alpha }B\cup BwB\text{.}$ (E) Proof. (a) $BwB·B{w}_{\alpha }B=Bw{𝔛}_{\alpha }{𝔛}_{P-\left\{\alpha \right\}}H{w}_{\alpha }B=Bw{𝔛}_{\alpha }{w}^{-1}w{w}_{\alpha }{w}_{\alpha }^{-1}{𝔛}_{P-\left\{\alpha \right\}}{w}_{\alpha }{w}_{\alpha }^{-1}H{w}_{\alpha }B=Bw{w}_{\alpha }B$ (for $w{𝔛}_{\alpha }{w}^{-1}\subseteq B,$ ${w}_{\alpha }^{-1}{𝔛}_{P-\left\{\alpha \right\}}{w}_{\alpha }\subseteq B$ and ${w}_{\alpha }^{-1}H{w}_{\alpha }\subseteq B\text{).}$ (b) If $w\alpha >0$ (a) gives the result. If $w\alpha <0$ set $w\prime =w{w}_{\alpha }\text{.}$ Then $w\prime \alpha >0$ and $w=w\prime {w}_{\alpha }\text{.}$ By (a) $BwB·B{w}_{\alpha }B=Bw\prime {w}_{\alpha }B·B{w}_{\alpha }B=Bw\prime B·B{w}_{\alpha }B·B{w}_{\alpha }B=Bw\prime B$ $\left(B\cup B{w}_{\alpha }B\right)$ (by Lemma 24) $=Bw\prime B\cup Bw\prime {w}_{\alpha }B=BwB\cup Bw{w}_{\alpha }B\text{.}$ $\square$

Corollary: If $w\in W$ and $w={w}_{\alpha }{w}_{\beta }\dots$ is an expression of minimal length of $w$ as a product of simple reflections then $BwB=B{w}_{\alpha }BB{w}_{\beta }B\dots \text{.}$

Lemma 26: Let $G$ be the Chevalley group $\left(G=⟨{𝔛}_{\alpha } | \text{all} \alpha ⟩\right)\text{.}$ Then $G$ is generated by all ${𝔛}_{\alpha },$ ${\omega }_{\alpha }$ for $\alpha$ a simple root. (E) Proof. We have ${\omega }_{\alpha }{𝔛}_{\beta }{\omega }_{\alpha }^{-1}={𝔛}_{{w}_{\alpha }\beta }\text{.}$ Since the simple reflections generate $W$ and every root is conjugate under $W$ to a simple root the result follows. $\square$

Theorem 4: (Bruhat, Chevalley)

 (a) $\underset{w\in W}{\cup }BwB=G\text{.}$ (E) (b) $BwB=Bw\prime B⇒w=w\prime \text{.}$ (U)
Thus any system of representatives for $N/H$ is also a system of representatives for $B\G/B\text{.}$ Proof. (a) By Lemma 26 $\underset{w\in W}{\cup }BwB$ contains a set of generators for $G\text{.}$ Since $\underset{w\in W}{\cup }BwB$ is closed under multiplication by these generators (by Lemma 25) and reciprocation it is equal to $G\text{.}$ (b) Suppose $BwB=Bw\prime B$ with $w,w\prime \in W\text{.}$ We will show by induction on $N\left(w\right)$ that $w=w\prime \text{.}$ (Here $N\left(w\right)$ is as in the Appendix II.) If $N\left(w\right)=0$ then $w=1$ so $w\prime \in B\text{.}$ Then $w\prime B{w\prime }^{-1}=B$ so $w\prime P=P$ and $w\prime =1$ (see Appendix II.23). Assume $N\left(w\right)>0$ and choose $\alpha$ simple so that $N\left(w{w}_{\alpha }\right) Then $w{w}_{\alpha }\in Bw\prime BB{w}_{\alpha }B\subseteq Bw\prime B\cup Bw\prime {w}_{\alpha }B=BwB\cup Bw\prime {w}_{\alpha }B\text{.}$ Hence by induction $w{w}_{\alpha }=w$ or $w{w}_{\alpha }=w\prime {w}_{\alpha }\text{.}$ But $w{w}_{\alpha }=w$ implies ${w}_{\alpha }=1$ which is impossible. Hence $w{w}_{\alpha }=w\prime {w}_{\alpha }$ so $w=w\prime \text{.}$ $\square$

Remark: The groups $B,N$ form a $B-N$ pair in the sense of J. Tits (Annals of Math. 1964). We shall not axiomatize this concept but adapt certain arguments, such as the last one, to the present context.

