Zeros of weakly holomorphic modular forms of level 4
ZZEROS OF WEAKLY HOLOMORPHIC MODULAR FORMS OF LEVEL 4
ANDREW HADDOCK AND PAUL JENKINS
Abstract.
Let M (cid:93)k (4) be the space of weakly holomorphic modular forms of weight k andlevel 4 that are holomorphic away from the cusp at ∞ . We define a canonical basis forthis space and show that for almost all of the basis elements, the majority of their zerosin a fundamental domain for Γ (4) lie on the lower boundary of the fundamental domain.Additionally, we show that the Fourier coefficients of the basis elements satisfy an interestingduality property. Introduction
In recent years there has been a great deal of interest in the question of locating the zerosof modular forms. Much of this work stems from results of F. Rankin and Swinnerton-Dyer [15], who showed that the zeros of the classical Eisenstein series E k , for even k ≥ ( Z ). Similar resultshave been obtained for Eisenstein series for a number of other congruence subgroups in thepapers [10, 12, 14, 17, 18].In contrast, the zeros for Hecke eigenforms of level 1 become equidistributed in the funda-mental domain, rather than congregating on its boundary, as the weight increases (see [16]and [13]). Even considering this equidistribution, though, some zeros still appear on theboundary; Ghosh and Sarnak [11] showed that for these eigenforms, the number of zeroson the boundary and center line of the fundamental domain for SL ( Z ) also grows with theweight.The second author, with Duke [5] and with Garthwaite [9], studied the zeros of weaklyholomorphic modular forms of level N = 1 , ,
3. Specifically, they showed that for a two-parameter family of weakly holomorphic modular forms that is a canonical basis for thespace, almost all of the basis elements have all (if N = 1) or most (if N = 2 ,
3) of theirzeros on a lower boundary of a fundamental domain for Γ ( N ). Thus, most or all of thezeros of these families of weakly holomorphic forms are the “real zeros” studied by Ghoshand Sarnak.In this paper, we examine the zeros of a canonical basis for the space of weakly holomorphicmodular forms of level 4 which are holomorphic away from the cusp at ∞ , and show thatmost of the zeros lie on a circular arc along the lower boundary of a fundamental domain.Additionally, we give a generating function for this canonical basis and show that the Fouriercoefficients of the basis elements satisfy several interesting properties.2. Definitions and Statement of Results
We let M k (4) be the space of holomorphic modular forms of even integer weight k for thegroup Γ (4) = { ( a bc d ) ∈ SL ( Z ) : c ≡ } , and we let M ! k (4) be the space of weakly Date : October 31, 2018.2010
Mathematics Subject Classification. a r X i v : . [ m a t h . N T ] M a y ANDREW HADDOCK AND PAUL JENKINS holomorphic modular forms, or modular forms that are holomorphic on the upper half planeand meromorphic at the cusps. In this paper, we specifically examine the space M (cid:93)k (4), whichis the subspace of M ! k (4) that is holomorphic away from the cusp at infinity. Atkinson [3]studied modular forms in M ! k (4) and M (cid:93)k (4), giving explicit descriptions of the action of adifferential operator, recurrences for their coefficients, and an identity for the exponents oftheir infinite product expansions. This generalized work of Bruinier, Kohnen, and Ono [4]in level 1 and of Ahlgren [1] for levels 2 , , ,
7, and 13.Any fundamental domain for Γ (4) has three cusps, which can be taken to be ∞ , 0, and . The fundamental domain we use is bordered by the lines Re( z ) = ± and the semicirclesdefined by ± + e iθ for θ ∈ [0 , π ]. As shown in Figure 1, instead of taking a single cusp at , we take a symmetric fundamental domain which meets the real line at 0 and ± . The Figure 1.
