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Here's a question that (I hope) may seem very trivial for you, and I hope one of you may provide me with a reference answering it (unless it's a trivial colloquial knowledge).

Let $f$ be an indefinite ternary quadratic form that is anisotropic (does not represent 0).

By Dickson's theorem a universal (representing all positive and negative numbers) ternary $f$ is isotropic, and here we see that $f$ cannot be universal.

Question: can it happen that for any such $f$ there is a \textbf{negative} number $d$ which $f$ does not represent?

Or: is there an indefinite ternary quadratic form $f$ such that $f$ represents all negative integers, but is not universal (thus avoids some positive number $d$)?

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    $\begingroup$ Of course there is. In addition, you can say that infinitely many of them. You need to specify their form. Write down the quadratic form more specifically using undetermined coefficients. $\endgroup$ – individ Feb 26 '17 at 8:49
  • $\begingroup$ @individ: I'm interested to know if a form $f = a x^2 + b y^2 + c z^2$ with $a, b > 0$ and $c < 0$ can be "negatively universal" but avoid instead some positive integer. In principal, I need ternary quadratic forms with rational coefficients and look at which numbers they represent over rationals, although it seems my question won't change much in this setting. $\endgroup$ – SashaKolpakov Feb 26 '17 at 8:56
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    $\begingroup$ Once the coefficients, we can ask, then you can choose what prefer. They will be infinitely many. For example. Some solutions can be through the Pell equation ask. artofproblemsolving.com/community/c3046h1048910___2 Reduced ternary form to some Pell equation. Formula however get bulky.... There is another approach. Using the solution $aX^2+bY^2=cZ^2$ To build the needed solutions. $\endgroup$ – individ Feb 26 '17 at 9:17
  • $\begingroup$ That sounds interesting, although the quadratic form you brought up is isotropic and thus universal. My concern is that "negative universality" implies may be "total universality", but I fail to find a concrete example showing the opposite. $\endgroup$ – SashaKolpakov Feb 26 '17 at 13:56
  • $\begingroup$ Then write down the parameterization of the solutions of Diophantine equations in General. $aX^2+bXY+cY^2-jZ^2=q$ Then it will become clear how the factors are interrelated and pick the right. $\endgroup$ – individ Feb 26 '17 at 14:14
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Two parts. If the form is anisotropic, there is a specific prime (actually an even number of primes, by the product formula for the Hilbert Norm Residue symbol) $p$ for which the form is anisotropic. The problem is that the form does not integrally represent anything in the $p$-adic squareclass of the discriminant of the form. This follows from just a few pages in Cassels, rational Quadratic Forms, which I recommend; mostly pages 58-59.

Next, it is possible to have exceptions that do not directly arise from congruences. The famous 1951 example of Siegel is a genus with two forms, $$ x^2 - 2 y^2 + 64 z^2, $$ $$ (2x+z)^2 - 2 y^2 + 16 z^2. $$ Note that the binary $4x^2 + 4 xz + 17 z^2$ is indeed in the same genus as $x^2 + 64 y^2.$ However, while $x^2 + 64 y^2$ is a fourth power in the class group, $4x^2 + 4 xz + 17 z^2$ is a square but not a fourth power. Now, $(2x+z)^2 - 2 y^2 + 16 z^2$ fails to represent any $m^2,$ where all prime factors of $m$ are $\pm 1 \pmod 8.$ Now, with $ x^2 - 2 y^2 + 64 z^2, $ we can take $x=y=8, z=1$ to get $0.$ Isotropic. At the same time, neither form represents any $\pm 3 \pmod 8.$

A more recent example of Schulze-Pillot and Xu is $$ x^2 + 100 y^2 - 5 z^2, $$ $$ 4 x^2 + 25 y^2 - 5 z^2. $$ Note that the binary $4x^2 + 25 y^2$ is in the same genus as $x^2 + 100 y^2.$ However, while $x^2 + 100 y^2$ is a fourth power in the class group, $4x^2 + 25 y^2$ is a square but not a fourth power. Now, $4 x^2 + 25 y^2 - 5 z^2$ fails to represent any $m^2,$ where all prime factors of $m$ are $\pm 1 \pmod 5.$ Now, with $ x^2 + 100 y^2 - 5 z^2, $ we can take $x=5, y=1, z=5$ to get $0.$ Isotropic. At the same time, neither form represents any $\pm 3 \pmod 5.$

For the moment, I think that we cannot have spinor exceptional integers (as in the two examples) without congruence obstructions, even when all is isotropic. I do not have an immediate proof. Well, this seems to do it: the genus and the spinor genus coincide unless the discriminant is divisible by $64$ or by $p^3$ for odd prime $p.$ Furthermore, they still coincide unless the diagonalization in $\mathbb Q_p$ is $a x^2 + bp y^2 + c p^2 z^2,$ whereupon we see that half the values $\pmod p$ are not represented, all $n$ with Legendre $(n,p) = -(a,p).$ More messy for $2,$ generally need $\pmod 8.$

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  • $\begingroup$ Dear Will, thank you for this explanation. If I may, I would like to ask a more precise question: is there a reference to an example of a ternary quadratic form that represents all negative integers over Q, but avoids some positive integer (I need forms over Q mostly). If no reference exists, is it possible to come up with a concrete example? In my mind representing every negative integer (over Q, not integrally) is quite a strong property to be checked. $\endgroup$ – SashaKolpakov Feb 27 '17 at 3:55
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    $\begingroup$ @SashaKolpakov I do not believe that to be possible. That's what I was talking about in the last paragraph, beginning with "Well, this seems to do it." $\endgroup$ – Will Jagy Feb 27 '17 at 4:13
  • $\begingroup$ Dear Will, thank you again! I will need however some more information about the link between genera and spinor genera with representability (you consider the case when they don't coincide, but what happens if they do?) I'm sure you may provide me with a reference for further reading which will help understanding things and finishing the "impossibility" proof (I do need to write a proof to this fact and thus need to learn!) Although your help has already been great. Thank you! $\endgroup$ – SashaKolpakov Feb 28 '17 at 10:53
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    $\begingroup$ @SashaKolpakov I put a few dozen pdfs at zakuski.math.utsa.edu/~kap For indefinite ternaries, the big result of Eichler was that equivalence class and spinor genus coincide. If spinor genus and genus coincide, the numbers represented are specified entirely by congruences; your condition on negative numbers says there are no congruence obstructions. $\endgroup$ – Will Jagy Feb 28 '17 at 17:08
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    $\begingroup$ @SashaKolpakov meanwhile, it seems possible to me that the methods of Dickson suffice for this problem. So, unless John Voight takes an interest, I suggest you write to Gonzalo Tornaria, who has been helpful to me. cmat.edu.uy/~tornaria As far as Dickson (you did not state his result correctly) you mostly want his Studies (about 1930) and Modern Elementary (1939) $\endgroup$ – Will Jagy Feb 28 '17 at 18:49
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Representation of a number we write.

$$aX^2+bY^2=cZ^2+q$$

I think that the only way to record the desired polynomial is to use the solutions of any equation.

$$ax^2+by^2=cz^2$$

Knowing the solutions of this equation and substituting them into the linear Diophantine equation.

$$axs+byp-czk=1$$

$(s;p;k) - $ variables which are solutions of this equation. Then the solution of the first equation can be written as.

$$X=\frac{x}{2}(ck^2-as^2-bp^2+q)+s$$

$$Y=\frac{y}{2}(ck^2-as^2-bp^2+q)+p$$

$$Z=\frac{z}{2}(ck^2-as^2-bp^2+q)+k$$

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