Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

[This question arises from trying to understand an incompletely formulated earlier question here.]

Let $G$ be a semisimple algebraic group over an algebraically closed field $K$ of characteristic $p \geq 0$, with some fixed maximal torus $T$ of dimension $r$ (the rank of $G$). The adjoint quotient $\pi:G \rightarrow X$ involves an irreducible affine variety $X$ of dimension $r$ associated to the algebra of invariants $K[G]^G$ (Chevalley). Here $X$ identifies with the orbit space of $T$ under the action of the Weyl group.

Each fiber of $\pi$ is the union of one or more conjugacy classes, but only the semisimple classes are actually closed. Call $\sigma:X \rightarrow G$ a section of $\pi$ if $\pi \sigma = \text{id}_X$. Then $S:= \sigma(X)$ is an example of a cross-section: a closed irreducible subset of $G$ meeting each fiber of $\pi$ in a single point. Call cross-sections $S$ and $S'$ "conjugate" if there is a morphism $\gamma:X \rightarrow G$ (possibly constant) such that $\gamma(s) s \gamma(s)^{-1} \in S'$ for each $s \in S$. .

General problem: Classify all cross-sections of $\pi$, up to conjugacy.

In his 1965 Publ. Math. IHES paper on regular elements, available via Numdam here, Steinberg assumed that $G$ is simply connected and constructed an explicit section $\sigma$ of $\pi$. Its image $S \subset G$ in fact consists entirely of regular points; each fiber is the closure of a regular class and contains a unique (closed) semisimple class. In the simply connected case, the affine variety $X$ can be identified naturally with $\mathbb{A}^r$, via the characters of fundamental representations (for positive roots relative to $T$.)

Example: When $G=\mathrm{SL}_n(K)$, Steinberg's cross-section involves a version of the classical rational canonical form. If $n=3$ one gets essentially the companion matrices of the characteristic (=minimal) polynomials $x^3 - a x^2 + bx -1$:

$$ \begin{pmatrix} a & b & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} $$

In a 2003 AMS conference proceedings, Friedman and Morgan here work exclusively over $\mathbb{C}$. Their main theorem shows that (for $G$ still simply connected) all cross-sections are conjugate to the Steinberg cross-section. But they use methods of complex algebraic geometry which go well beyond Steinberg's relatively elementary group-theoretic framework.

(!) If $p=0$, is there a more direct proof of the Friedman-Morgan theorem in the algebraic group setting, valid over any algebraically closed field of characteristic 0?

The $p=0$ assumption is actually necessary here, as shown more recently by Vladimir L. Popov in Cross-sections, quotients, and representation rings of semisimple algebraic groups in Transformtion Groups 16 (2011), 827-856. Popov works over our field $K$, where if $p=0$ he observes that all cross-sections arise from sections of $\pi$. (But if the natural map from a simply connected covering group to $G$ fails to be bijective, there are no cross-sections, so he studied instead "rational" sections over a dense open subset of $X$ for any $p$.)

It's worse when $p>0$. By replacing the 0 in the center of the above matrix by $a^{p^d}-a$ for $d>0$, Popov gets cross-sections which don't come from sections and are not all conjugate (for separability reasons).

(2) If $p>0$, can one classify the cross-sections up to conjugacy? (Is the Steinberg section somehow special?)

share|improve this question
    
The Friedman-Morgan proof looks completely algebraic (i.e., it isn't clear what "analytic flavor" you're referring to). In a few places they write $\mathbf{C}$, but only in places where it could just as well be any algebraically closed field of char. 0. Please clarify why the proof by Friedman and Morgan doesn't satisfy your requirements. –  user30180 Dec 29 '12 at 23:07
    
@ayanta: I've tried to clarify (1), which is less important than (2); but a different approach to (1) might shed light on where Steinberg's construction of a section fits into the more complicated prime characteristic picture. –  Jim Humphreys Dec 30 '12 at 13:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.