Basics of binary quadratic forms and Gauss composition

Basics of binary quadratic forms and Gauss composition Andrew Granville Université de Montréal SMS summer school: Counting arithmetic objects Monday June 3rd, 014, 3:30-5:00 pm 0

Sums of two squares 1 Any prime p 1 (mod 4) can be written as the sum of two squares Geometry of numbers type proof Since p 1 (mod 4) = i Z : i 1 (mod p). Idea: Find smallest non-zero integer lattice point (x, y) Z : x iy (mod p)

Sums of two squares Since p 1 (mod 4) = i Z : i 1 (mod p). Consider now the set of integers {m + ni : 0 m, n [ p]} # pairs m, n is ([ p] + 1) > p, so by the pigeonhole principle, two are congruent mod p; say that m + ni M + Ni (mod p) where 0 m, n, M, N [ p] and (m, n) (M, n).

Sums of two squares 3 Since p 1 (mod 4) = i Z : i 1 (mod p). Consider now the set of integers {m + ni : 0 m, n [ p]} # pairs m, n is ([ p] + 1) > p, so by the pigeonhole principle, two are congruent mod p; say that m + ni M + Ni (mod p) where 0 m, n, M, N [ p] and (m, n) (M, n). Let r = m M and s = N n so that r is (mod p) where r, s [ p] < p, and r and s are not both 0. Now r + s (is) + s = s (i + 1) 0 (mod p) and 0 < r + s < p + p = p. The only multiple of p between 0 and p is p, and therefore r + s = p.

Sums of two squares 4 What integers can be written as the sum of two squares? (a + b )(c + e ) = (ac + be) + (ae bc).

Sums of two squares 5 What integers can be written as the sum of two squares? (a + b )(c + e ) = (ac + be) + (ae bc). Generalization: (a + db )(c + de ) = (ac + dbe) + d(ae bc).

Sums of two squares 6 What integers can be written as the sum of two squares? (a + b )(c + e ) = (ac + be) + (ae bc). Generalization: (a + db )(c + de ) = (ac + dbe) + d(ae bc). Gauss s view: A binary quadratic form is of the shape f(x, y) := ax + bxy + cy. Here we take f(x, y) = x + dy and f(a, b)f(c, e) = f(ac + dbe, ae bc) The latter values in f, namely ac + dbe and ae bc, are bilinear forms in a, b, c, e. Does this generalize to other such multiplications?

Pell s equation Are there integer solutions x, y to x dy = 1? Pell s equation 7

Pell s equation 8 Pell s equation Are there integer solutions x, y to x dy = 1? Can always be found using continued fraction for d. (Brahmagupta, 68 A.D.; probably Archimedes, to solve his Cattle Problem one needs to find a solution to u 609 7766v = 1. The smallest solution has about 10 6 digits!)

Pell s equation 9 Pell s equation Are there integer solutions x, y to x dy = 1? Can always be found using continued fraction for d. (Brahmagupta, 68 A.D.; probably Archimedes, to solve his Cattle Problem one needs to find a solution to u 609 7766v = 1. The smallest solution has about 10 6 digits!) Solution to Pell s Equation Let d be a non-square integer. x, y Z for which x dy = 1, with y 0. If x 1, y 1 smallest positive solution, then all others given by x n + dy n = (x + dy) n

Pell s equation 10 x dy = 1? Solution to Pell s Equation Let d be a non-square integer. x, y Z for which x dy = 1, with y 0. If x 1, y 1 smallest positive solution, then all others given by. x n + dy n = (x + dy) n Better to look for solutions to x dy = ±4, Understanding when there is solution with is a difficult question (great recent progress by Fouvry and Kluners).

Pell s equation 11 Theorem Any quadratic irrational real number has a continued fraction that is eventually periodic. Here are some examples of the continued fraction for d: = [1, ], 3 = [1, 1, ], 5 = [, 4], 6 = [,, 4], 7 = [, 1, 1, 1, 4], 8 = [, 1, 4], 10 = [3, 6], 11 = [3, 3, 6], 1 = [3,, 6], 13 = [3, 1, 1, 1, 1, 6],... If p k /q k are the convergents for d then p n 1 dq n 1 = ( 1)n.

Pell s equation 1 Longest continued fractions and the largest fundamental solutions = [1, ], 1 1 = 1 3 = [1, 1, ], 3 1 = 1 6 = [,, 4], 5 6 = 1 7 = [, 1, 1, 1, 4], 8 7 3 = 1 13 = [3, 1, 1, 1, 1, 6], 18 13 5 = 1 19 = [4,, 1, 3, 1,, 8], 170 19 39 = 1 = [4, 1,, 4,, 1, 8], 197 4 = 1 31 = [5, 1, 1, 3, 5, 3, 1, 1, 10], 150 31 73 = 1 43 = [6, 1, 1, 3, 1, 5, 1, 3, 1, 1, 1], 348 43 531 = 1 46 = [6, 1, 3, 1, 1,, 6,, 1, 1, 3, 1, 1], 4335 46 3588 = 1 76 = [8, 1,, 1, 1, 5, 4, 5, 1, 1,, 1, 16], 57799 76 6630 = 1

Pell s equation 13 Length of longest cont fracts and fundl solutions 16 : 14395 94 1064 = 1 16 : 460799 14 414960 = 1 16 : 588599 133 4460 = 1 18 : 7756350 139 657889 = 1 0 : 178148040 151 140634693 = 1 : 170090565 166 1301564 = 1 6 : 78354373650 11 1916705353 = 1 6 : 69535918995 14 47533775646 = 1 6 : 588339537695 301 3391131083 = 1 34 : 785589801443970 331 15310986634573 = 1 37 : 440444569681418 41 146497463530785 = 1 40 : 8405609154695933775 56 3665019757349553 = 1 4 : 18114355061630786130 571 75798183506898587 = 1

Pell s equation 14 Length of fundamental solutions The length of the continued fractions here are around d, and the size of the fundamental solutions 10 d. How big is the smallest solution? We believe that the smallest solution is typically of size C d but not much proved. Understanding the distribution of sizes of the smallest solutions to Pell s equation is an outstanding open question in number theory.