A fundamental domain for Γ (4)bottom left semicircle is equivalent to the bottom right semicircle under the action of thematrix ( ), and the left border is equivalent to the right border by using the matrix ( ).We recall that the valence formula for level 4 is given by v ∞ ( f ) + v ( f ) + v ( f ) + (cid:88) z ∈ Γ (4) \H v z ( f ) = k , so that a modular form in M ! k (4) has a weighted sum of exactly k zeros and poles.We next define several useful modular forms of level 4. Let q = e πiz , as usual. It is wellknown that θ ( z ) ∈ M (4), where θ ( z ) = ∞ (cid:88) n = −∞ q n = 1 + 2 q + 2 q + 2 q + 2 q + . . . EROS OF WEAKLY HOLOMORPHIC MODULAR FORMS OF LEVEL 4 3 is the classical theta function. We note that θ ( z ) has its single zero at the cusp at . Wealso define F ( z ) = (cid:88) odd n> σ ( n ) q n = q + 4 q + 6 q + 8 q + 13 q + . . . = − E ( z ) + 3 E (2 z ) − E (4 z )24 ∈ M (Γ (4)) , where E ( z ) is the classical Eisenstein series of weight 2. The valence formula shows that F has a simple zero at ∞ , and vanishes nowhere else in the fundamental domain. With thesefunctions, we can define a Hauptmodul for Γ (4) by ψ ( z ) = θ ( z ) F ( z ) = q − + 8 + 20 q − q + 216 q − q + 1636 q + . . . ∈ M (cid:93) (4) . The function ψ ( z ) has a pole at ∞ and vanishes at the cusp at . We can define similarweight 0 Hauptmoduln by modifying the constant term; specifically, we define ψ ∞ ( z ) = ψ ( z ) − q − + 20 q − q + . . . , which has zero constant term and vanishes at − + i , and ψ ( z ) = ψ ( z ) −
16 = q − − q − q + . . . , which vanishes at the cusp at 0.We now define a canonical basis for M (cid:93)k (Γ (4)) similar to the bases defined for levels 1 , k = 2 (cid:96) , the elements f (4) k,m ( z ) of this basis have a Fourierexpansion of the form f (4) k,m ( z ) = q − m + ∞ (cid:88) n = (cid:96) +1 a (4) k ( m, n ) q n , where m ≥ − (cid:96) is the order of the pole at infinity. These basis elements can be constructedrecursively: the first basis element f (4) k, − (cid:96) ( z ) is given by F (cid:96) ( z ), and f (4) k,m ( z ) is constructedfrom f (4) k,m − ( z ) by multiplying by ψ ( z ) and subtracting off integral linear combinations ofpreviously constructed basis elements f (4) k,n ( z ) with n < m to obtain the appropriate gap in theFourier expansion. In general, these basis elements take the form f (4) k,m ( z ) = F (cid:96) ( z ) · P ( ψ ( z )),where P ( x ) ∈ Z [ x ] is a generalized Faber polynomial of degree (cid:96) + m (see [7, 8]). From thevalence formula, since f (4) k,m has a pole of order m at ∞ and no other poles, it must have m + (cid:96) zeros elsewhere in the fundamental domain or at the cusps 0 and . Since ψ givesa bijection between the fundamental domain and the Riemann sphere, these zeros all lie onthe bottom arc of the fundamental domain if and only if the zeros of the polynomial P ( x )are real and lie in the interval [0 , Theorem 1.
Let f (4) k,m ( z ) be defined as above. If (cid:96) ≥ and m ≥ (cid:96) + 16 , or if (cid:96) < and m ≥ | (cid:96) | + 16 , then at least (cid:106) √ m + k (cid:107) of the m + k nontrivial zeros of f (4) k,m ( z ) in thefundamental domain for Γ (4) lie on the lower boundary of the fundamental domain. ANDREW HADDOCK AND PAUL JENKINS
We note that this bound is not sharp. Computations of Faber polynomials show that mostof these basis elements appear to have more zeros on this arc. However, in some cases, wedo not expect all of the zeros to be on the arc. For instance, if the Faber polynomial P [ x ] isof very small degree, then for large enough (cid:96) its zeros may be computed to be outside of theinterval [0 , M (cid:93)k (Γ (4)) consisting of forms that vanishat the cusps 0 and . We denote the basis elements as g (4) k,m ( z ) = q − m + ∞ (cid:88) n = (cid:96) − b (4) k ( m, n ) q n for m ≥ − (cid:96) + 2. The g (4) k,m ( z ) can be constructed in a manner similar to the f (4) k,m ( z ); thefirst basis element is given by F (cid:96) ( z ) ψ ( z ) ψ ( z ), and they have the general form g (4) k,m ( z ) = F (cid:96) ( z ) ψ ( z ) ψ ( z ) Q ( ψ ( z )), where Q ( x ) ∈ Z [ x ] is a Faber polynomial of degree (cid:96) + m −
2. Theyhave a correspondingly smaller gap in their Fourier expansions. The proof of Theorem 1 canbe modified to show that the majority of the zeros of the g (4) k,m ( z ) in a fundamental domainfor Γ (4) also lie on the lower boundary.It turns out that the f (4) k,m and the g (4) k,m are closely related and share several interestingproperties. For instance, in examining the Fourier expansions of several basis elements ofweight 6, we see that f (4)6 , − ( z ) = q + 12 q + 66 q + 232 q + 627 q + 1452 q + . . .f (4)6 , − ( z ) = q + 32 q + 244 q + 1024 q + 3126 q + 7808 q + 16808 q + . . .f (4)6 , − ( z ) = q + 198 q + 704 q + 2685 q + 8064 q + 17006 q + . . .f (4)6 , ( z ) = 1 − q − q − q + . . . , while the corresponding basis elements in weight − g (4) − , ( z ) = q − − q − + 504 − q + 40996 q − q + . . .g (4) − , ( z ) = q − − q − − q − + 7032 q − q + 1017684 q + . . .g (4) − , ( z ) = q − − q − + 88902 q − q + 22573848 q + . . .g (4) − , ( z ) = q − − q − − q − − q + 1947264 q − q + . . . In these examples, the basis elements have Fourier expansions with exponents that are alleven or all odd, and the coefficient of q n in f (4) k,m ( z ) is the negative of the coefficient of q m in g (4)2 − k,n . These properties hold in general, and we will prove the following theorems. Theorem 2.
