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), 345348, 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 nontrivial 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.uniheidelberg.de/~hb3/ \\ Bhargava, High composition laws and applications, http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_13.pdf

$\begingroup$ For some interesting discussion related to your question: icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_13.pdf $\endgroup$– Frank ThorneCommented 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. $\endgroup$– Samuel HambletonCommented Oct 7, 2011 at 2:24

$\begingroup$ Bhargava, sorry. $\endgroup$– Samuel HambletonCommented 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. $\endgroup$– Samuel HambletonCommented 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 JagyCommented Oct 19, 2011 at 5:58
1 Answer
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.

$\begingroup$ 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 nontrivial kernel. I want to $\endgroup$ Commented 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. $\endgroup$ Commented 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.numbertheory, so more people will read it. $\endgroup$ Commented Oct 17, 2011 at 10:55

$\begingroup$ Thanks again Dr. Lemmermeyer. I now have a few things to think about. $\endgroup$ Commented 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 nontrivial 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. $\endgroup$ Commented Oct 21, 2011 at 3:52