By Ronald S. Irving

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It remains to consider the case that ys £ 5 W . 2 of [De 1], there is a reflection s' E S with ys = s'y. lilZy lZyws(l). ws(\) = 1, then by induction there is a reflection t with y = tws. Thus tw = (tws)s = (tws)(tws)s = (tws)ys = (tws)sfy. We find that T/ = Hy. The same argument yields the converse. This proves the Proposition for 11' w(l) and completes the proof. 3. A Jantzen sum formula The main result of this subsection is proved using duality polynomials, although the statement makes no mention of them.

Proof of equivalence of (i) and (iii). Under either the assumptions of (i) or (iii), the polynomials Qy,w(

The following well-known and trivial Lemma summarizes the information we need. 1. Let x and w be elements ofW. Let s E B. (i) r preserves length; r permutes the set B of simple reflections and the set R of reflections. (ii) Given x and w in VV, we have jo(wx) = (in) In particulary ifws > w, then J0(W)T(S) jo(w)r(x). < jo(w). 0V) jo ^ an order-reversing involution ofW as poset under the Bruhat order. (v) If sw > w, then sjo(w) < jo(w). 2. Let w E 5 W . Then wsjo(w) E 5VV. Proof. Since w is the shortest element in the coset Wsw, we have sw > w for all s E S.