User ncr - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:13:01Z http://mathoverflow.net/feeds/user/16419 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123634/reference-on-generators-of-subgroups-of-symplectic-groups/123658#123658 Answer by ncr for Reference on generators of subgroups of symplectic groups ncr 2013-03-05T19:26:45Z 2013-03-06T02:17:38Z <p>I think you might be interested in the article "Simple Graded Rings of Siegel Modular Forms, Differential Operators and Borcherds Products" by H. Aoki and T. Ibukiyama published in International Journal of Mathematics, Vol. 16, No. 3 (2005) 249–279. In Lemma 6.2 on pg. 265 they prove: "For any natural number $N$, the group $\Gamma_0(N)$ is generated by the above four kinds of matrices":</p> <p><code>$T = \begin{pmatrix} 0 &amp; 1 &amp; 0 &amp; 0\\ 1 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1\\ 0 &amp; 0 &amp; 1 &amp; 0 \end{pmatrix}$</code></p> <p><code>$u(x)=\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0\\ x &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; -x\\ 0 &amp; 0 &amp; 0 &amp; 1 \end{bmatrix}, x\in \mathbb{Z}$</code></p> <p><code>$u(S) = \begin{bmatrix} 1_2 &amp; S\\ 0 &amp; 1_2 \end{bmatrix}, S={}^tS\in M_2(\mathbb{Z})$</code></p> <p><code>$C(a,b,c,d)=\begin{bmatrix} a &amp; 0 &amp; b &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 0\\ cN &amp; 0 &amp; d &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1 \end{bmatrix}$</code> where <code>$\begin{bmatrix} a &amp; b \\ cN &amp; d\end{bmatrix}\in\Gamma_0^{(1)}(N)$</code></p> <p>(Sorry for the laTexing; I couldn't get matrices to work).</p> http://mathoverflow.net/questions/108846/modularity-of-higher-dimensional-abelian-varieties/108880#108880 Answer by ncr for Modularity of higher dimensional abelian varieties ncr 2012-10-05T02:06:21Z 2012-10-05T02:06:21Z <p>There has been recent work in some concrete cases. There's a paper by <a href="http://arxiv.org/abs/0912.0049" rel="nofollow">Poor and Yuen</a> that gives computational evidence for a special case of the so-called Paramodular Conjecture. This Conjecture is described as "a precise and testable modularity conjecture for rational abelian surfaces $\mathcal{A}$ with trivial endomorphisms, $End_\mathbb{Q} \mathcal{A} = \mathbb{Z}$ in the abstract of a paper by <a href="http://arxiv.org/abs/1004.4699" rel="nofollow">Brumer and Kramer</a>. To the best of my knowledge, this is the most precise version of a general prediction made by Yoshida as described in</p> <p>H. Yoshida, On generalization of the Shimura-Taniyama conjecture I and II, Siegel Modular Forms and Abelian Varieties, Proceedings of the 4-th Spring Conference on Modular Forms and Related Topics, 2007, pp. 1-26.</p> http://mathoverflow.net/questions/102575/are-there-cusp-forms-for-the-full-modular-group-sp2-z-and-representations-det3/102613#102613 Answer by ncr for Are there cusp forms for the full modular group Sp(2,Z) and representations det^3 \otimes Sym^2j(\rho_standard) ncr 2012-07-19T02:48:58Z 2012-07-19T02:48:58Z <p>Though I haven't seen it, I've heard that Tomoya Kiyuna, a student at Kyushu University, has done this for the case $j=4$ as part of his master's thesis. In particular, he finds eighteen explicit generators of the module of vector-valued Siegel modular forms of the symmetric tensor 8. </p> http://mathoverflow.net/questions/115128/what-happens-to-zetas-when-all-its-im-rho-n-are-scaled-linearly Comment by ncr ncr 2012-12-02T03:56:48Z 2012-12-02T03:56:48Z For more zeta zeros check out <a href="http://www.lmfdb.org/zeros/zeta/" rel="nofollow">lmfdb.org/zeros/zeta</a> There are over 36 million there. http://mathoverflow.net/questions/102575/are-there-cusp-forms-for-the-full-modular-group-sp2-z-and-representations-det3/102613#102613 Comment by ncr ncr 2012-07-23T13:12:46Z 2012-07-23T13:12:46Z I wrote to Kiyuna and he tells me that there are 18 generators: 6 of them are theta series (presumably products of theta constants) and the remaining 12 are a kind of Rankin-Cohen construction. I don't know much more but he tells me there will be a preprint in the next couple of weeks. If you'd like to send me an email (see my website for my email address), I can give you his email address.