Let f (4) k,m ( z ) and g (4) k,m ( z ) be defined as above. Then for all integers k, m, n with k even, we have the duality of coefficients a (4) k ( m, n ) = − b (4)2 − k ( n, m ) . Theorem 3.
Let f (4) k,m ( z ) be defined as above. If n (cid:54)≡ m (mod 2) , then a (4) k ( m, n ) = 0 . EROS OF WEAKLY HOLOMORPHIC MODULAR FORMS OF LEVEL 4 5
Theorem 2 establishes the duality of Fourier coefficients between the basis elements f (4) k,m of weight k and the basis elements g (4)2 − k,m of weight 2 − k , while Theorem 3 shows thatthe exponents appearing in the Fourier expansion of f (4) k,m ( z ) are either all even or all odd.The duality results are analogous to similar dualities first proved in [5] for integral weightmodular forms of level 1 and in [19] for half integral weight modular forms of level 4.The remainder of the paper proceeds as follows. In section 3 we give a generating functionfor the f (4) k,m ( z ) and prove Theorems 2 and 3. In section 4 we use this generating functionto approximate the f (4) k,m by trigonometric functions along the lower boundary arc of thefundamental domain and prove Theorem 1. In section 5 we give details of the computationsused in bounding the error term of this approximation.3. The Generating Function and Coefficients
In this section, we prove Theorem 2 and give a generating function for the basis elements f (4) k,m ( z ) and g (4) k,m ( z ). We also prove Theorem 3. We rely heavily on a theorem of El-Guindy [6],which gives duality results and generating functions for modular forms in any level for whichthere exist modular forms of weights 0 and 2 with properties similar to that of the f (4)0 ,m ( z )and the g (4)2 ,m ( z ). Ahlgren [1] proved that the levels given by the genus zero primes 2 , , , , p for which the modular curve X ( p ) is hyperelliptic.We first note that the constant term of G ( z ) = f (4) k,m ( z ) g (4)2 − k,n ( z ) = ( q − m + O ( q (cid:96) +1 ))( q − n + O ( q − (cid:96) ) is given by a (4) k ( m, n ) + b (4)2 − k ( n, m ). This form G ( z ) must be a modular form of weight2 that vanishes at both 0 and and has poles only at ∞ . By the valence formula, sucha form must have at least one pole, so the subspace of M (cid:93) (4) consisting of such forms isgenerated by the derivatives q ddq f (4)0 ,m ( z ) for m ≥
1. Thus, the constant term of G ( z ) must be0, and we obtain the duality a (4) k ( m, n ) = − b (4)2 − k ( n, m ), proving Theorem 2.To obtain a generating function for weight k = 0, we follow the proof of [6, Theorem 1.1],replacing El-Guindy’s ψ (cid:96),n with the modular form f (4)0 ,(cid:96) and his Φ (cid:96) with g (4)2 , . The reader mayverify that these forms satisfy the appropriate conditions for the argument. This gives usthe generating function (cid:88) m ≥ f (4)0 ,m ( τ ) q m = g (4)2 , ( z ) f (4)0 , ( z ) − f (4)0 , ( τ ) . By the duality above, this sum is also equal to − (cid:80) g (4)2 ,n ( z ) e πinτ . Note that the denominator f (4)0 , ( z ) − f (4)0 , ( τ ) is equal to ψ ( z ) − ψ ( τ ) = ψ ( z ) − ψ ( τ ) = ψ ∞ ( z ) − ψ ∞ ( τ ) . With these generating functions for weights 0 and 2, we see the conditions of [6, Theorem1.2] are satisfied for level 4, and we apply this theorem to obtain generating functions for allof the f (4) k,m ( z ) and g (4) k,m ( z ) by setting F ( z ) = f (4) k, − (cid:96) ( z ) for each even weight k . Specifically, weobtain the formula(3.1) (cid:88) m ≥− (cid:96) f (4) k,m ( τ ) q m = f (4) k, − (cid:96) ( τ ) g (4)2 − k,(cid:96) +1 ( z ) f (4)0 , ( z ) − f (4)0 , ( τ ) = − (cid:88) m ≥ (cid:96) +1 g (4) k,m ( z ) e πimτ . ANDREW HADDOCK AND PAUL JENKINS
