TITLE: How do I use the fact that $\langle \gamma, \lambda \rangle > 0$?
QUESTION [9 upvotes]: If I post all the details of my question up to the point where I am stuck, no one is going to want to read all that.  So I hope there is someone reading this question who is familiar with Weyl chambers and systems of positive roots will immediately know the answer.  The book I'm using is Humphreys, Linear Algebraic Groups.
$G$ is a connected linear algebraic group with maximal torus $T$.  A regular cocharacter of $T$ is a cocharacter whose image is contained in exactly the same Borel subgroups as those which contain $T$.  If $I(T)$ is the identity component of the intersection of all Borel subgroups containing $T$, then $\mathscr L(I(T))$ is $\textrm{Ad } T$ stable, so there is an $\textrm{Ad } T$-stable complement $$\bigoplus\limits_{\alpha} \mathfrak g_{\alpha}'$$ in $\mathfrak g$ for various nontrivial characters $\alpha$.  The set of such characters is denoted $\Psi$.

This is page 157 of Humphreys' book.  Why is it the case that $\langle \alpha, \lambda \rangle > 0$ if and only if $B(\lambda) \cap Z_{\alpha} = B_{\alpha}$?  This is true by definition when $\lambda = \lambda_0$, but even with the Weyl group equivariance ($w.B(\lambda) = B(w.\lambda)$) and whatnot I don't understand why his claim suddenly follows.  I understand everything in the argument up to that point.

REPLY [4 votes]: Here is the answer to my question in my own words.
Lemma 1: Let $G \times X \rightarrow X$ be a group action which can be realized as the action of a closed subgroup of $\textrm{GL}(V)$ on a closed orbit in projective space $\mathbb{P}(V)$, $\lambda$ a cocharacter of $T$, and $P \in X$.  The morphism $k^{\ast} \rightarrow X$, $x \mapsto \lambda(x)P$ extends (uniquely) to a morphism $\mathbb{P}^1 \rightarrow X$.  The points $\lambda(0)P$ and $\lambda(\infty)P$ are fixed points of $\textrm{Image } \lambda$.
Theorem: Let $\lambda$ be a regular cocharacter of $T$.  Let $\mathcal B$ be the projective variety of Borel subgroups of $G$.

There exists a Borel subgroup $B(\lambda) \in \mathcal B$, together with an open neighborhood $U$ of $B(\lambda)$, such that $\lambda(\infty)B' = B(\lambda)$ for all $B' \in U$.
The group $B(\lambda)$ is unique, and it contains $T$.
For a given regular cocharacter $\lambda$, there exists a finite collection of nontrivial characters $\beta_i$, which are trivial on $R(G) \cap T$, and have the property that if $\lambda'$ is a regular cocharacter, then $B(\lambda') = B(\lambda)$ if and only if $\langle \beta_i, \lambda' \rangle > 0$ for all $i$.
If $w \in W(G,T)$, then $w.B(\lambda) = B(w.\lambda)$.

If the Weyl group of $G$ has order two, then the theorem immediately tells us that the "finite collection" of nontrivial characters $\beta_i$ can be replaced by a single character $\beta$.
The Weyl chamber of a point $B \in \mathcal B^T$ is $$\textrm{wc}(B) = \{ \lambda \in Y(T)_{\textrm{reg}} : B(\lambda) =  B\}$$
The last point of the theorem, combined with the fact that $W(G,T)$ acts on $\mathcal B^T$ simply transitively, tells us that the function $\lambda \mapsto B(\lambda)$ is a $W(G,T)$-equivariant map of $Y(T)_{\textrm{reg}}$ onto the finite set $\mathcal B^T$, and that the Weyl chambers are just the fibers of this function.
If $\beta_i$ are the characters of $T$ assigned to a regular cocharacter $\lambda$ as in the statement of the theorem, then the condition that $\lambda'$ be in $\textrm{wc}(B(\lambda))$ can be rephrased as the condition that $\langle \beta_i, \lambda' \rangle$ be $ > 0$ for all $i$.
Now fix an $\alpha \in \Psi$, and let $T_{\alpha} = (\textrm{Ker } \alpha)^0$.  Then $T_{\alpha}$ is a singular subtorus of $T$ of codimension one.  We can talk about Weyl chambers and whatnot in the group $Z_G(T_{\alpha})$, whose Weyl group has order two.  Let $B_0(\lambda)$ be the Borel subgroup assigned to $\lambda$ in the group $Z_G(T_{\alpha})$.
Lemma 2: If $B$ is a Borel subgroup of $G$ containing $T$, then $B \cap Z_G(T_{\alpha})$ is a Borel subgroup of $Z_G(T_{\alpha})$ which also contains $T$, and $\textrm{wc}(B) \subseteq \textrm{wc}(B \cap Z_G(T_{\alpha}))$.
To prove this, we identify $\mathcal B_G$ with $G/B$, and $\mathcal B_{Z_G(T_{\alpha})}$ with $Z_G(T_{\alpha})/Z_G(T_{\alpha}) \cap B$, and use the uniqueness of $B_0(\lambda)$ given in the theorem.
Now to put all this together, choose a regular cocharacter $\lambda_0$ with $\langle \alpha, \lambda_0 \rangle > 0$, and let $B_{\alpha} = B(\lambda_0) \cap Z_G(T_{\alpha})$.  By the second lemma, $B_{\alpha} = B_0(\lambda_0)$, and since the Weyl group has order two, there exists a single nontrivial character $\beta$, trivial on $T \cap R(Z_G(T_{\alpha}))$ (hence trivial on $T_{\alpha}$), with the property that $\langle \beta, \lambda \rangle > 0$ if and only if $B_0(\lambda) = B_{\alpha}$.
Now, the fact that $\beta$ is trivial on $T_{\alpha}$ tells us that $T_{\alpha} \subseteq \textrm{Ker } \beta$, whence $T_{\alpha} = T_{\beta}$.  Some commutative algebra argument gives us that $a \alpha = b \beta$ for some integers $a, b$, with $b$ positive.  I claim that $a$ is also positive: since $B_0(\lambda_0) = B_{\alpha}$, we must have $\langle \beta, \lambda_0 \rangle > 0$.  Multiplying both sides by $b$, the fact that $\langle \alpha, \lambda_0 \rangle > 0$ shows that $a$ must also be positive.
So we can just replace $\beta$ by $\alpha$ here; they serve the same function.  Thus $B_0(\lambda) = B_{\alpha}$ if and only if $\langle \alpha, \lambda \rangle > 0$.
Main result: $B(\lambda) \cap Z_G(T_{\alpha}) = B_{\alpha}$ if and only if $\langle \alpha, \lambda \rangle > 0$.
Now this follows immediately from the second lemma and the statement right above, because $B(\lambda) \cap Z_G(T_{\alpha}) = B_0(\lambda)$.