Isotropic ternary forms

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..

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.

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)?

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 $$ M^T G M = -380 H $$

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