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 $\Psi$.

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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.


2 Answers 2


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 $B(\lambda) \cap Z_G(T_{\alpha}) = B_0(\lambda)$.

  • $\begingroup$ Nice. I need over 10 characters here. +1 $\endgroup$
    – Brian Ng
    Jul 31, 2016 at 16:09

The argument in 25.4 is typically somewhat dense, and I think your problem is partly due to the quantifiers involved at each step. Once you work out an assignment of regular $1$-psg's to the finitely many Borel subgroups containing the fixed maximal torus $T$, the rest of the argument involves a fixed $\alpha \in \Psi$. Here, there are in fact pairs $\pm\alpha$ (using additive notation), related by a sort of "reflection" called here $\sigma_\alpha$. Now, the $W$-equivariance is applied to this element of the Weyl group (always relative to the fixed $T$), so "half" of the $\lambda$ get assigned to one Borel subgroup of $Z_\alpha$ and their images under the reflection get assigned to the second.

My problem might be with quantifiers. It seems like we first pick a particular regular cocharacter $\lambda_0$ such that $\langle\alpha, \lambda_0\rangle > 0$ for all $\alpha \in \Psi$, and then we set $B_\alpha$ to be the Borel subgroup $B(\lambda_0) \cap Z_g((\text{Ker}\,\alpha)^0)$.

No, the first step is just to assign each $\lambda$ one of the finitely many Borel subgroups containing $T$. It does not make sense to say as you do "for all $\alpha \in \Psi$", since such roots and their negatives give opposite values here. Keep in mind that the special set $\Psi$ of roots has no particular structure at this point, but does consist of pairs $\alpha$, $-\alpha$. Such a pair determines a rank $1$ centralizer group with $2$ Borel subgroups, but it is not possible to assign those Borel subgroups of $Z$ at first to one or the other root. (Later on, the set $\Psi$ gets more precise structure as a root system.)

Then we fix an $\alpha \in \Psi$, and the claim is that for every regular cocharacter $\lambda$, that $B(\lambda) \cap Z_G((\text{Ker}\,\alpha)^0) = B_\alpha$ if and only if $\langle\alpha, \lambda\rangle > 0$. That is what I don't understand. Like how is $\langle \alpha, \lambda\rangle > 0$ related to $\langle \alpha, \lambda_0\rangle > 0$?

When you fix $\alpha$, you need to choose names for the $2$ Borel subgroups containing $T$ here. For this, take any regular cocharacter $\lambda_0$ with value at $\alpha$ positive. This just allows you to label the $2$ Borels. (Heuristically, one should visualize a sort of Euclidean space with a reflecting hyperplane for each pair of roots, so the regular points lie in various Weyl chambers determined by the complement of the hyperplanes. Here, you have assigned to each given pair of roots a positive and negative side of its hyperplane.) In turn, you have now assigned half of the regular cocharacters to the "positive" Borel you chose, the other half to the "negative" one. (Heuristically, half of the regular $\lambda$ lie on the plus side of the hyperplane and half on the minus.)

Thanks for answering. I still don't get it. You're saying the following.

  • Step 1: Fix $\alpha \in \Psi$.
  • Step 2: Choose $\lambda_0 \in Y(T)_{\text{reg}}$ with $\langle \alpha, \lambda_0\rangle > 0$, and set $B_\alpha := B(\lambda_0) \cap Z_G((\text{Ker}\,\alpha)^0)$.
  • Step 3: If $\lambda$ is any other regular cocharacter of $T$ with $\langle\alpha, \lambda\rangle > 0$, then also $B(\lambda) \cap Z_G((\text{Ker}\,\alpha)^0) = B_\alpha$.

Why is Step 3 true?

There is a crucial Step 0 which you need before you can carry out Steps 1-3. Section 25.4 (Weyl chambers) of Humphreys is mostly devoted to this: Chevalley's assignment of each regular $\lambda$ to one of the $|W|$ Borel subgroups containing $T$. Humphreys's early paragraph there is just motivational, while the end of the secion summarizes it but gives you trouble.

Chevalley's main idea here is to rely on the way tori act on a flag variety, which substitutes often for traditional use of Lie algebras in Lie group theory. After you assign the regular $\lambda$ to Weyl chambers, permuted simply transitively by $W$, it is possible to carry out your Steps 1-3. In Step 3, the geometric intuition is that the "hyperplane" made up of singular $\lambda$ for which the pairing with your fixed $\alpha$ is $0$ divides the regular $\lambda$ or Borel subgroups as indicated. (Maybe the discussion in Borel's original notes, reproduced in his Springer text, would be more useful. The ideas are the same, but the organization of arguments is different in each source.) Keep in mind that the reflection for $\alpha$ switches the two half-spaces across that hyperplane, at the same time it interchanges the two Borel subgroups of the smaller centralizer $Z_\alpha$. (Here, half of the Weyl chambers give one of these groups, the other half give the other. But the labels for these two subgroups depend on the arbitrary convention you made by choosing $\lambda_0$ on the positive side of the hyperplane.)

