So, suppose given a root system {\Phi} in a euclidean space {E}, which arises from a semisimple Lie algebra and a Cartan subalgebra as before. The first goal of this post is to discuss the “splitting”

\displaystyle \Phi = \Phi^+ \cup \Phi^-

(disjoint union) in a particular way, into positive and negative roots, and the basis decomposition into simple roots. Here {\Phi^- = - \Phi^+}.

To do this, choose {v \in E} such that {(v, \alpha) \neq 0} for {\alpha \in \Phi}. Then define {\Phi^+} to be those roots {\alpha} with {(v,\alpha)>0} and {\Phi^-} those with {(v,\alpha) < 0}. This was easy. We talked about positive and negative roots before using a real-valued linear functional, which here is given by an inner product anyway.

Bases

OK. Next, I claim it is possible to choose a linearly independent set {\Delta \subset \Phi^+} such that every root is a combination

\displaystyle \alpha = \sum k_i \delta_i, \quad \delta_i \in \Delta, \ k_i \in \mathbb{Z}

with all the {k_i \geq 0} or all the {k_i \leq 0}.

Then {\Delta} will be called a base. It is not unique, but I will show how to construct this below.

An amusing application of the condition on the simple roots is that {(\alpha, \beta) \leq 0} when {\alpha, \beta \in \Delta, \alpha \neq \beta}; otherwise, if {(\alpha,\beta)>0} we would have

\displaystyle s_{\alpha}(\beta) = \beta - 2\frac{ (\beta, \alpha)}{(\alpha, \alpha)} \alpha \in \Phi

but this has {\beta \in \Delta} occurring with coefficient {1} and {\alpha} occuring with negative coefficient. This contradicts the hypothesis on simple roots.

Anyway, to construct {\Delta}, first choose {v} as above and the corresponding splitting {\Phi = \Phi^+ \cup \Phi^-}. Let {\Delta} denote the set of positive roots that cannot be written as the sum of two positive roots—in other words, the indecomposable ones. I claim {\Delta} spans {E}. Indeed, it is enough to check that each positive root is a sum of indecomposable ones. If {\alpha \in \Phi^+} is not, we can choose {\alpha} such that {(\alpha, v)} is minimal among roots which are not sums of indecomposable ones; then, however, {\alpha} is itself not indecomposable, so we can write {\alpha = \beta + \beta'} for {\beta, \beta' \in \Phi^+}. Then {\beta, \beta'} are by the inductive hypothesis (since {(\beta, v), (\beta', v) < (\alpha, v)}) sums of indecomposable ones; thus so is {\alpha}, contradiction.

Next, I claim that {\Delta} as constructed above via indecomposable roots satisfies {(\alpha, \beta) \leq 0} for distinct {\alpha, \beta \in \Delta}, which we showed was necessary for a construction of simple roots. If {(\beta, \alpha)>0} then {\beta - \alpha} is a root (cf. the discussion of maximal strings here). If {\beta - \alpha \in \Phi^+}, then {\beta = (\beta - \alpha) + \alpha} is not indecomposable. Similarly if {\beta - \alpha \in \Phi^-}, then {\alpha = (\alpha -\beta) + \beta} is not indecomposable. This will let us prove linear independence. Indeed, if we had an expression

\displaystyle \sum a_i \delta_i = \sum b_j \delta'_j

for {\delta_i, \delta'_j \in \Delta} distinct sets and {a_i, b_j \geq 0}, then taking inner products with itself yields

\displaystyle \sum_{i,j} a_i b_j (\delta_i, \delta'_j) \geq 0

which is a contradiction by {(\delta_i, \delta'_j) \leq 0} unless all the {a_i, b_j} are zero, which is impossible since taking inner products with {v} gives something positive.

Incidentally, this resembles the argument in Sylvester’s theorem on quadratic forms.

Proposition 1 {\Delta} as just defined is a base.

 

We will denote it by {\Delta(v)} to emphasize the dependence on {v \in E}.

It is in fact the case that any base {\Delta} can be obtained in such a fashion. To see this, we will construct {v} such that {(\alpha, v) > 0} for all {\alpha \in \Delta} (and consequently {v} is orthogonal to no member of {\Phi}). I claim then that {\Delta = \Delta(v)}. Indeed, {\Delta} leads to a decomposition {\Phi^+ \cup \Phi^-} where {\Phi^+} consists of roots that are positive linear combinations of {\Delta}, and {\Phi^-} those that are negative linear combinations of {\Delta}. It is clear that the splitting {\Phi = \Phi^+ \cup \Phi^-} is also obtained by taking inner products with {v} as at the beginning of this post—i.e., {\Phi^+} consists of those roots that bracket positively with {v}. Now it is clear that {\Delta \subset \Phi^+} is indecomposable because it is a basis, and the set of indecomposable elements of {\Phi^+} is itself a basis by the above arguments of Proposition 1. If a basis is contained in another, the two are equal though. So {\Delta = \Delta(v)}.

It now remains to construct {v} from {\Delta}, and in fact we can do it for any basis of {E}. For each {\alpha \in \Delta}, one defines {v_{\alpha}} to be the projection of {\alpha} into the space spanned by {\Delta - \{ \alpha \}}. Then {v_{\alpha}} is orthogonal to {\Delta - \{ \alpha \}} but brackets positively with {\alpha}. So {\sum v_{\alpha}} brackets positively with all of {\Delta}.

Proposition 2 Any basis can be written as {\Delta(v)} for some {v \in E} orthogonal to no element of {\Phi}.

 

Next, there is another result:

Proposition 3 Given {\alpha \in \Phi^+}, we can write {\alpha = \delta_1 + \dots + \delta_r} where each partial sum is a positive root.

 

We can write {\alpha} as a sum {\sum_{\delta \in \Delta} k_{\delta} \delta} where {k_{\delta} \in \mathbb{Z}_{\geq 0}}. Induct on the degree {\sum k_{\delta}}; if it is one, then {\alpha \in \Delta} already. If not, then {\alpha} is at least not orthogonal to all of {\Delta}. Now if {(\alpha, \delta) \leq 0} for all {\delta \in \Delta} then it is easily checked that {\{ \alpha \} \cup \Delta} would be linearly independent, so we can choose {\delta \in \Delta } with

\displaystyle (\alpha, \delta) > 0.

As a result {\alpha - \delta \in \Phi}, and it is moreover a positive root. We can apply the inductive hypothesis to {\alpha - \delta} and use {\alpha = (\alpha - \delta) + \delta}.

Weyl chambers

Consider {Y = \{ v\in E: (v,\alpha) \neq 0 \ \forall \alpha \in \Phi \}}; this is the complement of the union of several hyperplanes. Each {v \in Y} corresponds to a basis {\Delta}, and it is clear that {\Delta(v) = \Delta(v')} if {v,v'} are sufficiently close to one another. In particular, the connected components of {Y} are in one-to-one correspondence with the bases of {Y}.

These connected components are called Weyl chambers. Next, we’ll define a group that permutes the Weyl chambers.

I learned this from Shlomo Sternberg’s book.

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