This generating function and duality theorem are analogous to theorems in [5] for level1, which showed that the f (1) k,m are dual to the f (1)2 − k,n , and in [9] for genus zero prime levels N , which showed that the f ( N ) k,m and the g ( N )2 − k,n are dual. Because Γ (4) has three cusps,we find additional modular forms in M (cid:93)k (4) which satisfy similar properties. Let h (4) k,m ( z ) bethe unique modular form in M (cid:93)k (4) that vanishes at the cusp 0 and has Fourier expansionbeginning q − m + O ( q (cid:96) ), and let i (4) k,m ( z ) be the unique modular form in M (cid:93)k (4) that vanishesat the cusp and has Fourier expansion beginning q − m + O ( q (cid:96) ). These forms may also begenerated recursively in much the same way as the f (4) k,m . El-Guindy’s Theorem 1.2, with F ( z ) = h (4) k, − (cid:96) +1 ( z ), now gives a generating function (cid:88) m ≥− (cid:96) +1 h (4) k,m ( τ ) q m = h (4) k, − (cid:96) +1 ( τ ) i (4)2 − k, − (cid:96) ( z ) h (4)0 , ( z ) − h (4)0 , ( τ ) . Note that the denominator of the generating function has not changed, since h (4)0 , = ψ , andthat a similar duality theorem between the coefficients of the h (4) k,m and the i (4)2 − k,n may beobtained by a similar argument as above or via El-Guindy’s theorem. Previously, Atkinson [3]obtained this generating function in level 4 for the the h (4)2 ,m ( z ) and the i (4)0 ,m ( z ), but this isnot quite sufficient to apply El-Guindy’s theorem and obtain generating functions for allweights.To prove Theorem 3, we recall the definition of several operators. For a matrix γ = ( a bc d ) ∈ GL +2 ( Q ) and a modular form f of weight k , we have the usual slash operator( f | [ γ ])( z ) = (det γ ) k − ( cz + d ) − k f (cid:18) az + bcz + d (cid:19) . Note that f | [ γ ] | [ γ ] = f | [ γ γ ] and that for a modular form of level N , we have f | [ γ ] = f when γ ∈ Γ ( N ). With this notation, the standard U p and V p operators on modular formscan be written as U p f ( z ) = p − (cid:88) i =0 f | (cid:2)(cid:0) i p (cid:1)(cid:3) ,V p f ( z ) = p − k f | (cid:2)(cid:0) p
00 1 (cid:1)(cid:3) . Recall (see [2] for details) that U p maps a modular form f ( z ) = (cid:80) a ( n ) q n ∈ M ! k ( N ) tothe modular form (cid:80) a ( pn ) q n , which is in M ! k ( N p ) if p (cid:45) N , in M ! k ( N ) if p | N , and in M ! k ( Np )if p | N . The operator V p maps f ( z ) = (cid:80) a ( n ) q n ∈ M ! k ( N ) to (cid:80) a ( n ) q pn ∈ M ! k ( N p ). Thecombination U p V p is the identity map, while the operator V p U p sends the form (cid:80) a ( n ) q n tothe form (cid:80) a ( pn ) q pn , retaining only the terms whose exponents are 0 (mod p ).We next prove that V U preserves the space M (cid:93)k (4). Suppose that a modular form f ∈ M ! k (4) is in the subspace M (cid:93)k (4), so that f | [ γ ] is holomorphic at ∞ for all γ ∈ SL ( Z ) \ Γ (4).Since V U f ( z ) = ( f ( z )+ f ( z + )), we need only show that f | [( )] | [ γ ] is holomorphic at ∞ for γ ∈ SL ( Z ) \ Γ (4). Since this can be written as f | [ α ] | [( ∗ ∗ ∗ )] for some α ∈ SL ( Z ) \ Γ (4),the result follows.Note that Theorem 3 can be restated as V U f (4) k,m ( z ) = 0 if m is odd, EROS OF WEAKLY HOLOMORPHIC MODULAR FORMS OF LEVEL 4 7 V U f (4) k,m ( z ) = f (4) k,m ( z ) if m is even,since V U f (4) k,m ( z ) must be an element of M (cid:93)k (4) and contains only the terms with evenexponent in the Fourier expansion of f (4) k,m ( z ). If k ≥ m is odd, then examining theFourier expansion of V U f (4) k,m ( z ) shows that it must be holomorphic. Since it vanishes at ∞ to order at least (cid:96) + 1, it must be identically zero. If k ≥ m is even, then taking thedifference V U f (4) k,m ( z ) − f (4) k,m ( z ) cancels the pole at ∞ , giving a holomorphic form in M k (4)vanishing at ∞ to order greater than (cid:96) , which must thus be zero. For k <
0, the forms F − (cid:96) ( z ) · V U f (4) k,m ( z ) and F − (cid:96) ( z ) · ( V U f (4) k,m ( z ) − f (4) k,m ( z )) are holomorphic modular forms in M (4) which vanish at ∞ and must therefore be identically zero. This proves Theorem 3.4. Integrating the Generating Function