Theorem 4': For a fixed $w\in W$ choose ${\omega }_{w}$ representing $w$ in $N\text{.}$ Set $Q=P\cap {w}^{-1}\left(P\right),$ $R=P\cap {w}^{-1}P$ (as before $P$ denotes the set of positive roots). Write ${U}_{w}$ for ${𝔛}_{Q}\text{.}$ Then:

 (a) $BwB=B{\omega }_{w}{U}_{w}\text{.}$ (E) (b) Every element of $BwB$ has a unique expression in this form. (U) Proof. (a) $BwB=Bw{𝔛}_{R}{𝔛}_{Q}H$ by Lemma 17 and Lemma 21) $=Bw{𝔛}_{R}{w}^{-1}w{𝔛}_{Q}H=Bw{𝔛}_{Q}H$ (since $w{𝔛}_{R}{w}^{-1}\subseteq B\text{)}=B{\omega }_{w}{𝔛}_{Q}\text{.}$ (b) If $b{\omega }_{w}x=b\prime {\omega }_{w}x\prime$ then ${b}^{-1}b\prime ={\omega }_{w}x{x\prime }^{-1}{\omega }_{w}^{-1}\text{.}$ Relative to an appropriate basis this is both superdiagonal and subdiagonal unipotent and hence $=1\text{.}$ Thus $b=b\prime ,$ $x=x\prime \text{.}$ $\square$

Exercise: (a) Prove $B$ is the normalizer in $G$ of $U$ and also of $B\text{.}$ (b) Prove $N$ is the normalizer in $G$ of $H$ if $k$ has more than 3 elements.

Examples: Let $ℒ={s\ell }_{n}$ so that $G={SL}_{n},$ and $B,H,N$ are respectively the superdiagonal, diagonal, monomial subgroups, and $W$ may be identified with the group of permutations of the coordinates. Going to $G={GL}_{n}$ for convenience, we get from Theorem 4: $\left(*\right)$ the permutation matrices ${S}_{n}$ form a system of representatives for $B\G/B\text{.}$ We shall give a simple direct proof of this. Here $k$ can be any division ring. Assume given $x\in G\text{.}$ Choose $b\in B$ to maximize the total number of zeros at the beginnings of all of the rows of $bx\text{.}$ These beginnings must all be of different lengths since otherwise we could subtract a multiple of some row from an earlier one, i.e., modify $b,$ and increase the total number of zeros. It follows that for some $w\in {S}_{n},$ $wbx$ is superdiagonal, whence $x\in B{w}^{-1}B\text{.}$ Now assume $BwB=Bw\prime B$ with $w,w\prime \in {S}_{n}\text{.}$ Then ${w}^{-1}bw\prime$ is superdiagonal for some $b\in B\text{.}$ Since $w,w\prime$ are permutation matrices and the matric positions where the identity is nonzero are included among those of $b,$ we conclude that ${w}^{-1}w\prime$ is superdiagonal, whence $w=w\prime ,$ which proves $\left(*\right)\text{.}$ Next we will give a geometric interpretation of the result just proved. Let $V$ be the underlying vector space, A flag in $V$ is an increasing sequence of subspaces ${V}_{1}\subset {V}_{2}\subset \dots \subset {V}_{n},$ where $\text{dim} {V}_{i}=i\text{.}$ Associated with the chosen basis $\left\{{v}_{1},\dots ,{v}_{n}\right\}$ of $V$ there is a flag ${F}_{1}\subset \dots \subset {F}_{n}$ defined by ${F}_{i}=⟨{v}_{1},\dots ,{v}_{i}⟩$ called the standard flag. Now $G$ acts on $V$ and hence on flags. $B$ is the stabilizer of the standard flag, so $B\G/B$ is in one-to-one correspondence with the set of $G\text{-orbits}$ of pairs of flags. Define a simplex to be a set of points $\left\{{p}_{1},\dots ,{p}_{n}\right\}$ of $V$ such that $\text{dim} ⟨{p}_{1},\dots ,{p}_{n}⟩=n\text{.}$ A flag ${V}_{1}\subset \dots \subset {V}_{n}$ is said to be incident with this simplex if ${V}_{i}=⟨{p}_{\pi 1},\dots ,{p}_{\pi i}⟩$ for some $\pi \in {S}_{n}\text{.}$ Hence there are $n!$ flags incident with a given simplex.

