Thursday: here is an example I proved in full detail, that illustrates the use of the mappings in one direction, along with the possible intricacy of the difference between finding all rational null vectors and successfully finding all primitive integer null vectors:
All solutions to $$ 2(x^2 + y^2 + z^2) - 113 (yz+zx+xy) = 0 $$ with $$ \gcd(x,y,z) = 1 $$ can be written as one of four recipes, with the understanding that we sort by absolute value and possibly multiply through by $-1$ so as to demand $x \geq |y| \geq |z|,$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
37 u^2 + 51 uv + 8 v^2 \\
8 u^2 -35 uv -6 v^2 \\
-6 u^2 + 23 uv + 37 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
32 u^2 + 61 uv + 18 v^2 \\
18 u^2 -25 uv -11 v^2 \\
-11 u^2 + 3 uv + 32 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
38 u^2 + 45 uv + 4 v^2 \\
4 u^2 -37 uv -3 v^2 \\
-3 u^2 + 31 uv + 38 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
29 u^2 + 63 uv + 22 v^2 \\
22 u^2 -19 uv -12 v^2 \\
-12 u^2 -5 uv + 29 v^2
\end{array}
\right)
$$
In all four cases we simply discard occurrences when the resulting $x,y,z$ have a common factor.
The four are all of the form $X = R U,$ where
$$ X =
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right)
$$
and
$$ U =
\left(
\begin{array}{r}
u^2 \\
uv \\
v^2
\end{array}
\right).
$$
Clearly we take $\gcd(u,v) = 1.$ We can also take $u,v \geq 0.$ This is an artifact of the extreme symmetry of the ternary form and the extremely special form of the four matrices $R$ that I chose.
jagy@phobeusjunior:~$ ./isotropy_just_ordered 2 113 1200
29 22 -12 % B lambda 0 / B lambda 1 = 1
32 18 -11 % B lambda 0 / B lambda 1 = 1
37 8 -6 % B lambda 0 / B lambda 1 = 1
38 4 -3 % B lambda 0 / B lambda 1 = 1
188 171 -86 % B lambda 0 / B lambda 49 = 7^2
211 144 -82 % B lambda 0 / B lambda 49 = 7^2
226 123 -76 % B lambda 0 / B lambda 49 = 7^2
243 94 -64 % B lambda 0 / B lambda 49 = 7^2
246 88 -61 % B lambda 0 / B lambda 49 = 7^2
258 59 -44 % B lambda 0 / B lambda 49 = 7^2
264 38 -29 % B lambda 0 / B lambda 49 = 7^2
268 11 -6 % B lambda 0 / B lambda 49 = 7^2
396 262 -151 % B lambda 0 / B lambda 169 = 13^2
432 209 -134 % B lambda 0 / B lambda 169 = 13^2
472 129 -94 % B lambda 0 / B lambda 169 = 13^2
489 76 -58 % B lambda 0 / B lambda 169 = 13^2
516 458 -233 % B lambda 0 / B lambda 361 = 19^2
526 447 -232 % B lambda 0 / B lambda 361 = 19^2
628 311 -198 % B lambda 0 / B lambda 361 = 19^2
656 262 -177 % B lambda 0 / B lambda 361 = 19^2
671 232 -162 % B lambda 0 / B lambda 361 = 19^2
692 183 -134 % B lambda 0 / B lambda 361 = 19^2
726 47 -32 % B lambda 0 / B lambda 361 = 19^2
727 36 -22 % B lambda 0 / B lambda 361 = 19^2
804 787 -382 % B lambda 0 / B lambda 961 = 31^2
894 688 -373 % B lambda 0 / B lambda 961 = 31^2
953 946 -456 % B lambda 0 / B lambda 1369 = 37^2
1034 492 -317 % B lambda 0 / B lambda 961 = 31^2
1062 443 -296 % B lambda 0 / B lambda 961 = 31^2
1102 363 -256 % B lambda 0 / B lambda 961 = 31^2
1123 314 -228 % B lambda 0 / B lambda 961 = 31^2
1159 1046 -528 % B lambda 0 / B lambda 1849 = 43^2
