Nonabelian Cohen-Lenstra Moments for the Quaternion Grou (University of Wisconsin-Madison) Featuring Joint Work with Jack Klys (University of Toronto) October 14, 2017
Cohen-Lenstra heuristics For a field k, let Cl(k) denote the class grou of the field. Write Cl (k) for the Sylow -subgrou of Cl(k), called the -art of the class grou. Conjecture (Cohen-Lenstra [4]) Fix an odd rime and a finite abelian -grou A. The robability that Cl(Q( ±d)) is isomorhic to A as ±d varies over negative (resectively ositive) quadratic discriminants is Prob(Cl (Q( d)) = A) = c 1 #Aut(A) Prob(Cl (Q( d)) = A) = c + 1 #A#Aut(A)
Nonabelian Cohen-Lenstra heuristics For a air of grous G and G G S 2 Melanie Wood gave a generalization of Cohen-Lenstra heuristics by defining the Cohen-Lenstra Moment E ± (G, G ) to be #{L/Q( d) unram. : Gal(L/Q( d)) = G, Gal( L/Q) = G } lim X 0<±d<X if the limit exists. Conjecture (Wood [10]) For an admissible, good air (G, G ) 0<±d<X 1 E (G, G ) = #H 2(G, c)[2] #Aut G (G) E + (G, G ) = #H 2(G, c)[2] #c#aut G (G) and for an admissible, not good air (G, G ), E ± (G, G ) =.
Quaternion Grou H 8 = a, b, z : a 2 = b 2 = [a, b] = z, z 2 = [a, z] = [b, z] = 1 For any ositive integer k, there exists a unique grou G k Hk 8 S 2 u to isomorhism with [G k, Hk 8 ] = 2 making (Hk 8, G k ) an admissible air G k = H k 8 σ where σ acts comonentwise by σ(a) = az, σ(b) = bz, and σ(z) = z. This air is not good. Theorem (A. (k=1), A.-Klys) E ± (H k 8, H k 8 σ ) =
Quaternion Grou Theorem (Lemmermeyer [8]) There exists an unramified extension M/Q( d) with Gal(M/Q( d)) = H 8 and Gal(M/Q) = H 8 σ if and only if there is a factorization d = d 1 d 2 d 3 satisfying d 1, d 2, d 3 are corime quadratic discriminants, at most one of which is negative. ( ) ( ) ( ) d 1d 2 d 3 = 1d 3 d 2 = 2d 3 1 = 1 for rimes i d i Moreover, for each factorization there are exactly 2 ω(d) 3 such extensions. The number of unramified H 8 extensions M/Q( d) Galois over Q can be exressed by 1 ( ( )) ( ( )) ( ( )) d1 d 2 d1 d 3 d2 d 3 8 d=d 1d 2d 3 d
k=1 Fix d 1 and d 2 and vary d 3 = m to find the exected value of a d1,d 2,m = 1 ( ( )) ( ( )) ( ( )) d1 d 2 d1 m d2 m 8 d=d 1d 2m d Lemma m sqf a d1,d 2,mm s has a meromorhic continuation to C with no oles or zeroes in an oen neighborhood of Re(s) 1 excet for a simle ole at s = 1 with an exlicit residue. d 1d 2m<X a d1,d 2,m c d 1,d 2 d 1 d 2 X
k=1 E ± (H 8, H 8 σ ) = lim N d 1,d 2<N lim N d 1,d 2<N = c d1,d 2 d 1 d 2 ( ( L 1, )) d 1d 2 d 1 d 2 by a result due to Goldfeld-Hoffstein [6]. When we consider H8 k for k > 1, this method gives more comlicated roducts of secial values of L-functions. Checking that the corresonding sum diverges becomes increasingly difficult for large k, and Goldfeld-Hoffstein s result was already nontrivial.
Counting Function For an admissible air (G, G ) and a quadratic discriminant define f G,G (d) = #{L/Q( d) unram. : Gal(L/Q( d)) = G, Gal( L/Q) = G } Define the numerator of the fraction in E ± (G, G ) to be N ± (G, G ; X ) = f G,G (d) 0<±d<X and denote the exected value in general to be g(d) E ± (g(d)) = lim X This way E ± (G, G ) = E ± (f G,G (d)). 0<±d<X 0<±d<X 1
Arbitrary k Theorem (A.-Klys [3]) ( ) k ) (fh8,h E 8 σ (d) = 1 3 ω(d) 32 k E + ( (fh8,h 8 σ (d) 3 ω(d) ) k ) = 1 192 k Theorem (A.-Klys [3]) N (H k 8, H k 8 σ ; X ) N + (H k 8, H k 8 σ ; X ) 1 4 k #Aut σ (H8 k) 0< d<x 1 24 k #Aut σ (H8 k) 0<d<X 3 kω(d) 3 kω(d) This imlies E ± (H k 8, Hk 8 σ ) = as a corollary
Sketch of Proof Use the same sieve as Fouvry-Klüners [5] did when finding the average value of the 4-rank of the class grou of quadratic fields. f (d) = 1 8 d=d 1d 2d 3 d ( ( )) ( d1 d 2 ( )) ( d1 d 3 ( )) d2 d 3 Take the k th ower and exand the sum so that, for u {0, 1,..., 5} k, f (d) k = d= D u u,v ( ) Φk (u,v) Du BIG IDEA: the main term only comes from those terms which cancel out comletely by quadratic recirocity, i.e. if D u, D v 1 then Φ k (u, v) = Φ k (v, u) so that the main term is d= D u D v ( 1) Φ k (u,v) Du 1 Dv 1 2 2
What else has been done? By methods related to those resented in this talk: (D 4, D 4 C 2 ) for D 4 the dihedral grou of order 8. (A. [1]) (G, G ) for [G : G] = 2 and G a central C 2 extension of C2 n not containing a C 2 D 4 with G (C 2 D 4 ) = C 2 C 4. (Klys [7]) (A, A C 2 ) for A a finite abelian 2-grou of exonent 8, and in a more recent rerint any finite abelian 2-grou (Smith [9]) In a recent rerint [2], I generalize Lemmermeyer s key theorem classifying unramified H 8 -extensions of Q( d) by certain factorizations d = d 1 d 2 d 3 to classify unramified G-extensions M/K of an abelian extension K/Q with Gal(M/K) = G for airs (G, G ) such that [G, G ] Z(G ). This result can be used to do the case: (Ucoming) (G, G ) for [G : G] = 2 and G any central C m 2 extension of C n 2
Possible Future Directions Exected numbers of G-extensions M/K as K varies over abelian number fields with Galois grou A with Gal(M/K) = G. My generalization of Lemmermeyer s conditions aly in this case for admissible airs (G, G ) with [G, G ] Z(G ), suggesting that a similar sieve will work here too. Investigating the ossibility that Smith s methods for (A, A C 2 ) for A an abelian 2-grou can be alied to (G, G ) for G any finite 2-grou. What should we exect the main term of N ± (G, G ; X ) to look like for not good airs? This would be an exansion of Wood s conjecture.
Acknowledgements I would like to thank my PhD advisor Nigel Boston, my coauthor Jack Klys, and Melanie Matchett Wood for many helful conversations and contributions. I would also like to thank the organizers of the Maine-Québec Number Theory Conference, esecially Andrew Knightly, for having me here to give this resentation.
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