Pell s equation 15 Descent on solutions of x dy = n, d > 0 Let ɛ d = x 1 + y 1 d, the smallest solution x1, y 1 in positive integers to x 1 dy 1 = 1. Given a solution of with x, y 0, let x dy = n α := x + y d > n. If nɛ k d α < nɛ k+1 d let β := αɛ k d = u + dv so that n β < nɛd with u, v 1 and u dv = n.

Binary quadratic forms 16 Representation of integers by binary quadratic forms What integers are represented by binary quadratic form f(x, y) := ax + bxy + cy? That is, for what N are there coprime m, n such that N = am + bmn + cn? WLOG gcd(a, b, c) = 1. Complete the square to obtain 4aN = (am + bn) dn where discriminant d := b 4ac, so d 0 or 1 (mod 4). When d < 0 the right side can only take positive values... easier than when d > 0. If a > 0 then positive definite binary quadratic form.

Binary quadratic forms 17 x + y represents the same integers as X + XY + Y If N = m + n then N = (m n) + (m n)n + n, If N = u + uv + v then N = (u + v) + v.

Binary quadratic forms 18 x + y represents the same integers as X + XY + Y If N = m + n then N = (m n) + (m n)n + n, If N = u + uv + v then N = (u + v) + v. ( ) x y = M ( ) X Y where M = ( ) 1 1 0 1 transforms x +y into X +XY +Y, and the transformation is invertible, since det M = 1.

Binary quadratic forms 19 x + y represents the same integers as X + XY + Y If N = m + n then N = (m n) + (m n)n + n, If N = u + uv + v then N = (u + v) + v. ( ) x y = M ( ) X Y where M = ( ) 1 1 0 1 transforms x +y into X +XY +Y, and the transformation is invertible, since det M = 1. Much more generally define {( ) } α β SL(, Z) = : α, β, γ, δ Z and αδ βγ = 1 γ δ Then ax + bxy + cy represents ( the ) same ( integers ) as AX + BXY + CY x X whenever = M with y Y M SL(, Z). These quadratic forms are equivalent..

Binary quadratic forms 0 Equivalence ax + bxy + cy is( equivalent ) ( to) AX + BXY + CY x X if equal whenever = M with M SL(, Z). y Y This yields an equivalence relation and splits the binary quadratic forms into equivalence classes. Write ax + bxy + cy = ( x y ) ( ) ( ) a b/ x b/ c y ( ) a b/ Discriminant(f) = det. We deduce that b/ c AX +BXY +CY = ( X Y ) M T ( a b/ b/ c ) M ( ) X, Y so A = aα + bαγ + cγ and C = aβ + bβδ + cδ as ( ) ( ) A B/ = M T a b/ M. B/ C b/ c Hence two equivalent bqfs have same discriminant.

Binary quadratic forms 1 Equivalence classes of binary quadratic forms 9X + 8XY + 58Y is equivalent to x + y

Binary quadratic forms Equivalence classes of binary quadratic forms 9X + 8XY + 58Y is equivalent to x + y Gauss: Every equivalence class of bqfs (with d < 0) contains a unique reduced representative, defined as a < b a c, and b 0 whenever a = c.

Binary quadratic forms 3 Equivalence classes of binary quadratic forms 9X + 8XY + 58Y is equivalent to x + y Gauss: Every equivalence class of bqfs (with d < 0) contains a unique reduced representative, defined as a < b a c, and b 0 whenever a = c. If so, d = 4ac ( b ) 4a a a = 3a and hence a d /3. Therefore, for given d < 0, finitely many a, and so b (as b a), and then c = (b d)/4a is determined; so only finitely many (h(d), the class number, the number of equivalence classes) reduced bqfs of discrim d.

Binary quadratic forms 4 Equivalence classes of binary quadratic forms 9X + 8XY + 58Y is equivalent to x + y Gauss: Every equivalence class of bqfs (with d < 0) contains a unique reduced representative, defined as a < b a c, and b 0 whenever a = c. If so, d = 4ac ( b ) 4a a a = 3a and hence a d /3. Therefore, for given d < 0, finitely many a, and so b (as b a), and then c = (b d)/4a is determined; so only finitely many (h(d), the class number, the number of equivalence classes) reduced bqfs of discrim d. In fact h(d) 1 since we always have the principal form: { x (d/4)y when d 0 (mod 4), x + xy + (1 d) 4 y when d 1 (mod 4).

Binary quadratic forms 5 Gauss s reduction Theorem Every positive definite binary quadratic form is properly equivalent to a reduced form. Proof. A sequence of equivalent forms; algorithm terminates when we reach one that is reduced. Given (a, b, c): ( ) ( ) ( ) x 0 1 x i) If c < a the transformation = y 1 0 y, yields (c, b, a) which is properly equivalent to (a, b, c).

