Why is the opposite Borel subgroup used? 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?
 A: 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.
