Modular Forms of Weight One - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:45:30Z http://mathoverflow.net/feeds/question/61263 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61263/modular-forms-of-weight-one Modular Forms of Weight One Peter Humphries 2011-04-11T05:10:26Z 2013-05-08T17:29:21Z <p>Consider the space of newforms $S^{\mathrm{new}}_k(\Gamma_1(q))$ of weight $k$ and level $q$ for the congruence subgroup $\Gamma_1(q)$ of $\mathrm{SL}_2(\mathbb{Z})$; for simplicity's sake, let's assume that $q$ is prime. Then for $k \geq 2$, it is known via Riemann-Roch that $$\dim S^{\mathrm{new}}_k(\Gamma_1(q)) = \frac{k - 1}{24} (q^2 - 1) + E(q,k)$$ for an error term $E(q,k)$. This error term can be calculated explicitly (though not particularly neatly): see Theorem 13 of <a href="http://www.math.ubc.ca/~gerg/papers/downloads/DSCFN.pdf" rel="nofollow">http://www.math.ubc.ca/~gerg/papers/downloads/DSCFN.pdf</a>. So for $k \geq 2$, it is certainly possible to determine $\dim S^{\mathrm{new}}_k(\Gamma_1(q))$ precisely.</p> <p>For $k = 1$, on the other hand, no such precise equations seem to exist, as the method used to prove the $k \geq 2$ case breaks down. Instead, it is conjectured (see Conjecture 2.1 of <a href="http://arxiv.org/pdf/0906.4579v1" rel="nofollow">http://arxiv.org/pdf/0906.4579v1</a>) that $$\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = \frac{q - 2}{2} h(K_q) + O_{\varepsilon}(q^{\varepsilon}),$$ for any $\varepsilon > 0$ with the error term is uniform in $q$, and where $h(K_q)$ is the class number of $\mathbb{Q}(\sqrt{-q})$; here the leading term comes from the dihedral modular forms, while the error term is due to the others (icosahedral etc.).</p> <p>Now note that the leading term in the formula for $S^{\mathrm{new}}_k(\Gamma_1(q))$ for $k \geq 2$ vanishes when $k = 1$, so if that formula where to be valid for $k = 1$, we would be left with the error term $E(q,k)$, which we can explicitly compute.</p> <p><strong>Question</strong>: Is there a reason why we should not expect $\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = E(q,1)$? Obviously a quick check on Magma or Sage should prove that this is not the case, but unfortunately I don't have either installed.</p> <p>If not, is there any chance that we will one day find a closed form for $\dim S^{\mathrm{new}}_1(\Gamma_1(q))$?</p> http://mathoverflow.net/questions/61263/modular-forms-of-weight-one/61266#61266 Answer by Emerton for Modular Forms of Weight One Emerton 2011-04-11T06:45:07Z 2011-04-11T06:45:07Z <p>The formula for the dimension of $S_k$ when $k \geq 2$ can be thought of as a Riemann--Roch calculation, applied to an appropriately chosen line bundle on the modular curve. The point is that when $k \geq 2$, this line bundle is positive, so positive that the $H^1$-term in Riemann--Roch vanishes. Thus the dimension of the space of cuspforms coincides with the degree of the line bundle ($+ 1 - g$, where $g$ is the genus of the modular curve), which is linear in $k$.</p> <p>On the other hand, when $k = 1$, the $H^1$ term in the analogous Riemann--Roch formula need not vanish. In fact, what one finds is that the $H^0$ and $H^1$ terms essentially cancel each other, but one gains no information about whether $H^0$ is actually non-zero. It is whatever it is. This is the basic reason that it is hard to give a formula for the dimension of spaces of weight one forms.</p> http://mathoverflow.net/questions/61263/modular-forms-of-weight-one/130101#130101 Answer by Marc Palm for Modular Forms of Weight One Marc Palm 2013-05-08T17:29:21Z 2013-05-08T17:29:21Z <p>Another natural reason is that there is no pseudo matrix coefficient for limit of discrete series representations, since they are not square integrable. If so, you could simply plugin this pseudo coefficient times the characteristic function associated to the congruence subgroup into the Arthur trace formula, and compute the geometric side.</p>