# Root in positive Weyl chamber

Let $\mathfrak{g}$ be a complex simple Lie algebra. We fix a Cartan subalgebra $\mathfrak{t}\subset \mathfrak{g}$. Let $R\subset \mathfrak{t}^*$ the set of roots. We fix $\Pi\subset R$ the set of simple roots, which induces the set of positive roots $R^+$. Let $W$ be the Weyl group which generator by the reflection $s_\alpha$ of simple roots.

Let $K\subset \mathfrak{t}^*$ be the Weyl chamber, that is (for some scalar product on $\mathfrak{t}^*$) $$K=\{X\in \mathfrak{t}^*: (X,\alpha)>0, \hbox{ for all } \alpha\in \Pi\}.$$

Question : can we show that, except the case $A_1,A_2$, $R\cap K$ is empty?

I'm almost sure about this result. For $A_n,B_n,C_n,D_n, G_2$, the explicit calculation gives us the result. I have not done the calculation for the type $E$ and $F$. I think a direct proof (without much calculations, for example the Dykin diagram contains enough information to conclude) should exist. Thank you!

Add: As pointed by Sasha, where are 2 possible length of root.

Add 2 and answer: Thank to Sasha, I can complete the proof as short as possible.

We know that $W$ acts on $R$. Fact: every two elements $\alpha,\beta$ connected by a single line in Dykin diagram is in the same $W$-orbit, since $s_\alpha s_\beta(\alpha)=\beta$.

From this, we can get that: in the (resp. non-)simple-lanced case, we have only one (resp. two) $W$-orbit. This means $R\cap \overline{K}$ has one (resp. two) element. To show the element of $\alpha \in R\cap \overline{K}$ on the Wall, by Chevalley's lemma, it is equivalent to the stabilizer $W_\alpha$ is non trivial. This is consequence of $|W|>|R|$ (resp. $|W|>|R|-1$).

• It is easy to classify $R \cap \overline{K}$. In the simply-laced case this set consists just of one root (the highest root). In the non-simply-laced case it consists of two roots (the highest long root and the highest short root). However, typically both lie on a wall of $K$. – Sasha Mar 27 '15 at 19:35
• Dear @Sasha, thank you for the comment. Is there some reference? What means simple-lanced? I am sorry, Lie algebra is far away my area... – shu Mar 27 '15 at 19:40
• If Lie algebras are far away your area, why do you need it ? – Dietrich Burde Mar 27 '15 at 19:49
• @DietrichBurde, to learn and try to use it for calculating some explicit example or to construct some contraexample (like symmetric space.) This question comes from the Rigidity of Witten genus for homogeneous space. I do not know why you ask this question? – shu Mar 27 '15 at 20:07
• @shu: Your question isn't clear to me. The set $R \cap K$ is empty to begin with (as Sasha comments, only one or two roots can be dominant weights, and they don't lie in the interior of $K$). On the other hand, $\rho$ isn't a root but does lie in the (dominant) open Weyl chamber $K$. What are you trying to find a proof for without using classification? – Jim Humphreys Mar 27 '15 at 20:52

## 3 Answers

Rather than prolong the tangled comments, I'll try to provide a straightforward answer to the current formulation of the question. As noted already, there are two small cases of irreducible root systems (belonging to simple Lie algebras) in which the dominant (open) Weyl chamber denoted $K$ here contains a root: type $A_1$ and $2\rho = \alpha$ (the sole positive root), type $A_2$ and $\rho = \alpha_1 + \alpha_2$.

Beyond these cases, there are no others. Possibly there is a somewhat more conceptual proof of this, but at the moment I can only verify case-by-case that the highest root $\tilde{\alpha}$ and (if there are two root lengths) the highest short root $\alpha_0$ lie in the closure of $K$ but not in $K$ itself. In the background is the fact that $\tilde{\alpha}$ and (if it exists) $\alpha_0$ are the only roots which are dominant weights: highest weights of irreducible finite dimensional representations. (This fact does have conceptual origins, which I won't go into here.)

