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79 Pages

### MODULAR FORMS-page59

Course: MATH 300, Fall 2009
School: Stony Brook University
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Word Count: 359

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We 55 6.4 know that any elliptic curve is isomorphic to a Hesse cubic curve. Let us give another cubic equation for an elliptic curve, called a Weierstrass equation. Its coecients will give us new examples of modular forms. Recall that dim Th(k, )ab = k. Let use &lt;, &gt; to denote the linear span. We have Th(1, ) 1 1 =&lt; 1 1 (z ; ) &gt;=&lt; T &gt;; 22 22 Th(2, ) =&lt; T 2 ,...

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We 55 6.4 know that any elliptic curve is isomorphic to a Hesse cubic curve. Let us give another cubic equation for an elliptic curve, called a Weierstrass equation. Its coecients will give us new examples of modular forms. Recall that dim Th(k, )ab = k. Let use <, > to denote the linear span. We have Th(1, ) 1 1 =< 1 1 (z ; ) >=< T >; 22 22 Th(2, ) =< T 2 , X >, Th(3, ) 1 1 =< T 3 , T X , Y >, 22 for some functions X Th(2, ), Y Th(3, ) 1 1 . Now the following seven func22 tions T 6, T 4X , T 2X 2, X 3, T 3Y , T X Y , Y 2 all belong to the space Th(6, ). They must be linearly dependent and we have aT 6 + bT 4 X 2 + cT 2 X 2 3 + dX + eT 3 Y + f T X Y + gY 2 = 0. (6.12) Assume g = 0, d = 0. It is easy to nd X = X + T 2 , Y = Y + XT + T 3 which reduces this expression to the form Y 2 T X 3 AXT 4 BT 6 = 0, (6.13) for some scalars A, B . Let (z ) = X/T 2 , 1 (z ) = Y /T 3 . Dividing (6.13) by T 6 we obtain a relation 1 (z )2 = (z + )3 A(z ) + B. (6.14) Since both X and T 2 belong to the same space Th(2, ) the functions (z ), 1 (z ) have periods Z + Z and meromorphic on C. As we shall see a little later, 1 (z ) = d . dz Consider the map E = C/ P2 , z (T (z )3 , T (z )X (z ), Y (z )). Since T (z )3 , T (z )X (z ), Y (z ) all belong to the same space Th(3, ) 1 , 1 this map is 22 well-dened and holomorphic. It diers from the map from Example 3.2 only by a composition with a translation on E and a linear change of the projective coordinates coordinates. This is because, for any f Th(k; ) we have 2 ei[a +2(z +a)b] f (z + b + a ) Th(k; )ab k (see Lecture 3). So it is an isomorphism onto its image. The relation (6.13) tells us that the image is the plane projective curve of degree 3 given by the equation y 2 t x3 Axt2 Bt3 = 0, (6.15) Now it is clear why we assumed that the coecients d, g in (6.12) are not equal to zero. If g = 0, we obtain an equation f (x, y, t) = 0 for the image of E in which y
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Stony Brook University - MATH - 300
54LECTURE 6. MODULAR FORMSHere we used that (T S ) = (ST )1 since (T 2 S 2 ) = 1 and similarly(T ST ) = (ST S )1 , (T ST S ) = (ST ST )1 ,(T ST ST ) = (ST ST S )1 .Also (ST ) = (ST ST )1 . Thus it is enough to verify that the elements S, ST, ST S,T
Stony Brook University - MATH - 300
536.3 Let us give some examples.Example 6.1. Let( ) = ( )24 .It is called the discriminant function. We know that ( ) satises (6.1) with k = 6with respect to the group = SL(2, Z). By (4.9)( ) =Since( ) = q1 1 1 8.(2 )8 2 2Y(1 q m )24m=1we s
Stony Brook University - MATH - 300
52LECTURE 6. MODULAR FORMSrepresented by a rational number x or the stabilizer group x is conjugate to asubgroup of SL(2, Z) . In fact, if g x = for some g SL(2, Z), theng x g 1 = .Since + = = 0, + we haveg x g 1 cfw_10, Z1Let h be the sma
Stony Brook University - MATH - 300
516.2 Suppose we have n + 1 linearly independent functions f0 , . . . , fn satisfying (6.1)(with the same number k). Then we can consider the mapf : H CPn , (f0 ( ), . . . , fn ( ).(6.7)When we replace with + , the coordinates of the image will all
