Product on representations of an integer by a quadratic form? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:00:59Z http://mathoverflow.net/feeds/question/105448 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105448/product-on-representations-of-an-integer-by-a-quadratic-form Product on representations of an integer by a quadratic form? Paul-Olivier Dehaye 2012-08-25T07:06:18Z 2012-08-26T23:15:30Z <p>Define the quadratic form $$Q(z_1,z_2,z_3,z_4) = 13 + \sum_{i=1}^4 (10+i)z_i +5 \sum_{1 \le i \le j \le 4} z_iz_j.$$ Then, $r_Q(n) := \left|{(z_1,z_2,z_3,z_4) \in \mathbb{Z}^4 : Q(z_1,z_2,z_3,z_4) = n }\right|$ is weakly multiplicative. I can prove this by using the generating function $\sum r_Q(n) q^n$ which is in the Eisenstein subspace of $\mathcal{M}_2(\chi_5)$, with $\chi_5$ the Legendre symbol. </p> <p>Because $r_Q(n)$ counts something, a likely alternative explanation is that there exists some product on solutions to $Q(\vec{z})=n$ that would explain this multiplicativity directly, hence the question:</p> <blockquote> <p>Does there exist a product $\times$ on solutions to $Q(\vec{z})=n$ such that, whenever $(m,n)=1$, </p> <ul> <li>$Q(\vec{x})=m$ and $Q(\vec{y})=n$ imply $Q(\vec{x}\times \vec{y})=mn$;</li> <li>$Q(\vec{z})=mn$ implies we can find unique $\vec{x}, \vec{y}$ with $Q(\vec{x})=m$, $Q(\vec{y})=n$ and $\vec{x} \times \vec{y} = mn$?</li> </ul> </blockquote> <p>I am interested in this because results of Garvan, Kim and Stanton give a bijection between the representations of $n$ by $Q$ and the number of $5-$core partitions of size $n-1$, which would lead to a product on 5-cores that I would like to understand. This multiplicativity has been used at 5 <em>only</em> by GKS to show combinatorially that $p(5n+4) \equiv 0 \mod5$. </p> <p><strong>Note 1:</strong> After the change of variables $v_i := 5z_i+i$ and the introduction of the fifth variable $v_0 := -v_1-v_2-v_3-v_4$, one can also define $Q$ by the more symmetric and homogeneous $$Q(v_0,v_1,v_2,v_3,v_4) = \frac{1}{10} \sum_{i=0}^4 v_i^2,$$ but we are not looking at all solutions then: we need $\sum_{i=0}^4 v_i = 0$ and $v_i \equiv i \mod 5$.</p> <p><strong>Note 2:</strong> For the multiplicativity, 5 is special at the moment, but could conceivably be replaced by 7 and 11 later, judging from the theory of Garvan, Kim and Stanton. However I am hoping that a combinatorial construction for the product on 5-cores could be generalized more widely.</p> <p><strong>UPDATE:</strong> I am sure the $r_Q(n)$ is not completely multiplicative. In fact, here is a list of the first 50 values, starting at 1: 1, 1, 2, 3, 5, 2, 6, 5, 7, 5, 12, 6, 12, 6, 10, 11, 16, 7, 20, 15, 12, 12, 22, 10, 25, 12, 20, 18, 30, 10, 32, 21, 24, 16, 30, 21, 36, 20, 24, 25, 42, 12, 42, 36, 35, 22, 46, 22, 43, 25,... ( it's at <a href="http://oeis.org/A053723" rel="nofollow">http://oeis.org/A053723</a> ). There are actually closed forms there for $r_Q(p^e)$, but I don't see how they help for my question.</p> <p>In particular, we have $r_Q(2) = 1$, and $r_Q(4) = 3$, corresondign respectively to the solutions $(0, -1, 0, -1)$ and $(-1, 0, -1, 0), (0, -1, -1, -1), (0, 0, 0, -1)$.</p>