(Workshop on Harmonic Analysis on symmetric spaces I.S.I. Bangalore : 9th July 2004) B.Sury

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1 Is e π 163 odd or even? (Worksho on Harmonic Analysis on symmetric saces I.S.I. Bangalore : 9th July 004) B.Sury e π 163 = The object of this talk is to exlain this amazing fact. The exlanation involes SL(, ZZ), ellitic curves, modular forms, class field theory and Artin s recirocity, among other things. 1 Quadratic forms We shall consider only ositive definite, binary quadratic forms over ZZ. Any such form looks like f(x, y) = ax + bxy + cy with a, b, c ZZ; it takes only values > 0 excet when x = y = 0. Two forms f and g are said to be equivalent (according to Gauss) if A = ( ) q SL(, (ZZ) such that f(x, y) = g(x + qy, rx + sy). Obviosly, r s equivalent forms reresent the same values. Indeed, this is the reason for the definition of equivalence. One defines the discriminant of f to be disc(f) = b 4ac. Further, f is said to be rimitive if (a, b, c) = 1. Note that if f is +ve-definite, the discriminant D must be < 0 (because 4a(ax + bxy + cy ) = (ax + by) Dy reresents +ve as well as -ve numbers if D > 0.) One has: Theorem 1.1 For any D < 0, there are only finitely many classes of rimi- 1

2 tive, +ve definite forms of discriminant D. [This is the class number h(d) of the field Q( D); an isomorhism is obtained by sending f(x, y) to the ideal azz + b+ D ZZ]. This is roved by means of reduction theory. The idea is to show that each form is equivalent to a unique reduced form. Reduced forms can be comuted - there are even algorithms which can be imlemented in a comuter which can determine h(d) and even the h(d) reduced forms of discriminant D. A rimitive, +ve definite, binary quadratic form f(x, y) = ax + bxy + cy is said to be reduced if b a c and b 0 if either a = c or b = a. These clearly imly D 0 < a 3. For examle, the only reduced form of discriminant D = 4 is x + y. The only two reduced forms of discriminant D = 0 are x + 5y x + xy + 3y. and The grou SL(, ZZ) is a discrete subgrou of SL(, IR) such that the quotient sace SL(, ZZ)\SL(, IR) is non-comact, but has a finite SL(, IR)-invariant measure. Reduction theory for SL(, ZZ) is (roughly) to find a comlement to SL(, ZZ) in SL(, IR); a nice comlement is called a fundamental domain. Viewing the uer half-lane h as the quotient sace SL(, IR)/SO(), {z h : Im(z) 3/, Re(z) 1/} is (the image in h) of a fundamental domain (see the accomanying figure) :

3 Fundamental domains can be very useful in many ways; for examle, they give even a resentation for SL(, ZZ). In this case, such a domain is written in terms of the Iwasawa decomosition of SL(, IR). One has SL(, IR) = KAN in the usual way. The, reduction theory for SL(, ZZ) says SL(, IR) = KA 3 N 1 SL(, ZZ). Here A t = {diag(a 1, a ) SL(, IR) : a i > 0 and a 1 a t} ( ) 1 x and N u = { N : x u}. 0 1 What does this have to with quadratic forms? Well, GL(, IR) acts on the sace S of +ve-definite, binary quadratic forms as follows: Each P S can be reresentated by a +ve-definite, symmetric matrix. For g GL(, IR), t gp g S. This action is transitive and the isotroy at I S is O(). In other words, S can be identified with GL(, IR)/O() i.e. S = { t gg : g GL(, IR)}. In general, this works for +ve-definite quadratic forms in n variables. It is easy to use the above identification and the reduction theory statement for SL(, ZZ) to show that each +ve definite, binary quadratic form is equivalent to a unique reduced form. Indeed, writing f = t gg and g = kanγ, t gg = t γ t na nγ with n U 1/ and a A 4/3 ; so t na n is a reduced form equivalent to f. To see how useful this is, let us rove a beautiful discovery of Fermat, viz., that any rime number 1 mod 4 is exressible as a sum of two squares. Since ( 1)! 1 mod and since ( 1)/ is even, it follows that ( 1!) 1 mod i.e., (( 1 )!) + 1 = q for some natural number q. Now the form x +( 1 )!xy+qy is +ve definite and has discriminant 4. Now, the only reduced form of discriminant 4 is x + y as it is trivial to see. Since each form is equivalent to a reduced form (by reduction theory), the forms x + 1!xy + qy and x + y must be equivalent. As the former form has as the value at (1, 0), the latter also takes the value for some integers x, y. 3

