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[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?)

[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 Regular elements of semi-simple algebraic groups (Numdam), 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, in Automorphism sheaves, spectral covers, and the Kostant and Steinberg sections, 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?)

[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?)

[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?)

[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 algebraic geometry having an analytic flavor.

(!) If $p=0$, is there a 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?)

[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?)