# Is the following construction of the 0-Hecke monoid (well) known?

Let W be a Coxeter group with Coxeter generators S. The corresponding 0-Hecke monoid H(W) has generating set S, the braid relations of W and the relations that each element of S is an idempotent. If one specializes the Hecke algebra associated to W to q=0 one gets the monoid algebra of H(W) (replace generators by their negatives to see this). It is also the monoid generated by foldings across the walls of the fundamental chamber of the Coxeter complex of W. It has been studied by a number of people and is sometimes called the Springer-Richardson monoid.

Margolis and I came across the following construction of it and I wanted to know if it is known. Let P(W) be the power set of W. It is a monoid with usual set product: $AB=\lbrace ab\mid a\in A, b\in B\rbrace$. Let $I(w)$ be the principal Bruhat ideal generated by $w\in W$, e.g., $I(s)=\lbrace 1,s\rbrace$ for $s\in S$. Then the principal Bruhat ideals form a submonoid of P(W) isomorphic to H(W).

Question: Does this construction appear explicitly in the literature and what is a reference?

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Since there hasn't been an answer yet - I'd just like to comment that I think it is likely that this is known, since I know of two different contexts where this might be a useful statement. Unfortunately, it may be buried as a lemma. I suggest trying Deodhar's papers on combinatorial aspects of Kazhdan--Lusztig elements/polynomials and the Kostant--Kumar papers on cohomology and K-theory of flag varieties. –  Alexander Woo Nov 24 '11 at 0:22
@Alexander, thanks I will take a look. –  Benjamin Steinberg Nov 24 '11 at 0:50
Following up on Alexander's comment, you might also find it interesting to take a look at work of A. Knutson and E. Miller on subword complexes where they study properties of the `Demazure product', which is exactly the 0-Hecke product. It also appears in work of Drew Armstrong on sorting orders and in work of mine on total positivity. I had given an answer along these lines, but decided it wasn't really an answer to your question to your question. –  Patricia Hersh Nov 10 '12 at 22:28
Why the down vote on this old question? –  Benjamin Steinberg Jan 29 '13 at 2:00