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There is something I am missing, or some misconception I have, and it would be nice if someone could clear it up for me:

I have read that the set of classes of binary quadratic forms of given discriminant forms a finite abelian group. Here is my issue:

Let's choose the discriminant $D=-56$, for example. Then the class number is three and the three reduced binary quadratic forms are:

$$a^2+14b^2 $$ $$2a^2+7b^2$$ $$3a^2+2ab+5b^2 $$

Now, 3 and 5 are both in the third class. If we multiply them to get 15, we see that 15 is in class 1 ($1^2+14 \cdot 1^2$) and in class 2 ($2 \cdot 2^2 +7 \cdot 1^2) $

But how is this possible if the classes form an abelian group... wouldn't that mean that the product of two classes must equal precisely one other class and that class only?

Again, I know I am missing something and some part of my argument must sound absurd, but I don't see which part....

Edit: @Brent Kerby: I understand what you mean by not multiplying point-wise, but even when considering the classes as a whole I have the same problem. Here is what I mean:

$$ (3a^2+2ab+5b^2)(3c^2+2cd+5d^2) = (3ac+ad+bc+5bd)^2 + 14(ad-bc)^2 $$

and also

$$ (3a^2+2ab+5b^2)(3c^2+2cd+5d^2) = 2(2ad+2bc+3bd-ac)^2 +7(ac+ad+bc-bd)^2 $$

Hence the product of the third class with itself gives both the first class and the second class. The case with $3 \cdot 5 = 15$ was only an example (with $a=1$, $b=0$, $c=0$, $d=1$)

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    $\begingroup$ I just saw your edit, and I've expanded my answer accordingly. $\endgroup$ Mar 5, 2015 at 15:12
  • $\begingroup$ Uh ... How is the class number for discriminant $-56$ three? Not four? $\endgroup$ Jul 12, 2020 at 0:06

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You are confusing classes of quadratic forms with the sets of integers that they represent. The class of a quadratic form consists of the quadratic forms which may be obtained from it by an appropriate linear change-of-variables. For example, in the form $3a^2+2ab+5b^2$, if we apply a change of variables $a\mapsto a+b$, $b\mapsto b$, then we obtain $$3(a+b)^2+2(a+b)b+5b^2 = 3a^2 + 8ab + 10b^2$$ So $3a^2+8ab+10b^2$ is a quadratic form in the class of $3a^2+2ab+5b^2$. Quadratic forms in the same class always represent the same sets of integers; for example, as the integer 5 is represented by $3a^2+2ab+5b^2$, namely with $(a,b)=(0,1)$, it is also represented by $3a^2+8ab+10b^2$, with $(a,b)=(-1,1)$.

The group operation takes two classes of quadratic forms and outputs another class of quadratic form; this operation is somewhat complicated and does not just consist of multiplying the two forms together pointwise (which would not result in a quadratic form, but rather a form of degree 4). The point is that the group operation is defined on the set of classes of quadratic forms, rather than on the integers that they represent.

In response to the edit: It looks like your definition of the product of quadratic forms $f$ and $g$ is $h$ where $$f(x,y)g(z,w)=h(B_1(x,y;z,w),B_2(x,y;z,w))$$ and $B_1$ and $B_2$ are bilinear forms, say, $$B_i(x,y;z,w) = a_ixz+b_ixw+c_iyz+d_iyw$$ The problem is that, under this definition, as your examples show, $h$ is not uniquely determined (even up to equivalence classes), so this does not give a well-defined operation. However, it does become a well-defined operation if we impose the restrictions $a_1b_2-a_2b_1=f(1,0)$ and $a_1c_2-a_2c_1=g(1,0)$. See, for example, Cox, Primes of the form $x^2+ny^2$, p. 43. In both of your examples, these restrictions are not satisfied; however, your second example can be modified to satisfy the restrictions by negating $B_1$: $$(3a^2+2ab+5b^2)(3c^2+2cd+5d^2)=2(-2ad-2bc-3bd+ac)^2+7(ac+ad+bc−bd)^2$$ Thus the product of the class of $3a^2+2ab+5b^2$ with itself is the class of $2a^2+7b^2$ and only this class.

By the way, in your list of reduced binary quadratic forms of discriminant $-56$, you're missing the form $3a^2-2ab+5b^2$. Although this form is equivalent to $3a^2+2ab+5b^2$ under the change-of-variables $(a,b)\mapsto (a,-b)$, such an equivalence is not a ``proper equivalence", because the change-of-variables matrix has determinant -1 instead of 1. In order for the group operation to be well-defined, classes of quadratic forms must be defined based on proper equivalence rather than equivalence. In this case, the result will be a group of order 4.

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    $\begingroup$ Nice answer, Brent. $\endgroup$
    – user02138
    Feb 28, 2015 at 0:15
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    $\begingroup$ I see, so if we split up the classes more, by considering only proper equivalence and not improper equivalence, thereby making a larger list of classes, then we do get a group structure, which we didn't get before when we were less strict on the definition of equivalence... thank you, that helped +1 and bounty $\endgroup$ Mar 5, 2015 at 16:01

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