In his seminal 2004 paper "Higher Composition Laws I" in the Annals of Mathematics, (doi:10.4007/annals.2004.159.217), Bhargava proves that for fixed $D \neq 0$, there is a bijective correspondence between the following two collections:

  • Primitive Bhargava cubes of discriminant $D$ modulo the action of $(\mathrm{SL}_2(\mathbb{Z}))^{\times 3}$, and
  • Triplets of oriented ideal classes $I_1, I_2, I_3$ in the narrow class group of the unique oriented quadratic ring of discriminant $D$ with the property that $I_1 \cdot I_2 \cdot I_3$ is the class of the unit ideal.

He then claims that, as a corollary of the above correspondence, one recovers the standard bijective correspondence due to Gauss between primitive integral binary quadratic forms and ideal classes in the narrow class group. Why is this true, that Gauss's correspondence is a corollary? I don't immediately see how this follows.

What I know: It's not hard to see explicitly that there is a Bhargava cube such that one of its three associated quadratic forms equals any given integral binary quadratic form. Thus, the map sending an ideal class to its norm form is surjective. I'm having trouble showing that this map is injective.

  • $\begingroup$ It seems that your confusion isn't really with Bhargava's cube law but rather why Gauss composition implies that $\text{SL}_2(\mathbb{Z})$-equivalence classes of binary quadratic forms parametrize ideal classes of quadratic orders. Do you understand the latter? $\endgroup$ – Stanley Yao Xiao May 30 '18 at 9:19
  • $\begingroup$ @StanleyYaoXiao I understand how to prove (independently of anything Bhargava does) the correspondence between $\mathrm{SL}_2(\mathbb{Z})$ equivalence classes of primitive integral binary quadratic forms of non-square discriminant $D$ and ideal classes of the narrow class group of the quadratic order of discriminant $D$. But the virtue of Bhargava's approach is that it works even in the case where $D$ is a square -- he seems to claim that the form--ideal-class correspondence can be deduced for all $D$ from the cube--triple-of-ideals correspondence. $\endgroup$ – Ashvin Swaminathan May 30 '18 at 14:23

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