In order to study the root systems individually, it's best to rely on the tables at the end of Bourbaki's Chapters 4-6 in Lie Groups and Lie Algebras. For the highest short roots, when they exist, it may be easiest to look at the table in my textbook on page 66. For the exceptional types, it's useful to refer to the explicit tables of positive roots arranged by height appended to Springer's IHES paper here. In all cases, what has to be checked is that some simple root is orthogonal to the dominant root in question, i.e., adding or subtracting it doesn't produce another root. (The dominance means that the addition part is automatic.)

ADDED: After all the edits and comments, maybe it's useful to outline a more detailed proof (based on the Bourbaki/Serre treatment of root systems). Note that only the axiomatic theory of root systems is actually involved here, leaving aside the origins in Lie algebras.

While developing basic facts about a (reduced) root system in a real euclidean space $E$, with Weyl group $W$, one needs to show that there exist simple systems of roots $\alpha_1, \dots, \alpha_n$ and resulting positive systems (all conjugate under $W$). In particular, $W$ is generated by the reflections $s_i$ corresponding to the $\alpha_i$, and the positive roots have a natural partial ordering $\beta \leq \alpha$.

The proofs show that in fact $W$ acts simply transitively on the systems of simple roots. When all hyperplanes orthogonal to roots are removed from $E$, the remaining open sets (Weyl chambers) are then permuted simply transitively by $W$. One of these is the positive (or dominant) Weyl chamber denoted here by $K$, defined by the condition that $(\lambda, \alpha_i) > 0$ for all $\lambda \in K$ and all $\alpha_i$. Then one shows that the closure $\overline{K}$ is a fundamental domain for $W$: every element of $E$ is $W$-conjugate to a unique element of $\overline{K}$.

With this information in place, one can say more special things about an irreducible root system (one coming from a simple Lie algebra):

1) Under the partial ordering, there is a unique maximal root $\tilde{\alpha}$.

2) $W$ acts irreducibly on $E$.

3) There are at most two root lengths (long, short), those of the same length being $W$-conjugate (so only one of a given length can lie in $\overline{K}$). (If all roots are of equal length, the system is usually called "simply-laced".)

4) If there are two root lengths, $\tilde{\alpha}$ is long. Comparing with the dual root system (where long and short are interchanged), one sees that in this case there is a unique maximal short root $\alpha_0$ in the partial ordering of the original system. Moreover, $\alpha_0$ lies in $\overline{K}$.

Only at this point does the classification of irreducible root systems by Dynkin diagrams come into play. As I've indicated above, one can identify case-by-case which highest long/short roots are orthogonal to particular simple roots; as noted, the only cases where $\tilde{\alpha} \in K$ are types $A_1, A_2$.

• @shu: Having gone back to the foundational material on root systems, I'm motivated to outline the full story more carefully. This is all fairly elementary, starting with the axioms, but it does need to be done systematically. Though some shortcuts are possible for the narrow question you've raised, the details in the outline are useful to know about. – Jim Humphreys Mar 31 '15 at 0:52

This is only a slight modification of the argument already given, but I liked it enough to type it in.

Since $W$ acts with no stabilizer on the open Weyl chamber $K$, for $K$ to contain a root $\beta$, this $\beta$ cannot be perpendicular to any other root. In particular the Dynkin diagram $D$ has to be a complete graph (but the triangle $K_3 = \widehat A_2$ is already infinite type, so as a graph $D$ has to be $K_1$ or $K_2$), and by inspection of $B_2,G_2$ we see that every root is perpendicular to some other root. So the only possibilities are $A_1,A_2$.

Instead of asking when $\tilde \alpha$ is regular, you can ask: what is the stabilizer of $\tilde \alpha$? Because $\tilde\alpha$ is dominant, this is the standard parabolic subgroup generated by the simple reflections fixing $\tilde \alpha$; those are associated to the simple roots which are orthogonal to $\tilde \alpha$, which you can read off from the affine Dynkin digram: they are the nodes that are not linked to the additional node (in the affine Dynkin diagram, one can interpret the additional node as $-\tilde\alpha$). Of course it is the same as knowing on which walls $\tilde \alpha$ lies.

So in type $A_n$ the stabilizer is trivial for $n \leq 2$ and of type $A_{n-2}$ for $n > 2$.

For the dominant short root, you can use the affinization of the dual Dynkin diagram.