Binary quadratic forms 6 Gauss s reduction Theorem Every positive definite binary quadratic form is properly equivalent to a reduced form. ( ) ( ) ( ) x 0 1 x i) If c < a the transformation = y 1 0 y, yields (c, b, a) which is properly equivalent to (a, b, c). ii) If b > a or b a let b be the least residue, in absolute value, of b (mod( a), ) so( a < ) b ( ) a, say b x 1 k x = b ka. Then let = y 0 1 y. The resulting form (a, b, c ) is properly equivalent to (a, b, c). iii) If( c = ) a and ( a < ) ( b < ) 0 then we use the transformation = x 0 1 x y 1 0 y yielding the form (a, b, a). If resulting form not reduced, repeat

Binary quadratic forms 7 Gauss s reduction Theorem Every positive definite binary quadratic form is properly equivalent ( to ) a reduced ( form. ) ( ) x 0 1 x i) If c < a then = y 1 0 y. ( ) ( ) ( ) x 1 k x ii) If b > a or b a then = y 0 1 y. ( ) ( ) ( ) x 0 1 x iii) If c = a and a < b < 0 then = y 1 0 y If resulting form not reduced, repeat The algorithm terminates after (iii), and since (ii) is followed by (i) or (iii), and since (i) reduces the size of a.

Binary quadratic forms 8 Gauss s reduction Theorem; examples (76, 17, 155) of discriminant 31, The sequence of forms is (76, 65, 14), (14, 65, 76), (14, 9, ), (, 9, 14), (, 1, 4), the sought after reduced form.

Binary quadratic forms 9 Gauss s reduction Theorem; examples (76, 17, 155) of discriminant 31, The sequence of forms is (76, 65, 14), (14, 65, 76), (14, 9, ), (, 9, 14), (, 1, 4), the sought after reduced form. (11, 49, 55) of discriminant 19, gives the sequence of forms (11, 5, 1), (1, 5, 11), (1, 1, 5).

Binary quadratic forms 30 Restriction on values taken by a bqf Suppose d = b 4ac with (a, b, c) = 1, and p is a prime. (i) If p = am + bmn + cn for some integers m, n then d is a square mod 4p. (ii) If d is a square mod 4p then there exists a binary quadratic form of discriminant d that represents p.

Binary quadratic forms 31 Restriction on values taken by a bqf Suppose d = b 4ac with (a, b, c) = 1, and p is a prime. (i) If p = am + bmn + cn for some integers m, n then d is a square mod 4p. (ii) If d is a square mod 4p then there exists a binary quadratic form of discriminant d that represents p. Proof. (i) If p ad and p = am +bmn+cn. Therefore 4ap = (am + bn) dn and so dn is a square mod 4p. Now p n else p 4ap + dn = (am + bn) so that p am which is impossible as p a and (m, n) = 1. We deduce that d is a square mod p. (ii) If d b (mod 4p) then d = b 4pc for some integer c, and so px + bxy + cy is a quadratic form of discriminant d which represents p = p 1 + b 1 0 + c 0.

Binary quadratic forms 3 Class number one Theorem Suppose h(d) = 1. Then p is represented by the form of discrim d if and only if d is a square mod 4p.

Binary quadratic forms 33 Class number one Theorem Suppose h(d) = 1. Then p is represented by the form of discrim d if and only if d is a square mod 4p. ( Fundamental discriminants: If q d then q = and d 8 or 1 (mod 16).) The only fundamental d < 0 with h(d) = 1 are d = 3, 4, 7, 8, 11, 19, 43, 67, 163. (Heegner/ Baker/ Stark)

Binary quadratic forms 34 Class number one Theorem Suppose h(d) = 1. Then p is represented by the form of discrim d if and only if d is a square mod 4p. ( Fundamental discriminants: If q d then q = and d 8 or 1 (mod 16).) The only fundamental d < 0 with h(d) = 1 are d = 3, 4, 7, 8, 11, 19, 43, 67, 163. (Heegner/ Baker/ Stark) Euler noticed that the polynomial x + x + 41 is prime for x = 0, 1,,..., 39, and some other polynomials. Rabinowiscz s criterion We have h(1 4A) = 1 for A if and only if x + x + A is prime for x = 0, 1,,..., A.

Binary quadratic forms 35 Class number one Rabinowiscz s criterion We have h(1 4A) = 1 for A if and only if x + x + A is prime for x = 0, 1,,..., A. If p d then p is rep d by x + y if and only if ( 1/p) = 1, p is rep d by x + y if and only if ( /p) = 1, p is rep d by x + xy + y if and only if ( 3/p) = 1, p is rep d by x + xy + y if and only if ( 7/p) = 1, p is rep d by x + xy + 3y if and only if ( 11/p) = 1, p is rep d by x + xy + 5y if and only if ( 19/p) = 1, p is rep d by x + xy + 11y if and only if ( 43/p) = 1, p is rep d by x + xy + 17y if and only if ( 67/p) = 1, p is rep d by x + xy + 41y if and only if ( 163/p) = 1.

Binary quadratic forms 36 Class number not one What about when the class number is not one? First example, h( 0) =, the two reduced forms are x + 5y and x + xy + 3y. p is represented by x +5y if and only if p = 5, or p 1 or 9 (mod 0); p is represented by x + xy + 3y if and only if p =, or p 3 or 7 (mod 0).