It can be shown, by induction on $n$ (see Steinberg, T.A.M.S. 1951), that $\left(*\right)$ given any two flags there is a simplex incident with both. Thus associated to each pair of flags there is an element of ${S}_{n},$ the permutation which transforms one to the other. Hence $B\G/B$ corresponds to ${S}_{n}\text{.}$ Thus $\left(*\right)$ is the geometric interpretation of the Bruhat decomposition.

(b) Consider $ℒ=\left\{X\in {s\ell }_{m} | XA+A{X}^{t}=0\right\}$ where $A$ is fixed and nonsingular (i.e., consider the invariants of the automorphism $X\to A\left(-{X}^{t}\right){A}^{-1}\text{).}$ If $m=2n$ and $A$ is skew this gives an algebra of type ${C}_{n},$ if $m=2n$ and $A$ is symmetric this gives an algebra of type ${D}_{n},$ and if $m=2n+1$ and $A$ is symmetric this gives an algebra of type ${B}_{n}\text{.}$ If we take

$A= [ 1⋰1 1⋰1 ]$

in the first case and

$A=[1⋰1]$

in the second and third cases an element $y\in ℒ$ is superdiagonal if and only if $\text{ad} y$ preserves $\underset{\alpha >0}{\Sigma }{ℒ}_{\alpha }$ (with the usual ordering of roots). Exponentiating we get the invariants of $X\stackrel{\sigma }{\to }A{X}^{-1t}{A}^{-1}$ (that is $XA{X}^{t}=A\text{).}$ In the first case we get $G\subseteq {Sp}_{m},$ in the second and third cases $G\subseteq {SO}_{m}$ (relative to a form of maximal index). (For the proof that equality holds see Ree, T.A.M.S. 1957.) The automorphism $\sigma$ above preserves the basic ingredients $B,H,N$ of the Bruhat decomposition of ${SL}_{m}\text{.}$ From this a Bruhat decomposition for ${Sp}_{m}\text{.}$ and ${SO}_{m}$ can be inferred. By a slight modification of the procedure, we can at the same time take care of unitary groups.

(c) If $ℒ$ is of type ${G}_{2}$ it is the derivation algebra of a split Cayley algebra. The corresponding group $G$ is the group of automorphisms of this algebra.

Since the results labelled (E) depend only on the relations (R) (which are independent of the representation chosen) we may extract from the discussion so far the following result.

Proposition: Let $G\prime$ be a group generated by elements labelled ${x}_{\alpha }^{\prime }\left(t\right)$ $\left(\alpha \in \Sigma ,t\in k\right)$ such that the relations (R) hold and let $U\prime ,H\prime ,\dots$ be defined as in $G\text{.}$

 (1) Every element of $U\prime$ can be written in the form $\underset{\alpha >0}{\Pi }{x}_{\alpha }^{\prime }\left({t}_{\alpha }\right)\text{.}$ (2) For each $w\in W,$ write $w={w}_{\alpha }{w}_{\beta }\dots$ a product of reflections. Define ${\omega }_{w}^{\prime }={\omega }_{\alpha }^{\prime }{\omega }_{\beta }^{\prime }\dots$ (where ${\omega }_{\alpha }^{\prime }={w}_{\alpha }^{\prime }\left(1\right)\text{).}$ Then every element of $G\prime$ can be written $u\prime h\prime {\omega }_{w}^{\prime }v\prime$ (where $u\prime \in U\prime ,$ $h\prime \in H\prime ,$ $v\prime \in {U}_{w}^{\prime }\text{).}$