The proof of Theorem 1 follows that of the main theorem of [9], which proves a similarresult for levels 2 and 3. We integrate the generating function for the basis elements f (4) k,m ( z )to approximate them by a real-valued trigonometric function along the lower arc of thefundamental domain. This approximation is good enough that to prove that most of thezeros must lie along this arc.We first write the generating function (3.1), with r = e πiτ and q = e πiz , as ∞ (cid:88) m = − (cid:96) f k,m ( z ) r m = F (cid:96) ( z ) F (cid:96) ( τ ) g (4)2 , ( τ ) f (4)0 , ( τ ) − f (4)0 , ( z ) . We multiply by r − m − and integrate around r = 0 to obtain2 πif (4) k,m ( z ) = (cid:73) F (cid:96) ( z ) F (cid:96) ( τ ) g (4)2 , ( τ ) f (4)0 , ( τ ) − f (4)0 , ( z ) r − m − dr. Changing variables from r to τ , we see that2 πif (4) k,m ( z ) = (cid:90) + iA − + iA F (cid:96) ( z ) F (cid:96) ( τ ) g (4)2 , ( τ ) f (4)0 , ( τ ) − f (4)0 , ( z ) e − πimτ πidτ, where A ≥ is a real number.We write g (4)2 , ( τ ) as a derivative, since it can be checked that that ddτ f (4)0 , ( τ ) = − πig (4)2 , ( τ ).We therefore simplify and obtain f (4) k,m ( z ) = (cid:90) + iA − + iA F (cid:96) ( z ) F (cid:96) ( τ ) ddτ ( f (4)0 , ( τ ) − f (4)0 , ( z )) f (4)0 , ( τ ) − f (4)0 , ( z ) e − πimτ − πi dτ. We now assume that z is on the left semicircular border of our fundamental domain, so that z = − + e iθ for some 0 ≤ θ ≤ π . We move the contour of integration downward, noting thatwe pick up a term of − πi times the residue of the integrand exactly when f (4)0 , ( τ ) − f (4)0 , ( z )is zero, and that this expression has a simple zero exactly when τ is equivalent to z underthe action of Γ (4). The first residues occur when τ = z and when τ = z z +1 . In computingthe residues, we note that the integral of the logarithmic derivative of f (4)0 , ( τ ) − f (4)0 , ( z ) will ANDREW HADDOCK AND PAUL JENKINS give a factor of 1, which is multiplied by the remaining terms in our integral to give residuesof F (cid:96) ( z ) F (cid:96) ( γz ) e − πin ( γz ) − πi , where γ = ( ) or ( ).Since F ∈ M (4), we know that F (cid:96) ( γz ) = ( cz + d ) k F (cid:96) ( z ), and the residue becomes1 − πi ( cz + d ) − k e − πin ( γz ) . Substituting in both values of γ , we obtain f k,m ( z ) − e − πimz − (4 z + 1) − k e − πim ( z z +1 ) = (cid:90) C F (cid:96) ( z ) F (cid:96) ( τ ) ddτ ( f , ( τ ) − f , ( z )) f , ( τ ) − f , ( z ) e − πimτ − πi dτ. Since z = − + e iθ , we find that − e − πimz − (4 z + 1) − k e − πim ( z z +1 ) is equal to e πm sin θ e − ikθ (cid:18) kθ πm − πm θ (cid:19) , and, multiplying by e ikθ e − πm sin θ , we obtain(4.1) e ikθ e − πm sin θ f k,m (cid:18) −
14 + 14 e iθ (cid:19) − (cid:18) kθ πm − πm θ (cid:19) = e ikθ e − πm sin θ (cid:90) C F (cid:96) ( z ) F (cid:96) ( τ ) g , ( τ ) f , ( τ ) − f , ( z ) e − πimτ dτ. Let f ( z ) = (cid:80) c ( n ) e πiz be a Fourier series with real coefficients c ( n ). It is clear that f ( a + bi ) = (cid:80) c ( n ) e πi ( a + bi ) = (cid:80) c ( n ) e − πia e − πb = (cid:80) c ( n ) e πi ( − a + bi ) = f ( − a + bi ), so that f ( z ) = f ( − z ). When z = − + e iθ for 0 < θ < π , so that z is on the bottom boundaryof our fundamental domain, then ( ) z = z z +1 = − + e iθ − e iθ +1 = − e − iθ = − z . If f is amodular form for Γ (4), it follows that f ( z ) = (4 z + 1) − k f ( z z +1 ) = e − ikθ f ( z ). Multiplying by e ikθ , we find that e ikθ f ( z ) = e − ikθ f ( z ) = e ikθ f ( z ). Thus, for any f ∈ M ! k (4), the normalizedmodular form e ikθ f ( z ) is real-valued on the lower boundary of this fundamental domain.By the argument above, we know that the left-hand side of (4.1) is a real-valued functionof θ for 0 ≤ θ ≤ π . When the argument inside the cosine function takes on a value of mπ for m ∈ Z , the cosine term will be ±
2. Since this argument ranges from 0 at θ = 0 to kπ + πm at θ = π , the cosine term is ± m + 1 + k times. Thus, if we can bound the right-handside of (4.1) in absolute value by 2, then by the Intermediate Value Theorem there must beat least m + k zeros of the modular form f (4) k,m ( z ) on this arc.The valence formula predicts that f (4) k,m ( z ) has exactly k zeros in the fundamental domain;the pole of order m at ∞ means that k + m zeros remain. These must be simple and lieon the lower boundary if the weighted modular form is close enough to the cosine function.Unfortunately, it is difficult to move the contour down far enough to capture all of thezeros on the bottom arc without picking up additional residues or increasing the difficultyof bounding the integral. We settle for showing that the majority of the zeros are on the arcby picking a contour of a fixed height that is low enough to capture most zeros, but not lowenough to pick up extra images of z under the action of Γ (4). EROS OF WEAKLY HOLOMORPHIC MODULAR FORMS OF LEVEL 4 9
