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?



Possible Borel subgroups of GL_n?
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.)