Corollary 1: Suppose $G\prime$ is as above and $\phi$ is a homomorphism of $G\prime$ onto $G$ such that $\phi \left({x}_{\alpha }^{\prime }\left(t\right)\right)={x}_{\alpha }\left(t\right)$ for all $\alpha$ and $t\text{.}$ Then:

 (a) Uniqueness of expression holds in (1) and (2) above. (b) $\text{ker} \phi \subseteq \text{center of} G\prime \subseteq H\prime \text{.}$ Proof. (a) Suppose $\Pi {x}_{\alpha }^{\prime }\left({t}_{\alpha }\right)=\Pi {x}_{\alpha }^{\prime }\left({\stackrel{\sim }{t}}_{\alpha }\right)\text{.}$ Applying $\phi$ we get $\Pi {x}_{\alpha }\left({t}_{\alpha }\right)=\Pi {x}_{\alpha }\left({\stackrel{\sim }{t}}_{\alpha }\right)$ and by Lemma 17 ${t}_{\alpha }={\stackrel{\sim }{t}}_{\alpha }$ for all $\alpha \text{.}$ Hence $\phi |U\prime$ is an isomorphism. Now if $u\prime h\prime {\omega }_{w}^{\prime }v\prime =\stackrel{\sim }{u}\prime \stackrel{\sim }{h}\prime {\omega }_{w}^{\prime }\stackrel{\sim }{v}\prime$ by applying $\phi$ we get $\phi \left(u\prime \right)\phi \left(h\prime \right){\omega }_{w}\phi \left(v\prime \right)=\phi \left(\stackrel{\sim }{u}\prime \right)\phi \left(\stackrel{\sim }{h}\prime \right){\omega }_{w}\phi \left(\stackrel{\sim }{v}\prime \right)\text{.}$ By Theorem 4' and Lemma 21 $\phi \left(u\prime \right)=\phi \left(\stackrel{\sim }{u}\prime \right)$ and $\phi \left(v\prime \right)=\phi \left(\stackrel{\sim }{v}\prime \right)\text{.}$ Hence $u\prime =\stackrel{\sim }{u}\prime$ and $v\prime =\stackrel{\sim }{v}\prime$ so $h\prime {\omega }_{w}^{\prime }=\stackrel{\sim }{h}\prime {\omega }_{w}^{\prime }$ so $h\prime =\stackrel{\sim }{h}\prime \text{.}$ (b) Let $x\prime =u\prime h\prime {\omega }_{w}^{\prime }v\prime \in \text{ker} \phi \text{.}$ Then $1=\phi \left(u\prime \right)\phi \left(h\prime \right){\omega }_{w}\phi \left(v\prime \right)\in UH{\omega }_{w}{U}_{w};$ so $w=1,$ ${\omega }_{w}^{\prime }=1,$ $\phi \left(u\prime \right)=1,$ $\phi \left(v\prime \right)=1\text{.}$ Hence $u\prime =v\prime =1$ so $x\prime =h\prime =\Pi {h}_{\alpha }^{\prime }\left({t}_{\alpha }\right)\text{.}$ Then $x\prime {x}_{\beta }^{\prime }\left(u\right){x\prime }^{-1}={x}_{\beta }^{\prime }\left(\underset{\alpha }{\Pi }{t}_{\alpha }^{⟨\beta ,\alpha ⟩}u\right)$ by (R8). Applying $\phi$ we see that $\underset{\alpha }{\Pi }{t}_{\alpha }^{⟨\beta ,\alpha ⟩}=1\text{.}$ Hence $x\prime$ commutes with $x\beta \prime \left(u\right)$ for all $\beta$ and $u,$ so is in center of $G\prime \text{.}$ To complete the proof it is enough to show that center of $G\subset H$ (for we have shown $\text{ker} \phi \subset H\prime \text{).}$ If $x=uh{\omega }_{w}v\in \text{center of} G$ and $w\ne 1$ then there exists $\alpha >0$ such that $w\alpha <0\text{.}$ Then $x{x}_{\alpha }\left(1\right)={x}_{\alpha }\left(1\right)x$ which contradicts Theorem 4'. Hence $w=1$ so $x=uh\text{.}$ Let ${w}_{0}$ be the element of $W$ making all positive roots negative. Then $x={\omega }_{{w}_{0}}x{\omega }_{{w}_{0}}^{-1}$ is both superdiagonal and subdiagonal. Since $h$ is diagonal, $u$ is diagonal, and also unipotent. Hence $u=1$ and $x=h\in H\text{.}$ $\square$

