Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

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:

Does there exist a product $\times$ on solutions to $Q(\vec{z})=n$ such that, whenever $(m,n)=1$,

  • $Q(\vec{x})=m$ and $Q(\vec{y})=n$ imply $Q(\vec{x}\\times \vec{y})=mn$;
  • $Q(\vec{z})=mn$ implies we can find unique $\vec{x}, \vec{y}$ with $Q(\vec{x})=m$, $Q(\vec{y})=n$ and $\vec{x} \times \vec{y} = mn$?

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 only by GKS to show combinatorially that $p(5n+4) \equiv 0 \mod5$.

Note 1: 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 $v_i \equiv i \mod 5$.

Note 2: 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.

UPDATE: 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 http://oeis.org/A053723 ). There are actually closed forms there for $r_Q(p^e)$, but I don't see how they help for my question.

In particular, we have $r_Q(2) = 1$, and $r_Q(4) = 3$, corresondign respectively to the solutions $(0, -1, 0, -1)$ and $(-1, 0, -1, 0), (0, -1, -1, -1), (0, 0, 0, -1)$.

share|improve this question
    
Are you sure your $r_Q(n)$ is not completely multiplicative? If not, there should be very small $m,n$ with $\gcd(m,n) \neq 1$ and $r_Q(mn) \neq r_Q(m) r_Q(n).$ –  Will Jagy Aug 25 '12 at 20:58
    
Now that I think of it, it might be enough to ask how $r_Q(n^2)$ and $r_Q(n)^2$ compare. –  Will Jagy Aug 25 '12 at 21:09
    
It is not completely multiplicative. I think the two conditions you wrote are equivalent. Anyways, it fails $r_Q(4) = r_Q(2)^2$, as shown in my update. –  Paul-Olivier Dehaye Aug 26 '12 at 3:39
    
Presumably you should look for a multiplication identity in the quaternions $v_1+iv_2+jv_3+kv_4$ that preserves the congruence conditions $v_i\equiv i\pmod 5$. Also, your original polynomial is a quadratic polynomial, but not a quadratic form, since it has linear and constant terms. –  Greg Martin Aug 27 '12 at 7:23
    
Yes, a potential way to get the product is by showing that $Q$ behaves like a norm on an order of a particular quaternion algebra. I am hoping invariants of the form can help narrow down where to look. –  Paul-Olivier Dehaye Aug 27 '12 at 8:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.