Let $f,g, u,v \in \mathbb{Z}[x,y]$ be binary quadratic forms with co-prime coefficients. We say that the pair $(f,g)$ and $(u,v)$ are $\operatorname{GL}_2(\mathbb{Z})$-equivalent if there exists $T = \left(\begin{smallmatrix} t_1 & t_2 \\ t_3 & t_4 \end{smallmatrix} \right) \in \operatorname{GL}_2(\mathbb{Z})$ such that

$$\displaystyle f(t_1 x + t_2 y, t_3 x + t_4 y) = u(x,y).$$ and $$\displaystyle g(t_1 x + t_2 y, t_3 x + t_4 y) = v(x,y).$$

Note that if the pair $(f,g)$ is equivalent to $(u,v)$, then $f$ is equivalent to $u$ and $g$ equivalent to $v$ as individual binary quadratic forms, but the converse need not be true, since it is not clear that there exists a common matrix $T$ that sends $f$ to $u$ and $g$ to $v$.

Aside from the discriminants $\Delta(f), \Delta(g)$ of the individual forms $f$ and $g$, the pair $(f,g)$ has one more invariant which can be taken to be

$$\displaystyle \Delta(f,g) = 2 f_2 g_0 - f_1 g_1 + 2 f_0 g_2,$$

where $f(x,y) = f_2 x^2 + f_1 xy + f_0 y^2$ and $g(x,y) = g_2 x^2 + g_1 xy + g_0 y^2$.

Denote by $h(d_1, d_2, d_3)$ to be the class number of $\operatorname{GL}_2(\mathbb{Z})$-equivalence classes of pairs of binary quadratic forms $(f,g)$ with $\Delta(f) = d_1, \Delta(g) = d_2$, and $\Delta(f,g) = d_3$. Are there any estimates for $h(d_1, d_2, d_3)$ in terms of $d_1$ and $d_2$, or average estimates with $0 < \max\{|d_1|, |d_2|,|d_3|\} \leq X$?

I have consulted an old paper by Dickson (https://www.jstor.org/stable/2370100?seq=1#page_scan_tab_contents) but he only treated the easier case of $\operatorname{GL}_2(\mathbb{Q})$-equivalence.

Any help or reference would be appreciated!

  • $\begingroup$ Did my answer help ? $\endgroup$ – few_reps Apr 12 '17 at 7:02
  • $\begingroup$ @few_reps: It did provide some insights, but I am really looking for an average with $d_1, d_2, d_3$ varying simultaneously $\endgroup$ – Stanley Yao Xiao Apr 12 '17 at 14:48

We can show that $ h(1,1,n) $ increases like $ n $-th powers of 2 : indeed, in the half-plane presentation of the set of positive definite quadratic forms, the euclidean form is identified to $ i $ and the other unimodular forms are the points of the Serre Tree that stand at the middle of the edges. The trace ( $d_3 $ ) of a form is then the distance between that form and the Euclidean form on this tree. Say $ A_n $ is the set of those forms of trace $ n $ . The cardinal of $ A_n $ is $ 2^n $ and the cardinal of $ A_n /\mathrm{SO}_2(\mathbf Z) $ is something we might call $ sh(1,1,n) $: the number of proper equivalence classes. Thus $ h(1,1,n )$ is somewhere between $ 2^{n-4} $ and $ 2^n $ .

In the general case , I guess the behaviour is similar (when defined). At least if $ d_3 $ measures this kind of distances ... Indeed, in this case, if we take two forms $ A $ and $ B $ in the usual fundamental domain of the action of $ \mathrm{ Sl}_2(\mathbf Z) $, there is a unique representant of $ B $ in each tile and we can count the number of these representants which are at distance n in this tesselation from $ A $ .

Thus probably at $ d_1 $ and $ d_2 $ fixed, $ h(d_1,d_2,d_3) $ grows like some $ \gamma^n$ , with $ \gamma \leq 2 $ .

| cite | improve this answer | |
  • $\begingroup$ Yes you are right, there is a joint invariant of the pair that must also remain the same. I will edit the question. Thank you! $\endgroup$ – Stanley Yao Xiao Apr 8 '17 at 22:05

The exact class number formula was given by Jorge Morales in the following paper:

Jorge Morales. The classification of pairs of binary quadratic forms. Acta Arith., 59(2):105–121, 1991.

In particular, he showed that

$$\displaystyle h(d_1, d_2, d_3) = m \sum_{n | 4(d_3^2 - d_1 d_2)} \left(\frac{d_1}{n}\right),$$

where $m = 1$ if $d_3^2 - d_1 d_1 \geq 1$ and $m = 2$ if $d_3^2 - d_1 d_1 < 0$, and $\left(\frac{\cdot}{n}\right)$ is the Jacobi symbol.

The case when computing the class number $h(d_1, d_2, d_3)$ of pairs $(f,g)$ with $\Delta(f) = d_1, \Delta(g) = d_2, \Delta(f,g) = d_3$ and $f,g$ are both positive definite was resolved by an earlier paper by K. Hardy and K. Williams.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.