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Given a character $\chi$ and $k$ odd how can one compute a basis for the space of modular forms $M_\frac{k}{2}(\Gamma_0(4),\chi)$. By compute a basis I mean, finding the beginning of the Fourier expansions. I am looking for computer programs, which can do that for me.

I have heard of the package SAGE, which seems to do the job for integral weight modular forms. There is even the function http://www.sagemath.org/doc/reference/sage/modular/modform/half_integral.html but the examples all have q-expansions starting with q, so I guess this is not really a basis for the space of all modular forms but only cusp forms. MAGMA does not seem to include this functionality, either.

So, are there any packages which can do this? Since I have not found a package, I have some doubts that there is really an algorithm working in general. If there is no algorithm known to handle this, what methods are available in order to compute a basis "by hand"?

Thanks.

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Did you look at Cohen-Oesterle? I think it has explicit formulae for the Eisenstein series, or at least their dimension, but I'm not at work to check. –  Kevin Buzzard Apr 7 '10 at 18:23
    
@Kevin: I suppose you mean "Dimensions des espaces de formes modulaires". There are no explicit formulae for Eisenstein series nor are there any proofs. Or is there another paper by both authors which I am not aware of. –  wood Apr 7 '10 at 23:17
    
@wood: I did mean the latter paper. My memory was that they used Riemann-Roch to compute both the dimension of the spaces of modular forms and of cusp forms (the point being that that they work in both integer and half-integer weight); perhaps I was over-optimistic in hoping that they gave an explanation of the difference! –  Kevin Buzzard Apr 8 '10 at 7:13
    
@Kevin: I asked Cohen about the lack of proofs in their paper, and he told me that both he and Oesterle had come up with the same formulas independently by different methods, so published a paper together with no proofs. Cohen suggested that anybody who could follow his proof could probably come up with it just as easily. I complained to Jordi Quer about this and he wrote journals.impan.gov.pl/aa/Inf/145-4-4.html, which does only integral weight. –  William Stein Feb 8 '11 at 19:53
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4 Answers

up vote 7 down vote accepted

Edit: Here's a rather silly method that should work if SAGE is just giving you cusp forms: $\Gamma_0(4)$ has a single normalized cusp form of weight 6, given by $\eta(2\tau)^{12} = q - 12q^3 + 54q^5 - \dots$, so take your basis of cusp forms of weight $k/2 + 6$, and divide each element by this form to get a basis of the space of modular forms of weight $k/2$.

Edit in response to Buzzard: Thanks for pointing out that I should make this argument. Here is a proof that the cusp form has minimal vanishing at all cusps. $\Gamma_0(4)$ is conjugate to $\Gamma(2)$ by $\tau \mapsto 2\tau$, so it suffices to check that $\Delta(\tau)$, the square of $\eta(\tau)^{12}$, vanishes to twice the minimum order at each cusp of $\Gamma(2)$. The quotient $\Gamma(1)/\Gamma(2) \cong S_3$ acts transitively on the cusps of $X(2)$ with stabilizers of order 2, so the quotient map to $X(1)$ has ramification degree 2 at each cusp. $\Delta(\tau)$ is invariant under the weight 12 action of $\Gamma(1)$, and $\Delta(\tau)$ has minimal vanishing at infinity on $X(1)$.

Old answer: If you have a cusp form of weight $k/2$ for $\Gamma_0(4)$ (e.g., given to you by SAGE), you can multiply it by the modular function $\frac{\eta(\tau)^8}{\eta(4\tau)^8} = q^{-1} - 8 + 20q - 62q^3 + 216q^5 - \dots$ to get a modular form of the same weight, that is nonvanishing at one of the three cusps and vanishing at the other two. If you want a form that is nonzero at one of the other cusps, you can multiply by $\frac{\eta(4\tau)^8}{\eta(\tau)^8}$ (has a pole at zero) or by $\frac{\eta(\tau)^{16}\eta(4\tau)^8}{\eta(2\tau)^{24}}$ (pole at $1/2$). [Constant term $-8$ added Sept. 23, in response to an email correction from Michael Somos.]

