5
$\begingroup$

Let $\Delta = \sigma + 4 m$ be the fundamental discriminant of a quadratic field, where $\sigma \in \{ 0, 1 \}$. The binary quadratic form $Q(x, y) = A x^2 + B x y + C y^2$ of discriminant $\Delta$ belongs to the identity class (principal) of the narrow class group of forms, under substitution by $(x,y) \mapsto M (x', y')$ with $M \in \text{SL}_2(\mathbb{Z})$, if there exist rational integers $x, y$ satisfying $Q(x, y) = 1$, or equivalently, if $Q(x, y)$ is equivalent to $Q_0(x, y) = x^2 + \sigma x y - m y^2$. Davenport and Heilbronn, On the density of discriminants of cubic fields (I), Bull. Lond. Math. Soc. 1, (1969), 345--348, for example, considered binary cubic forms $a x^3 + b x^2 y + c x y^2 + d y^3$ of discriminant $D = 18 a b c d + b^2 c^2 - 4 a c^3 - 4 b^3 d - 27 a^2 d^2$, and their quadratic covariant binary quadratic forms $(b^2 - 3 a c) x^2 + (b c - 9 a d) x y + (c^2 - 3 b d) y^2$, of disc $\Delta = - 3 D$. M. Bhargava has described the composition of binary cubic forms in which $3 \mid b, c$. I would have thought, probably mistakenly, that a group law on the former binary cubic forms would consist of composing quadratic covariants and finding a binary cubic form for which the covariant belongs to the class of the composed form. I would like to know which class of binary cubic forms to consider as the neutral element of a class group of forms. For example, the binary cubic form $C = (39820040, 28889459, 6986439, 563185)$ of discriminant $D = -3299$ has quadratic covariant $x^2 - 99 x y - 24 y^2$ of discriminant $9897$. Does $C$ belong to an identity class of a class group of binary cubic forms? \\ Lemma : Let $\mathcal{P} : x^2 + \sigma x y - m y^2 = 1$ be a Pell conic with $\Delta = - 3 D = \sigma + 4 m$, $\sigma \in \{ 0, 1 \}$, and let $(x, y) = (b, - 3 a) \in \mathcal{P}(\mathbb{Z})$ be the least non-trivial point such that $3 \mid y$. Let $c = \frac{b^2 - 1}{3 a}$, and $d = \frac{b c - \sigma }{9 a}$. Then the binary cubic form $(a, b, c, d)$ of discriminant $D$ has quadratic covariant equal to $Q_0(x, y)$. \\ Proof : Since $b^2 - 3 \sigma a b - 9 m a^2 = 1$, $c = \frac{b^2 - 1}{3 a} = \sigma b + 3 a m \in \mathbb{Z} $. Also, $$d = \frac{b c - \sigma }{9 a} = \frac{b (\sigma b + 3 a m ) - \sigma }{9 a} = \frac{\sigma (b^2 - 1) + 3 a b m }{9 a} = \frac{3 \sigma a b + 9 \sigma a^2 m + 3 a b m }{9 a} = \frac{\sigma - D}{4} b + \sigma a m ,$$ which is clearly an integer. The following identity shows that $(a, b, c, d)$ is a binary cubic form of discriminant $D$. $$18 a b c d + b^2 c^2 - 4 a c^3 - 4 b^3 d - 27 a^2 d^2 = \frac{4 \sigma b - 3 \sigma a - 4 \frac{b^2 - 1}{3 a}}{9 a} = -\frac{\sigma + 4 m }{3} = D .$$ The quadratic covariant of $C(x ,y) = (a, b, c, d)$ is $$( b^2 - 3 a c , b c - 9 a d , c^2 - 3 b d ) = ( b^2 - b^2 + 1, b c - b c + \sigma , \frac{(b^2 - 1)^2 - b^2 (b^2 - 1) + 3 \sigma a b}{9 a^2} ),$$ equal to $(1, \sigma , - m )$. Should $C(x, y)$ be called principal? When $\Delta > 0$, what is a different $\text{SL}_2(\mathbb{Z})$ class of binary cubic form with principal quadratic covariant? \\ References : \\ Lemmermeyer, Conics A poor mans elliptic curves http://www.rzuser.uni-heidelberg.de/~hb3/ \\ Bhargava, High composition laws and applications, http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_13.pdf

