Parity of the rank of a Demushkin group #
When every cup square on Hยน(G, ๐ฝ_p) vanishes, the cup form of a Demushkin group
(LinearMap.cupForm) is a nondegenerate alternating form, so the Demushkin rank is even. At an odd
prime this is automatic by graded commutativity (LinearMap.isAlt_cupForm_of_ne_two); in particular
the rank cannot be one there. This is the parity constraint on the odd-prime normal forms. At
p = 2 the vanishing of the cup squares is decided by Labute's invariant q, in
TauCeti.Topology.Algebra.Group.Profinite.Demushkin.CupSquare.
Main results #
TauCeti.IsDemushkin.even_demushkinRank_of_forall_cupFp_self_eq_zero: the rank is even when every cup square vanishes.TauCeti.IsDemushkin.even_demushkinRank_of_ne_two: the rank is even at an odd prime.TauCeti.IsDemushkin.demushkinRank_ne_one_of_ne_two: rank one occurs only atp = 2.
References #
- J. P. Labute, Classification of Demushkin groups, Canad. J. Math. 19 (1967), 106โ132.
A Demushkin group on which every cup square vanishes has even rank: the cup form is then
a nondegenerate alternating form on Hยน(G, ๐ฝ_p), whose dimension is the rank.
At an odd prime, the rank of a Demushkin group is even: every cup square vanishes, because the
cup product is graded-commutative and 2 is invertible.
A Demushkin group at an odd prime cannot have rank one.