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116 P.E.T. JØRGENSEN, D.P. PROSKURIN, AND YU.S. SAMO ĬLENKO
where σ (n+1)
0 is the unique element of Sn+1 with maximal possible length of
the reduced decomposition.
Remark 3. 1. The element σ (n+1)
0
σ (n+1)
0
Set Un = φ(σ (n+1)
0 ): Then
of the group Sn+1 has the form
= (σ1 · · · σn)(σ1 · · · σn−1) · · · (σ1σ2)σ1.
Un = (T1T2 · · · Tn)(T1T2 · · · Tn−1) · · · (T1T2)T1.
2. It is easy to see that the operator Un is selfadjoint, and, taking adjoints,
we can rewrite (6) in the following form:
(−1) |J| P (DJ) ∗ = (−1) n+1 (7)
1 + Un − Pn+1.
J⊂S, J=∅
3. Note also that, for all J ⊂ S, the group WJ is isomorphic to Sk for
some k < n, or to the direct product of some such groups.
2. In what follows we shall use the following properties of the operator Un.
Proposition 3. ker Pn+1 is invariant with respect to the action of Un.
Proof. First we show that for all J ⊂ S,
P (D ∗ J): ker Pn+1 ↦→ ker Pn+1.
It can be easily obtained from the equality
Then by (7), we have
Pn+1P (DJ) ∗ = P (DJ)P (WJ)P (DJ) ∗ = P (DJ)Pn+1.
Un − (−1) n 1: ker Pn+1 ↦→ ker Pn+1.
Proposition 4. Let operators {Ti, i = 1, . . . , n} satisfy the braid condition
TiTi+1Ti = Ti+1TiTi+1, i = 1, . . . , n − 1, and TiTj = TjTi, |i − j| ≥ 2. Then
(8)
TkUn = UnTn+1−k, ∀ k = 1, . . . n.
Proof. 1. For n = 1 the equality is evident.
2. Suppose that (8) holds for any n ≤ m. Note that
Um+1 = T1T2 · · · Tm+1Um.