$\endgroup$
6
  • $\begingroup$ For some interesting discussion related to your question: icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_13.pdf $\endgroup$ Oct 6, 2011 at 16:23
  • $\begingroup$ Thank you Dr. Thorne. I am still reading Brargava's interesting articles on composition. I find `triplicate central coefficients' problematic for related questions I have on binary cubic forms. I thought that a group law on classes of BCFs (binary cubic forms) should be to compose $C_1 \cdot C_2$, $C_j = (a_j, b_j, c_j, d_j)$, compose for $j = 1, 2$, $Q_j = b_j^2 - 3 a_j c_j, b_j c_j - 9 a_j d_j , c_j^2 - 3 b_j d_j$, find a BCF for which the quadratic covariant is $Q_3$, the composed quadratic form. Pepin says more than one class of BCF corresponds to the pricipal class of BQF. I am confused. $\endgroup$ Oct 7, 2011 at 2:24
  • $\begingroup$ Bhargava, sorry. $\endgroup$ Oct 7, 2011 at 2:25
  • $\begingroup$ I've ticked Franz Lemmermeyer's Answer correct since I'm sure that after thinking about this answer, my confusion will be resolved, but more answers are also welcome. $\endgroup$ Oct 19, 2011 at 5:23
  • $\begingroup$ I have asked them to merge the two registered accounts of yours that I know about, 17053 and 18487. $\endgroup$
    – Will Jagy
    Oct 19, 2011 at 5:58

1 Answer 1

3
$\begingroup$

I'm not sure that I will be answering your question, so let me first recall the background. The equivalence classes of primitive binary cubic forms with discriminant $\Delta$ correspond to $SL_2(\mathbb Z)$-equivalence classes of triply symmetric Bhargava cubes. There is a natural homomorphism from this group to the strict class group of binary quadratic forms with discriminant $\Delta$, and in fact the image has order dividing three for (almost trivial) reasons. The kernel of this homomorphism has order $(U:U^3)$, where $U$ is the unit group of the quadratic order with discriminant $\Delta$.

The group structure on the set of binary cubic forms is defined via the group structure on the group of $\Gamma$-equivalence classes of Bhargava cubes, where $\Gamma$ is the direct product of three copies of $SL_2(\mathbb Z)$. Given a binary quadratic form $Q$ with discriminant $\Delta$ it is easy to see that there is exactly one $\Gamma$-equivalence class of triply symmetric cubes belonging to $(Q,Q,Q)$.

Your problem seems to come from confusing the $\Gamma$-equivalence classes of cubes in the second paragraph with the $SL_2(\mathbb Z)$-equivalence classes of cubes from the first one, and I think that this is what you asked in your comment - observe that an element $S \in SL_2(\mathbb Z)$ acts on a cube via the action of $(S,S,S)$.

I do not understand your question in the numerical example; perhaps you can clarify this part by editing the question.

$\endgroup$
6
  • $\begingroup$ Thank you Dr. Lemmermeyer. I will think about it for a little while to sort out my confusion and then edit the question, which could have been more clear. I would like to understand the group law on classes of binary cubic forms $(a, b, c, d)$ without requiring $b$ and $c$ to be divisible by $3$. I have used Davenport and Heilbronn's quadratic covariant from "On the density of discriminants of cubic fields (I)" $D = 18 a b c d + b^2 c^2 - 4 a c^3 -4 b^3 d - 27 a^2 d^2$ and the quadratic covariant's disc is $\Delta = - 3 D$. When $D < 0$ there should be a non-trivial kernel. I want to $\endgroup$ Oct 16, 2011 at 3:05
  • $\begingroup$ ... know which class of binary cubic forms to consider the neutral element of the group. Again, I'll think about your answer. $\endgroup$ Oct 16, 2011 at 3:07
  • $\begingroup$ Perhaps math.ubc.ca/~mantilla/pruebas.pdf will help. If you should edit the question, please add a tag nt.number-theory, so more people will read it. $\endgroup$ Oct 17, 2011 at 10:55
  • $\begingroup$ Thanks again Dr. Lemmermeyer. I now have a few things to think about. $\endgroup$ Oct 19, 2011 at 5:27
  • $\begingroup$ You say that there is exactly one $\Gamma$-equivalence class of triply symmetric cubes belonging to $(Q_0, Q_0, Q_0)$, but that the kernel of the homomorphism is non-trivial when $\Delta > 0$. Clearly I am confusing something especially when trying to apply it to a slightly different object. Would you be able to clarify? Thanks. $\endgroup$ Oct 21, 2011 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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