Isotropic ternary forms

It is well known that some questions about isotropic ternary forms reduces to the study of the special case $f_0(X)=xz-y^2, X=(x,y,z)$, see page 301 of Cassel's "Rational quadratic forms" (Dover, 2008). In page 303 it is said that given any other general integral isotropic form $f(X)$, there is an integer $m\neq0$ and a nonsingular $3\times 3$ integral matrix $M$ such that $$mf(X)=f_0(MX).$$ The question is how to compute explicitly $M$. When $f(X)=ax^2+y^2-z^2$ it is quite easy to find $M$. In general, for $f(X)=ax^2+by^2-z^2$, $a>b>1$, it becomes more difficult. Of course, since the form is isotropic, by Legendre's Theorem, $a$ and $b$ satisfy the three relations: $$(i) \,\,\, aRb \, , \,\,\,\,\,(ii) \,\,\, bRa \, , \,\,\,\,\,(iii) \,\,\, (-ab)/d^2Rd$$ where $R$ is the equivalence relation $mRn$ if, and only if, $m$ is a square modulo $n$ and $d=d(a,b)=g.c.d(a,b)$. Any general formula to obtain $M$ for a specific isotropic ternary $f(X)$ should take this relations into consideration.

For example, the ternary form $f(X)=19x^2+5y^2-z^2$ is isotropic since $19+45=64$ and $76+5=81$ (this ternary form was considered in Integral orthogonal group for indefinite ternary quadratic form). What is $M$ for this particular ternary form?

• Use stereographic projection. – Daniel Loughran Jun 1 '15 at 15:33
• I suppose it would help to clarify matters if the OP indicates if we are meant to assume a rational point is given to us (in which case Daniel L's simple suggestion seems most efficient) or if we are just given the local congruence conditions; in effect, how do we know isotropicity holds? – grghxy Jun 2 '15 at 1:07
• I wrote my own software for this task. There is no easy answer to what $m$ or $M$ needs to be. I recently did a rather large project on $f(X) = A (x^2 + y^2 + z^2) - B(yz+zx+xy)$ for integers $B > A>0,$ which is isotropic when both $B+2A$ and $B-A$ can be written as $u^2 + 3 v^2.$ Final, the methods are all in Fricke and Klein (1897). – Will Jagy Jun 3 '15 at 1:04

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..

Gauss' Disquisitiones gives a completely algorithmic solution to this question in his development of reduction theory for ternary quadratic forms over the integers (via artful systematic use of his binary reduction theory applied to the xy-part, yz-part, and xz-part in appropriate order, along with those quadratic congruence conditions). Is there a reason his algorithm is insufficient (other than that it seems nobody has ever "coded it up", for reasons I do not know)?

• Thank you for your comments. But I still do not know how to obtain $M$ even for the the example. All I have is the congruence conditions from the fact that $f$ is isotropic. @grghxy, could you please be more specific about your reference to Gauss' Disquisitiones? – user148455 Jun 2 '15 at 10:59
• @user148455: It is classical via projection from a rational point (as Daniel Loughran suggests) to build a birational isomorphism between a conic with a rational point and the projective line, so composing one such with the inverse of another gives the desired isomorphism between conics. Since you have rational points in your case of interest, that is what you should do. If all you have is passage of congruence tests, then follow Gauss; look at the table of contents and flip the pages of an actual copy of the book and you'll find what is needed spread over many sections. – grghxy Jun 3 '15 at 2:50
• @Will Jagy, Thanks a lot for your answer. I have tried to prove the formula in Cassel's book but I have failed. I am also trying to locate the conference proceedings edited by Olga Taussky you mentioned. Of course, a specific example (apart the trivial ones when one of the positive coefficients is a square) would be very interesting and helpful. – user148455 Jun 3 '15 at 19:05

Wednesday morning, June 3.

I will be back home tonight, probably tomorrow I can provide the specific example requested.

