TheLemma.tex

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\begin{document}
\title{The Lemma}

Suppose $k$ is a finite field, $\G$ is a connected reductive group over $k$ and $\T$ is a maximal torus in $G$.  Let $\N$ be the normalizer of $\T$ and $\W$ the absolute Weyl group $\N / \T$.  The absolute Galois group $\Galk = \Gal(\bar{k}/k)$ acts on everything; write $G = \G^\Galk$, $T = \T^\Galk$ and $W = \W^\Galk$.  Set $\hatT = \Hom(T, \CC^\times)$.  We have that $W$ acts on $T$ and $\hat{T}$; we call an element or character \emph{regular} if it is not fixed by any nontrivial element of $W$ (this condition is sometimes referred to as strongly regular).  Write $\Treg$ for the set of regular elements in $T$, $\Threg$ for the set of regular characters, $\Tirr$ for the set of irregular elements and $\Thirr$ for the set of irregular characters.  Note that in general neither the regular nor the irregular elements of $T$ or $\hatT$ form a subgroup.

\begin{problem}
Identify the set $Q_{\T}$ of $\alpha \in \hatT$ with the following property:
\begin{itemize}
\item For every $\chi \in \Threg$ there is a $w \in W$ with $\frac{\chi}{w(\chi)} = \alpha$.
\end{itemize}
\end{problem}

\section{Facts}

\begin{proposition} \label{pigeonhole}
Suppose $\#\Threg > (\# W - 1) \cdot \# \Thirr$.  Then $Q_{\T} = \{ 1 \}$.
\end{proposition}
\begin{proof}
For $w \in W$, set
$$S_w = \{\chi \in \Threg \st  \frac{\chi}{w(\chi)} = \alpha\}.$$
Note that if $S_1$ is nonempty then we get $\alpha = 1$ immediately, so we may assume the contrary.  Then by the pigeonhole principle, there is a $w \in W$ with $\# S_w > \# \Thirr$.  Pick $\chi \in S_w$; since $\# S_w > \#\Thirr$ there is some $\chi' \in S_w$ with $\frac{\chi}{\chi'}$ regular.  We now have
$$\frac{\chi}{w(\chi)} = \alpha = \frac{\chi'}{w(\chi')}$$
and therefore $\frac{\chi}{\chi'}$ is fixed by $w$.  Since $\frac{\chi}{\chi'}$ is regular, we must have $w = 1$ and thus 
$$\alpha = \frac{\chi}{\chi} = 1.$$
\end{proof}

Recall that Frobenius acts on $X^*(\T)$ via an endomorphism $F = qF_0$, where $F_0$ is an automorphism of finite order.  So it makes sense to vary $q$: we fix $F_0$ and consider the tori dual to the $\Galk$-modules with Frobenius acting through $qF_0$.

\begin{corollary}[{c.f. \cite[Lemma 8.4.2]{carter}}]
Consider the family of tori $\T_q$ with the same $F_0$.  Then for sufficiently large $q$, $Q_{\T_q} = \{ 1 \}$ (regardless of the $\G$ in which $\T_q$ is embedded).
\end{corollary}
\begin{proof}
We will write $\T$ for a general torus in the family and $r$ for the common dimension.  For $w \in W$ with $w \ne 1$ the centralizer $\Z_{\T}(w)$ is a proper $F$-stable subgroup of $\T$, and thus $\dim(\Z_{\T}(w)) \le r - 1$.  By \cite[3.3.5]{carter} $\# T$ is a polynomial in $q$ of degree $r$ and $\# \Z_{\T}(w)^\Galk = \# \Z_T(w)$ is a polynomial in $q$ of degree at most $r-1$.  Thus the ratio
$$\frac{\# \Treg}{\# \Tirr} = \frac{\# T - \sum_{1 \ne w \in W} \# \Z_T(w)}{\sum_{1 \ne w \in W} \# \Z_T(w)}$$
grows without bound as $q$ does.  Applying this result to the dual torus $\hatT$ yields the corollary.
\end{proof}

\begin{proposition} \label{irr-sub}
The set $Q_{\T}$ is a subgroup of $\hatT$, contained within $\Thirr$ and stable under the action of $W$.
\end{proposition}
\begin{proof}
If $\alpha \in \Threg \cap Q_{\T}$ then there is some $w \in W$ with $\frac{\alpha}{w(\alpha)} = \alpha$, so $\alpha = 1$ which is not regular.

