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### MODULAR FORMS-page9

Course: MATH 300, Fall 2009
School: Stony Brook University
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the 5 is upper half-plane. Let us see how the group GL(2, Z) acts on the both sets. We have M f = a((x + y ) z (x + y ))((x + y ) z (x + y )) = a(x( z1 ) y ( + z ))(x( z ) y ( + z )) = a| z |2 (x + z + z y )(x y ). z z Let us consider the action of GL(2, Z) on C \ R by fractional-linear transformations (also called Moebius transformations) dened by the formula z = z + . z +...

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the 5 is upper half-plane. Let us see how the group GL(2, Z) acts on the both sets. We have M f = a((x + y ) z (x + y ))((x + y ) z (x + y )) = a(x( z1 ) y ( + z ))(x( z ) y ( + z )) = a| z |2 (x + z + z y )(x y ). z z Let us consider the action of GL(2, Z) on C \ R by fractional-linear transformations (also called Moebius transformations) dened by the formula z = z + . z + (1.5) Notice that Im M z = Im z + (z + )( z + ) = Im z. = Im z + |z + |2 |z + |2 (1.6) This explains why the transformation is well-dened on C \ R. Also notice that M 1 = det M . Thus the root z is transformed to the root z = M 1 z and we obtain, for any M GL(2, 1 Z), M f = a|z + |2 (x M z )(x M z ). 1.4 Until now we considered binary forms up to the equivalence dened by an invertible integral substitution of the variables. We say that two binary forms are properly equivalent if they dier by a substitution with determinant equal to 1. In other words, we restrict ourselves with with the subgroup SL(2, Z) of GL(2, Z). Since 10 GL(2, Z) = SL(2, Z) SL(2, Z) 0 1 10 (ax2 + 2bxy + cy 2 ) = ax2 2bxy + cy 2 we obtain that each f is 0 1 properly equivalent to a form ax2 + 2bxy + cy 2 , where (a, b, c) and and = {(a, b, c) R3 : |2b| c a, a, ac b2 > 0}. Denition. We shall say that f = ax2 + 2bxy + cy 2 is properly reduced if (a, b, c) .
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Stony Brook University - MATH - 300
85Denition. Let X = H /. A point x = is called an elliptic point of order 2(resp, of order 3) if (1) i (resp. (1) ) and = 1.Theorem 8.4. The genus of H / is equal tog =1+r2r3r,12432where is the index of / (1) in (1)/(1), r2 is the number of e
Stony Brook University - MATH - 300
84LECTURE 8. THE MODULAR CURVEThis is called the Hurwitz formula. The number n here is called the degree of themeromorphic function f . Formula (8.21) says that this number is equal to #f 1 (y )for almost all y P1 (C).We shall dene the triangulation
Stony Brook University - MATH - 300
83Proof. Replacing D with D + div(f ), we do not change the dimensions of the spacesL(D) and L(KX D) but change deg(D) by deg(D + div(f ) = deg D + deg(div(f ).It follows from Riemann-Roch that deg(div(f ) = 0.Corollary 8.2.deg KX = 2g 2.Proof. Take
Stony Brook University - MATH - 300
82LECTURE 8. THE MODULAR CURVEdtwhen restricted to U U . Here dtU is the derivative of the function gU,U = tU t1 :UUtU (U U ) tU (U U ).Two meromorphic dierentials are said be equal if they coincide when restrictedto the subcover formed by interse
Stony Brook University - MATH - 300
81zero (resp. the order of pole) of f at x. We have the following easily veried propertiesof x (f ):Lemma 8.3. Let x X and f, g be two meromorphic functions on X . Then(i) x (f g ) = x (f ) + x (g );(ii) x (f + g ) = mincfw_x (f ), x (g ) if f + g =
Stony Brook University - MATH - 300
80LECTURE 8. THE MODULAR CURVE. The natural inclusion U H factors through the map U/x H /x . Taking csmall enough and using (8.8) we see that this map is injective. Its image is an openneighborhood U of the cusp c H /. Let h be the index of the cusp.
