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Let G is a linear algebraic group over algebraic closed field, B is an Borel subgroup of G. Does there exist g$\in$G which is only in a finite numbers of conjugates of B (they are also Borel subgroups) ?

I choose this version of condition from the book:Tauvel, Patrice, and W. T. Rupert. Lie algebras and algebraic groups 28.2.1.

It appears in the lemma before the density theorem of Borel subgroups, but I do not see any book do this for Borel subgroups directly, they all do it for Cartan subgroups, choosing a semisimple regular element of the unique maximal torus, but Borel subgroups are bigger, maybe they have more intersections.

The similar question is at here with no answer: https://math.stackexchange.com/questions/3113958/element-in-finite-number-of-borel-subgroups

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    $\begingroup$ Regular semisimple elements have this property; the only Borel subgroups containing them will be Weyl group conjugates of a fixed one. $\endgroup$
    – Venkataramana
    Commented Mar 8, 2019 at 7:33
  • $\begingroup$ Thanks. Weyl groups appear later in many books, maybe the question should be changed, can we get the density theorem from this direct way, without vicious circle? $\endgroup$
    – Strongart
    Commented Mar 8, 2019 at 14:18
  • $\begingroup$ More generally, regularity (without assuming semisimplicity) can be defined as belonging to only a finite number of Borel subgroups (or, equivalently, having minimal-dimensional centraliser). $\endgroup$
    – LSpice
    Commented Jan 30, 2020 at 20:11

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First of all, it's probably intended that "linear algebraic groups" are semisimple or at least reductive (and connected). For example, a solvable group might have no semisimple elements except 1 (etc.) and Venkatarama's comment won't apply. The basic object of study is a connected semisimple group, for which Steinberg's 1965 IHES 25 paper extended Kostant's Lie algebra ideas in characteristic 0 to such a group over an algebraically closed field of arbitrary characeristic. (Chapters 2 and 4 of my 1995 AMS monograph Conjugacy Classes in Semisimple Algebraic Grooups expose all of this theory.but Steinberg'spaper is freely available online from numdam.org.)

In this context, a regular semisimple element lies in only finitely many Borel subgroups, and a regular unipotent element lies in a unique such subgroup. But the situation is more complicated for an arbitrary regular element, defined to be one whose centralizer is of minimal dimension (equal to the rank of the group).

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    $\begingroup$ There's no reason to assume $G$ semisimple/reductive. Indeed, Borel subgroups contain the solvable radical $R$, and more precisely are the inverse images of Borel subgroups of the corresponding semisimple quotient $G/R$. Hence the question immediately reduces to the semisimple case. $\endgroup$
    – YCor
    Commented Mar 8, 2019 at 23:28
  • $\begingroup$ Agreed. but my "assumption" was meant to be a reply to Venkataramana's comment. $\endgroup$
    – Jim Humphreys
    Commented Mar 12, 2019 at 19:56

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