Corollary 2: Center $G\subseteq H\text{.}$

Corollary 3: The relations (R) and those in $H$ on the ${h}_{\alpha }\left(t\right)$ form a defining set of relations for $G\text{.}$ Proof. If the relations in $H$ are imposed on $H\prime$ then $\phi$ in Corollary 1 becomes an isomorphism by (b). $\square$

Corollary 4: If $G\prime$ is constructed as $G$ from $ℒ,k,\dots$ but using a perhaps different representation space $V\prime$ in place of $V,$ then there exists a homomorphism $\phi$ of $G\prime$ onto $G$ such that $\phi \left({x}_{\alpha }^{\prime }\left(t\right)\right)={x}_{\alpha }\left(t\right)$ if and only if there exists a homomorphism $\theta :H\prime \to H$ such that $\theta {h}_{\alpha }^{\prime }\left(t\right)={h}_{\alpha }\left(t\right)$ for all $\alpha$ and $t\text{.}$ Proof. Clearly if $\phi$ exists then $\theta$ exists. Conversely assume $\theta$ exists. Matching up the generators of $H\prime$ and $H,$ we see that the relations in $H\prime$ form a subset of those in $H\text{.}$ By Corollary 3 and the fact that the relations (R) are the same for $G\prime$ and $G,$ the relations on ${x}_{\alpha }^{\prime }\left(t\right),\dots$ in $G\prime$ form a subset of those on ${x}_{\alpha }\left(t\right),\dots$ in $G\text{.}$ Thus $\phi$ exists. $\square$

So far the structure of $H$ has played a minor role in the proceedings. To make the preceding results more precise we will now determine it.

We recall that $H$ is the group generated by all ${h}_{\alpha }\left(t\right)$ $\left(\alpha \in \Sigma ,t\in k\right)$ and $\left(*\right)$ ${h}_{\alpha }\left(t\right)$ acts on the weight space ${V}_{\mu }$ as multiplication by ${t}^{⟨\mu ,\alpha ⟩}\text{.}$ Also, we recall that by Theorem 3(e), a linear function $\mu$ on $ℋ$ is the highest weight of some irreducible representation provided $⟨\mu ,\alpha ⟩=\mu \left({H}_{\alpha }\right)\in {ℤ}^{+}$ for all $\alpha >0\text{.}$ Clearly, it suffices that $⟨\mu ,{\alpha }_{i}⟩\in {ℤ}^{+}$ for all simple roots ${\alpha }_{i}\text{.}$ Define ${\lambda }_{i},i=1,2,\dots ,\ell$ by $⟨{\lambda }_{i},{\alpha }_{j}⟩={\delta }_{ij}\text{.}$ We see that ${\lambda }_{i}$ occurs as the highest weight of some irreducible representation, and we call ${\lambda }_{i}$ a fundamental weight.

