# Siegel's Mass Formula for ternary indefinite quadratic forms

In his paper "On the theory of indefnite quadratic forms", Siegel gives the formula (Thm. 1) $$\mu(S,T)=\prod_p \alpha_p(S,T),$$ where

• $S$ is an $m\times m$ non singular integral symmetric matrix of signature $(r,m-r)$,
• $T$ is an $n\times n$ integral symmetric matrix,
• $\mu(S,T)$ is the measure of the representation of $T$ by $S$,
• $\alpha_p(S,T)$ is the $p$-adic density of the representation of $T$ by $S$ ($p$ over the rational primes),

and $$n\leq r,\qquad n\leq m-r,\qquad 2(n+1) < m.$$

In my case, $n=1$ thus $T=t$ is just a (non zero) integer number, and $S$ has signature $(m-1,1)$ (or $(1,m-1)$). Hence, the theorem holds for $$m>4.$$ However, in the fourth page, he added that the formula also holds for $m=4$.

I known that when $m=3$ ($S$ is a ternary quadratic form), the formula doesn't holds in general and strange things happen (see BOROVOI ).

My question is: Does the formula works for the following particular case?:

$$S= \begin{pmatrix} A & \\ &-a \end{pmatrix}$$

where

• $A$ is a $2\times2$ integral positive definite symmetric matrix,
• $a$ is a positive integer, and
• $t$ is a negative integer.

I known that this is true for $A=I_2$ and $a=1$ (see this paper).

Borovoi does not claim what you seem to think. Siegel's formalism discusses, for positive forms, the number of representations of an integer by an entire genus of positive ternary forms, each form weighted according to the number of its integral automorphs. For indefinite forms, the automorph groups are infinite and one must discuss the number of essentially inequivalent representations of an integer by a form, that is representations that cannot be taken to each other by automorphs.

That being said, most indefinite ternary forms are in a genus with only one integer equivalence class of forms. In such cases, Siegel's answers for a genus agree with those for the single form.

Borovoi simply repeats one of the more famous examples of a genus of more than one class. In the first edition (1988) of SPLAG by Conway and Sloane, this is on pages 404-405. The two classes are represented by $$x^2 - 2 y^2 + 64 z^2$$ and $$-9 x^2 + 2 x y + 7 y^2 + 2 z^2.$$ In at least three variables, with indefinite forms, spinor genera coincide with equivalence classes. That is, every indefinite ternary form is spinor regular. There is, as there is here, a finite set of squareclasses of "spinor exceptional integers" that may fail to be represented by a form, even though another form in the genus represents them. Borovoi does not mention it in this manner (he takes only my $\; s=1$), but here we go:

Let us define a (positive) integer $s$ such that all prime factors $q$ of $s$ satisfy $$q \equiv \pm 1 \pmod 8,$$ with the consequence that also $s \equiv \pm 1 \pmod 8.$ Then, in notation going back to Jones and Pall (1939), Borovoi's form does not integrally represent $s^2,$ as in $$-9 x^2 + 2 x y + 7 y^2 + 2 z^2 \neq s^2.$$ Borovoi gives all necessary steps. I should say already that every example I know of spinor exceptional integers can be proved, after the fact, by very simple factoring arguments.

Proof (Zagier): ASSUME $$-9 x^2 + 2 x y + 7 y^2 + 2 z^2 = s^2$$ in integers, where
$$q | s \; \; \Rightarrow q \equiv \pm 1 \pmod 8$$ Rewrite as $$2 z^2 - s^2 = 8(x^2 - y^2) + (x-y)^2.$$ The left hand side is odd, the right side is also odd, so $x-y$ is odd, and the right side is $1 \pmod 8.$ If $z$ were even, the left side would be $7 \pmod 8,$ so we find that $z$ is odd. As a result, $z^2 \equiv 1 \pmod 8$ and $2 z^2 \equiv 2 \pmod {16}.$ As $s \equiv \pm 1 \pmod 8,$ we get $s^2 \equiv 1 \pmod {16}.$ Put together, we get $$2 z^2 - s^2 = 1 \pmod {16}.$$ But $8 (x^2-y^2) \equiv 8 \pmod {16}.$ Now we find $$(x-y)^2 \equiv 9 \pmod {16}.$$ Backing this up we get $$x-y \equiv \pm 3 \pmod 8.$$ Thus there is some prime $r \equiv \pm 3 \pmod 8$ such that $r | (x-y).$ But, from $2 z^2 - s^2 = 8(x^2 - y^2) + (x-y)^2,$ we see $$(x-y) | (2 z^2 - s^2).$$ Put those together, we have a prime $r$ such that $$r | (2 z^2 - s^2) \; \; \mbox{with} \; \; r \equiv \pm 3 \pmod 8.$$ By a standard application of quadratic reciprocity, we find that $$r | s$$ which is a contradiction of the assumption. Just for completeness, if $r$ does not divide $s,$ then $s$ has a multiplicative inverse $\pmod r.$ So $2 z^2 - s^2 \equiv 0 \pmod r$ becomes $2 z^2 \equiv s^2 \pmod r,$ then $4 z^2 \equiv 2 s^2 \pmod r,$ finally $\left( \frac{2z}{s} \right)^2 \equiv 2 \pmod r.$ However, $2$ is not a quadratic residue $\pmod r.$