The maximum height of the other images of the bottom arc of the fundamental domainunder Γ (4) is . We pick the contour τ = u + i , which has a height of . Fixing theheight of the contour for τ means that we must also limit z so that it does not cross belowthat height, giving a zero in the denominator. We limit z = − + e iθ by picking θ ∈ [ π , π ],where we find that Im( z ) = sin θ ≥ √ > . > for θ in the given interval. Thisrestriction on θ also determines the number of zeros we can prove are on the arc. When θ = π , the term in the cosine function takes on the value kπ + πm − πm √ ; when θ = π , thisterm takes on the value kπ + πm + πm √ . We take the difference of these terms and findthat (cid:98) k + m √ (cid:99) zeros must lie on the arc. Therefore, bounding the integral by 2 will proveTheorem 1.To obtain this bound, we note that in absolute value, the right-hand side of (4.1) is e − πm (sin θ − ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) − F (cid:96) ( − + e iθ ) F (cid:96) ( u + i ) g , ( u + i ) f , ( u + i ) − f , ( − + e iθ ) e − πimu du (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . It is clear that this is bounded above by e − πm (sin θ − ) (cid:90) − (cid:12)(cid:12)(cid:12)(cid:12) F ( − + e iθ ) F ( u + i ) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) (cid:12)(cid:12)(cid:12)(cid:12) g , ( u + i ) f , ( u + i ) − f , ( − + e iθ ) (cid:12)(cid:12)(cid:12)(cid:12) du. The dependence on m in the integral has been removed.Unfortunately, the quotient of F s takes on values greater than 1, so a naive bound, re-placing each term with its maximum value over the appropriate ranges of u and θ , givesexponential growth in (cid:96) . However, for θ ∈ [ π , π ], the term outside the integral gives expo-nential decay in m , so for m large enough with respect to (cid:96) , this decay dominates and theintegral must be less than 2. All that remains is to bound the integral and determine theconditions for the size of m with respect to (cid:96) .We rewrite our basis elements in terms of F , θ , and ψ , noting that τ = u + i for − ≤ u ≤ and z = − + e iθ for θ ∈ [ π , π ]. We also drop the subscript on ψ to makethe notation less cumbersome, and find that our integral is bounded above by e − πm (sin θ − ) (cid:90) − (cid:12)(cid:12)(cid:12)(cid:12) F ( z ) F ( τ ) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) (cid:12)(cid:12) θ ( τ ) − F ( τ ) (cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ψ ( τ ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) du. Thus, we must find upper and lower bounds for F ( z ) and F ( τ ), as well as upper bounds forthe remaining terms.We find the following inequalities to hold for θ ∈ [ π , π ] and − ≤ u ≤ :0 . ≤ | F ( z ) | ≤ . , . ≤ | F ( τ ) | ≤ . ,θ ( τ ) ≤ . ,θ ( z ) ≤ . . Therefore, we have that | θ ( τ ) − F ( τ ) | ≤ . , (cid:12)(cid:12)(cid:12)(cid:12) F ( z ) F ( τ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ . , (cid:12)(cid:12)(cid:12)(cid:12) F ( τ ) F ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ . . Additionally, it is true that e − π (sin θ − ) ≤ . , and that the quotient of ψ -functions can contribute at most 80 . (cid:96) ≥ e − πm (sin θ − ) (cid:90) − (cid:12)(cid:12)(cid:12)(cid:12) F ( z ) F ( τ ) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) (cid:12)(cid:12) θ ( τ ) − F ( τ ) (cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ψ ( τ ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) du ≤ (0 . m (4 . (cid:96) (50 . . . We compute that (0 . m (50 . . ≤ m ≥
16, and that (0 . m (4 . ≤ m ≥
4. Therefore, when (cid:96) ≥