Binary quadratic forms 37 Class number not one What about when the class number is not one? First example, h( 0) =, the two reduced forms are x + 5y and x + xy + 3y. p is represented by x +5y if and only if p = 5, or p 1 or 9 (mod 0); p is represented by x + xy + 3y if and only if p =, or p 3 or 7 (mod 0). Cannot always distinguish which primes are represented by which bqf of discriminant d by congruence conditions. Euler found 65 such idoneal numbers. No more are known at most one further idoneal number.

Structure of Ideals 38 Ideals in quadratic fields Any ideal I in a quadratic ring of integers: R := {a + b d : a, b Z} is generated by elements.

Structure of Ideals 39 Ideals in quadratic fields Any ideal I in a quadratic ring of integers: R := {a + b d : a, b Z} is generated by elements. If I Z then principal. Else r + s d I with s 0, wlog s > 0. Select s minimal.

Structure of Ideals 40 Ideals in quadratic fields Any ideal I in a quadratic ring of integers: R := {a + b d : a, b Z} is generated by elements. If I Z then principal. Else r + s d I with s 0, wlog s > 0. Select s minimal. Claim: If u + v d I then s divides v (else if ks + lv = g := gcd(s, v) then (kr + lu) + g d = k(r + s d) + l(u + v d) I #)

Structure of Ideals 41 Ideals in quadratic fields Any ideal I in a quadratic ring of integers: R := {a + b d : a, b Z} is generated by elements. If I Z then principal. Else r + s d I with s 0, wlog s > 0. Select s minimal. Claim: If u + v d I then s divides v (else if ks + lv = g := gcd(s, v) then (kr + lu) + g d = k(r + s d) + l(u + v d) I #) Let v = ms, so that (u+v d) m(r +s d) = u mr. Therefore I = {m(r + s d) + n : m Z, n I Z}. Now I Z is an ideal in Z so principal, = g say hence I = r + s d, g Z.

Structure of Ideals 4 Any ideal I R := {a + b d : a, b Z} has the form I = r + s d, g Z.

Structure of Ideals 43 Any ideal I R := {a + b d : a, b Z} has the form I = r + s d, g Z. More: d R, so g d I and sd + r d I, and so s divides both g and r.

Structure of Ideals 44 Any ideal I R := {a + b d : a, b Z} has the form I = r + s d, g Z. More: d R, so g d I and sd + r d I, and so s divides both g and r. Therefore r = sb and g = sa. Also s(b d) = (r + s d)(b d) I Z and so s(b d) is a multiple of g = sa; hence a divides b d. Therefore I = s b + d, a Z for some integers s, a, b where a divides b d.

Structure of Ideals 45 Binary quadratic forms and Ideals I = s a, b + d Z If f(x, y) = ax + bxy + cy then ( af(x, y) = ax + b + ) ( d y ax + b d so we see that af(x, y) is the Norm of y ) ( ax + b+ d y ).

Structure of Ideals 46 Binary quadratic forms and Ideals I = s a, b + d Z If f(x, y) = ax + bxy + cy then ( af(x, y) = ax + b + ) ( d y ax + b ) d y ( so we see that af(x, y) is the Norm of ax + b+ d y So the set of possible values of f(x, y) with x, y Z is in 1-to-1 correspondence with the elements of a, b+ d Z. ).

Structure of Ideals 47 Equivalence of ideals Any two equivalent bqfs can be obtained from each other by a succession of two basic transformations: x x+y, y y gives a, b + d Now a, b+ d Z = a, a+b+ d Z Z a, a + b + d Z

Structure of Ideals 48 Equivalence of ideals Any two equivalent bqfs can be obtained from each other by a succession of two basic transformations: x x+y, y y gives a, b + d Now a, b+ d Z = a, a+b+ d Z x y, y x gives a, b + d Z a, a + b + d Z Z c, b + d Z. Since b+ d b+ d = d b 4 = ac, and therefore b + d a, b + d Z = a b + d, c Z.

Structure of Ideals 49 Equivalence of ideals Any two equivalent bqfs can be obtained from each other by a succession of two basic transformations: x x+y, y y gives a, b + d Now a, b+ d Z = a, a+b+ d Z x y, y x gives a, b + d Z a, a + b + d Z Z c, b + d Z. Since b+ d b+ d = d b 4 = ac, and therefore b + d a, b + d Z = a b + d, c Z. So, equivalence of forms, in setting of ideals, gives: For ideals I, J of Q( d), we have that I J if and only there exists α Q( d), such that J = αi.

Structure of Ideals 50 For ideals I, J of Q( d), we have that I J if and only there exists α Q( d), such that J = αi.

Structure of Ideals 51 For ideals I, J of Q( d), we have that I J if and only there exists α Q( d), such that J = αi. This works in any number field; moreover then one has finitely many equivalence classes, and i bounds for the smallest element of each class.

Structure of Ideals 5 For ideals I, J of Q( d), we have that I J if and only there exists α Q( d), such that J = αi. This works in any number field; moreover then one has finitely many equivalence classes, and i bounds for the smallest element of each class. Any ideal I = a, b+ d with d < 0 then we plot Z- linear combinations on the complex plane and they form a lattice, Λ = a, b+ d geometry of lattices. Equivalence: Two lattices Λ, Λ are homothetic if there exists α C such that Λ = αλ, and we write Λ Λ. Divide through by a, every such lattice is homothetic to 1, τ where τ = b+ d a, in the upper half plane.