Update. Maybe I should add a more algebraic formulation, implicit in the development, since my previous comments have emphasized more the geometric intuition and motivation (needed for this subtle argument due to Chevalley). After the study of rank $1$ centralizers of singular tori in 25.3, one takes a fixed root $\alpha \in \Psi$ and the corresponding group $Z = Z_\alpha$, which has two Borel subgroups $B_\alpha$ and $B_\alpha'$ interchanged by a reflection $s_\alpha$ generating the Weyl group of $Z$. This is a subgroup of the big Weyl group $W$ and has a representative $n$ in $Z$ for $s_\alpha$. Thus, conjugation by $n$ takes $B_\alpha$ to $B_\alpha'$, and vice versa since $s_\alpha$ has order $2$.

Having proved the main assertion of 25.4 about assignment of regular $\lambda$ to the $|W|$ Borel subgroups containing $T$, one returns to this rank $1$ setup. Using an arbitrary $\lambda_0$ for which the pairing with $\alpha$ is positive, you designate (say) $B_\alpha$ as the Borel subgroup of $Z$ gotten by intersecting $Z$ with $B(\lambda_0)$. In turn, $W$-equivariance means that conjugating by $n$ a typical $B(\lambda)$ whose pairing with $\alpha$ is positive yields $B(s_\alpha \lambda)$. In turn, this intersects $Z = nZn^{-1}$ in $nB_\alpha n^{-1} = B_\alpha'$.

This yields what you need in the statement of Humphreys's summary in Proposition 25.4. It is all somewhat complicated to write down, but it is near the core of Chevalley's program. So far of course you only look at one root at a time, but for his full classification theorem you need to deal with independent pairs of roots. (Humphreys followed his treatment, but Springer and others later used a more elegant argument based on Serre relations, due to Takeuchi. In any case, the key idea here is that everything follows from rank $1$ and $2$ relations, just as in the finite reflection group $W$.)

  • $\begingroup$ My problem might be with quantifiers. It seems like we first pick a particular regular cocharacter $\lambda_0$ such that $\langle \alpha, \lambda_0 \rangle > 0$ for all $\alpha \in \Psi$, and then we set $B_{\alpha}$ to be the Borel subgroup $B(\lambda_0) \cap Z_G((\textrm{Ker } \alpha)^0)$. $\endgroup$
    – D_S
    Jul 28, 2016 at 18:39
  • $\begingroup$ Then we fix an $\alpha \in \Psi$, and the claim is that for every regular cocharacter $\lambda$, that $B(\lambda) \cap Z_G((\textrm{Ker } \alpha)^0) = B_{\alpha}$ if and only if $\langle \alpha, \lambda \rangle > 0$. That is what I don't understand. Like how is $\langle \alpha, \lambda \rangle > 0$ related to $\langle \alpha, \lambda_0 \rangle > 0$? $\endgroup$
    – D_S
    Jul 28, 2016 at 18:40
  • $\begingroup$ Thanks for answering. I still don't get it. You're saying Step 1: Fix $\alpha \in \Psi$. Step 2: Choose $\lambda_0 \in Y(T)_{\textrm{reg}}$ with $\langle \alpha, \lambda_0 \rangle > 0$, and set $B_{\alpha} := B(\lambda_0) \cap Z_G((\textrm{Ker } \alpha)^0)$. Step 3: If $\lambda$ is any other regular cocharacter of $T$ with $\langle \alpha, \lambda \rangle > 0$, then also $B(\lambda) \cap Z_G((\textrm{Ker } \alpha)^0) = B_{\alpha}$. Why is Step 3 true? $\endgroup$
    – D_S
    Jul 29, 2016 at 15:05
  • $\begingroup$ Thanks very much for your extended explanation, I am still trying to understand it. How in the world did you have time to learn all this stuff as an undergraduate? I've been trying to learn the basic theory of linear algebraic groups for almost two years. $\endgroup$
    – D_S
    Jul 30, 2016 at 1:53
  • $\begingroup$ @D_S I read a lot. Alternatively, you can use void salts and imp stool to enchant your weapons for like $+10$ intelligence, I highly suggest it. $\endgroup$
    – Brian Ng
    Jul 30, 2016 at 3:36

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