0, if m ≥ (cid:96) + 16, it follows that the integral is boundedby 2 and Theorem 1 is true. When (cid:96) <
0, we just switch (cid:12)(cid:12)(cid:12) F ( z ) F ( τ ) (cid:12)(cid:12)(cid:12) (cid:96) for (cid:12)(cid:12)(cid:12) F ( τ ) F ( z ) (cid:12)(cid:12)(cid:12) | (cid:96) | ≤ . | (cid:96) | . Theresult holds for m ≥ | (cid:96) | + 16, proving Theorem 1.We note that this argument may be easily modified to prove that the majority of the zerosof the g (4) k,m also lie on the lower boundary of the fundamental domain, as the generatingfunction is almost identical due to (3.1).5. Computations
In this section, we justify the numerical bounds given in section 4, using ideas from [9].All computations were verified using Sage. In this section, we write a modular form G ( z ) = (cid:80) a ( n ) q n as ˜ G ( z ) + RG ( z ), where ˜ G ( z ) = (cid:80) n ≤ N a ( n ) q n is the truncation of G ( z ) (we take N = 50 throughout this section) and RG ( z ) is the tail of the series.For θ ∈ [ π , π ], the maximum value of | q | is t = e − π √ . For modular forms in τ = u + i ,we have | e πiτ | = s = e − π .Recall that F ( z ) = (cid:80) odd n> σ ( n ) q n . It is clear that σ ( n ) ≤ n + n ≤ n + n for integers n ≥
1. Using this with derivatives of the geometric series, we can bound | F ( z ) | as follows. | F ( z ) | ≤ ∞ (cid:88) odd n> | σ ( n ) || q n | ≤ ˜ F ( z ) + ∞ (cid:88) odd n = N +1 ( n + n ) t n ≤ ˜ F ( z ) + 2 t (1 − t ) − N (cid:88) n =1 ( n + n ) t n . We can directly evaluate this last term. Letting N = 50, we compute an upper bound for | RF ( z ) | . | RF ( z ) | ≤ t (1 − t ) − (cid:88) n =1 ( n + n ) t n ≤ . · − EROS OF WEAKLY HOLOMORPHIC MODULAR FORMS OF LEVEL 4 11
An upper bound for ˜ F ( z ) is given by the calculation | ˜ F ( z ) | ≤ (cid:80) n> σ ( n ) t n ≤ . | F ( z ) | ≤ . θ .An upper bound for | F ( τ ) | is calculated similarly, replacing | q | with | r | = s = e − π toobtain | RF ( τ ) | ≤ . · − and | F ( τ ) | ≤ . . We will also need to find lower bounds for | F ( z ) | and | F ( τ ) | . We do so by finding an upperbound for the absolute values of the derivatives of F ( z ) = F ( − + e iθ ) with respect to θ and F ( τ ) = F ( u + i ) with respect to τ . We then calculate values of ˜ F at equally spacedpoints in the regions θ ∈ [ π , π ] , u ∈ [ − , ] and use the bounds on the derivatives and thetails to compute a lower bound for the value of the function.The derivative of F ( z ) with respect to θ is given by ddθ F ( z ) = ddθ (cid:32) ∞ (cid:88) odd n> σ ( n ) e πi ( − + e iθ ) (cid:33) = ∞ (cid:88) odd n> − πn σ ( n ) e πni (cos θ − − πn sin θ (cos θ + i sin θ )Taking absolute values and again using derivatives of the geometric series, we find that (cid:12)(cid:12)(cid:12)(cid:12) ddθ F ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ π (cid:32) (cid:88) odd n> nσ ( n ) t n + 2 t (1 − t ) + 6 t (1 − t ) − (cid:88) n =1 ( n + n ) t n (cid:33) , which we calculate yields (cid:12)(cid:12)(cid:12)(cid:12) ddθ F ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ . . If we evaluate | F ( z ) | at the points θ = π + n for 0 ≤ n ≤ π , the spacing between thepoints is small enough that on the entire interval, | F ( z ) | cannot be below 1 . · = 0 . | ˜ F ( z ) | on these pointsis at least 0 . | RF ( z ) | is negligible, so the smallest that | F ( z ) | can be on π ≤ θ ≤ π is 0 . F ( τ ) = F ( u + i ), where − ≤ u ≤ . We calculatethe derivative of F with respect to τ , arriving at ddτ F ( τ ) = ddτ (cid:32) (cid:88) odd n> σ ( n ) e πinτ (cid:33) = (cid:88) odd n> πinσ ( n ) e πinτ . We use the same derivatives of geometric series to bound the tail, and obtain the bound (cid:12)(cid:12)(cid:12)(cid:12) ddτ F ( τ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ π (cid:32) (cid:88) odd n> nσ ( n ) s n + 2 s (1 − s ) + 6 s (1 − s ) − (cid:88) n =1 ( n + n ) s n (cid:33) ≤ . . For integers n ∈ [0 , (cid:12)(cid:12)(cid:12) ˜ F ( u + i ) (cid:12)(cid:12)(cid:12) at the points u = − + n ,and find that the minimum value is 0 . | RF ( τ ) | is negligible, and sincethe derivative is at most 31 .