We now show that $Q_{\T}$ is a group.  Certainly $1 \in Q_{\T}$.  Suppose $\alpha, \alpha' \in Q_{\T}$ and $\chi \in \Threg$.  Then there are $w, w' \in W$ with
\begin{align*}
\frac{\chi}{w(\chi)} &= \alpha, \\
\frac{w(\chi)}{w'(w(\chi))} &= \alpha'.
\end{align*}
Multiplying the two relations yields $\frac{\chi}{w'w(\chi)} = \alpha\alpha'$, so $\alpha\alpha' \in Q_{\T}$.  Since $Q_{\T}$ is finite and contained within $\hatT$ it is thus closed under inversion.

Finally, suppose $\tau \in W$.  Given $\chi \in \Threg$ with $\alpha = \frac{\chi}{w(\chi)}$ we have
$$\tau(\alpha) = \frac{\tau(\chi)}{\tau w(\chi)} = \frac{\tau(\chi)}{w' \tau(\chi)}$$
for some $w' \in W$.  Since $\tau$ permutes the regular characters we get that $\tau(\alpha) \in Q_{\T}$.
\end{proof}

\begin{proposition} \label{orderdiv}
If $\alpha \in Q_{\T}$ has order $m$ and $\chi \in \Threg$ has order $k$ then $m$ divides $k$.
\end{proposition}
\begin{proof}
There is a $w \in W$ with
$$\frac{\chi}{w(\chi)} = \alpha.$$
Since $w(\chi)$ also has order $k$, raising both sides to the $k$th power  yields $\alpha^k = 1$.
\end{proof}

\section{Rank 1 and 2 tori}

Write $k_r$ for the degree $r$ extension of $k$, $\T_r$ for the restriction of scalars torus $\Res_{k_r/k} \Gm$ and $\T_r^s$ for $\{x \in \Res_{k_r/k} \Gm \st \Nm_{k_r/k_s} x = 1\}$.

In rank one there are two tori:
\begin{itemize}
\item $\Gm$
\item $\T_2^1$
\end{itemize}

In rank two there are 
\begin{itemize}
\item $\T_1 \times \T_1$
\item $\T_1 \times \T_2^1$
\item $\T_2^1 \times \T_2^1$
\item $\T_2$
\item $\T_3^1$
\item $\T_4^2$
\item $\T_6^3 \cap \T_6^2 \simeq \T_6^3 / \T_2^1$
\end{itemize}

There are two rank 1 semisimple groups over $k$: $\SL_2$ and $\PGL_2$.  In both cases, for both the split and anisotropic tori, the Weyl group acts as inversion, so the irregular characters are precisely those of order $1$ or $2$.  When $q = 2^k$ neither torus has any nontrivial irregular characters so $Q_{\T} = \{1\}$.  When $q = 3$ the split torus has no regular characters and thus $Q_{\T} = \Thirr$ has order $2$.  The anisotropic torus has $4$ elements, $2$ regular and $2$ irregular characters.  Again $Q_{\T}$ has order $2$ since the Weyl group exchanges the two regular characters and their quotient is the nontrivial irregular character.  For $q = 5$ the split torus has the same behavior as the anisotropic torus for $q = 3$: $Q_{\T}$ has order $2$.  The anisotropic torus has regular characters of order $3$ and $6$ so $Q_{\T} = \{1\}$ by Proposition \ref{orderdiv}; the same holds for the split torus when $q = 7$.  For the anisotropic torus ($q = 7$) and all larger $q$, $Q_{\T}$ is trivial by Proposition \ref{pigeonhole}.