Structure of Ideals 53 Fundamental discriminants and orders A square class of integers, like 3, 1, 7, 48,... gives same field Q( 3n ) = Q( 3) minimal one? Candidate: The only one that is squarefree? However, from theory of bqfs need discriminant 0 or 1 (mod 4). Divisibility by 4 correct price to pay. The fundamental discriminant of a quadratic field to be the smallest element of the square class of the discriminant which is 0 or 1 (mod 4). For d squarefree integer, the fundamental discriminant D is { d if d 1 (mod 4) D = 4d if d or 3 (mod 4). The ring of integers is Z [ D+ D ] or Z[ω] = 1, ω Z, ω := { 1+ d if d 1 (mod 4) d = D/ if d or 3 (mod 4).

Composition 54 Gauss s Composition Law The product of any two values of a principal form gives a third value of that quadratic form: (a + db )(c + de ) = (ac + dbe) + d(ae bc).

Composition 55 Gauss s Composition Law The product of any two values of a principal form gives a third value of that quadratic form: (a + db )(c + de ) = (ac + dbe) + d(ae bc). Gauss: if f and g are bqfs discrim d, then bqf h of discrim d, such that any f(a, b)g(c, e) = h(m, n), m = m(a, b, c, e), n = n(a, b, c, e) are bilinear forms.

Composition 56 Gauss s Composition Law The product of any two values of a principal form gives a third value of that quadratic form: (a + db )(c + de ) = (ac + dbe) + d(ae bc). Gauss: if f and g are bqfs discrim d, then bqf h of discrim d, such that any f(a, b)g(c, e) = h(m, n), m = m(a, b, c, e), n = n(a, b, c, e) are bilinear forms. Gauss showed this explicitly via formulae; e.g., for three bqfs of discrim 71, m + mn + 9n = (4a + 3ab + 5b )(3c + ce + 6e ). with m = ac 3ae bc 3be and n = ac+ae+bc be. Gauss called this composition.

Composition 57 m + mn + 9n = (4a + 3ab + 5b )(3c + ce + 6e ). Gauss showed composition stays consistent under the equivalence relation. Allows us to find a group structure on the classes of quadratic forms of given discriminant, the class group. Gauss s proof is monstrously difficult, even in the hands of the master the algebra involved is so overwhelming that he does not include many details.

Composition 58 m + mn + 9n = (4a + 3ab + 5b )(3c + ce + 6e ). Gauss showed composition stays consistent under the equivalence relation. Allows us to find a group structure on the classes of quadratic forms of given discriminant, the class group. Gauss s proof is monstrously difficult, even in the hands of the master the algebra involved is so overwhelming that he does not include many details. Gauss s student Dirichlet found several ways to simplify composition. The first involved finding forms that are equivalent to f and g that are easier to compose:

Composition 59 Dirichlet s composition of forms For any given integer w there exist integers m, n with (am + bmn + cn, w) = 1. Given quadratic forms f and g, find f f such that (f (1, 0), g(1, 0)) = 1. There exists F f and G g such that F (x, y) = ax +bxy +cy and G(x, y) = Ax +bxy +Cy with (a, A) = 1. If f and g have the same discriminant then there exist h such that F (x, y) = ax + bxy + Ahy and G(x, y) = Ax + bxy + ahy with (a, A) = 1. d = b 4aAh. If H(x, y) = aax + bxy + hy then H(ux hvy, auy + Avx + bvy) = F (u, v)g(x, y)

Composition 60 Dirichlet s composition of ideals Dirichlet simplified by defining ideals: To multiply two ideals, IJ = {ij : i I, j J}. m + mn + 9n = (4a + 3ab + 5b )(3c + ce + 6e ).

Composition 61 Dirichlet s composition of ideals Dirichlet simplified by defining ideals: To multiply two ideals, IJ = {ij : i I, j J}. m + mn + 9n = (4a + 3ab + 5b )(3c + ce + 6e ). (4, 3+ 71 ) corresponds to 4a + 3ab + 5b, and (3, 1+ 71 ) corresponds to 3c + ce + 6e.

Composition 6 Dirichlet s composition of ideals Dirichlet simplified by defining ideals: To multiply two ideals, IJ = {ij : i I, j J}. m + mn + 9n = (4a + 3ab + 5b )(3c + ce + 6e ). (4, 3+ 71 ) corresponds to 4a + 3ab + 5b, and (3, 1+ 71 ) corresponds to 3c + ce + 6e. Then ) ( 3, 1 + ) 71 = ( 4, 3 + 71 ( 1, 5 + 71 which corresponds to 1x 5xy +y, also of disc 71, ),

Composition 63 Dirichlet s composition of ideals Dirichlet simplified by defining ideals: To multiply two ideals, IJ = {ij : i I, j J}. m + mn + 9n = (4a + 3ab + 5b )(3c + ce + 6e ). (4, 3+ 71 ) corresponds to 4a + 3ab + 5b, and (3, 1+ 71 ) corresponds to 3c + ce + 6e. Then ) ( 3, 1 + ) 71 = ( 4, 3 + 71 ( 1, 5 + 71 which corresponds to 1x 5xy +y, also of disc 71, but not reduced. Reduction then yields: (1, 5, ) (, 5, 1) (, 1, 9) ),

Composition 64 Comparing Dirichlet s compositions If F = ax + bxy + Ahy, G = Ax + bxy + ahy then H(ux hvy, auy + Avx + bvy) = F (u, v)g(x, y) for H(x, y) = aax + bxy + hy.

