Extending Zagier s Theorem on Continued Fractions and Class Numbers

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1 Extending Zagier s Theorem on Continued Fractions and Class Numbers Colin Weir University of Calgary Joint work with R. K. Guy, M. Bauer, M. Wanless West Coast Number Theory December 2012

2 The Story of How 3 People Spent a Summer Failing to Answer Questions from Richard Guy Colin Weir University of Calgary Joint work with R. K. Guy, M. Bauer, M. Wanless West Coast Number Theory December 2012

3 In the beginning... Zagier s Theorem Let p be a prime such that p 3 (mod 4). Let [b 0 ; b 1, b 2,..., b r ] be the negative continued fraction expansion of p. If h(p) = 1 then ( r ) h( p) = 1 3 i=1 b i 3r

4 In the beginning... Zagier s Theorem Let p be a prime such that p 3 (mod 4). Let [b 0 ; b 1, b 2,..., b r ] be the negative continued fraction expansion of p. If h(p) = 1 then ( r ) h( p) = 1 b i 3r 3 i=1 }{{} Magic Number

5 In the beginning... Zagier s Theorem Let p be a prime such that p 3 (mod 4). Let [b 0 ; b 1, b 2,..., b r ] be the negative continued fraction expansion of p. If h(p) = 1 then ( r ) h( p) = 1 b i 3r 3 i=1 }{{} Magic Number Richard Guy s Questions: 1. Why is the magic number divisible by 3? Is this always true? 2. What happens when p = 1 (mod 4)? 3. What happens when p is not prime? 4. What if h(p) 1?

6 In the beginning... Zagier s Theorem Let p be a prime such that p 3 (mod 4). Let [b 0 ; b 1, b 2,..., b r ] be the negative continued fraction expansion of p. If h(p) = 1 then ( r ) h( p) = 1 b i 3r 3 i=1 }{{} Magic Number Richard Guy s Questions: 1. Why is the magic number divisible by 3? Is this always true? 2. What happens when p = 1 (mod 4)? 3. What happens when p is not prime? 4. What if h(p) 1?... See Richard Guy s talk at the Joint Meetings for a more complete list of questions we failed to answer.

7 In the beginning... Zagier s Corollary Let p be a prime such that p 3 (mod 4). Let [b 0 ; b 1, b 2,..., b r ] be the negative continued fraction expansion of p. If h(p) = 1 then ( r ) h( p) = 1 b i 3r 3 i=1 }{{} Magic Number Richard Guy s Questions: 1. Why is the magic number divisible by 3? Is this always true? 2. What happens when p = 1 (mod 4)? 3. What happens when p is not prime? 4. What if h(p) 1?... See Richard Guy s talk at the Joint Meetings for a more complete list of questions we failed to answer.

8 Zagier s Theorem Soit K un corps quardratique réel avec discriminant D dont l unité fondamentale est totalement positive. Soit χ un caractère sur les idéaux de K tel que l on a χ((α)) = sign(n(α)) pour un idéal principal (α), c est-à-dire, χ est un caractère sure le group des classes d idéaux au sens restreint qui n est pas induit per un caractère sure les groupe des classes d idéaux au sens large. Supposons maintenant que χ est réel. Il est connu que ces caractères correspondent à toutes les décompositions possibles (à l ordre près) D = D 1 D 2 du discriminant de K en produit de deux discriminants de corps. On a alors: h(d 1 )h(d 2 ) = 1 6 C Cl(K ) χ(c)m(c) (avec les conventions h( 3) = 1/3, h( 4) = 1/2)

9 Zagier s Theorem en anglais Let K be a real quadratic field with discriminant D and fundamental unit of positive norm. Let χ be a character (to be defined in English later) and suppose that χ is real. If D = D 1 D 2 for two fundamental discriminants D 1,D 2 of imaginary quadratic fields then h(d 1 )h(d 2 ) = 1 6 C Cl(K ) χ(c)m(c) (with the conventions h( 3) = 1/3, h( 4) = 1/2)

