The following seems to be true: if $|W\_q| := \sum {q^{l(w)}}$, where the sum is taken over the elements $w$, then $|W\_q| = \prod {(1 + q +...+ q^{e\_i})}$, where the product is taken over the exponents $e\_i$.
In other words, if $V$ is the root space for $W$ and the polynomials $f\_k$ are the basis of $S[V]$ sur $S^W[V]$, then the set of $deg(f\_k)$ is the same with $l(w)$.
I couldn't find any proof for this. Yes, I tried Bourbaki.







For a Weyl group, what is the connection between its exponents and lengths of its elements?
I would leave this as a comment but I don't appear to have enough reputation points for that. Just to add to Philippe's answer that you will also find this as Theorem 10.2.3 in Carter's "Simple Groups of Lie Type", (it appears even earlier than this in Steinberg's Lecture Notes on Chevalley Groups, see Theorem 26 - pg. 131). Indeed this formula is key to giving factorisations of the orders of finite reductive groups as polynomials in $q$, (see Carter's "Finite Groups of Lie Type: Conjugacy Classes and Complex Characters" - Section 2.9).