1179 118 -88 % B lambda 0 / B lambda 961 = 31^2
1188 19 2 % B lambda 0 / B lambda 961 = 31^2
1199 1002 -524 % B lambda 0 / B lambda 1849 = 43^2
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
hi
jagy@phobeusjunior:~$ ./isotropy_binaries_combined 2 113 1200 | sort -n
x y z recipe u v
29 22 -12 < 29, 63, 22 > 1 0
32 18 -11 < 32, 61, 18 > 1 0
37 8 -6 < 37, 51, 8 > 1 0
38 4 -3 < 38, 45, 4 > 1 0
188 171 -86 < 37, 51, 8 > 1 2
211 144 -82 < 38, 45, 4 > 1 2
226 123 -76 < 32, 61, 18 > 1 2
243 94 -64 < 29, 63, 22 > 1 2
246 88 -61 < 38, 45, 4 > 2 1
258 59 -44 < 37, 51, 8 > 2 1
264 38 -29 < 29, 63, 22 > 2 1
268 11 -6 < 32, 61, 18 > 2 1
396 262 -151 < 37, 51, 8 > 1 3
432 209 -134 < 38, 45, 4 > 1 3
472 129 -94 < 29, 63, 22 > 3 1
489 76 -58 < 32, 61, 18 > 3 1
516 458 -233 < 38, 45, 4 > 2 3
526 447 -232 < 37, 51, 8 > 2 3
628 311 -198 < 38, 45, 4 > 3 2
656 262 -177 < 32, 61, 18 > 2 3
671 232 -162 < 37, 51, 8 > 3 2
692 183 -134 < 29, 63, 22 > 2 3
726 47 -32 < 32, 61, 18 > 3 2
727 36 -22 < 29, 63, 22 > 3 2
804 787 -382 < 32, 61, 18 > 1 5
894 688 -373 < 29, 63, 22 > 1 5
953 946 -456 < 38, 45, 4 > 3 4
1034 492 -317 < 37, 51, 8 > 1 5
1062 443 -296 < 29, 63, 22 > 5 1
1102 363 -256 < 38, 45, 4 > 1 5
1123 314 -228 < 32, 61, 18 > 5 1
1159 1046 -528 < 32, 61, 18 > 1 6
1179 118 -88 < 38, 45, 4 > 5 1
1188 19 2 < 37, 51, 8 > 5 1
1199 1002 -524 < 29, 63, 22 > 1 6
well, then. I put in some blank lines..
Here is a necessary and sufficient condition for $G^{-}(f)$ to be non-empty, taken from [1, Exercise 6.21]:
Let $S = \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix}$ with $a,b$ and $c \in \mathbb{Z}$. Then the following are equivalent:
$(1)$ There exists $A \in \text{GL}_2(\mathbb{Z})$ with $\det(A)= -1$ such that $^tASA = S$.
$(2)$ The matrix $S$ is $\text{SL}_2(\mathbb{Z})$-equivalent to $\begin{pmatrix} a & -b/2 \\ -b/2 & c \end{pmatrix}$.
$(3)$ There exists $S' = \begin{pmatrix} a' & b'/2 \\ b'/2 & c' \end{pmatrix}$ with $a', b'$ and $c' \in \mathbb{Z}$ which is $\text{SL}_2(\mathbb{Z})$-equivalent to $S$ such that $b'$ is divisible by $a'$. (Here we may take $b' = a'$ or $b' = 0$.)
Hint: If $\,^tASA = S$ with $\det(A)= -1$, then show that $\text{Tr}(A) = 0$ and $A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
Then reduce the case when $A$ is upper triangular.
If the above conditions hold, the $\text{SL}_2(\mathbb{Z})$-class to which such an $S$ belongs is called an ambig class (sic). This is also what John Robertson calls an ambiguous class in [2]. From this source, we get
Let $D \in \mathbb{Z}$ which is not a square and such that $D \equiv 0$ or $D \equiv 1 \,(\text{mod } 4)$. Then the number of ambiguous classes is the order of the genus group of $D$, that is, the order of $H^{+}(D) \otimes \mathbb{Z}/2\mathbb{Z}$ where $H^{+}(D)$ is the strict class group of $D$.
[1] T. Ibukiyama, M. Kaneko, "Quadratic Forms and Ideal Theory of Quadratic Fields" in Bernoulli Numbers and Zeta Functions, pp 75-93, 2014.
[2] J. Robertson, "Computing in Quadratic Orders", 2009.
Best Answer
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.