The fact in Cassels is correct. It goes back to the seminal book by Fricke and Klein in 1897. It is the first volume of two on automorphic forms. Volume I is about group theory. There is quite a lot. One major thread is the relation between $SL_2 \mathbb Z$ and the integer automorphism group of a form that is already known to be isotropic over $\mathbb Q,$ with integer coefficients.

Let $H$ be the Hessian matrix of second partial derivatives of your $f_0$ by the chosen variables $x,y,z.$ Let $G$ be the Hessian of $f.$ The matrix $M$ and nonzero integer $m$ are to satisfy $$M^t H M = m G.$$ The theorem is that such exist. One constraint is automatic, $m$ must lie in a fixed integer squareclass. If we write $t = \det M,$ $h = \det H,$ $g = \det G,$ we arrive at $$h t^2 = m^3 g.$$ If some $t_0$ is the integer of smallest absolute value for which there is an integer solution to $h t_0^2 = m^3 g,$ we know that any successful $t$ must be of the form $t = t_0 w^2.$

There is something special for isotropic ternaries here: if $\gcd(x,y,z) = 1$ and $xz-y^2 = 0$ and either $x \geq 0$ or $z \geq 0,$ then there are integers $u,v,$ not both zero and with $\gcd(u,v) = 1,$ such that $$x = u^2; \; \; y = uv; \; \; z = v^2.$$ It is for this reason that finding the transition matrix $M$ leads immediately to all rational isotropic vectors of $f,$ just plug in $u,v.$ The hard part, and it is hard, is going from rational solutions to primitive integer solutions. Ummm, I may have lost track of the direction here, the useful direction for what i am describing is $H = r N^T G N,$ where the $H$ refers to $xz-y^2$ and $r$ is rational. One thing this explains completely is why the integer null cone for an isotropic ternary always comes out in terms of binary quadratic forms in some $u,v,$ and the hard part is then "What is the gcd of the three entries $x,y,z?$"

All this, and a good deal more, is in Fricke and Klein in German, in Wilhelm Magnus, Noneuclidean Tessellations and their Groups, and in a conference proceedings edited by Olga Taussky, with major article by Zassenhaus. I cannot remember all details on the Taussky book, I will check later.

Ternary Quadratic Forms and Norms edited by Olga Taussky (1982). Pages 5-30 is William Plesken, Automorphs of Ternary Quadratic Forms. The word automorph is one of the traditional terms for what would now be called a member of the (integer) automorphism group of the form, or rotation group, or orthogonal group.

Since $xz - y^2$ has the mixed coefficient $1,$ we need to double it to get an integral matrix, and this is also the Hessian matrix,

$$H = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & -2 & 0 \\ 1 & 0 & 0 \end{array} \right)$$

This becomes annoying for diagonal forms, where we still double the diagonal entries: $$G = \left( \begin{array}{rrr} 38 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & -2 \end{array} \right)$$

In this direction, take $$N = \left( \begin{array}{rrr} 38 & 30 & 16 \\ 19 & -25 & -7 \\ -38 & 5 & 9 \end{array} \right)$$ for one of infinitely many solutions to $$N^T H N = -95 G$$

give me a few more minutes...

In the other direction, take $$M = \left( \begin{array}{rrr} 2 & 2 & -2 \\ 3 & -8 & -1 \\ 11 & -4 & 9 \end{array} \right)$$ then $$M^T G M = -380 H$$

• Thank you for this computation. About double the matrices in order to get integer matrices consider the family $f=ax^2+by^2-z^2$. If, say, $b$ is a square, the form becomes (variables change) $g=ax^2+y^2-z^2$, with $a>1$. Surely, it is isotropic. Denote $H'$ the matrix of $f_0=xz-y^2$ (without double). Then, $N=\left(\begin{array}{ccc} 0 & a & a \\ a & 0 & 0 \\ 0 & -1 & 1 \end{array} \right)$ verifies $N^{t}H'N=-aG$. – user148455 Jun 4 '15 at 17:00

The mention of another quadratic form. You can use the standard approach.