Composition 65 Comparing Dirichlet s compositions If F = ax + bxy + Ahy, G = Ax + bxy + ahy then H(ux hvy, auy + Avx + bvy) = F (u, v)g(x, y) for H(x, y) = aax + bxy + hy. ( The two quadratic forms F and G correspond to ( ) ( and. The product is A, b+ d aa, b+ d a, b+ d ), so the composition of F and G must be aax + bxy + hy. )

Composition 66 Comparing Dirichlet s compositions If F = ax + bxy + Ahy, G = Ax + bxy + ahy then H(ux hvy, auy + Avx + bvy) = F (u, v)g(x, y) for H(x, y) = aax + bxy + hy. ( The two quadratic forms F and G correspond to ( ) ( and. The product is A, b+ d aa, b+ d a, b+ d ), so the composition of F and G must be aax + bxy + hy. Identity of ideal class group: principal ideas. Inverses: ( a, b + d ) ( a, b d ) = ( a, a b + d a (a, b, c) = (a), ), a b d, b d 4 So an ideal and its conjugate are inverses in class group. )

Pre-homogenous vector spaces 67 A more general set up Let G(Z) be SL(, Z), an algebraic group ; V (Z) the space of bqfs over Z, a representation. Seen that: The G(Z)-orbits parametrize the ideal classes in the associated quadratic rings. Do other such pairs exist? That is an algebraic group G and associated representation V such that G(Z) V (Z) parametrizes something interesting? Eg rings, modules etc of arithmetic interest.

Pre-homogenous vector spaces 68 A more general set up Let G(Z) be SL(, Z), an algebraic group ; V (Z) the space of bqfs over Z, a representation. Seen that: The G(Z)-orbits parametrize the ideal classes in the associated quadratic rings. Do other such pairs exist? That is an algebraic group G and associated representation V such that G(Z) V (Z) parametrizes something interesting? Eg rings, modules etc of arithmetic interest. - In our example there is just one orbit over C: A pre-homogenous vector space is a pair (G, V ) where G is an algebraic group and V is a rational vector space representation of G such that the action of G(C) on V (C) has just one Zariski open orbit.

Pre-homogenous vector spaces 69 A pre-homogenous vector space is a pair (G, V ) where G is an algebraic group and V is a rational vector space representation of G such that the action of G(C) on V (C) has just one Zariski open orbit. Bhargava s program centres around study of G(Z) V (Z) for pre-homogenous vector spaces (G, V ).

Pre-homogenous vector spaces 70 A pre-homogenous vector space is a pair (G, V ) where G is an algebraic group and V is a rational vector space representation of G such that the action of G(C) on V (C) has just one Zariski open orbit. Bhargava s program centres around study of G(Z) V (Z) for pre-homogenous vector spaces (G, V ). There are just 36 of them (Sato-Kimura, 1977), but they have proved yo be incredibly rich in structure of interest to number theorists.

Pre-homogenous vector spaces 71 Bhargava composition Recently Bhargava gave a new insight into the composition law. Note: If IJ = K then IJK is principal.

Pre-homogenous vector spaces 7 Bhargava composition We begin with a -by--by- cube. a, b, c, d, e, f, g, h. Six faces, can be split into three parallel pairs. To each consider pair of -by- matrices by taking the entries in each face, with corresponding entries corresponding to opposite corners of the cube, always starting with a. Hence we get the pairs ( ) ( ) a b e f M 1 (x, y) := x + y, c d g h ( ) ( ) a c b d M (x, y) := M 3 (x, y) := x + e g ( ) a b x + e f y, f h ( ) c d y, g h where we have, in each added the dummy variables, x, y. The determinant, Q j (x, y), of each M j (x, y) gives rise to a quadratic form in x and y.

Pre-homogenous vector spaces 73 ( ) ( ) a b e f M 1 (x, y) := x + y, c d g h ( ) ( ) a c b d M (x, y) := x + y, e g f h ( ) ( ) a b c d M 3 (x, y) := x + y, e f g h Q j (x, y) = det M j (x, y), a bqf Now apply( an SL(, ) Z) transformation in one direction. α β That is, if SL(, Z) then we replace the face γ δ ( ) ( ) ( ) a b a b e f by α + β c d c d g h and ( ) e f g h by ( ) a b γ + c d ( ) e f δ. g h

If and ( ) α β γ δ M 1 (x, y) := Pre-homogenous vector spaces 74 ( ) a b x + c d ( ) e f y, g h SL(, Z) then we replace the face ( ) a b c d ( ) e f g h by by ( ) a b α + c d ( ) a b γ + c d ( ) e f β g h ( ) e f δ. g h Then M 1 (x, y) gets mapped to {( ) ( ) } {( ) a b e f a b α + β x+ γ + c d g h c d that is M 1 (αx + γy, βx + δy). Therefore ( ) } e f δ y, g h Q 1 (x, y) = det M 1 (x, y) gets mapped to Q 1 (αx + γy, βx + δy). which is equivalent to Q 1 (x, y).