\begin{tabular}{l|l|l}
Group & Torus & Weyl Group Action \\
\hline && \\
$\SL_3,$ & $\T_1 \times \T_1$ & Full $S_3$ action \\
$\PGL_3$ & $\T_2$ & $\ZZ/2$: Galois action \\
& $\T_3^1$ & $\ZZ/3$: Galois action \\
\hline && \\
$\SU_3$ & $\T_6^3 \cap \T_6^2$ & trivial Weyl group \\
& $\T_2$ & $\ZZ/2$: Galois action \\
$\PU_3$ & $\T_6^3 / \T_2^1$ & trivial Weyl group \\
& $\T_2$ & $\ZZ/2$: Galois action \\
\hline && \\
$\SO_5,$ & $\T_1 \times \T_1$ & Full $D_8$ action \\
$\Sp_4$ & $\T_1 \times \T_2^1$ & $\ZZ/2 \times \ZZ/2$: inversion on each component \\
& $\T_4^2$ & $\ZZ/4$: Galois action \\
\hline && \\
$\SO_4$ & $\T_1 \times \T_1$ & $\ZZ / 2 \times \ZZ / 2$: inversion on each component \\
& $\T_1 \times \T_2^1$ & $\ZZ / 2 \times \ZZ / 2$: inversion on each component \\
& $\T_2^1 \times \T_2^1$ & $\ZZ / 2 \times \ZZ / 2$: inversion on each component \\
& $\T_4^2$ & $\ZZ / 2$: subgroup of Galois \\
\hline && \\
$\SL_2 \times \SL_2,$ & $\T_1 \times \T_1$ & $\ZZ / 2 \times \ZZ / 2$: inversion on each component \\
$\SL_2 \times \PGL_2,$ & $\T_1 \times \T_2^1$ & $\ZZ / 2 \times \ZZ / 2$: inversion on each component \\
$\PGL_2 \times \PGL_2$ & $\T_2^1 \times \T_2^1$& $\ZZ / 2 \times \ZZ / 2$: inversion on each component
\end{tabular}

First note that $Q_{\T}$ is automatically trivial when the rational Weyl group is trivial.  For the tori with cyclic rational points of order $m$ where the Weyl group induces the Galois action, a character is irregular if and only if it maps into the $q-1$ roots of unity (or $q^2-1$ in the case $\T_4^2 \subset \SO_4$).
\begin{itemize}
\item In $\T_2$, $m=q^2 - 1$ and thus there are $q-1$ irregular characters and $q^2 - q$ regular characters so Proposition \ref{pigeonhole} implies $Q_{\T}$ is trivial.
\item In $\T_3^1$, $m = q^2 + q + 1 = (q - 1)(q + 2) + 3$ so there are $3$ irregular characters if $q \equiv 1 \pmod{3}$ and $1$ otherwise.  Again Proposition \ref{pigeonhole} does the job for all $q$.
\item In $\T_4^2$, $m = q^2 + 1 = (q-1)(q+1) + 2 = (q^2 - 1) + 2$ so there are $2$ irregular characters if $q$ is odd and $1$ otherwise (in both the $\SO_4$ and $\SO_5$ cases).  Proposition \ref{pigeonhole} works for all $q$.
\end{itemize}

For the tori that decompose as a product of rank 1 tori with a corresponding decomposition of the Weyl group, $Q_{\T}$ will split up analogously.  So we get some examples with nontrivial $Q_{\T}$ in the $q=3$ and $q=5$ cases described in the rank 1 discussion.

