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Let $G$ be a connected split reductive group over a finite field $k$. Suppose $G$ has connected centre. Let $T$ be a maximal split torus with Weyl group $W$. Note that $W$ acts on the finite group $T(k)$; thus, it acts on characters of $T(k)$.

Let $\theta: T(k) \rightarrow \mathbb{C}^\times$ be a $W$-invariant character.

Question: Does $\theta$ extend to a character of $G(k)$?

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    $\begingroup$ Isn't the order $2$ character of the split torus of $SL_2(\mathbb F_p)$ a counterexample? $\endgroup$
    – Will Sawin
    Jun 9, 2022 at 23:08
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    $\begingroup$ But the centre of SL_2 is disconnected (in characteristic not equal to 2). $\endgroup$
    – Dr. Evil
    Jun 10, 2022 at 0:25
  • $\begingroup$ What about $\textbf{PGL}_2$, then? $\endgroup$
    – Jason Starr
    Jun 10, 2022 at 1:57
  • $\begingroup$ I think the order 2 character of maximal torus of PGL_2 actually extends to all of PGL_2(k). (Note abelianization of PGL_2(k) is nontrivial) $\endgroup$
    – Dr. Evil
    Jun 10, 2022 at 6:28
  • $\begingroup$ @Dr.Evil, indeed, for $p \ne 2$ (otherwise there is no non-trivial character!) the desired extension is $g \mapsto \operatorname{sgn}_k(\det(g))$. $\endgroup$
    – LSpice
    Jun 12, 2022 at 13:06

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Edit. The answer below is incorrect. The correct computation is the determinant of the Cartan matrix, which happens to equal $1$ for $E_8$.

I believe the answer is "no" for $G$ equal to the split form of $E_8$ over a finite field $k$ of odd characteristic. The set of rational points is a simple finite group. The center of $E_8$ is trivial. For a maximal split torus $T$, I believe the action of the Weyl group $W$ on $T(k)/(T(k))^2$ is trivial, since $W$ is generated by simple reflections arising from copies of $\textbf{SL}_2$ in the group (the "root groups"), and the action of the Weyl group is trivial for $\textbf{SL}_2$. So this would mean that every nontrivial character of $T(k)/(T(k))^2$ gives a counterexample.

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    $\begingroup$ Thanks. I'm not following the argument that $W$ acts trivially $T(k)/(T(k))^2$. I would appreciate more details. $\endgroup$
    – Dr. Evil
    Jun 13, 2022 at 0:04

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