Exercise 3.5.13 ($H(d) = \sum_{r>0,\, r^2 \mid d} h(d/r^2)$)

Proof. Let $f(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$ be a form of discriminant $d=b^2-4\mathit{ac}$ and $r=\gcd (a,b,c)$. We suppose that $f\ne 0$, so that $r\ne 0$.

Then $a=\mathit{rA},b=\mathit{rB},c=\mathit{rC}$, where $A,B,C$ are integers satisfying $\gcd (A,B,C)=1$, because $r=a\wedge b\wedge c=\mathit{rA}\wedge \mathit{rB}\wedge \mathit{rc}=r(A\wedge B\wedge C)$, so $A\wedge B\wedge C=1$. Then $g(x,y)=(a∕r)x^2+(b∕r)\mathit{xy}+(c∕r)y^2=A{x}^2+\mathit{Bxy}+C{y}^2$ is a form of discriminant $D=B^2-4\mathit{AC}=d∕r^2$, and this form is primitive since $A\wedge B\wedge C=1$.

For every integer $d$ such that $d$ is not a perfect square, let $\mathrm{Cl}(d)$ denote the set of classes of of forms of discriminant $d$, and let $P(d)$ denote the set of classes of primitive forms of discriminant $d$ (these forms are supposed definite positive if $d<0$).

Let $C_r$ be the set of classes of forms $f(x,y)=a{x}^2+\mathit{bx}+\mathit{cy}$ of discriminant $d$ such that $a\wedge b\wedge c=r$ (this makes sense because the discriminant and the g.c.d. of the coefficients does not depend on the choice of a representative of the class by Problem 11).

Then

this union being a disjoint union. If $r=a\wedge b\wedge c$, then $r>0$, and $r\mid a,r\mid b,r\mid c$, thus $r^2\mid d=b^2-4\mathit{ac}$. So $C_r=\varnothing$ if these conditions are not satisfied. This gives

Consider now, for $r>0,r^2\mid d$, the map

which sends the class $\bar{f}$ of $f(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$ on the class $\bar{g}$ of $g(x,y)=(a∕r)x^2+(b∕r)\mathit{xy}+(c∕r)y^2$, where $r=a\wedge b\wedge c$.

We show first that this map is well defined. If $f\stackrel{+}{\sim }{f}^{\prime }$, where $f^{\prime }(x,y)=a^{\prime }{x}^2+b^{\prime }\mathit{xy}+c^{\prime }{y}^2$, then $r=a\wedge b\wedge c=a^{\prime }\wedge b^{\prime }\wedge c^{\prime }$ (Problem 11). There is a matrix $A=\left(\begin{array}{cc}\hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill \gamma \hfill & \hfill \delta \hfill \end{array}\right)\in \Gamma$ such that $f^{\prime }=f\cdot A$, i.e.

If $g^{\prime }(x,y)=(a^{\prime }∕r)x^2+(b^{\prime }∕r)\mathit{xy}+(c^{\prime }∕r)y^2$, then

Therefore $g^{\prime }=g\cdot A\stackrel{+}{\sim }g$, thus $\bar{g}$ does not depend on the choice of the representative $f$ of $\bar{f}$, and so $\phi$ is well defined.

Moreover, $\phi$ is a bijection, whose reciprocal bijection sends the class of the primitive form $A{x}^2+\mathit{Bxy}+C{y}^2$ on the class of $\mathit{rA}{x}^2+\mathit{rBxy}+\mathit{rC}{y}^2$ in $C_r$. This shows that

Then the equality (1) gives

So

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