The only remaining cases are the split tori.  Consider $\PGL_3$ first, and let $g$ be a generator of $k^\times$ and $\zeta$ a $(q-1)^{\mathrm{st}}$ root of unity.  Characters of the split torus are of the form
$$\chi_{a, b} \colon \begin{pmatrix}
g^x & & \\
& g^y & \\
& & 1 \\
\end{pmatrix}
\mapsto \zeta^{ax + by},$$
where $a$ and $b$ are specified modulo $q-1$.  The Weyl group acts as follows:
\begin{align*}
w_{1,2}: \chi_{a,b} &\mapsto \chi_{b, a}, \\
w_{2,3}: \chi_{a,b} &\mapsto \chi_{a,-a-b}, \\
w_{3,1}: \chi_{a,b} &\mapsto \chi_{-a-b,b}, \\
w_{1,2,3}: \chi_{a,b} &\mapsto \chi_{b,-a-b}, \\
w_{1,3,2}: \chi_{a,b} &\mapsto \chi_{-a-b,a}. \\
\end{align*}
A character is irregular if $a=b$ (fixed by $w_{1,2}$) or $a = -2b$ (fixed by $w_{2,3}$) or $b = -2a$ (fixed by $w_{3,1}$).  The first two conditions intersect when $a = b = -2b$, which occurs $3$ times if $q \equiv 1 \pmod{3}$ and once otherwise; first and third is similar.  The last two conditions intersect when $a = 4a = -2b$, which occurs with the same frequency.  In fact, this same set consists of the intersection of all three conditions (and the characters fixed by $w_{1,2,3}$ and $w_{1,3,2}$).  So by inclusion-exclusion we have

\vspace{0.2in}

\begin{tabular}{lll}
$3q-9$ irregular characters & $q^2-5q+10$ regular characters & if $q \equiv 1 \pmod{3}$, \\
$3q-5$ irregular characters & $q^2-5q+6$ regular characters & if $q \not\equiv 1 \pmod{3}$.
\end{tabular}

\vspace{0.2in}

Note that the number of regular characters is always a multiple of $6$, which is forced by the fact that the Weyl group acts simply on $\Threg$.  For $q = 2, 3$ there are no regular characters, so $Q_{\T} = \Thirr$.  Proposition \ref{pigeonhole} is effective when
\begin{align*}
q^2 - 5q + 10 > 5(3q - 9) \Leftrightarrow q \ge 19 & q \equiv 1 \pmod{3} \\
q^2 - 5q + 6 > 5(3q-5) \Leftrightarrow q \ge 23 & q \not\equiv 1\pmod{3}
\end{align*}
so we need to check $q=4,5,7,8,9,11,13,16,17$.  Consider $\frac{\chi}{w(\chi)}$ as $w$ ranges over nontrivial Weyl group elements.  We get
\begin{align*}
\chi_{a-b,b-a} \\
\chi_{0,2b-a} \\
\chi_{2a-b,0} \\
\chi_{a-b,2b-a} \\
\chi_{2a-b,b-a}
\end{align*}
We're looking for an irregular character $\chi_{c,d}$ that is hit by one of these options for every pair $(a,b)$.  Suppose first that $c=d$.  If $a-b = c = d = b-a$ then $2a = 2b$ so $(a,b) = (\frac{q-1}{2}, 0)$ or $(0,\frac{q-1}{2})$.  In either case $(c, d) = (\frac{q-1}{2}, \frac{q-1}{2})$.  If $a - b = 2b - a = \frac{q-1}{2}$ then again $2a = 2b$ so $a = a - b$ so $b = 0$ and $a = \frac{q-1}{2}$.  The same analysis from $b - a = 2a - b = \frac{q-1}{2}$ gives $a = 0$ and $b = \frac{q-1}{2}$.  None of these give regular $\chi_{a,b}$.

