Abstract
The Ish arrangement was introduced by Armstrong to give a new interpretation of the $q; t$-Catalan numbers of Garsia and Haiman. Armstrong and Rhoades showed that there are some striking similarities between the Shi arrangement and the Ish arrangement and posed some problems. One of them is whether the Ish arrangement is a free arrangement or not. In this paper, we verify that the Ish arrangement is supersolvable and hence free. Moreover, we give a necessary and sufficient condition for the deleted Ish arrangement to be free L’arrangement Ish a été introduit par Armstrong pour donner une nouvelle interprétation des nombres $q; t$-Catalan de Garsia et Haiman. Armstrong et Rhoades ont montré qu’il y avait des ressemblances frappantes entre l’arrangement Shi et l’arrangement Ish et ont posé des conjectures. L’une d’elles est de savoir si l’arrangement Ish est un arrangement libre ou pas. Dans cet article, nous vérifions que l’arrangement Ish est supersoluble et donc libre. De plus, on donne une condition nécessaire et suffisante pour que l’arrangement Ish réduit soit libre.
Highlights
Let K be a field of characteristic 0 and {x1, . . . , x } a basis for the dual space (K )∗ of the -dimensional vector space K
The Shi arrangement originally defined over R was introduced by J.Y
Theorem 1.1 ([Armstrong (2013); Headley (1997)]) The characteristic polynomial of the Shi arrangement and the Ish arrangement are given by χ(Shi( ), t) = χ(Ish( ), t) = t(t − ) −1
Summary
Let K be a field of characteristic 0 and {x1, . . . , x } a basis for the dual space (K )∗ of the -dimensional vector space K. Theorem 1.1 ([Armstrong (2013); Headley (1997)]) The characteristic polynomial of the Shi arrangement and the Ish arrangement are given by χ(Shi( ), t) = χ(Ish( ), t) = t(t − ) −1. Note that the implications (2) ⇒ (3) ⇒ (4) are general properties for arrangements [Orlik and Terao (1992)] This theorem asserts that there are no differences among these properties for N -Ish arrangements. |N |), where |Nj| denotes the cardinality of Nj. Corollary 1.5 The cone over the Ish arrangement c(Ish( )) is free with exponents exp(c(Ish( ))) =
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