In equation $$aX^2+bY^2+cZ^2=qXY+dXZ+tYZ$$

$a,b,c,q,d,t$ integer coefficients which specify the conditions of the problem.

For a more compact notation, we introduce a replacement.

$k=(q+t)^2-4b(a+c-d)$

$j=(d+t)^2-4c(a+b-q)$

$n=t(2a-t-d-q)+(2b-q)(2c-d)$

Then the formula in the general form is:

$$X=(2n(2c-d-t)+j(q+t-2b\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{n})ps+$$

$$+(2b-q-t\pm\sqrt{k})s^2$$

$$Y=(2n(2c-d-t)+j(2(a+c-d)-q-t\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{ n })ps+$$

$$+(q+t+2(d-a-c)\pm\sqrt{k})s^2$$

$$Z=(j(q+t-2b\pm\sqrt{k})-2n(2(a+b-q)-d-t))p^2+$$

$$+2((2(a+b-q)-d-t)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2$$

And more.

$$X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+$$

$$+(d+t-2c\pm\sqrt{j})s^2$$

$$Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2+$$

$$+2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2$$

$$Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps+$$

$$+(2(a+b-q)-d-t\pm\sqrt{j})s^2$$

$p,s$ are integers and are given us.

Solution when there is at least one root of a whole. For this it is necessary to consider the possible equivalent forms. This is done by changing the type $X \longrightarrow X+kY$

The desired coefficients $k$ can be found by solving a simpler equation. As a rule, either the Pell equation or representation of a number as a sum of squares. The equation is much simple than the original.

Not entirely clear where one detail which is not mentioned. Ternary quadratic form always amounts to a Pell equation. For example if you take a fairly simple equation. And set some conditions for the coefficients.

All of numbers can be any character.In Equation: $qX^2+Y^2=Z^2+a$

If the ratio is factored so: $a=(b-c)(b+c)$

Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$

where: $f=(q+1)k^2-2kt-(q-1)t^2$

Then the solutions are of the form:

$$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$$

$$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$$

$$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$$

All of numbers can be any character.

For the equation: $qX^2+Y^2=Z^2+j$

In the case where a square: $a=\sqrt{\frac{j}{q}}$

Using equation Pell: $p^2-(q+1)s^2=1$

Then the solution can be written:

$$X=2s(s\pm{p})L\pm{ap^2}+2aps\pm{a(q+1)s^2}=bL+af$$

$$Y=(p^2\pm2ps+(1-q)s^2)L\pm{ap^2}+2aps\pm{a(q+1)s^2}=cL+af$$

$$Z=(p^2\pm2ps+(q+1)s^2)L\pm{ap^2}+2a(q+1)ps\pm{a(q+1)s^2}=fL+at$$

$L$ - any integer number given by us.

The most interesting thing is that these numbers are solutions of equations:

$qb^2+c^2=f^2$

$t^2-(q+1)f^2=\pm{q}$

If we use the equation Pell: $p^2-(q+1)s^2=k$

And substituting the solutions in the upper formula, we have solutions of the following equations.

$qb^2+c^2=f^2$

where: $c-b=k$

$t^2-(q+1)f^2=\pm{qk^2}$

True, I use this formula in reverse order. Find solutions of Pell's equation is much more complicated than the simple equations like Pythagorean triples. So find them and then have solutions of Pell's equation. The most interesting thing is that the solution of Pell related to Pythagorean triples.

You can also write a more General formula. In this case it will be necessary to consider all the possible equivalent forms. In this case, anyway. The problem is reduced to solving a Pell equation. I think it's easier to solve.

This approach makes it quite easy to prove that the curves for triangular numbers is always possible to write the solution. If the coefficients of the 1-St degree is not equal to 0. And the coefficients of the 2nd degree don't create a trivial situation. The formula there. https://math.stackexchange.com/questions/794510/curves-triangular-numbers