When reading some papers on flag varieties, I sometimes find some remarks mentioning opposite Borel subgroup. It seems that people do so when they consider algebraic group. To my understanding, it is just a convention but it's very confusing to me because I don't understand the motivation.

For instance, the following is from the page 3 of Line bundles on Bott-Samelson varieties by Lauritzen and Thomsen. Here, $G$ is a connected semisimple, simply connected linear algebraic group over an algebraically closed field, and $B$ is a Borel subgroup.

It is well known, that $\mathcal{L}_{G/B}(\lambda)$ is globally generated exactly when $\lambda$ is dominant with respect to the Borel subgroup opposite to $B$ (or equivalently $\langle \lambda, \alpha^\vee \rangle \geq 0$ for all simple roots $\alpha \in S$).

I thought that $\lambda$ is dominant if and only if $\langle \lambda, \alpha^\vee \rangle \geq 0$. But why respect to the Borel subgroup opposite to $B$? Is this because of the convention regarding the opposite Borel subgroups, or is it just a typo?

I am more familiar with Lie groups and practically know nothing about algebraic groups. Is there a basic reference which explains this kind of convention in detail? Why is such convention necessary or has advantage?

  • 2
    $\begingroup$ Indeed, that is the definition of $\lambda$ being dominant. But choosing a set of simple roots and choosing a Borel subgroup are the same thing. So either determines which weights are dominant. And in general for algebraic groups, it turns out to be convenient to fix the simple roots, and then work with the Borel subgroup corresponding to the negative roots. The general reason for this is essentially that the induction functor one considers for algebraic groups is in some (not very correct) sense "dual" to the one for Lie algebras. $\endgroup$ Apr 24, 2018 at 9:14

1 Answer 1


The best reason I have seen for needing both Borels is that it makes induction work properly. Suppose that $B^+$ is the Borel we measure weights against, where all of the $1 + e_i$ elements live, and $B^{-}$ is the opposite Borel, with all the $1 + f_i$'s. Here is a pretty sketchy idea of why induction through the opposite Borel "does the right thing":

Let $\lambda$ be any weight, form the one-dimensional representation $k_\lambda$ of $B^-$, and induce this up to $G$. Recall that in the algebraic category, induction through $B^-$ is something about taking sections of the associated bundle to the principal $B^-$-bundle $G \to G/B^-$, but after unwinding this definition, we find that the induced representation is a subspace of the space of regular maps $G \to k_\lambda$ satisfying some conditions:

$$ \mathrm{Ind}_{B^-}^G k_\lambda = \{f: G \to k_\lambda \mid f(bg) = bf(g) \text{ for all } b \in B^{-} \}$$

Any highest-weight vector $f$ (by definition) satisfies $f(u) = (u \cdot f)(1) = f(1)$ for all $u \in U^+$, the unipotent radical of $B^+$. Because $B^- U^+$ is dense in $G$ (this is the crucial point, and is certainly not true for $B^+ U^+$), all values of a highest-weight vector $f$ are determined by $f(1)$, since $f(bu) = \lambda(b) f(u) = \lambda(b) f(1)$ for all $b \in B^-$, $u \in U^+$.

And so we see that the space of highest-weight vectors is at most one-dimensional, generated by any $f \in \mathrm{Ind}_{B^-}^G k_\lambda$ with $f(1) \neq 0$. Furthermore, the action of $B^-$ fully determines the values of $f$ everywhere, and gives us a method to construct $f$, if only we knew lots of stuff about the geometry of $G/B^-$. It turns out this can be done precisely when $\lambda$ is dominant. (for a somewhat high-brow reference, see http://www.math.harvard.edu/~lurie/papers/bwb.pdf).

How I think of this is that induction through $B^-$ "pulls down" the highest weight vectors into a representation, while induction through $B^+$ "pulls up" the lowest weight vectors, so reverses the dominance condition.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .