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I am trying to understand the interaction between Borel subgroups of $GL_n$ and its roots. Is it correct to say that for any choice of roots among each pair of reciprocal roots there is a Borel subgroup containing those root subgroups? (Meaning that exactly those roots appear on the decompsition of its Lie algebra) If not, what is the precise condition on the choice of roots that is needed for such a Borel to exist?

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Not every collection of choices between roots $\alpha$ and $-\alpha$ is "allowed". Yes, there is a partition of the set of all roots into positive $\Phi$ and negative $-\Phi$, but, also, $\Phi$ must be closed under addition. In the case of $GL(n)$, the Weyl group (permutation matrices, if you like) acts simply-transitively on the set of such choices, and/so there are exactly $n!$ such choices, rather than the $2^{n(n-1)/2}$ choices of sign.

[Edit:] In response to @Brad H-D's comment/query: yes, I was a bit glib... Indeed, it is not the case that all possible (iterated) sums of positive roots are roots, since this would (erroneously) require that there be infinitely-many, etc. Rather, as Brad H-D leadingly-asked, it means that either the sum of two "positive" roots is a "positive" root, or is not a root at all. (Thx, Brad H-D.)

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  • $\begingroup$ When you say "$\Phi$ must be closed under addition" do you mean instead that the sum of two positive roots is either a positive root or not a root? $\endgroup$
    – Brad Hannigan-Daley
    Commented Nov 21, 2011 at 0:52
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The question itself seems too elementary for this site, since it just involves the standard axiomatic treatment of root systems as in Bourbaki Groupes et algebres de Lie, VI.1.7. The question is really about an arbitrary reductive algebraic group (with nontrivial derived group) over an algebraically closed field, along with its Borel subgroups in natural bijection with systems of positive roots relative to a fixed maximal torus. Such a torus $T$ lies in exactly $|W|$ Borel subgroups, where $W=N_G(T)/T$ is the Weyl group. At this point the axiomatic theory takes over and provides straightforward criteria for a given "closed" set of roots to be the positive roots for some choice of simple roots: the set has to be disjoint from its negative and together with its negative exhaust all roots (Prop. 20, Cor. 1).

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    $\begingroup$ With respect, there is no way this question is too elementary for this site. It is a reasonable question for a graduate student to ask -- or a research mathematician in another field, who might not know the standard references. I agree that the confusion is straightforward and easily cleared up, but I would say this shows that the question is the right amount of elementary for this site! $\endgroup$
    – Tom Church
    Commented Nov 21, 2011 at 2:18
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    $\begingroup$ I agree with Jim. However I am not voting to close this question. $\endgroup$
    – Bruce Westbury
    Commented Nov 21, 2011 at 5:00
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    $\begingroup$ @Tom: I don't mean to imply that the structure theory of reductive groups is itself easy or elementary, just that the question asked needs a systematic reference like Bourbaki rather than an ad hoc answer online. The Borel-Chevalley theory was an impressive development of Cartan's earlier work on Lie groups, while the 1965 IHES paper by Borel-Tits allowed arbitrary ground fields and was integrated with Bourbaki's axiomatization of root systems. Aside from that, the question needs more context: what sources dealing with Borel subgroups is the questioner starting from? $\endgroup$
    – Jim Humphreys
    Commented Nov 21, 2011 at 13:39
  • $\begingroup$ I think we (collectively) want both: the facile, possibly ad-hoc also tangible example/data-point, and the systematic reference to neo-classical literature. That is, examples are good, and over-arching concepts are good, and neither supplants the other. The pseudo-joke I tell my students about the definition of "parabolic" is its definition "G/P is complete", with the caveat that "complete" does not necessarily mean "projective", and so on. Yes, the classical projective imbeddings of Grassmannians and such are interesting, but I think it's ok to just give the example, too. Best to all... $\endgroup$
    – paul garrett
    Commented Nov 22, 2011 at 0:08

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