4 Class field theory/recirocity One way to motivate recirocity is as follows. A rime is of the form x + y ( 1 ) = 1 (i.e., 1 is a square mod ). A rime is of the form x + 7y is a cube mod and 1 mod 3. A rime is of the form x + 64y is a 4th ower mod and -1 is a square mod. The oint of quadratic recirocity is that one can exress a condition of the form ( a ) = 1 in terms of congruences for. For instance, ( 3 ) = 1 ±1 mod 1. ( 5 ) = 1 ±1, ±11 mod 0. ( 7 ) = 1 ±1, ±3, ±9 mod 8. The quadratic recirocity law (QRL) says: q odd rimes ( q ) = 1 q ±d mod 4 for some odd d. Abelian class field theory and Artin s recirocity law in articular - QRL corresonds to the secial case of quadratic extensions - tells us when a rime slits comletely in a finite abelian extension of Q, in terms of congruences. Here slits comletely in Q(α) if the minimal olynomial of α over Q slits into linear factors when viewed modulo. For e.g. in Q(e πi/n ), a rime slits comletely 1 mod n. In any finite extension field K of Q, one can do algebra as in ZZ and Q, exceting the fact that unique factorisation is absent, in general. Fortunately, a finite 4

5 grou (called the class grou of K) measures the deviation from this roerty holding good. For K = Q( D) with D < 0, the order h(d) of the class grou of K gives the number of +ve-definite, rimitive, reduced, binary, quadratic forms. Class Field Theory has two arts - one consists of the recirocity law and the other is an existence theorem of a certain field called the Hilbert class field corresonding to any field K. The latter is the maximal, unramified, abelian extension of K. For examle, the Hilbert class field of Q( 14) is Q( 14)( 1). One has: Theorem.1 Let n > 0 be square-free and 3 mod 4. Then, an odd rime can be exressed as x +ny if, and only if, slits comletely in the Hilbert class field of Q( n). Remark There is an analogous version when n 3(4). In that case one looks at rimes exressible as x + xy + ( 1+n 4 )y and one considers the so-called ring class field of ZZ[ n]. Of course, ( n) = 1 imlies that divides x + ny for some integers x, y. Unlike the case of n = 1 (and the cases n =, 3, 4, 7), there are many (as many as h( 4n) ) reduced forms (among which is the form x +ny ) and the condition ( n ) = 1 only imlies that is reresented by one of these forms. When do we know that is reresented by x + ny itself? Now, the revious theorem can be used to determine the rimes exressible in the form x + ny rovided one can determine the Hilbert class field of Q( n). Indeed, if L = Q( n)(α) is the Hilbert class field (actually the ring class field of ZZ[ n] and f n (X) is the minimal olynomial of α (where α O L ), then for a rime with n, disc.f n, we have: = x + ny ( n ) = 1 and f n(x) 0 mod for some x ZZ. As before, there is an analogous version for n 3 (mod 4). 5

6 3 The modular function For τ h, the uer half-lane, consider the lattice ZZ+ZZτ and the functions g (τ) = 60 g 3 (τ) = 140 m,n m,n ( ( 1 = (π)4 1 + (m + nτ) 4 1 ( ( 1 = (π)6 1 + (m + nτ) 6 1 )) σ 3 (n)e πinτ n=1 )) σ 5 (n)e πinτ. n=1 [Note that (z) = 4(z) 3 g (τ)(z) g 3 (τ) where the Weierstrass -function on ZZ + ZZτ is given by (z) = ( 1 ).] z w (z w) w It can be shown that (τ) d = g (τ) 3 7g 3 (τ) 0. The ellitic modular function j : h C is defined by j(τ) = 1 3 g(τ) 3 (τ). The adjective modular accomanies the j-function because of the invariance roerty: { ( j(τ) = j(τ ) τ SL(, ZZ)(τ) = d aτ + b a cτ + d : c In fact, we have: ) } b SL(, ZZ). d Theorem 3.1 (i) j is holomorhic on h. (ii) j has the invariance roerty above. (iii) j : h C is onto. The roof of (iii) needs the fundamental domain of SL(, ZZ) we referred to earlier. That fact that satisfies the equation ( ) = 4 3 g g 3 imlies, by the above theorem, that the j-function, gives an isomorhism from the set SL(, ZZ)\h to the set all comlex ellitic curves C/ZZ + ZZτ. 6

