System of $ n $ simultaneously diagonal real quadrics in $ n+1 $ variables has all solutions real

Question

A single isotropic real quadric in 2 variables always has 2 (projective) real solutions. See Zero set of system of two real quadratic forms for the explicit form.

I've noticed that a system of 2 simultaneously diagonal isotropic real quadrics in 3 variables always seems to have 4 (projective) real solutions.

By a diagonal quadric I mean something with no cross terms, for example $ x^2+y^2-z^2 $

Is it true in general that a system of $ n $ simultaneously diagonal isotropic real quadrics in $ n+1 $ variables always has $ 2^n $ (projective) real solutions?

Note: I know this isn't strictly true. For example $ x^2+y^2-z^2,x^2-y^2+z^2 $ is a system of 2 simultaneously diagonal isotropic real quadrics in 3 variables that only has 2 (projective) real solutions. However, somehow this doesn't seem to happen in the generic case? Perhaps this is "non-generic" in the sense the ideal generated by these quadrics contains a non-isotropic quadric. In particular, the Groebner basis includes the definite form $ x^2 $. In fact $ x^2 $ is even just a linear combination. Maybe there is some restriction to be mentioned here about the pencil generated by the quadrics not including any non-isotropic quadrics? Or would it be too strong to assume that? almost like assuming the conclusion

So perhaps a more refined question is:

Under what additional hypotheses on the quadrics can we conclude that a system of $ n $ simultaneously diagonal isotropic real quadrics in $ n+1 $ variables always has $ 2^n $ (projective) real solutions?

Answer 1

Suppose we have a system of $n$ diagonal quadratic equations along with a normalization condition in $n+1$ variables: \begin{align*} \sum_{j=1}^{n+1} \alpha_{ij} x_{j}^2 &= 0, \qquad i=1, \cdots, n \\ \sum_{j=1}^{n+1} x_j^2 &= 1. \end{align*} We can convert this into a linear system by defining the variable $y_j = x_j^2$. Thus consider the following auxiliary system of $n+1$ linear equations in $n+1$ variables: \begin{align*} \sum_{j=1}^{n+1} \alpha_{ij} y_j &= 0, \qquad i=1, \cdots, n \\ \sum_{j=1}^{n+1} y_j &= 1. \end{align*} This is a matrix equation $A y = b$ where $(A)_{ij} = \alpha_{ij}$ for $i=1,\cdots, n$ and $(A)_{n+1,j} = 1$ and $b = (0, \cdots, 0, 1)^T$.

If you assume the quadratic forms are linearly independent then $\det(A) \neq 0$ so there will be a unique solution to the linear system. But because $y_j = x_j^2$ then $x_j = \pm \sqrt{y_j}$ so that there are in general $2^{n+1}$ complex solutions to the quadratic system (counting multiplicity of zeroes). In particular this means there are $2^n$ complex projective solutions which is the result of Bezout's theorem.

However, we can also determine if the solutions are real by making the following observations: if $y_j < 0$ then both of the corresponding $x_j$'s will be imaginary, if $y_j = 0$ then $x_j = 0$ with multiplicity 2, and if $y_j > 0$ then both of the corresponding $x_j$'s are real. Thus we see that all of the $y_j$'s must be non-negative in order to guarantee real solutions.

Thus we arrive at the final result. If $y \geq 0$ and if $m$ is the number of (strictly) positive $y_j$'s in the solution to the linear system, then there are $2^{m-1}$ projective real solutions (each occurring with multiplicity $2^{n+1-m}$).

In your example $x^2 + y^2 - z^2 = 0$ and $x^2 - y^2 + z^2 = 0$ we have the linear system $$ \underbrace{\begin{pmatrix}1 & 1 & -1 \\ 1 & -1 & 1 \\ 1 & 1 & 1 \end{pmatrix}}_A \underbrace{\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} }_y = \underbrace{\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}}_b. $$ We have $\det(A) = 4$ and the solution is $$ y_1 = 0, \quad y_2 = 1/2, \quad y_3 = 1/2. $$ We see that $y \geq 0$ so the solution to the system of quadratic equations is real. Moreover, there are $2$ positive solutions and so there are $2^{2-1} = 2$ projective real solutions to the corresponding quadratic system as you noticed.

So I guess the reduction of your question becomes: when does a linear system $A y = b$ have a solution such that $y > 0$?