Lemma 27:

 (a) The additive group generated by all the weights of all representations forms a lattice ${L}_{1}$ having $\left\{{\lambda }_{i}\right\}$ as a basis. (b) The additive group generated by all roots is a sublattice ${L}_{0}$ of ${L}_{1}\text{.}$ Moreover, $\left(⟨{\alpha }_{i},{\alpha }_{j}⟩\right)$ $i,j=1,2,\dots ,\ell$ is a relation matrix for ${L}_{1}/{L}_{0},$ which is thus finite. (c) The additive group generated by all weights of a faithful representation on $V$ forms a lattice ${L}_{V}$ between ${L}_{0}$ and ${L}_{1}\text{.}$ Proof. Part (a) is immediate from the definition of the fundamental weights. (b) If ${\alpha }_{i}$ is a simple root and ${\alpha }_{i}=\Sigma {c}_{ij}{\lambda }_{j}$ $\left({c}_{ij}\in ℂ\right)$ then $⟨{\alpha }_{i},{\alpha }_{k}⟩={c}_{ik}$ and ${\alpha }_{i}=\Sigma ⟨{\alpha }_{i},{\alpha }_{j}⟩{\lambda }_{j}\text{.}$ (c) If $\alpha$ is a root, then since ${X}_{\alpha }\ne 0$ there exists $0\ne v\in {V}_{\mu }$ for some weight $\mu$ with $0\ne {X}_{\alpha }v\in {V}_{\mu +\alpha }\text{.}$ Hence $\alpha =\left(\mu +\alpha \right)-\mu \in {L}_{V}$ and ${L}_{0}\subseteq {L}_{V}\subseteq {L}_{1}\text{.}$ $\square$

Remark: All lattices between ${L}_{0}$ and ${L}_{1}$ can be realized as in Lemma 27 (c) by an appropriate choice of $V\text{.}$ In particular, ${L}_{V}={L}_{0}$ if $V$ corresponds to the adjoint representation, and ${L}_{V}={L}_{1}$ if $V$ corresponds to the sum of the representations having the fundamental weights as highest weights.

Lemma 28 (Structure of $H\text{):}$

 (a) For each $\alpha ,$ ${h}_{\alpha }\left(t\right)$ is multiplicative as a function of $t\text{.}$ (b) $H$ is an Abelian group generated by the ${h}_{i}\left(t\right)\text{'s}$ (with ${h}_{i}\left(t\right)={h}_{{\alpha }_{i}}\left(t\right)\text{).}$ (c) $\underset{i=1}{\overset{\ell }{\Pi }}{h}_{i}\left({t}_{i}\right)=1$ if and only if $\underset{i=1}{\overset{\ell }{\Pi }}{t}_{i}^{⟨\mu ,{\alpha }_{i}⟩}=1$ for all $\mu \in {L}_{V}\text{.}$ (d) The center of $G=\left\{\underset{i=1}{\overset{\ell }{\Pi }}{h}_{i}\left({t}_{i}\right) | \underset{i=1}{\overset{\ell }{\Pi }}{t}_{i}^{⟨\beta ,{\alpha }_{i}⟩}=1 \text{for all} \beta \in {L}_{0}\right\},$ hence is finite. Proof. (a), (b), and (c) follow from $\left(*\right)$ above. (a) and (c) are immediate and (b) results from ${t}^{⟨\mu ,\alpha ⟩}={t}^{\mu \left({H}_{\alpha }\right)}={t}^{\mu \left(\Sigma {n}_{i}{H}_{i}\right)}={t}^{\Sigma {n}_{i}⟨\mu ,{\alpha }_{i}⟩}$ if ${H}_{\alpha }=\Sigma {n}_{i}{H}_{i}\text{.}$ For (d), we note that $\underset{i=1}{\overset{\ell }{\Pi }}{h}_{i}\left({t}_{i}\right)$ commutes with ${x}_{\beta }\left(u\right)$ if and only if $\underset{i=1}{\overset{\ell }{\Pi }}{t}_{i}^{⟨\beta ,{\alpha }_{i}⟩}=1$ by Lemma 19(c). $\square$

Corollary:

 (a) If ${L}_{V}={L}_{1},$ then every $h\in H$ can be written uniquely as $h=\underset{i=1}{\overset{\ell }{\Pi }}{h}_{i}\left({t}_{i}\right),$ ${t}_{i}\in {k}^{*}\text{.}$ (b) If ${L}_{V}={L}_{0},$ then $G$ has center 1.