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@Scott: for the edited method to work, you need to know that the weight 6 cusp form has a zero of the smallest possible order at each cusp. Are you asserting that this is the case and if so how does one see this easily? –  Kevin Buzzard Apr 7 '10 at 17:32
    
Thank you. This seems very promising. –  wood Apr 7 '10 at 23:21
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"MAGMA does not seem to include this functionality, either."

Basis(HalfIntegralWeightForms(DirichletGroup(4).1^2,11/2));

[ 1 - 88*q^3 - 330*q^4 - 4224*q^7 - 7524*q^8 - 30600*q^11 + O(q^12),

q + 4*q^3 + 56*q^4 + 132*q^5 + 224*q^6 + 512*q^7 + 912*q^8 + 1525*q^9 + 
2752*q^10 + 4044*q^11 + O(q^12),

q^2 + 6*q^3 + 20*q^4 + 56*q^5 + 130*q^6 + 256*q^7 + 472*q^8 + 800*q^9 + 
1266*q^10 + 1970*q^11 + O(q^12)

]

Basis(HalfIntegralWeightForms(DirichletGroup(112).1^2,3/2));

[ 1 + 2*q^16 + 2*q^28 + O(q^30),

q - q^21 + 2*q^29 + O(q^30),

...

]

http://magma.maths.usyd.edu.au/calc

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2  
Your answer would be a bit more helpful if you added something like, "MAGMA in fact has such functionality. Here are some examples." –  S. Carnahan Apr 10 '10 at 16:25
    
Thanks, this is what I was looking for. When I searched MAGMA documentation, if missed that and only found Weight Half Forms WeightOneHalfData(M) : ModFrm -> List thinking that MAGMA was only able to find half integral weight forms of weight exactly 1/2. Thanks again –  wood Apr 12 '10 at 13:44
    
Actually, we can more easily use in MAGMA the command > ModularForms(DirichletGroup(4).1^2,11/2); and this doesn't need the "HalfIntegralWeight" phrase at all. –  Junkie Apr 21 '10 at 12:15
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You might also try Eichler and Zagier's book on the theory of Jacobi forms. For example, they show how to compute half-integer weight mf's of weight k+1/2 and level N from Jacobi forms when k is odd.

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Here is a standard approach: One has the Jacobi $\theta$-function $\sum_{n = -\infty}^{\infty} e^{2 \pi i n^2 \tau}$, which is weight $1/2$ on $\Gamma_1(4)$. Thus multiplication by $\theta$ induces an embedding $M_{k+\frac{1}{2}}(\Gamma_1(4N)) \hookrightarrow M_{k+1}(\Gamma_1(4 N)),$ for any integer $N$. It is not too hard to determine the image: given an element $f$ in $M_{k+1}(\Gamma_1(4 N))$, one must determine if $f/\theta$ is holomorphic in the upper half-plane, and at the cusps. This is just a question of $f$ having zeroes at the location of the zeroes of $\theta$. One can use this condition to compute the dimension of the image, and with more effort one should be able to find an actual basis of the image (although I have never tried to implement this latter step myself, and I don't know how hard it is in practice).

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I think one should add to Emerton's answer that the only zeroes of $\theta$ are at cusps, because one has $$\theta(\tau) = \prod (1-q^{2k})(1+q^{2k-1})^2,$$ with $q=e^{2 \pi i \tau}$. So the method Emerton proposes does not require any transcendental computations. –  David Speyer Apr 6 '10 at 23:08
    
But it will be a pain to do at level $4N$ because you will have to isolate the subspace of $M_{k+1}(\Gamma_1(4N))$ consisting of forms which vanish at all cusps to at least the degree that $\theta$ vanishes. If $N>1$ then the degeneracy maps are typically ramified at the cusps so this will be a pain to do: the problem is that a computer algebra package typically computes modular forms as $q$-expansions at infinity, and seeing what's going on at the middle cusp will be a bit of a pain. –  Kevin Buzzard Apr 7 '10 at 17:34
    
Dear Kevin, I believe you. I've used the algorithm I described to compute dimensions of spaces of half-integral weight forms, but I don't think I've used it to actually compute a basis. On the other hand, I think in his thesis Nick Ramsey used this description, or some variant of it, to $p$-adically interpolate half-integral weight forms. So it has some theoretical value, which might make up for its computational inadequacy! –  Emerton Apr 7 '10 at 18:03
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