26, the function decreases by at most 31 . · = 0 . | F ( τ ) | ≥ . − . . > . θ ( z ) and θ ( τ ), we note that the coefficients of θ are againmultiples of sigma functions, so we use similar bounding techniques as with F to computethat | θ ( τ ) | ≤ . , | θ ( z ) | ≤ . τ and z .These computations give us easy bounds for the first two terms in our integral; it is moredifficult to bound the Hauptmodul quotient. We can bound Rψ ( τ ) and Rψ ( z ) by computingthat Rψ ( τ ) = ψ ( τ ) − ˜ ψ ( τ ) = θ ( τ ) F ( τ ) − ˜ ψ ( τ )= (cid:32) ˜ ψ ( τ ) + θ ( τ ) − ˜ ψ ( τ ) ˜ F ( τ )˜ F ( τ ) (cid:33) ˜ F ( τ ) F ( τ ) − ˜ ψ ( τ ) ≤ | ˜ ψ ( τ ) | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ F ( τ ) F ( τ ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + | ˜ θ ( τ ) − ˜ ψ ( τ ) ˜ F ( τ ) | + | Rθ ( τ ) || F ( τ ) |≤ | ˜ ψ ( τ ) | (cid:12)(cid:12)(cid:12)(cid:12) RF ( τ ) F ( τ ) (cid:12)(cid:12)(cid:12)(cid:12) + | ˜ θ ( τ ) − ˜ ψ ( τ ) ˜ F ( τ ) | + | Rθ ( τ ) || F ( τ ) | . We compute explicit bounds on all of these terms, and find that Rψ ( τ ) ≤ (62 . · (6 . · − ) + 2 . · − + 1 . · − . ≤ . · − . Similarly, Rψ ( z ) ≤ . · − . We also bound the derivatives of the truncations of ψ ( z ) and the real and imaginary partsof ψ ( τ ). Doing so allows us to evaluate the functions at equally spaced points as before toget maximum and minimum values for ψ ( z ) and the real and imaginary parts of ψ ( τ ).We take the derivative of ˜ ψ ( τ ) with respect to u , and for both the real and imaginaryparts we achieve a bound of (cid:12)(cid:12)(cid:12)(cid:12) ddu Re( ˜ ψ ( τ )) (cid:12)(cid:12)(cid:12)(cid:12) , (cid:12)(cid:12)(cid:12)(cid:12) ddu Im( ˜ ψ ( τ )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:88) n = − πn | a (1 , n ) | s n ≤ . . The bound on the derivative of ˜ ψ ( z ) with respect to θ is quite manageable as well; we have (cid:12)(cid:12)(cid:12)(cid:12) ddθ ˜ ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:88) n = − | a (1 , n ) | nπ t n ≤ . . With this, we find that 0 . ≤ ψ ( z ) ≤ . ψ ( z ) is real-valued on the lower boundary of the fundamental domain. EROS OF WEAKLY HOLOMORPHIC MODULAR FORMS OF LEVEL 4 13
We next need to bound (cid:12)(cid:12)(cid:12)(cid:12) ψ ( τ ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ψ ( z ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) ψ ( z ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) . We let D ( z, τ ) = ψ ( z ) ψ ( τ ) − ψ ( z ) = ψ ( z )Re( ψ ( τ )) − ψ ( z ) + i Im( ψ ( τ )) , noting that its numerator is real-valued. We will bound the numerator and denominatorseparately, using the bounds on Rψ ( τ ) and Rψ ( z ) above, and find maximum values of | D ( z, τ ) | for values of u on each of several subintervals of [ − , ] to bound the originalintegral. Computations are made easier by noting thatIm (cid:18) ψ (cid:18) u + i (cid:19)(cid:19) = − Im (cid:18) ψ (cid:18) − u + i (cid:19)(cid:19) . Using the bound on Rψ ( τ ), we find that for u ∈ [ − . , − . ψ ( τ )) <
0, Re( ψ ( τ )) >
32, or | Im( ψ ( τ )) | >
16. All of these imply thatin this interval, | D ( z, τ ) | <
1. Thus, over this interval, the Hauptmodul quotient is less than2, and we have a bound of 2 · (cid:82) − . − . du ≤ . u ∈ [ − . , − . | Re( ψ ( τ )) | ≤ . ψ ( τ ) here is quite small, so we consider it to be 0. Asits absolute value is ≥
0, it actually increases our minimum value calculations if consid-ered. We calculate the minimum value of Re( ψ ( τ )) − ψ ( z ) to be 0 . ψ ( τ )) < ψ ( z ) for all values of τ and z , so we take the maximum of Re( ψ ( τ )) and theminimum of ψ ( z ) to get a minimum difference. These bounds then give us (cid:90) − . − . (cid:12)(cid:12)(cid:12)(cid:12) ψ ( τ ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) du ≤ (cid:90) − . − . (cid:18) . . (cid:19) du ≤ ( . − . (cid:18) . . (cid:19) ≤ . . Multiplying this quantity by 2 to include the interval [ . , . . u ∈ [ − . , . ≤ | Re( ψ ( τ )) | . We proceed similarly,and compute that | Re( ψ ( τ )) − ψ ( z ) | ≥ . (cid:90) . − . (cid:12)(cid:12)(cid:12)(cid:12) ψ ( τ ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) du ≤ · (cid:90) − . (cid:18) . . (cid:19) du ≤ . . Over the entire interval, the Hauptmodul fraction gives a maximum value of (cid:90) . − . (cid:12)(cid:12)(cid:12)(cid:12) ψ ( τ ) ψ ( τ ) − ψ ( z ) (cid:12)(cid:12)(cid:12)(cid:12) du ≤ . . . . , completing our proof of Theorem 1. References
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Department of Mathematics, Brigham Young University, Provo, UT 84602
E-mail address : [email protected] Department of Mathematics, Brigham Young University, Provo, UT 84602
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