Pre-homogenous vector spaces 75 Now M (x, y) gets mapped to ( ) aα + eβ cα + gβ x + aγ + eδ cγ + gδ ( ) α β = M γ δ (x, y); ( bα + fβ dα + hβ bγ + fδ dγ + hδ ) y hence the determinant, Q (x, y), is unchanged. An analogous ( ) calculation reveals that M 3 (x, y) gets mapped to α β M γ δ 3 (x, y) and its det, Q 3 (x, y) also unchanged. Therefore we can act on our cube by such SL(, Z)- transformations, in each direction, and each of the three quadratic forms remains in the same equivalence class.

Pre-homogenous vector spaces 76 Another prehomogenous vector space We can act on our cube by such SL(, Z)-transformations, in each direction, and each of the three quadratic forms remains in the same equivalence class.

Pre-homogenous vector spaces 77 Another prehomogenous vector space We can act on our cube by such SL(, Z)-transformations, in each direction, and each of the three quadratic forms remains in the same equivalence class. Bhargava s cubes can be identified as a e 1 e 1 e 1 + b e 1 e e 1 + c e e 1 e 1 +d e e e 1 + e e 1 e 1 e + f e 1 e e +g e e 1 e + h e e e with the representation Z Z Z of the group SL(, Z) SL(, Z) SL(, Z).

Pre-homogenous vector spaces 78 Another prehomogenous vector space We can act on our cube by such SL(, Z)-transformations, in each direction, and each of the three quadratic forms remains in the same equivalence class. Bhargava s cubes can be identified as a e 1 e 1 e 1 + b e 1 e e 1 + c e e 1 e 1 +d e e e 1 + e e 1 e 1 e + f e 1 e e +g e e 1 e + h e e e with the representation Z Z Z of the group SL(, Z) SL(, Z) SL(, Z). This pair is also a prehomogenous vector space

Pre-homogenous vector spaces 79 Reducing a Bhargava cube Simplify entries using the following reduction algorithm: We select the corner that is to be a so that a 0. Transform cube to ensure a divides b, c and e. If not, say a does not divide e, n select integers α, β so that aα + eβ = (a, e). Let γ = e/(a, e), δ = a/(a, e). In transformed matrix a = (a, e), e = 0 and 1 a a 1. If a does not divide b or c, repeat the process. Each time we reduce a, so a finite process. Transform cube to ensure b = c = e = 0. Select α = 1, β = 0, γ = e/a, δ = 1, so that e = 0, b = b, c = c. We repeat this in each of the three directions to ensure that b = c = e = 0.

Pre-homogenous vector spaces 80 Reducing a Bhargava cube, II Replacing a by a, we have that the three matrices are: ( ) ( ) a 0 0 f M 1 = x + y, so Q 0 d g h 1 = adx + ahxy + fgy ; ( ) ( ) a 0 0 d M = x + y, so Q 0 g f h = agx + ahxy + dfy ; ( ) ( ) a 0 0 d M 3 = x + y, so Q 0 f g h 3 = afx + ahxy + dgy. All discrim (Q j ) = (ah) 4adfg,

Pre-homogenous vector spaces 81 Reducing a Bhargava cube, II Replacing a by a, we have that the three matrices are: ( ) ( ) a 0 0 f M 1 = x + y, so Q 0 d g h 1 = adx + ahxy + fgy ; ( ) ( ) a 0 0 d M = x + y, so Q 0 g f h = agx + ahxy + dfy ; ( ) ( ) a 0 0 d M 3 = x + y, so Q 0 f g h 3 = afx + ahxy + dgy. All discrim (Q j ) = (ah) 4adfg, and Q 1 (fy x 3 +gx y 3 +hy y 3, ax x 3 dy y 3 ) = Q (x, y )Q 3 (x 3, y 3 ) x 1 = fy x 3 + gx y 3 + hy y 3 and y 1 = ax x 3 dy y 3.

Pre-homogenous vector spaces 8 Reducing a Bhargava cube, II Replacing a by a, we have that the three matrices are: ( ) ( ) a 0 0 f M 1 = x + y, so Q 0 d g h 1 = adx + ahxy + fgy ; ( ) ( ) a 0 0 d M = x + y, so Q 0 g f h = agx + ahxy + dfy ; ( ) ( ) a 0 0 d M 3 = x + y, so Q 0 f g h 3 = afx + ahxy + dgy. All discrim (Q j ) = (ah) 4adfg, and Q 1 (fy x 3 +gx y 3 +hy y 3, ax x 3 dy y 3 ) = Q (x, y )Q 3 (x 3, y 3 ) x 1 = fy x 3 + gx y 3 + hy y 3 and y 1 = ax x 3 dy y 3. Dirichlet: a = 1. So Includes every pair of bqfs of same discriminant.

Pre-homogenous vector spaces 83 SL(, Z)-transformations. Forms-Ideals-Transformations ( ) ( ) 1 1 0 1 Generators of SL(, Z),. 0 1 1 0 Let z f := b+ d a in the upper half plane. z, z equivalent ( ) if( M) SL(, Z) such that z = u/v u z where = M. Hence z z +1 and z 1/z. v 1 Now a, b + d = a(1, z f ) so equivalent.

Pre-homogenous vector spaces 84 ( ) ( ) 1 1 0 1 Generators of SL(, Z),. correspond to two basic ops in Gauss s reduction algorithm 0 1 1 0 The first is x x + y, y y, so that f(x, y) g(x, y) := f(x+y, y) = ax +(b+a)xy+(a+b+c)y. Note that I g = (a, (b + a) + d) = I f, and z g = b a+ d a = z f 1.

