Are there any known criteria for quadratic mapping from R^n to R^n being surjective?

Question

Let quadratic mapping be the function from $\mathbb{R}^n$ to $\mathbb{R}^n$, where each coordinate is a quadratic form of $n$ variables. Are there any known criteria for it being surjective? May somebody give relevant references too?

Answer 1

If $n=2$ the quadratic map $\mathbb{R}^2\to\mathbb{R}^2$ with $(x_1,x_2)\mapsto (x_1^2-x_2^2,2x_1x_2)$ is surjective. This follows because the map $\mathbb{C}\to\mathbb{C}$ with $z\mapsto z^2$ is surjective. Hence there exist real quadratic maps $\mathbb{R}^{2m}\to\mathbb{R}^{2m}$ for all even values of $n$. (Identify $\mathbb{R}^{2m}$ and $\mathbb{C}^m$ and consider $(z_1,\dots,z_m)\mapsto(z_1^2,\dots,z_m^2)$.) For $n=1$ a quadratic map $\mathbb{R}\to\mathbb{R}$ is not surjective. The question can thus be rephrased: for which values of $m\geq1$ is there a real quadratic surjective map $\mathbb{R}^{2m+1}\to\mathbb{R}^{2m+1}$?

Answer 2

Some comments.

Check this question and the comments.

A paper from the answer

Proposition 6. No algorithm is possible that, given a polynomial mapping $f : \mathbb{R}^n \to \mathbb{R}^n$ with computable coefficients, decides whether this mapping is surjective.

On the other hand another answer gives relatively efficient criterion for deciding if the map is bijective over $\mathbb{Q}$ (this implies surjective).

Answer 3

I do not know the answer in general but, just for fun, I tried to analyze the small dimensions. The results call for an obvious conjecture. This may be known, but I haven't seen this in the literature.

A quadratic map (given by homogeneous quadratic forms) $\varphi\colon\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ is essentially the same as an $n$-dimensional linear system $L_\varphi$ of quadrics in $\mathbb{R}\mathrm{p}^n$. We also have the projectivization, which is a rational map $\bar\varphi\colon\mathbb{R}\mathrm{p}^n\to\mathbb{R}\mathrm{p}^n$. For the sake of simplicity, let us assume that $L_\varphi$ has no basepoints, so that $\bar\varphi$ is regular (we do not need to blow up anything).

The first observation is that, assuming $n\ge1$, the original map $\varphi$ is surjective if and only if so is $\bar\varphi$. Indeed, we can restrict/normalize $\varphi$ to the unit spheres, and the key is the fact that the double covering $S^n\to\mathbb{R}\mathrm{p}^n$ is nontrivial.

Here is the common approach to the description of the topology of $L_\varphi$ (after Dixon, Agrachev, etc.) The spectral variety is the set $C\subset L_\varphi\cong\mathbb{R}\mathrm{p}^n$ of singular quadrics. Typically, it is a hypersurface of degree $n+1$ (possibly nonreduced), although in some very degenerate cases it may coincide with $L_\varphi$. Passing from quadrics to quadratic forms (and normalizing), we get a nontrivial double covering $S^n\to L_\varphi$ and the pull-back $\tilde C\subset S^n$ of $C$. On the complement $S^n\setminus\tilde C$ one has the index function, sending a point to the negative inertia index of the corresponding nondegenerate quadratic form. It takes values between $0$ and $n+1$, is symmetric in an appropriate sense, and has a lot of other nice properties that I will skip here.

Theorem. A necessary condition for a quadratic map $\varphi$ to be surjective is that the index function $\iota\colon S^n\setminus\tilde C\to\mathbb{Z}$ should not take value $0$. If $n\le2$ and $\varphi$ is generic, this condition is also sufficient.

Proof The necessity follows from the fact that the members of $L_\varphi$ are the pull-backs of hyperplanes in $\mathbb{R}\mathrm{p}^n$, and quadrics of index $0$ are empty.

The case $n=0$ is trivial: $\varphi$ is never surjective and $\iota$ always takes value $0$.

The case $n=1$ is also easy. Indeed, there are two projective classes of generic pencils, with the spectral curve $C\subset L_\varphi$ empty or not. In the former case, $\varphi$ is surjective; in the latter case, it is not and $\iota$ does take value $0$. (I can refer to my paper, but the fact is trivial and well known.)

Let $n=2$. A point $p$ in the target $\mathbb{R}\mathrm{p}^n$ defines a pencil $\ell\subset L_\varphi$, and the pull-back $\varphi^{-1}(p)$ is the base locus of $\ell$. Generic pencils of plane conics admit a simple projective classification (see my paper again); they are characterized by their basepoints. The index function of the only pencil with all four basepoints non-real does take value $0$; hence, such pencils do not occur if $\iota>0$. QED

Remark. Taking care of a few technicalities, I believe that the proof can be adjusted to non-generic linear systems. The big question is whether the sufficiency holds as well in higher dimensions.

Answer 4

I'll consider the question posed in Glasby's answer. Take $n=3$ and $\pi(x,y,z)=( (x-1)^2 - y^2, 2(x-1)y, xz )$. To see that $(u,v,w)$ lies in the image, note that there is always at least one pair $(x_0,y_0)$ such that $((x_0-1)^2-y_0^2,2(x_0-1)y_0) = (u,v)$ and $x_0 \neq 0$. To see this note that $(0,y_0) \mapsto (1-y_0^2,-2y_0)$ and $(2,-y_0) \mapsto (1-y_0^2,-2y_0)$. Thus $\pi(x_0,y_0,w/x_0) = (u,v,w)$.

Thus there exist surjective quadratic maps for all $n \neq 1$. This doesn't really answer the original request for a criterion though.

Answer 5

Similar questions were studied in the paper

A V Arutyunov, S E Zhukovskii, "Properties of real surjective quadratic mappings", SB MATH, 2016, 207 (9), DOI:10.1070/SM8611 .

In particular, for $n=3$ there are necessary and sufficient conditions for a quadratic mapping to be surjective. The paper is written in Russian. Probably it will be translated this year. If it is interesting, I can write a short summary.

Answer 6

A necessary condition is that if the mapping is $Q:x\mapsto(q_1(x),\cdots,q_n(x))$ where $x=(x_1,\cdots,x_n)\in\mathbb{R}^n$, then $q_1,\cdots,q_n$ are not simultaneously diagonalizable. In fact if not after a change base, we may assume that $Q(x_1,\cdots,x_n)=A\begin{bmatrix}x_1^2\\ \vdots\\ x_n^2\end{bmatrix}$, where $A\in M_n(\mathbb{R})$. But this map is obviously not surjective.

Using above we get the following necessary and sufficient condition for the case $n=2$.

$Q=(q_1,q_2)$ is surjective iff $q_1$ and $q_2$ are both regular and not simultaneously diagonalizable.

(The example of Glasby manifests this).