Corollary 5 (To Theorem 4'): Let $G$ be a Chevalley group as usual and let $G\prime$ be another Chevalley group constructed from the same $ℒ$ and $k$ as $G$ but using $V\prime$ in place of $V\text{.}$ If ${L}_{V\prime }\supseteq {L}_{V},$ then there exists a homomorphism $\phi :G\prime \to G$ such that $\phi \left({x}_{\alpha }^{\prime }\left(t\right)\right)={x}_{\alpha }\left(t\right)$ for all $\alpha ,t$ and $\text{ker} \phi \subseteq \text{Center of} G\prime ,$ where ${x}_{\alpha }^{\prime }\left(t\right)$ corresponds to ${x}_{\alpha }\left(t\right)$ in $G\prime \text{.}$ If ${L}_{V}={L}_{V\prime },$ then $\phi$ is an isomorphism. Proof. There exists a homomorphism $\theta :H\prime \to H$ such that $\theta {h}_{i}^{\prime }\left(t\right)={h}_{i}\left(t\right)$ by Lemma 23(c). If $\alpha$ is any root and ${H}_{\alpha }=\Sigma {n}_{i}{H}_{i},$ ${n}_{i}\in ℤ,$ then ${h}_{\alpha }\left(t\right)=\Pi {h}_{i}{\left(t\right)}^{{n}_{i}}$ and similarly for ${h}_{\alpha }^{\prime }\left(t\right)\text{.}$ Hence $\theta {h}_{\alpha }^{\prime }\left(t\right)={h}_{\alpha }\left(t\right)\text{.}$ By Corollary 4 to Theorem 4', $\phi$ exists. By Corollary 1, $\text{ker} \phi \subseteq \text{Center of} G\prime \text{.}$ If ${L}_{V}={L}_{V\prime },$ we have a homomorphism $\psi :G\to G\prime$ such that $\psi \left({x}_{\alpha }\left(t\right)\right)={x}_{\alpha }^{\prime }\left(t\right)\text{.}$ Hence, $\psi \circ \phi ={\text{id}}_{G\prime }:\phi \circ \psi ={\text{id}}_{G},$ and $\phi$ is an isomorphism. $\square$

We call the Chevalley groups ${G}_{0}$ and ${G}_{1}$ corresponding to the lattices ${L}_{0}$ and ${L}_{1}$ the adjoint group and the universal group respectively. If $G={G}_{V}$ is a Chevalley group corresponding to the lattice ${L}_{V},$ then by Corollary 5, we have central homomorphisms $\alpha$ and $\beta$ such that $\alpha :{G}_{1}\to {G}_{V}$ and $\beta :{G}_{V}\to {G}_{0}\text{.}$ We call $\text{ker} \alpha$ the fundamental group of $G$, and we see $\text{ker} \beta =\text{center of} G\text{.}$

Exercise: The center of the universal group, i.e., the fundamental group of the adjoint group is isomorphic to $\text{Hom}\left({L}_{1}/{L}_{0},{k}^{*}\right)\text{.}$ E.g., if $k=ℂ,$ the last group is isomorphic with ${L}_{1}/{L}_{0}\text{.}$ Also in this case the Center of ${G}_{V}\cong {L}_{1}/{L}_{V},$ and the fundamental group of ${G}_{V}\cong {L}_{1}/{L}_{V}\text{.}$

In the following table, we list some information known about the lattices and Chevalley groups of the various Lie algebras $ℒ\text{:}$

$Type of ℒ L1/L0 G0 GV G1 Aℓ ℤℓ+1 PSLℓ+1 SLℓ+1 Bℓ ℤ2 PSO2ℓ+1=SO2ℓ+1 Spin2ℓ+1 Cℓ ℤ2 PSp2ℓ Sp2ℓ D2n+1 ℤ4 PSO4n+2 SO4n+2 Spin4n+2 D2n ℤ2×ℤ2 PSO4n SO4n Spin4n E6 ℤ3 E7 ℤ2 E8 ℤ1 G0 = G1 F4 ℤ1 G0 = G1 G2 ℤ1 G0 = G1$