10 Zagier s Theorem en anglais Let K be a real quadratic field with discriminant D and fundamental unit of positive norm. Let χ be a character (to be defined in English later) and suppose that χ is real. If D = D 1 D 2 for two fundamental discriminants D 1,D 2 of imaginary quadratic fields then h(d 1 )h(d 2 ) = 1 6 C Cl(K ) χ(c)m(c) (with the conventions h( 3) = 1/3, h( 4) = 1/2) Zagier s Corollary Let p be a prime such that p 3 (mod 4). Let [b 0 ; b 1, b 2,..., b r ] be the negative continued fraction expansion of p. If h(4p) = 1 then ( r ) h( p) = 1 b i 3r 3 i=1 }{{} Magic Number

11 Extending Zagier s Theorem on Continued Fractions and Class Numbers Colin Weir University of Calgary Joint work with R. K. Guy, M. Bauer, M. Wanless West Coast Number Theory December 2012

12 Specializing Zagier s Theorem on Continued Fractions and Class Numbers Colin Weir University of Calgary Joint work with R. K. Guy, M. Bauer, M. Wanless West Coast Number Theory December 2012

13 The Remainder of the Talk Vague outline of Zagier s Proof What is χ? What is the Magic Number? The cases we could work out explicitly

14 What is χ? Definition Let χ be the character of the narrow class group (not induced from the wide class group) extending χ((α)) = sign(n(α)) on principal ideals (α), restricted to the wide class group. Lemma If χ is real and D = D 1 D 2 then L(s, χ) = L D1 (s)l D1 (s).

15 What is χ? Definition Let χ be the character of the narrow class group (not induced from the wide class group) extending χ((α)) = sign(n(α)) on principal ideals (α), restricted to the wide class group. Lemma If χ is real and D = D 1 D 2 then L(s, χ) = L D1 (s)l D1 (s). To apply the theorem we want: 1. Set D 1 (mod 4) so the f.u. of K has positive norm. 2. Want Cl(K ) no have no odd part. 3. We want Cl nar (K ) = Z/2Z Cl wide (K ) so that χ is real.

16 The Magic... Revealed! What is the magic number: Let U be any representative of the ideal class of C Cl(K ). Write a Z basis for U Then f.u. acts on the basis as an element A of SL(2, Z). SL(2, Z) acts on the upper half plane by F.L.T. Take the neg. continued fraction of the fixed point of A. Call it [b 0 ;b 1, b 2,..., b r ]. Then ( ) ( b1 1 b2 1 A = and the Dedekind sum m(c) = r i=1 (b i 3). ) ( br )

17 Our Zagier s Theorem Theorem Let K be the real quadratic field of discriminant D = 8p n i=1 q i such that p = 3 (mod 8) and q i = 1 (mod 8). Suppose that Cl(K ) = (Z/2Z) n. Then Cl(K ) = q i and h( p q i ) = 1 k j 2p q k 6 qj C= q j m

18 Our Zagier s Theorem Theorem Let K be the real quadratic field of discriminant D = 8p n i=1 q i such that p = 3 (mod 8) and q i = 1 (mod 8). Suppose that Cl(K ) = (Z/2Z) n. Then Cl(K ) = q i and h( p q i ) = 1 k j 2p q k 6 qj C= q j m Proof: 1. Show a 2 b 2 2p q i = q j has no solutions for a, b Z. 2. Use L(s, χ) = L D1 (s)l D1 (s) to show ( ) 2 χ(c) = = 1 qj 3. Use Zagier s Theorem.

19 A Question From Richard Guy Is it true/can anyone prove the following? Conjecture h( p)h( p) m( p) = { 0 (mod 16) h( p) 1, 7 (mod 8) 8 (mod 16) h( p) 3, 5 (mod 8)

20 A Question From Richard Guy Is it true/can anyone prove the following? Conjecture h( p)h( p) m( p) = { 0 (mod 16) h( p) 1, 7 (mod 8) 8 (mod 16) h( p) 3, 5 (mod 8) Please and Thank You

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