Errata for "Representations of finite Chevalley groups" by G.Lusztig
p. v, line -9 replace "Reginal" by "Regional"
p.5 in the last two displayed formulas (line -2,-4) replace $F^n$ by
$F^ng^{-1}$.
p.25,line 10 replace $i\ne 1$ by $i\ne 1,2$.
p.47, Ref.6: the author is C.T.Benson and C.W.Curtis
p.47, Ref 13: the year should be 1955
The text on p.18 starting with "One can show..." on line 11 contains several
errors. (I thank C. Bonnafe for pointing them out to me.) One should replace
it by the following text.
We show that the curves
(1) $xy^q-x^qy=z^{q+1}$.
(2) $x^{q+1}+y^{q+1}+z^{q+1}=0$
are equivalent over $F_{q^4}$. We make a change of coordinates $x=aX+bY$,
$y=cX+dY$ with $a,b,c,d$ to be determined. The equation (1) becomes
$(aX+bY)(c^qX^q+d^qY^q)-(a^qX^q+b^qY^q)(cX+dY)=z^{q+1}$,
$(ac^q-a^qc)X^{q+1}+(bd^q-b^qd)Y^{q+1}+(bc^q-a^qd)X^qY+(ad^q-b^qc)XY^q=z^{q+1}$.
We want to satisfy the conditions
(3) $ac^q-a^qc=-1, bd^q-b^qd=-1, bc^q=a^qd, ad^q=b^qc, ad-bc\ne0$.
Choose $v\in F_{q^2}-F_q, u=v^q$. Set $a=cu, b=dv$. The conditions (3) become
$c^{q+1}(u-u^q)=-1, d^{q+1}(v-v^q)=-1, cd(u-v)\ne0$.
Since $u\ne v$, the last condition is equivalent to $cd\ne0$.
Let $w=-1/(u-u^q)$. We have $w^q=-w$ hence $w^{q-1}=-1$. Hence
$w^{(q^4-1)/(q+1)}=(-1)^{q^2+1}=1$. Since $F_{q^4}-{0}$ is cyclic we can find
$c\in F_{q^4}-{0}$ such that $c^{q+1}=w$. Similarly we can find $d\in F_{q^4}-{0}$
such that $d^{q+1}=-1/(v-v^q)$. Thus equation (3) can be solved over $F_{q^4}$
and equation (1) can be transformed by a linear change of coordinates with
coefficients in $F_{q^4}$ into the equation (2).
From Theorem 3.23 for the unitary group $U_3(F_q)$ which acts on the curve (2)
we see that $F^2=-q$ on the first l-adic cohomology of (2) hence $F^4=q^2$ on the
first l-adic cohomology of (1). It follows that $F^4=q^2$ on $H_\epsilon$.
Note that $H_\epsilon$ may be identified with a subspace of the first l-adic
cohomology of the curve obtained from $C$ by taking orbits of the group
$\{\lambda\in H;\lambda^{{q+1)/2}=1\}$. The map $(x,y)\mapsto(u,z)$ where
$z=yx^{-1}$, $u=x^{-(q+1)/2}$ defines an isomorphism of this curve with
$C'=\{(u,z)\in k^*\times k;u^2=z^q-z\}$.
(This subspace is $H^-$, the $(-1)$-eigenspace of the involution induced by
$(u,z)\mapsto(-u,z)$.) Let $H^+$ be the $+1$-eigenspace of this involution
on the first l-adic cohomology of $C'$. It is easy to see that $\dim H^+=q$,
$\dim H^-=q-1$. The quotient of $C'$ by the involution above is
$\{z;z^q-z\ne0\}$. Hence $F$ acts as $1$ on $H^+$.
Now $H^-$ is a direct sum of two nonisomorphic irreducible $SL_2(F_q)$-modules
of dimension $(q-1)/2$ on which $F$ must acts by nonzero scalars say $a,b$. It
is enough to show that $a\ne b$. By the trace formula for Frobenius maps we have
$q-q1-(q-1)/2(a+b)=|C'{}^F}|$
and this is clearly 0. (If $(u,z)\in C'{}^F$ then $z^q=z$ hence $u=0$, absurd).
Thus $a+b=0$. Since $a,b$ are nonzero we see that $a\ne b$ as required.
Note that $a^4=b^4=q^2$. More precisely we show that $a^2=b^2=-q\e(-1)$.
Using again the trace formula for Frobenius maps we have
$q^2-q1-(q-1)/2(a^2+b^2)=|C'{}^{F^2}|=N$. Now
$N=|\{(u,z)\in F_{q^2}^*\times F_q;u^2=z^q-z\}|$. Also $z\to z^q-z$ is a
homomorphism from $F_{q^2}$ onto $\{x\in F_{q^2}|x^2+x=0\}$ with kernel
of size $q$. Hence
$N=q|\{(u,x)\in F_{q^2}^*\times F_{q^2};x^q+x=0,u^2=x\}|
=q|\{u\in F_{q^2}^*;u^{2q}+u^2=0\}|=q|\{u\in F_{q^2}^*;u^{2(q-1)}=-1\}|.
Now $u\to u^{q-1}$ is a homomorphism of $F_{q^2}^*$ onto
$P=\{v\in F_{q^2}^*;v^{q+1}=1\}$
with kernel of size $q-1$. Hence
$N=q(q-1)|\{v\in P;v^2=-1\}|=q(q-1)(1+\epsilon(-1))$.
We see that $q^2-q1-(q-1)a^2=q(q-1)(1+\epsilon(-1))$. Hence
$a^2=-q\epsilon(-1)=b^2$, as required.
Let $H_\epsilon^a$ (resp. $H_\epsilon^b$) be the $a$- (resp. $b$-) eigenspace
of $F$ on $H^-=H_\epsilon$. If $x,y\in H_\epsilon^a$ then
$q(x,y)=(Fx,Fy)=(ax,ay)=a^2(x,y)=-q\epsilon(-1)(x,y)$.
If $\epsilon(-1)=1$ it follows that $(x,y)=0$ so that
$H_\epsilon^a$ is an isotropic subspace of $H_\epsilon$ (necessarily maximal since
it has dimension $(q-1)/2$); similarly $H_\epsilon^b$ is a maximal isotropic subspace
of $H_\epsilon$.
If $\epsilon(-1)=-1$ then for $x\in H_\epsilon^a$, $y\in H_\epsilon^b$ we have
$q(x,y)=(Fx,Fy)=(ax,by)=ab(x,y)=q\epsilon(-1)(x,y)=-q(x,y)$. Hence $(x,y)=0$. It
follows that the restrictions of $(,)$ to $H_\epsilon^a$ and $H_\epsilon^b$ are
nondegenerate.