7 In fact, one has bijective corresondences between : (i) lattices L = ZZ + ZZτ C uto scalar multilication, (ii) comlex ellitic curves C/L uto isomorhism, (iii) the numbers j(τ), and (iv) Riemann surfaces of genus 1 uto comlex analytic isomorhism. As a matter of fact, SL(, ZZ)\h is the (coarse) moduli sace of ellitic curves over C. In general, various subgrous of SL(, ZZ) describe other moduli roblems for ellitic curves. This descrition has been vastly exloited by Shimura et al. in modern number theory. For instance, comlex saces like Γ 0 (N)\h have algebraic models over Q called Shimura varieties. The Taniyama-Shimura-Weil conjecture (which imlies Fermat s Last Theorem) says that any ellitic curve over Q admits a surjective, algebraic ma defined over Q from a rojectivised model of Γ 0 (N)\h onto it. The oint of this is that functions on Γ 0 (N)\h or even on SL(, ZZ)\h with nice analytic roerties are essentially modular forms and conjectures like Taniyama-Shimura-Weil and, more generally, those which come under the so-called Langlands Program say essentially that geometric objects over Q come from modular forms. As j : h C is SL(, ZZ) - invariant, one has j(τ + 1) = j(τ). So j(τ) is a holomorhic function in the variable q = e πiτ, in the region 0 < q < 1. Thus, j(τ) = c n (n < 0) vanish. n= In fact, j(τ) = 1 q c n q n is a Laurent exansion i.e., all but finitely many n 1 c n q n with c n ZZ n. (c 1 = , c = , c 3 = etc.) We shall kee this q-exansion of j in mind. 7

8 4 Comlex multilication We defined the j-function on h. One can think of j as a function on lattices ZZ + ZZτ. In articular, if O is an order in an imaginary quadratic field Q( n), it can be viewed as a lattice in C. In fact, any roer, fractional O-ideal I can be -generated i.e, is a free ZZ-module of rank i.e., is a lattice in C. Then, it makes sense to talk about j(i). Using basic roerties of ellitic functions, it is quite easy to show: Proosition: j(i) is an algebraic number of degree class number of O. In fact, a much stronger result holds and, it is : The First main theorem of Comlex multilication : Let O be an order in an imaginary quadratic field K. Let I O be a factional O-ideal. Then, j(i) is an algebraic integer and K(j(I)) is the Hilbert (ring) class field of O. In articular, K(j(O K )) is the Hilbert class field of K. We have almost come back where we started from. Indeed, it only remains to exlain the za of things now 1 A Corollary of the above theorem is: Proosition: Let O, K be as above and let I 1,..., I h be the ideal classes of O (i.e., h = [Hilbert class field of O : K] = [K(j(O)) : K]). Then, h (X j(i i )) is the minimal olynomial of any α such that K(α) = Hilbert i=1 class field of O. Note that α can be any j(i i ). Alying the theorem to j(τ) for τ imaginary quadratic, it follows that j(τ) is an algebraic integer of degree = class number of Q(τ) i.e, integers a 0,..., a h 1 such that j(τ) h + a h 1 j(τ) h a 0 = 0. Now, there are only finitely many imaginary quadratic fields Q(τ) = K which have class number 1. The largest D such that Q( D) has class number 1 A friend had confessed long ago that in his rimary school, he understood the tables but it took him a long time to understand the meaning of za in two two za four! 8

9 1 is 163. Since 163 3(4), the ring of integers is ZZ + ZZ( 1+i 163). Thus j( 1+i 163) ZZ. Now j(τ) = c q n q n with c n ZZ and n 1 q = e 1+i πi( 163 ) = e π 163. Thus e π e π e π = j(τ) ZZ. In other words, e π 163 integer = e π e π VOILA!!! 9

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