Pre-homogenous vector spaces 85 ( ) ( ) 1 1 0 1 Generators of SL(, Z),. correspond to two basic ops in Gauss s reduction algorithm 0 1 1 0 The first is x x + y, y y, so that f(x, y) g(x, y) := f(x+y, y) = ax +(b+a)xy+(a+b+c)y. Note that I g = (a, (b + a) + d) = I f, and z g = b a+ d a = z f 1. The second is x y, y x so that f(x, y) h(x, y) := f(y, x) = cx bxy + ay. Note that I h = (c, b + d), and z h = b+ d c. z f z h = b + d b + d a c that is z h = 1/z f. Then = d b 4ac = 1 I h (1, z h ) = (1, 1/z f ) (1, z f ) = (1, z f ) I f.

Pre-homogenous vector spaces 86 Since any SL(, Z)-transformation can be constructed out of the basic two transformation we deduce Theorem f f if and only if I f I f if and only if z f z f.

Invariants of number fields 87 The ring of integers of a quadratic field, revisited Integer solutions x, y to x + 19 = y 3? If so, y is odd else x 5 (mod 8) #. Also 19 y else 19 x = 19 x + 19 = y 3 0 (mod 19 ). Hence (y, 38) = 1.

Invariants of number fields 88 The ring of integers of a quadratic field, revisited Integer solutions x, y to x + 19 = y 3? If so, y is odd else x 5 (mod 8) #. Also 19 y else 19 x = 19 x + 19 = y 3 0 (mod 19 ). Hence (y, 38) = 1. Now (x + 19)(x 19) = y 3 and (x + 19, x 19) contains 19 and y 3, and so also (y 3, 38) = 1.

Invariants of number fields 89 The ring of integers of a quadratic field, revisited Integer solutions x, y to x + 19 = y 3? If so, y is odd else x 5 (mod 8) #. Also 19 y else 19 x = 19 x + 19 = y 3 0 (mod 19 ). Hence (y, 38) = 1. Now (x + 19)(x 19) = y 3 and (x + 19, x 19) contains 19 and y 3, and so also (y 3, 38) = 1. Hence the ideals (x + 19) and (x 19) are coprime Their product is a cube and so they are both cubes.

Invariants of number fields 90 Integer solutions x, y to x + 19 = y 3? The ring of integers of Q[ 19] has class number one. So every ideal is principal. Hence x + 19 = u(a + b 19) 3 where u is a unit.

Invariants of number fields 91 Integer solutions x, y to x + 19 = y 3? The ring of integers of Q[ 19] has class number one. So every ideal is principal. Hence x + 19 = u(a + b 19) 3 where u is a unit. Only units: 1 and 1. Change a, b, to ua, ub. Hence x + 19 = (a + b 19) 3 so that b(3a 19b ) = 1. = a(a 57b ) + b(3a 19b ) 19,

Invariants of number fields 9 Integer solutions x, y to x + 19 = y 3? The ring of integers of Q[ 19] has class number one. So every ideal is principal. Hence x + 19 = u(a + b 19) 3 where u is a unit. Only units: 1 and 1. Change a, b, to ua, ub. Hence x + 19 = (a + b 19) 3 = a(a 57b ) + b(3a 19b ) 19, so that b(3a 19b ) = 1. Therefore b = ±1 and so 3a = 19b ± 1 = 19 ± 1 which is impossible. We deduce: There are no integer solutions x, y to x + 19 = y 3.

Invariants of number fields 93 Integer solutions x, y to x + 19 = y 3? The ring of integers of Q[ 19] has class number one. So every ideal is principal. Hence x + 19 = u(a + b 19) 3 where u is a unit. Only units: 1 and 1. Change a, b, to ua, ub. Hence x + 19 = (a + b 19) 3 = a(a 57b ) + b(3a 19b ) 19, so that b(3a 19b ) = 1. Therefore b = ±1 and so 3a = 19b ± 1 = 19 ± 1 which is impossible. We deduce: There are no integer solutions x, y to x + 19 = y 3. However what about 18 + 19 = 7 3

Invariants of number fields 94 Integer solutions x, y to x + 19 = y 3? The ring of integers of Q[ 19] has class number one. So every ideal is principal. Hence x + 19 = u(a + b 19) 3 where u is a unit. Only units: 1 and 1. Change a, b, to ua, ub. Hence x + 19 = (a + b 19) 3 = a(a 57b ) + b(3a 19b ) 19, so that b(3a 19b ) = 1. Therefore b = ±1 and so 3a = 19b ± 1 = 19 ± 1 which is impossible. We deduce: There are no integer solutions x, y to x + 19 = y 3. However what about 18 + 19 = 7 3 The mistake: The ring of integers of Q[ 19] is not the set of numbers of the form a + b 19 with a, b Z. It is (a + b 19)/ with a, b Z and a b (mod ).

Invariants of number fields 95 Integer solutions x, y to x + 19 = y 3? The ring of integers of Q[ 19] has class number one. So every ideal is principal. Hence x + 19 = ( a+b 19 ) 3 8x + 8 19 = (a + b 19) 3 = a(a 57b ) + b(3a 19b ) 19, so that b(3a 19b ) = 8. Therefore b = ±1, ±, ±4 or ±8 and so 3a = 19 ± 8, 19 4 ± 4, 19 16 ± or 19 64 ± 1. The only solution is b = 1, a = ±3 leading to x = 18, y = 7, the only solutions.