For $c = d \ne \frac{q-1}{2}$ we still have either $a - b = 2b - a \Leftrightarrow 2a = 3b$ and $b = 2c$ or $2a-b = b-a \Leftrightarrow 3a = 2b$ and $a = 2c$.  For fixed $c$ the first choice yields two possibilities, as does the second.  For $q \ge 4$ there are more than four regular characters, so $\chi_{c,c}$ can't be hit by them all.

For $c = -2d$ we first consider the case $c = 0$ and $d = \frac{q-1}{2}$.  This case can be obtained through the second option for $a = 2b + \frac{q-1}{2}$, which is regular unless
$$b \in \{0, \frac{q-1}{2}, \frac{q-1}{8}, \frac{3(q-1)}{8}, \frac{5(q-1)}{8}, \frac{7(q-1)}{8}, \frac{q-1}{5}, \frac{2(q-1)}{5}, \frac{3(q-1)}{5}, \frac{4(q-1)}{5}\}.$$
None of the other options can produce $\chi_{c,d}$ from regular $\chi_{a,b}$.  There are $q-1$ choices for $b$, of which some are irregular depending on $b \pmod{40}$.

\begin{tabular}{ll}
$q$ & Number of regular $\chi_{a,b}$ \\
4 & 2 \\
5 & 2 \\
7 & 4 \\
8 & 6 \\
9 & 2 \\
11 & 5 \\
13 & 10 \\
16 & 10 \\
17 & 10
\end{tabular}

In no case can we obtain all regular characters in this form.

Next we consider the case that $c = -2d$ and both are nonzero.  If $a-b = -2(b-a)$ then $a=b$ so $\chi_{a,b}$ is not regular.  If $a-b = -2(2b-a)$ then $a = 3b$ which is regular unless 
$$b \in \{0, \frac{q-1}{2}, \frac{q-1}{7}, \frac{2(q-1)}{7}, \frac{3(q-1)}{7}, \frac{4(q-1)}{7}, \frac{5(q-1)}{7}, \frac{6(q-1)}{7}, \frac{q-1}{5}, \frac{2(q-1)}{5}, \frac{3(q-1)}{5}, \frac{4(q-1)}{5}\}.$$

\begin{tabular}{ll}
$q$ & Number of regular $\chi_{a,b}$ \\
4 & 2 \\
5 & 2 \\
7 & 4 \\
8 & 0 \\
9 & 6 \\
11 & 5 \\
13 & 10 \\
16 & 10 \\
17 & 14
\end{tabular}

If $2a-b = -2(b-a)$ then $b = 0$ and $a \in \{0, \frac{q-1}{2}\}$ so there are no other regular contributions.  Again in no case do we obtain all regular characters.

In summary, the split torus in $\PGL_3$ has $Q_{\T} = \{1\}$ for $q > 3$.

I won't go through the full analysis for $\SL_3$ but I wanted to point out that the action of the Weyl group is different on the split torus.

Characters of the split torus are of the form
$$\chi_{a, b} \colon \begin{pmatrix}
g^x & & \\
& g^y & \\
& & g^{-x-y} \\
\end{pmatrix}
\mapsto \zeta^{ax + by},$$
where $a$ and $b$ are specified modulo $q-1$.  The Weyl group acts as follows:
\begin{align*}
w_{1,2}: \chi_{a,b} &\mapsto \chi_{b, a}, \\
w_{2,3}: \chi_{a,b} &\mapsto \chi_{a-b,-b}, \\
w_{3,1}: \chi_{a,b} &\mapsto \chi_{-a,b-a}, \\
w_{1,2,3}: \chi_{a,b} &\mapsto \chi_{b-a,-a}, \\
w_{1,3,2}: \chi_{a,b} &\mapsto \chi_{-b,a-b}. \\
\end{align*}

\begin{thebibliography}{9}
\bibitem{carter}
  Roger Carter,
  \emph{Finite Groups of Lie Type: Conjugacy Classes and Complex Characters}.  Wiley \& Sons, Chichester, 1985.
\end{thebibliography}

\end{document}