Stony Brook University - MATH - 300
79where c is a positive real number. Since SL(2, Z) acts transitively on H \ H we cantake for a basis of open neighborhoods of each x Q the set of g -translates of the setsUc for all c &gt; 0 and all g SL(2, Z) such that g = x. Each g (Uc ) is equal to th
Stony Brook University - MATH - 300
78LECTURE 8. THE MODULAR CURVEExample 8.1. Let = (1). Let us show that there exists a holomorphic isomorphismH/SL(2, Z) C.=This shows that the set of isomorphism classes of elliptic curves has a natural structure3of a complex manifold of dimension
Stony Brook University - MATH - 300
Lecture 8The Modular Curve8.1 In this lecture we shall give an explicit formula for the dimension of the spacesMk (), where is any subgroup of nite index in SL(2, Z). For this we have to applysome techinique from algebraic geometry. We shall start wit
Stony Brook University - MATH - 300
4LECTURE 1. BINARY QUADRATIC FORMSamong all vectors not equal to v . I claim that (v , w ) forms a basis of .Assume it is false. Then there exists a vector x such that x = av + bwwhere one of the coecients a, b is a real number but not an integer. Aft
Stony Brook University - MATH - 300
75(i) Show thatrLC (2) = 48 + 16A4 ,rLC (4) = 28 A8 + 640A4 + 1104.(ii) Using (7.13) show that A8 = 759 4A4 .7.9 Let A = An be a commutative graded algebra over a eld F . Assume An=has no zero divisors, A0 = F 1 and dim AN &gt; 1 for some nN &gt; 0. Show
Stony Brook University - MATH - 300
74LECTURE 7. THE ALGEBRA OF MODULAR FORMS(i) Show that LC is an integral lattice if and only if for any x = ( 1 , . . . , n ) Cthe number wt(x) = #cfw_i : i = 0 (called the weight of x) is divisible by 4. Inthis case we say that C is a doubly even lin
Stony Brook University - MATH - 300
73Proof. Let be a normal subgroup of nite index in (1) which is contained in . Italways can be found by taking the intersection of conjugate subgroups g 1 g, g (1). We rst apply Lemma 7.3 to the case when B = M( ), A = M(1). SinceA C[T1 , T2 ] is nite
Stony Brook University - MATH - 300
72LECTURE 7. THE ALGEBRA OF MODULAR FORMS7.4 Our goal is to prove an analog of Theorem 7.1 for any subgroup of nite index of (1). Let be two such subgroups. Assume also that is normal in andlet G = / be the quotient group. The group G acts on Mk ( ) a
Stony Brook University - MATH - 300
71In fact there exists only one even unimodular lattice in R8 (up to equivalence oflattices). The lattice is the famous E8 lattice, the root lattice of simple Lie algebra oftype E8 .Fig.2Here the diagram describes a symmetric matrix as follows. All t
Stony Brook University - MATH - 300
70LECTURE 7. THE ALGEBRA OF MODULAR FORMS(e1 , . . . , en ). Then the pre-image of the standard lattice Zn = Ze1 + . . . + Zen is anintegral lattice L with the distance function Q.Let us dene the theta function of the lattice L by settingL ( ) =Xm=
Stony Brook University - MATH - 300
697.3 Let us give some examples.Example 7.1. We know that the Eisenstein series E2k is a modular form of weight22k with respect to (1). Since M4 (1) = Cg2 = CE4 , comparing the constantcoecients in the Fourier expansions we obtainE8 = (8) 2E4 .2
Stony Brook University - MATH - 300
68LECTURE 7. THE ALGEBRA OF MODULAR FORMSIn particular, this is true for g2 . For any other f M2 (1) we have f /g2 is (1)invariant and also holomorphic at (since g2 is not a cusp form). This shows thatf /g2 is constant andM2 (1) = Cg2 .Similar argum
