The q-invariant of the Demushkin normal forms #
Labute's classification puts a Demushkin group of rank n with q-invariant q in one of three
normal forms, presented on x₁, …, xₙ by the relator words
x₁^q (x₁, x₂)(x₃, x₄) ⋯ (x_{n-1}, x_n), forq ≠ 2;x₁² x₂^{2^f} (x₂, x₃)(x₄, x₅) ⋯ (x_{n-1}, x_n), forq = 2andnodd, together with its levelf = ∞, the wordx₁² (x₂, x₃)(x₄, x₅) ⋯ (x_{n-1}, x_n);x₁^{2+a} (x₁, x₂) x₃^{2^f} (x₃, x₄) ⋯ (x_{n-1}, x_n), forq = 2andneven,4 ∣ a.
This file checks that the parameter q of a normal form is the q-invariant of the group it
presents: whenever such a presented group is a Demushkin group, TauCeti.demushkinQ of it is q
for the first word (p^{v_p(q)} in general, so q itself for q = 0 or q a positive power of
p), and 2 for the dyadic words. The computation goes through the one-relator
abelianization structure theorem TauCeti.presentedProP.oneRelatorAbelianizationEquiv, in the form
TauCeti.demushkinQ_presentedProP_eq_pow_valuation: the exponent vector of the first word is
q e₁, that of the second is 2 e₁ + 2^f e₂ = 2 (e₁ + 2^{f-1} e₂), or 2 e₁ at level f = ∞,
and that of the third is
(2 + a) e₁ + 2^f e₃ = (2 + a)(e₁ + c e₃) with 2 + a = 2u for the unit u = 1 + a/2 of ℤ₂,
so the abelianizations are ℤ_p^{n-1} × ℤ_p ⧸ (q), ℤ₂^{n-1} × ℤ/2 and ℤ₂^{n-1} × ℤ/2. The
hypothesis 4 ∣ a in the third word is Labute's normalisation α ∈ 4ℤ₂, and it is what makes
1 + a/2 a unit: for a ≡ 2 mod 4 and f ≥ 2 the whole exponent vector is divisible by 4, so
the q-invariant would be at least 4. (For f = 1 the coordinate 2 at x₃ alone gives
q-invariant 2, whatever a is.)
Main results #
TauCeti.freeProP.toAdd_exponentSum_demushkinWordNeTwo,TauCeti.freeProP.toAdd_exponentSum_demushkinWordTwoOdd,TauCeti.freeProP.toAdd_exponentSum_demushkinWordTwoOddTop,TauCeti.freeProP.toAdd_exponentSum_demushkinWordTwoEven: the exponent vectors of the words on the free generators.TauCeti.demushkinQ_presentedProP_demushkinWordNeTwo_eq_zero_iff,TauCeti.demushkinQ_presentedProP_demushkinWordNeTwo,TauCeti.demushkinQ_presentedProP_demushkinWordNeTwo_of_eq_pow: theq-invariant of theq ≠ 2normal form vanishes exactly whenq = 0, isp^{v_p(q)}forq ≠ 0, and isqwhenqis a positive power ofp.TauCeti.demushkinQ_presentedProP_demushkinWordTwoOdd,TauCeti.demushkinQ_presentedProP_demushkinWordTwoOddTop,TauCeti.demushkinQ_presentedProP_demushkinWordTwoEven: the dyadic normal forms haveq-invariant2.
References #
- J. P. Labute, Classification of Demushkin groups, Canad. J. Math. 19 (1967), 106–132, p. 106 and Theorems 1–3.
- J. Neukirch, A. Schmidt, K. Wingberg, Cohomology of Number Fields, 2nd ed., Theorem 3.9.19.
Exponent vectors of the normal-form words #
The exponent vector of the q ≠ 2 word x₁^q (x₁, x₂) ⋯ (x_{n-1}, x_n) on the free generators
is q e₁.
The exponent vector of the q = 2, n odd word x₁² x₂^{2^f} (x₂, x₃) ⋯ (x_{n-1}, x_n) on
the free generators is 2 e₁ + 2^f e₂.
The exponent vector of the q = 2, n odd word at level f = ∞,
x₁² (x₂, x₃) ⋯ (x_{n-1}, x_n), on the free generators is 2 e₁.
The exponent vector of the q = 2, n even word
x₁^{2+a} (x₁, x₂) x₃^{2^f} (x₃, x₄) ⋯ (x_{n-1}, x_n) on the free generators is
(2 + a) e₁ + 2^f e₃.
The q-invariant of the normal forms #
The q-invariant of the q ≠ 2 normal form vanishes exactly when q = 0: for p ∣ q,
if ⟨x₁, …, xₙ ∣ x₁^q (x₁, x₂) ⋯ (x_{n-1}, x_n)⟩ is a Demushkin group, its q-invariant is 0
if and only if q = 0.
The q-invariant of the q ≠ 2 normal form is p^{v_p(q)}: for p ∣ q and q ≠ 0, if
⟨x₁, …, xₙ ∣ x₁^q (x₁, x₂) ⋯ (x_{n-1}, x_n)⟩ is a Demushkin group, its q-invariant is
p ^ padicValNat p q.
The q ≠ 2 normal form with parameter q = p^f, f ≥ 1, has q-invariant q: if
⟨x₁, …, xₙ ∣ x₁^q (x₁, x₂) ⋯ (x_{n-1}, x_n)⟩ is a Demushkin group, its q-invariant is q.
The q = 2, n odd normal form has q-invariant 2: for f ≥ 1, if
⟨x₁, …, xₙ ∣ x₁² x₂^{2^f} (x₂, x₃) ⋯ (x_{n-1}, x_n)⟩ is a Demushkin group, its q-invariant is
2.
The q = 2, n odd normal form at level f = ∞ has q-invariant 2: if
⟨x₁, …, xₙ ∣ x₁² (x₂, x₃) ⋯ (x_{n-1}, x_n)⟩ is a Demushkin group, its q-invariant is 2.
The q = 2, n even normal form has q-invariant 2: for 4 ∣ a and f ≥ 1, if
⟨x₁, …, xₙ ∣ x₁^{2+a} (x₁, x₂) x₃^{2^f} (x₃, x₄) ⋯ (x_{n-1}, x_n)⟩ is a Demushkin group, its
q-invariant is 2. The hypothesis 4 ∣ a makes 2 + a exactly divisible by 2.