Basis for modular forms of half-integral weight. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:06:46Z http://mathoverflow.net/feeds/question/20549 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20549/basis-for-modular-forms-of-half-integral-weight Basis for modular forms of half-integral weight. wood 2010-04-06T21:27:44Z 2010-09-23T07:30:13Z <p>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.</p> <p>I have heard of the package SAGE, which seems to do the job for integral weight modular forms. There is even the function <a href="http://www.sagemath.org/doc/reference/sage/modular/modform/half_integral.html" rel="nofollow">http://www.sagemath.org/doc/reference/sage/modular/modform/half_integral.html</a> 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.</p> <p>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"?</p> <p>Thanks.</p> http://mathoverflow.net/questions/20549/basis-for-modular-forms-of-half-integral-weight/20557#20557 Answer by Emerton for Basis for modular forms of half-integral weight. Emerton 2010-04-06T22:27:38Z 2010-04-06T22:27:38Z <p>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).</p> http://mathoverflow.net/questions/20549/basis-for-modular-forms-of-half-integral-weight/20595#20595 Answer by S. Carnahan for Basis for modular forms of half-integral weight. S. Carnahan 2010-04-07T05:44:37Z 2010-09-23T07:30:13Z <p><b>Edit:</b> Here's a rather silly method that should work if SAGE is just giving you cusp forms: <code>$\Gamma_0(4)$</code> has a single normalized cusp form of weight 6, given by <code>$\eta(2\tau)^{12} = q - 12q^3 + 54q^5 - \dots$</code>, 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 $k/2$.</p> <p><b>Edit in response to Buzzard:</b> Thanks for pointing out that I should make this argument. Here is a proof that the cusp form has minimal vanishing at all cusps. <code>$\Gamma_0(4)$</code> is conjugate to <code>$\Gamma(2)$</code> 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 <code>$\Gamma(2)$</code>. The quotient <code>$\Gamma(1)/\Gamma(2) \cong S_3$</code> 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 $X(1)$.</p> <p><b>Old answer:</b> If you have a cusp form of weight $k/2$ for <code>$\Gamma_0(4)$</code> (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.]</p> http://mathoverflow.net/questions/20549/basis-for-modular-forms-of-half-integral-weight/20659#20659 Answer by Kim Hopkins for Basis for modular forms of half-integral weight. Kim Hopkins 2010-04-07T20:20:33Z 2010-04-07T20:20:33Z <p>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. </p> http://mathoverflow.net/questions/20549/basis-for-modular-forms-of-half-integral-weight/20912#20912 Answer by Junkie for Basis for modular forms of half-integral weight. Junkie 2010-04-10T11:09:31Z 2010-04-10T11:09:31Z <p>"MAGMA does not seem to include this functionality, either."</p> <blockquote> <p>Basis(HalfIntegralWeightForms(DirichletGroup(4).1^2,11/2));</p> </blockquote> <p>[ 1 - 88*q^3 - 330*q^4 - 4224*q^7 - 7524*q^8 - 30600*q^11 + O(q^12),</p> <pre><code>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) </code></pre> <p>]</p> <blockquote> <p>Basis(HalfIntegralWeightForms(DirichletGroup(112).1^2,3/2)); </p> </blockquote> <p>[ 1 + 2*q^16 + 2*q^28 + O(q^30),</p> <pre><code>q - q^21 + 2*q^29 + O(q^30), </code></pre> <p>...</p> <p>]</p> <p><a href="http://magma.maths.usyd.edu.au/calc" rel="nofollow">http://magma.maths.usyd.edu.au/calc</a></p>