So, in fact, $$-9 x^2 + 2 x y + 7 y^2 + 2 z^2 \neq s^2.$$

As to the forms being in the same genus, write $$f(x,y,z) = x^2 - 2 y^2 + 64 z^2$$ and $$g(x,y,z) = -9 x^2 + 2 x y + 7 y^2 + 2 z^2.$$ Then $$g(3 v + 15 w, u + v + 7 w, u - 5 v - 32 w) = 9 f(u,v,w).$$ Also $$f(u + 7 v + 2 w, 13 u -5 v + 5 w, 2 u - v + w) = 9 g(u,v,w).$$

Other aspects: with prime $r \equiv \pm 3 \pmod 8,$ we do have $f(r,0,0) = r^2.$ Another factoring argument shows that $f$ does not primitively represent $r^2.$ That is, if $f(x,y,z) = r^2,$ then $\gcd (x,y,z) = r.$ Finally, I am less sure about this, but I believe that $g$ represents all such $r^2,$ as in $g(x,y,z) = r^2,$ in this case primitively as $g$ does not represent 1. Examples include $$g(0,1,1) = 9, \; g(0,1,3) = 25, \; g(1,2,7) = 121,\; g(0,1,9) = 169, \; g(0,7,3) = 361,$$ $$g(3,10,9) = 841, \; g(3,-14,9) = 1369, \; g(1,-16,7) = 1849,\; g(2,19,11) = 2809.$$ The shortest discussion on this is on page 352 of Rainer Schulze-Pillot, Exceptional Integers for Genera of Integral Ternary Positive Definite Quadratic Forms, Duke mathematical Journal, volume 102 (2000), pages 351-357. However, as in the title, he is talking about definite forms. There is also his survey, including Siegel for indefinite forms, Representation by integral quadratic forms-a survey, in Contemporary Mathematics volume 344 (2004) pages 303-321, the book is titled Algebraic and Arithmetic Theory of Quadratic Forms edited by Baeza, Hsia, Jacob, and Prestel.

• Will, it's a nice answer! Thank you very much. However, I want to check my conclusion from your answer in the next comment. Feb 4 '12 at 0:13
• Are you saying that if the ternary form $S=diag(A,-a)$ (with $A$ pos def and $a\in\mathbb N$) is in a genus with only one integer equivalence class, then Siegel's formula holds? Feb 4 '12 at 0:16
• @emiliocba, yes. What is your actual background as relates to this material? Feb 4 '12 at 1:46
• Thank you again. I want to known if these particular case holds generalizing the fact that it holds for $S=diag(1,1,-1)$ and $t<0$. Feb 4 '12 at 17:34

Let me settle the situation with $r^2$ when $r$ is a prime with $r \equiv \pm 3 \pmod 8.$ First, suppose we have some representation by the first form, $$x^2 - 2 y^2 + 64 z^2 = r^2.$$ Then $$x^2 - 2 y^2 = r^2 - 64 z^2 = (r + 8 z)(r - 8 z).$$ Now, as $r - 8 z \equiv \pm 3 \pmod 8,$ there is some prime $t \equiv \pm 3 \pmod 8$ with $$t^{2 k + 1} \parallel (r - 8 z).$$ However, we must have $$t^{2 m} \parallel (x^2 - 2 y^2),$$ so $$t^{2 l + 1} \parallel (r + 8 z).$$ As a result, $t | (2 r),$ so $t = r,$ so that $r | x$ and $r | y.$ We already had $t | (2z),$ so actually $r | z.$ Put these together, $x^2 - 2 y^2 + 64 z^2$ does represent $r^2$ but not primitively.

This next bit is unexpected, at least in terms of finding an explicit formula. We get a cheap trick out of $$-9 x^2 + 2 x y + 7 y^2 = (9x + 7 y)(-x+y).$$ If we take $y=x+1$ the result of the binary is $16 x + 7.$ Add on $2 z^2$ with $z=1$ we actually get $16 x + 9.$

Alright, back to prime $r \equiv \pm 3 \pmod 8.$ It follows that $r^2 \equiv 9 \pmod {16}.$ We have $$g(x,y,z) = -9x^2 + 2 x y + 7 y^2 + 2 z^2.$$ Take any number $n \equiv 9 \pmod {16}.$ Then we have an integral, and primitive, representation in $$g \left( \frac{n - 9}{16}, \; \frac{n + 7}{16}, \; 1 \right) = n.$$ You really do not get this sort of closed form answer with positive forms.

Though you should build a bark of dead men's bones,
And rear a phantom gibbet for a mast,
Stitch creeds together for a sail, with groans
To fill out, blood-stained and aghast;
Although your rudder be a dragon's tail
Long sever'd, yet still hard with agony
Your cordage large uprootings from the skull
Of bald Medusa, certes you would fail
To find the Melancholy - whether she
Dreameth in any isle of Lethe dull.

• Why do you quote a melancholic Ode, / And format it as though 'twere pseudocode? Feb 5 '12 at 3:39
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• @Noam, Excellent. I'll send you the book review and book pages I meant, 2 jpegs and a pdf. Feb 5 '12 at 23:13