Stony Brook University - MATH - 300
67Use the function q = e2i to map the segment cfw_ : |Re | 1 , Im = h onto the21circle C : |q | = e2h . When we move along the segment from the point 2 + ih to the1point 2 + ih the image point moves along the circle in the clockwise way. We have12
Stony Brook University - MATH - 300
66LECTURE 7. THE ALGEBRA OF MODULAR FORMSNote that when f is a modular form with respect to a group we haveg (f ) = (f ),g .For each H let8&gt;2&lt;m = 3&gt;:1if (1) i,if (1) e2i/3 ,otherwise.(7.3)Lemma 7.1. Let f ( ) be a modular form of weight k
Stony Brook University - MATH - 300
3and henceacbd=abcd.This can be also expressed by saying that the form f is obtained from the formf by using the change of variablesx x + y,y x + y.We write this in the formf = M f.According to Lagrange two binary quadratic forms f and g a
Stony Brook University - MATH - 300
Lecture 7The Algebra of ModularForms7.1 Let be a subgroup of nite index of (1). We setMk () = cfw_modular forms of weight k with respect to ,We also denote by Mk ()0 the subspace of cuspidal modular forms. It is clear thatMk () is a vector space ove
Stony Brook University - MATH - 300
63(iii) Show that L ( ) is almost modular form for the group0 (N ) = cfw_ SL(2, Z) : N |c,i.e.L ( + ) = ( + )k/2 (d)L ( ), + where (d) = ( (1)dk2D 0 (N ),) is the quadratic residue symbol.(iv) Prove that L ( ) is a modular form for 0 (2) w
Stony Brook University - MATH - 300
62LECTURE 6. MODULAR FORMS6.5 Show that0(z1 )det @ (z2 )r(z3 ) (z1 ) (z2 ) (z3 )111A = 01whenever z1 + z2 + z3 = 0. Deduce from this an explicit formula for the group law onthe projective cubic curve y 2 t = 4x3 g2 xt2 g3 t3 .6.6 (Weierstr
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,
Virginia Tech - CEE - 4674
AdvisoryCircularU.S. Departmentof TransportationFederal AviationAdministrationSubject: AIRPORT MASTER PLANSDate: July 29, 2005Initiated by: APP-400AC No: 150/5070-6BChange:_1. PURPOSE. This Advisory Circular (AC) provides guidance for the prep
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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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Virginia Tech - CEE - 4674
AC 150/5320-6D7l7195C.Muskeg. Muskeg is sometimes encountered in arctic areas. Muskeg is a highly organic soil depositwhich is essentially a swamp. Every effort should be made to avoid pavement construction on this material. Ifconstruction in areas o
Virginia Tech - CEE - 4674
AC150/5320-6D7l7l95C.Stabilized Subbase. Stabilized subbases also offer considerably higher strength to the pavement thanP-154. Recommended equivalency factors associated with stabilized subbase are presented in Table 3-7.-/TABLE 3-7. RECOMMENDED EQ
Virginia Tech - CEE - 4674
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
Virginia Tech - CEE - 4674
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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U.S. Departmentof TransportationFederal AviationAdministrationSubject: RUNWAY LENGTHREQUIREMENTS FOR AIRPORT DESIGNAdvisoryCircularDate: 7/1/2005Initiated by: AAS-100AC No: 150/5325-4BChange:1. PURPOSE. This Advisory Circular (AC) provides gui
Virginia Tech - CEE - 4674
U.S. Departmentof TransportationFederal AviationAdministrationSubject: RUNWAY LENGTHREQUIREMENTS FOR AIRPORT DESIGNAdvisoryCircularDate: 7/1/2005Initiated by: AAS-100AC No: 150/5325-4BChange:1. PURPOSE. This Advisory Circular (AC) provides gui