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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?

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    $\begingroup$ What is the connection with representation theory? $\endgroup$ – Douglas Zare Dec 31 '13 at 14:58
  • $\begingroup$ The problem is linked to representation theory via invariant theory. $\endgroup$ – Rampant_mouse Jan 2 '14 at 14:05
  • $\begingroup$ Nice question. But the two posted `answers' are really very far from answers and -- in my opinion -- should be deleted. $\endgroup$ – J. Martel Jan 2 '14 at 20:30
  • $\begingroup$ Criterion in terms of what? Foe $n=2$, there is a classification; it may serve as a source for further guessing: Alex Degtyarev. Quadratic transformations Rp2 --> Rp2. In: Topology of Real Algebraic Varieties and Related Topics, Amer. Math. Soc. Transl. (2), 173 (1996), 61-73. $\endgroup$ – Alex Degtyarev Jan 2 '14 at 22:54
  • $\begingroup$ @AlexDegtyarev: Could you be so generous as to elaborate in the form of an answer. I am unclear why the pull-back is quadric, why I must hope for degeneration, and what is a `double' hyperplane? $\endgroup$ – J. Martel Jan 3 '14 at 17:08
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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 $\mathbb{R}^{2m+1}\to\mathbb{R}^{2m+1}$?

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    $\begingroup$ This is a nice remark, but it does not settle the case of even dimension. The OP asks for a criterium... $\endgroup$ – Pietro Majer Dec 31 '13 at 12:02
  • $\begingroup$ Still I am curious to know whether the map can be surjective for $n$ odd $\geq 3$? $\endgroup$ – abx Dec 31 '13 at 15:13
  • $\begingroup$ @abx Are you asking if a surjective map exists for odd n >= 3 ? $\endgroup$ – joro Dec 31 '13 at 15:32
  • $\begingroup$ Yes (a quadratic map). $\endgroup$ – abx Dec 31 '13 at 15:34
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    $\begingroup$ y1 = 2 * x1 * x3; y2 = 2 * x2 * x3; y3 = x3^2 - x1^2 - x2^2. To see it is enough to write it in polar coordinates. For higher dimensions the situation is same yj = 2 * xj * xn; yn = xn^2 - squares of other variables $\endgroup$ – Rampant_mouse Jan 2 '14 at 14:33
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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).

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    $\begingroup$ 1) Here the polynomial mapping is polynomial, but of quite a special form. Is the quoted proposition true even in this very particular class? 2) The OP means a map whose coordinates are quadratic forms, i.e. homogeneous polynomials of degree 2. If we allow linear terms then it is very easy to build examples of invertible maps. $\endgroup$ – Pietro Majer Dec 31 '13 at 16:05
  • $\begingroup$ @PietroMajer thanks, I see, misread the question. $\endgroup$ – joro Dec 31 '13 at 16:11
  • $\begingroup$ @joro: in my opinion your `answer' needlessly clutters this particular question and deserves to be deleted. $\endgroup$ – J. Martel Jan 2 '14 at 20:45
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    $\begingroup$ Disagree with J. Martel, at least for me, joro's answer nice "sharing of knowledge" even if it does not precisely match the question $\endgroup$ – Alexander Chervov Jan 4 '14 at 12:11
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    $\begingroup$ The paper you quote has what I consider a misleading framing of the problem: Their map contains a computable number $a$ and the map will be surjective iff $a \neq 0$; there is no algorithm which takes a description of a computable number and determines whether it is zero. If you give your mapping $f$ in a more concrete manner (for example, with coefficients in $\mathbb{Q}$), then this will be computable by Tarski's Theorem mathworld.wolfram.com/TarskisTheorem.html . $\endgroup$ – David E Speyer Jan 13 '14 at 18:37
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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.

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  • $\begingroup$ In your case $n=1, 2$, which is the case $n=2,3$ in the original question, (1) would it be possible to reformulate your necessary and sufficient condition purely in terms of the original quadratic forms? (2) The example provided by @Glasby the quadratic forms $x_1^2-x_2^2$ ad $2x_1x_2$ are both of trivial inertia index, in the last paragraph before stating your theorem you are talking about the negative inertia index of which quadratic form? Thank you. $\endgroup$ – Name Jan 13 '14 at 11:42
  • $\begingroup$ In a sense, it is in terms of the original forms, as they span the linear system. But otherwise the answer is "no", as you can always choose those forms very close to each other, so that they'll be equivalent algebraically. It's the system that matters! Concerning the example, do not mix inertia index (number of negative squares in the diagonal form) and signature: in the example, both indices are equal to 1. A binary form of index 0 is $x_1^2+x_2^2$. $\endgroup$ – Alex Degtyarev Jan 13 '14 at 12:02
  • $\begingroup$ Let me illustrate the case $n=1$. All binary forms form a plane $\mathbb{R}\mathrm{p}^2$, and the discriminant is a conic in this plane. A pencil is a line, and what matters is whether this line intersects the conic. Obviously, any line is spanned by two points outside the conic, and separately any such form is equivalent to $x_1^2-x_2^2$. Changing the coordinates in the target, you can always say that the other form is arbitrary close to the first one; it's the direction that matters! $\endgroup$ – Alex Degtyarev Jan 13 '14 at 12:11
  • $\begingroup$ I was thinking about an explanation why the example provided by Glasby works, based on your necessary and sufficient condition for $n=1$. Thank you anyway, also for your thoughtful answer. $\endgroup$ – Name Jan 13 '14 at 15:17
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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.

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  • $\begingroup$ Just noticed that Rampant_mouse already answered this in a comment. $\endgroup$ – David Wehlau Jan 4 '14 at 11:22
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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.

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  • $\begingroup$ I think it is interesting that you state the main results here, if you can. $\endgroup$ – Amir Sagiv Sep 5 '16 at 13:56
  • $\begingroup$ Concerning quadratic mappings acting from $\mathbb{R}^3$ to $\mathbb{R}^3$ in the above mentioned paper it is proved that the following are equivalent: (i) quadratic mapping $Q$ is surjective; (ii) surjectivity of $Q$ is stable, i.e. any “close” to $Q$ quadratic mapping is surjective; (iii) $Q(x)\neq 0$ for each $x\neq 0$, and for each regular value $y$ of the mapping $Q$ the set $Q^{−1}(y)$ consists of 2 or 6 points. $\endgroup$ – Qwerty Sep 6 '16 at 8:11
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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).

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  • $\begingroup$ I don't think your criterion is correct. In particular, $q_1, q_2$ regular and not simultaneously diagonalizable does not imply that either of $q_1, q_2$ are indefinite (which certainly is necessary!). $\endgroup$ – J. Martel Jan 24 '14 at 15:01
  • $\begingroup$ @J.Martel Thank you for your comment. It does imply! If $q_1$ and $q_2$ are both regular, and at least one of them is definite then they are simultaneously diagonalizable (see Greub, Linear Algebra, 1967, Third edition, Ch. IX, p.256). $\endgroup$ – Name Feb 1 '14 at 8:06

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