Here ${G}_{V}$ is a Chevalley group other than ${G}_{0}$ and ${G}_{1},$ ${ℤ}_{n}$ is the cyclic group or order $n,$ ${SO}_{n}$ is the special orthogonal group, ${\text{Spin}}_{n}$ is the spin group, ${Sp}_{n}$ is the simplectic group, and $PG$ denotes the projective group of $G\text{.}$

To obtain the column headed by ${L}_{1}/{L}_{0}$ one reduces the relation matrix $\left(⟨{\alpha }_{i},{\alpha }_{j}⟩\right)$ to diagonal form. To show, for example, that ${SL}_{n}$ is the universal group of $ℒ={s\ell }_{n}$ of type ${A}_{n-1},$ we let ${\omega }_{i}$ be the weight ${\omega }_{i}:\text{diag}\left({a}_{1},\dots ,{a}_{n}\right)\to {a}_{i}\text{.}$ Then if ${\lambda }_{i}={\omega }_{1}+{\omega }_{2}+\dots +{\omega }_{i},$ $1\le i\le n-1,$ we have ${\lambda }_{i}\left({H}_{j}\right)={\lambda }_{i}\left({E}_{jj}-{E}_{j+1,j+1}\right)={\delta }_{ij}\text{.}$ Hence the fundamental weights are in the lattice associated with this representation. Since the center of ${SL}_{n}$ is generically cyclic of order $n,$ it follows that ${L}_{1}/{L}_{0}$ is isomorphic to ${ℤ}_{n}$ in this case.

Exercise: If $G$ is a Chevalley group, ${G}_{1},{G}_{2},\dots ,{G}_{r}$ subgroups of $G$ corresponding to indecomposable components of $\Sigma ,$ then:

 (a) Each ${G}_{i}$ is normal in $G$ and $G={G}_{1}{G}_{2}\dots {G}_{r}\text{.}$ (b) $G$ is universal (respectively adjoint) if and only if each ${G}_{i}$ is. (c) In each case in (b), the product in (a) is direct.

Corollary 6: If $\alpha$ is a root, then there exists a homomorphism ${\phi }_{\alpha }:{SL}_{2}\to ⟨{𝔛}_{\alpha },{𝔛}_{-\alpha }⟩$ such that ${\phi }_{\alpha }\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]={x}_{\alpha }\left(t\right),$ ${\phi }_{\alpha }\left[\begin{array}{cc}1& 0\\ t& 1\end{array}\right]={x}_{-\alpha }\left(t\right),$ ${\phi }_{\alpha }\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]={\omega }_{\alpha },$ and ${\phi }_{\alpha }\left[\begin{array}{cc}1& 0\\ 0& {t}^{-1}\end{array}\right]={h}_{\alpha }\left(t\right)\text{.}$ Moreover, $\text{ker} {\phi }_{\alpha }=\left\{1\right\}$ or $\left\{±1\right\}$ so that $⟨{𝔛}_{\alpha },{𝔛}_{-\alpha }⟩$ is isomorphic to either ${SL}_{2}$ or ${PSL}_{2}\text{.}$ Proof. Let ${ℒ}_{1}$ be of rank 1 spanned by $X,Y$ and $H$ with $\left[X,Y\right]=H,$ $\left[H,X\right]=2X$ and $\left[H,Y\right]=-2Y\text{.}$ Now ${ℒ}_{1}$ has a representation $X\to \left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],$ $Y\to \left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],$ $H\to \left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$ as ${s\ell }_{2}$ on a vector space $V\prime$ and a representation $X\to {X}_{\alpha },$ $Y\to {X}_{-\alpha },$ $H\to {H}_{\alpha }$ on the same vector space $V$ as the original representation of $ℒ\text{.}$ Since ${SL}_{2}$ is universal, the required homomorphism $\phi$ exists and has $\text{ker} \phi \subseteq \left\{±1\right\}$ by Corollary 5. $\square$

Exercise: If $G$ is universal, each ${\phi }_{\alpha }$ is an isomorphism.

## 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.