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79 Pages

### MODULAR FORMS-page66

Course: MATH 300, Fall 2009
School: Stony Brook University
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Word Count: 329

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6. 62 LECTURE MODULAR FORMS 6.5 Show that 0 (z1 ) det @ (z2 ) r(z3 ) (z1 ) (z2 ) (z3 ) 1 1 1A = 0 1 whenever z1 + z2 + z3 = 0. Deduce from this an explicit formula for the group law on the projective cubic curve y 2 t = 4x3 g2 xt2 g3 t3 . 6.6 (Weierstrass -function) It is dened by 1 Z (z ; ) = + z X \{0} 1 1 z + +2. Let = Z1 + Z2 . Show that (i) Z (z ) = (z ); (ii) Z (z + i ) = Z (z ) + i ,...

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6. 62 LECTURE MODULAR FORMS 6.5 Show that 0 (z1 ) det @ (z2 ) r(z3 ) (z1 ) (z2 ) (z3 ) 1 1 1A = 0 1 whenever z1 + z2 + z3 = 0. Deduce from this an explicit formula for the group law on the projective cubic curve y 2 t = 4x3 g2 xt2 g3 t3 . 6.6 (Weierstrass -function) It is dened by 1 Z (z ; ) = + z X \{0} 1 1 z + +2. Let = Z1 + Z2 . Show that (i) Z (z ) = (z ); (ii) Z (z + i ) = Z (z ) + i , i = 1, 2 where i = Z (i /2); (iii) 1 2 2 1 = 2i; (iv) Z (z ; ) = 1 Z (z ; ), where is any nonzero complex number. 6.7 Let (z ) be a holomorphic function satisfying (z ) /(z ) = Z (z ), (i) Show that (z ) = (z ); (ii) (z + i ) = ei (z+ i 2 (z ); (iii) (z ) = (z ), where (z ) is the Weierstrass -function. 6.8 Using the previous exercise show that the Weierstrass -function (z ) admits an in nite product expansion of the form Y z z 1z2 (z ) = z (1 )e + 2 ( ) \{0} which converges absolutely, and uniformly each in disc |z | R. 6.9 Let E be an elliptic curve and y 2 = 4x3 g2 x g3 be its Weierstrass equation. Show that any automorphism of E is obtained by a linear transformation of the variables (x, y ) which transforms the Weierstrass equation to the form y 2 = 4x3 c4 g2 x c6 g3 for some c = 0. Show that E is harmonic (resp. anharmonic) if and only if g3 = 0 (resp. g2 = 0). 6.10 Let k be an even integer and let L Rk be a lattice with a basis (e1 , . . . , ek ). Assume that ||v ||2 is even for any v L. Let D be the determinant of the matrix (ei ej ) and N be the smallest positive integer such that N ||v ||2 2Z for all v Rk satisfying v w Z for all w L. Dene the theta series of the lattice L by L ( ) = X #{v L : ||v ||2 = 2n}e2in . n=0 (i) Show that L ( ) = P v L 2 ei ||v|| ; (ii) Show that the functions (0, )k discussed in the beginning of Lecture 6 are special cases of the function L .
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Stony Brook University - MATH - 300
61Example 6.4. We apply the previous theorem to (z ; ) = (z ) and z =case = 0 (2) = cfw_ (1) : 2| .1.2In thisNow, replacing z with z/( + ) in (6.22), we get( + z;) = ( + )2 [z 2 + + + X(m,n)=(0,0)1].z m( + ) + n( + )Since Z + Z = Z( +
Stony Brook University - MATH - 300
60LECTURE 6. MODULAR FORMSIn particular, 120 (4) = (2 )4 /12,280 (6) = (2 )6 /216 and we can write117Y g2 = (2 )4 [+ 20X ], g3 = (2 )6.122163This gives32g2 27g3 = (2 )12 [(5X + 7Y )/12 + 100X 2 + 20X 3 42Y 2 ] =(2 )12 q + q 2 (. . .).3N
Stony Brook University - MATH - 300
59Setting t = e2iz , we rewrite the left-hand side as follows: cot(z ) = Xmeiz + eizt+1cos z= i iz) = i= i (1 2t ).sin ze eizt1m=0Dierentiating k 1 2 times in z , we getX(k 1)!X(z + m)k = (2i)kmZmk1 tm .m=1This gives us the needed
Stony Brook University - MATH - 300
58LECTURE 6. MODULAR FORMSas a holomorphic map from C \ cfw_e1 , e2 , e3 to (C/ )/(z z ). It can be shown thatit extends to a holomorphic isomorphism from the Weiertrass cubic y 2 = 4x3 g2 x g3onto (C/ ) \ cfw_0. This is the inverse of the map given
Stony Brook University - MATH - 300
57After all of these normalizations, the elliptic function (z ) with respect to is uniquely determined by the conditions (6.18) and (6.19). It is called called theWeierstrass function with respect to the lattice .One can nd explicitly the function (z
Stony Brook University - MATH - 300
56LECTURE 6. MODULAR FORMSenters only in the rst degree. Thus we can express y in terms of x, t and obtain thatE is isomorphic to P1 (C). If d = 0 we obtain that f could be chosen of degree 2.Again this is impossible. Note that we also have in (6.13)
Stony Brook University - MATH - 300
2LECTURE 1. BINARY QUADRATIC FORMSLet x = m1 v + m2 w . The length of x is given by the formulax|2 = |m1 v + m2 w|2 = (m1 , m2 )vvvwvwwwm1m2=am2 + 2bm1 m2 + cm2 ,12wherea = v v,b = v w,c = w w.(1.1)Let us consider the (binary) quadratic
Stony Brook University - MATH - 300
556.4 We know that any elliptic curve is isomorphic to a Hesse cubic curve. Let usgive another cubic equation for an elliptic curve, called a Weierstrass equation. Itscoecients will give us new examples of modular forms. Recall that dim Th(k, )ab =k.
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
Virginia Tech - CEE - 4674
U.S. Departmentof TransportationFederal AviationAdministrationSubject: STANDARDS FOR AIRPORT MARKINGS1. PURPOSE. This advisory circular (AC)contains the Federal Aviation Administration(FAA) standards for markings used on airportrunways, taxiways,
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U.S. Departmentof TransportationFederal AviationAdministrationAIRPORT DESIGN/ INCORPORATES CHANGES 1 THRU 16 /AC:Date:150/5300-139/29/89Advisory Circular3/28/07ParagraphAC 150/5300-13 CHG 11CONTENTSPageChapter 1. REGULATORY REQUIREMENTS AN
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AC 150/5320-6D7f7l95U! SS3NXlIHL EWIS82L-1011-100, 200nnnANNUAL DEPARTURESCONTACT AREA= 337 SQ. IN.DUAL SPACING= 52 IN.TANDEM SPACING = 70 IN.1,2006,0003,00025,00015,000-1719 -2018 -19-1617-1516-1415-1314-1213-10650-11600
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AC150/5320-6D7l7J95Increasing C, Factor. A value of C, lower than 0.75 represents a severely cracked base slab,(2)which would not be advisable to overlay without modification due to the likelihood of severe reflection cracking. Seeparagraph 406 f. In
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