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

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?

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 $$y > 0$$?

• can you say more about this? For example how do we know that there is any nonnegative solution at all (other than the $0$ solution)? Can we use some version of Farkas lemma en.wikipedia.org/wiki/Farkas%27_lemma? I certainly agree that if there is any real solution then there are indeed $2^{n+1}$ real solutions on the unit sphere, by varying signs, which in turn means $2^n$ real solutions in $\mathbb{R}P^n$ because of global sign. So in this case either every projective solution is real or no projective solution is real. Sep 19 at 18:09