# Compute the discriminant for reductive groups

Consider $G=GL_{2}$ and $F=k((\pi))$, and a diagonal matrix $t=\left(\begin{array}{cc}a&0\\0&b\end{array}\right)$.

The characteristic polynomial of $t$ is $X^{2}-(a+b)X+ab$, and the discriminant of the characteristic polynomial is $\Delta=(b-a)^{2}$.

In the theory of algebraic groups we also have another discriminant to check if it's regular semisimple: $\Delta'(t)=\prod\limits_{\alpha\in R}(\alpha(t)-1)=\frac{(b-a)^{2}}{ab}$.

For instance, we can obtain it as the valuation of the determinant of $Id-Ad(t)$ on $\mathfrak{gl}_{2}(F)/\mathfrak{t}(F)$, and so $val (\Delta(t))\neq val (\Delta'(t))$.

The reason is that somehow the second definition kills the center and is the right one only for semisimple.

My question is: do we have a group theoretic interpretation of $\Delta$ which gives us the right determinant?

Of course, I stated it for $GL_{2}$ but the same question holds more generally for reductive groups.

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What is "the same question" for more general reductive groups? That is, what are you putting in place of the discriminant of a quadratic polynomial? The setup looks rather specific to GL$_n$ (or SL$_n$). – Marguax Sep 26 '13 at 19:32
For a general reductive group G you have you have a finite flat surjective map map $T\rightarrow T/W$ and you can compute the discriminant of this map and compare to same $\Delta'$. – prochet Sep 27 '13 at 7:10
Have you looked at Steinberg's IHES article on regular semisimple elements of connected semisimple groups (where he identifies $T/W$ with the "quotient" of $G$ modulo conjugation when $G$ is simply connected, and also with an affine space)? – Marguax Sep 27 '13 at 10:03
it's only an affine space forr semisimple simply connected group – prochet Sep 27 '13 at 20:46
Yes, right, sorry I wasn't clear about that. – Marguax Sep 27 '13 at 23:15