Stony Brook University - MATH - 300
50LECTURE 6. MODULAR FORMSof in the space of holomorphic functions on H dened by(g )(z ) = |k g 1 .Note that we switched here to g1(6.6)in order to get(gg ) = (g ) (g ).It follows from the above that to check (6.1) for some subgroup it is enough
Stony Brook University - MATH - 300
Lecture 6Modular Forms6.1 We have seen already in Lecture 5 (5.2) and Corollary 5.3 that the functions( )4k = 00 (0; )4k (resp. ( )24 ) satisfy the functional equationf ( + 2) = f ( ),f (1/ ) = 2 f ( ),(resp.f ( + 1) = f ( ),f (1/ ) = 12 f ( ).In
Stony Brook University - MATH - 300
48LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONS5.7 Dene the Weierstrass -function by2z2z 2 ( 1 1 /61 1 1 ) 1 1 ( ; )1122 (z ; 1 , 2 ) = 1 e2222 1 1 (0).22Show that(i) (z ; 1 , 2 ) does not depend on the basis 1 , 2 of the lattice ;(ii) (
Stony Brook University - MATH - 300
47(iv) Show that the expressionQn1 =100 ( n ; ) 1 0 ( n ; )0 1 ( n ; )2200 (0; )n1 1 0 (0; )n1 0 1 (0; )n12.2does not change when is replaced with 2 .(v) Show thatQn1 =100 ( n ; ) 1 0 ( n ; )0 1 ( n ; )2200 (0; )n1 1 0 (0; )n1 0 1 (0; )n
Stony Brook University - MATH - 300
46LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONSExercises5.1 Show that the constant (M ) in (5.16) is equal to iodd. If is odd and is even, it is equal to0symbol, where we also set ( 1 ) = 1.ei/4 ( ). 12( | ) when is even and isHere ( x ) is the
Stony Brook University - MATH - 300
Lecture 1Binary Quadratic Forms1.1 The theory of modular form originates from the work of C.F. Gauss of1831 in which he gave a geometrical interpretation of some basic notions ofnumber theory.Let us start with choosing two non-proportional vectors in
Stony Brook University - MATH - 300
45Proof. We shall prove in the next lecture that it is enough to check this for generatorsof the group SL(2, Z). Also we shall show that the group SL(2, Z) is generated by thematrices M1 = ( 1 1 ) , M2 = ( 0 1 ) , I. We have from (4.14) and (4.15)011
Stony Brook University - MATH - 300
44LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONSNow take M = ( 0 1 ) . We have10eiz2/00 (z/ ; 1/ ) = B00 (z ; )for some B depending only on . Plugging in z = 0 and applying (4.2), we getB = i ,(5.10)where the square root takes positive values on
Stony Brook University - MATH - 300
43wherekk, a + b ).22Summarizing we obtain that, for any (z, ) Th(k; )ab ,(a , b ) = (a b +(5.4)2eik ( +)z ( + )z ; ) Th(k, )a b .(5.5)` Now let us replace with its inverse . We rewrite (5.13) and (5.14) as2eik ( +)z ( + )z ; ) Th(k, )aw
Stony Brook University - MATH - 300
42LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONSThus for any Th(k; )ab , we have(z ( + ) Th(e ; Z + Z),where= + , + em+n (z ) = e(m+n)( + ) (z (+ ).We have, using (5.1),e1 (z ) = e + (z ( + ) = e2i(ab ) eik(2z( +)+eik ( +)(z+1)22( + )z ) i
Stony Brook University - MATH - 300
Lecture 5Transformations of ThetaFunctions5.1 Let us see now that the theta constants ab and their derivatives ab satisfythe functional equation similar to (4.2). This will imply that certain powers of thetaconstants are modular forms. For brevity we
Stony Brook University - MATH - 300
40LECTURE 4. THETA CONSTANTS4.5 Prove the Jacobi triple product identity:Y11(1 q n )(1 + q n 2 t)(1 + q n 2 t1 ) =Xqr22tn .r Zn=14.6 Prove a doubling identity for theta constants:0 1 (2 )2 = 00 ( )0 1 ( ).22(see other doubling identities
Stony Brook University - MATH - 300
39This gives the Euler identityXY(1)r q r(3r+1)/2 =(1 q m ).r Z(4.23)m=1In particular, we get the following Fourier expansion for the Dedekinds function ( ):1 ( ) = q 24X(1)r q r(3r+1)/2 .r ZThe positive integers of the form n + (k 2) n(n2
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