So, suppose given a root system in a euclidean space , which arises from a semisimple Lie algebra and a Cartan subalgebra as before. The first goal of this post is to discuss the “splitting”
(disjoint union) in a particular way, into positive and negative roots, and the basis decomposition into simple roots. Here .
To do this, choose such that for . Then define to be those roots with and those with . 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.
OK. Next, I claim it is possible to choose a linearly independent set such that every root is a combination
with all the or all the .
Then 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 when ; otherwise, if we would have
but this has occurring with coefficient and occuring with negative coefficient. This contradicts the hypothesis on simple roots.
Anyway, to construct , first choose as above and the corresponding splitting . Let 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 spans . Indeed, it is enough to check that each positive root is a sum of indecomposable ones. If is not, we can choose such that is minimal among roots which are not sums of indecomposable ones; then, however, is itself not indecomposable, so we can write for . Then are by the inductive hypothesis (since ) sums of indecomposable ones; thus so is , contradiction.
Next, I claim that as constructed above via indecomposable roots satisfies for distinct , which we showed was necessary for a construction of simple roots. If then is a root (cf. the discussion of maximal strings here). If , then is not indecomposable. Similarly if , then is not indecomposable. This will let us prove linear independence. Indeed, if we had an expression
for distinct sets and , then taking inner products with itself yields
which is a contradiction by unless all the are zero, which is impossible since taking inner products with gives something positive.
Incidentally, this resembles the argument in Sylvester’s theorem on quadratic forms.
Proposition 1 as just defined is a base.
We will denote it by to emphasize the dependence on .
It is in fact the case that any base can be obtained in such a fashion. To see this, we will construct such that for all (and consequently is orthogonal to no member of ). I claim then that . Indeed, leads to a decomposition where consists of roots that are positive linear combinations of , and those that are negative linear combinations of . It is clear that the splitting is also obtained by taking inner products with as at the beginning of this post—i.e., consists of those roots that bracket positively with . Now it is clear that is indecomposable because it is a basis, and the set of indecomposable elements of is itself a basis by the above arguments of Proposition 1. If a basis is contained in another, the two are equal though. So .
It now remains to construct from , and in fact we can do it for any basis of . For each , one defines to be the projection of into the space spanned by . Then is orthogonal to but brackets positively with . So brackets positively with all of .
Proposition 2 Any basis can be written as for some orthogonal to no element of .
Next, there is another result:
Proposition 3 Given , we can write where each partial sum is a positive root.
We can write as a sum where . Induct on the degree ; if it is one, then already. If not, then is at least not orthogonal to all of . Now if for all then it is easily checked that would be linearly independent, so we can choose with
As a result , and it is moreover a positive root. We can apply the inductive hypothesis to and use .
Consider ; this is the complement of the union of several hyperplanes. Each corresponds to a basis , and it is clear that if are sufficiently close to one another. In particular, the connected components of are in one-to-one correspondence with the bases of .
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.