I have a field $K$ of transcendence degree two over $\mathbb{R}$, and elements $a_1,a_2,a_3\in K$. I would like to understand the set $$ Q = \{ u\in K^3 : \sum_i a_iu_i^2 = 1\} $$ In particular, I would like to know whether it is nonempty, and if so, I would like to find some examples of elements, and ideally some kind of formula for all elements. What general methods are known for this sort of question? My gut feeling is that $Q$ is likely to be empty for most choices of the parameters $a_i$; are there any theorems of that type?

Specifically, I am interested in the following case: \begin{align*} r &= y_2^3+(5y_1-9)y_2^2/5+(-20y_1+4y_1^2+24-16y_4)y_2/25-4(1-y_1)/25 \\ A &= \mathbb{R}[y_1,y_4,y_2]/(r) \\ K &= \text{field of fractions of } A \\ a_1 &= y_2(1-y_1-y_2) \\ a_2 &= y_4 \\ a_3 &= y_2. \end{align*}

Note that $A$ is a free module of rank three over the subring $A_1=\mathbb{R}[y_1,y_4]$, so $K$ is a free module of rank three over the purely transcendental subfield $K_1=\mathbb{R}(y_1,y_4)$. Using this, we could translate the question to one about a quadratic form in nine variables over $K_1$, which might or might not be more tractable.


Starting from Jason's observations, an obvious obstruction for the existence of a solution would be the existence of a point $M=(y_1,y_2)$ in $\mathbb{R}^2$ such that $b_1(M)>0$, $b_2(M)>0$, and $b_4(M)<0$. However, we have $$16 b_2=(4y_1^2)\,b_4-(5y_2-2)^2\,b_1$$ so whenever $b_1>0$ and $b_4<0$ we must have $b_2\leq0$ and this obstruction vanishes.

To put things differently, we have a smooth quadric fibration $f:Q\to U$ where $U$ is a suitable Zariski open subset of the real $(y_1,y_2)$-plane, and we are looking for a rational section of $f$. What I am saying is that, at least, for each $M\in U(\mathbb{R})$, the quadric $f^{-1}(M)$ has real points.

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    $\begingroup$ Wait, isn't that a solution? We're looking to solve $b_1^{\phantom2} v_1^2 + b_2^{\phantom2} v_2^2 + b_3^{\phantom2} v_3^2 = b_4^{\phantom2} v_4^2$ and the displayed identity says $(v_1,v_2,v_3,v_4) = (5y_2-2, 4, 0, 2y_1)$ works. $\endgroup$ – Noam D. Elkies Aug 31 '13 at 21:44
  • $\begingroup$ How could I miss that? $\endgroup$ – Laurent Moret-Bailly Sep 1 '13 at 7:00
  • $\begingroup$ Indeed, this gives a solution, thank you. Unfortunately I learn from it that my formulation of the problem was not optimal. Your solution gives $u_2=2/y_1$, and the denominator of $y_1$ prevents it from being useful for the intended application. I will have to think in more detail about which denominators would be harmless. Anyway, this at least gives me more insight into the structure of the problem. $\endgroup$ – Neil Strickland Sep 1 '13 at 13:15
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    $\begingroup$ Once you have one solution $(v_1,v_2,v_3,v_4)$, you can parametrize the others by the line joining any other solution $(w_1,w_2,w_3,w_4)$ to $(v_1,v_2,v_3,v_4)$. Often in such problems this gets you from a trivial or spurious solution to one that's relevant to your intended application. For starters it will probably give you a better handle on whether "useful" solutions exist. $\endgroup$ – Noam D. Elkies Sep 1 '13 at 15:02

There are some elementary observations. First of all, your equation $r=0$ allows you to eliminate the variable $y_4$. So the field $K$ is the same as the field $\mathbb{R}(y_1,y_2)$. Moreover, your quadratic form, $a_1u_1^2 + a_2u_2^2 + a_3u_3^2 - a_4u_4^2$ (where $a_4=1$) is not reduced. Multiply the equation by $y_2$, define $$v_1=y_2u_1,\ v_2 = u_2,\ v_3=y_2u_3,\ v_4=u_4.$$ Then your new defining polynomial is $f = b_1v_1^2 + b_2v_2^2 + b_3v_3^2 -b_4v_4^2 = 0$, where now $b_1$, $b_2$, $b_3$ and $b_4$ are actually integral elements in $\mathbb{R}[y_1,y_2]$, namely, $$b_1 = 1-y_1-y_2,\ b_2 = \frac{25}{16}(y_2^3+\frac{1}{5}(5y_1-9)y_2^2+\frac{4}{25}(y_1^2-5y_1+6)y_2 + \frac{4}{25}(y_1-1)), $$ $$b_3 = 1,\ b_4 = y_2.$$ Next, over the Zariski open $U=\text{Spec}\ \mathbb{R}[y_1,y_2][(b_1b_2b_3b_4)^{-1}]$, there is a finite, etale degree 2 morphism $\nu:V\to U$ and a relative Brauer-Severi variety $\pi:P_V\to V$ such that your quadric bundle over $U$ is the Weil restriction of $P_V$ via $\nu$. Thus existence of a $K$-rational point of your original quadric is equivalent to the triviality of the element $[P_V]$ in the Brauer group of $V$.

The Brauer group of a real quasi-projective surface has been studied, e.g., by Nikulin and Krasnov. First of all, there is the Brauer group of the corresponding complex surface, which has a local and a global part. If you want to prove non-existence, then the local part already may be sufficient.


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