Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Question : Does there exist a surjective function $F$ that maps $\mathbb{R}^n_+$ to $\mathbb{R}^n$ (where $\mathbb{R}^n_+$ denotes the set of vectors of length $n$ with only positive entries). The answer is yes by considering the function $F$: $$(x_1,\ldots,x_n)\to(\log x_1,\ldots,\log x_n)$$ It is easy to see that $F$ is surjective.

Now my real question is if there is such a function $F$ whose components are restricted to be polynomials of degree at most $n$ (and $F$ maps $\mathbb{R}^n_+$ to $\mathbb{R}^n$ and is surjective). For example: $$(x_1,x_2,x_3,x_4)\to(1-x_1,x_1\cdot x_2+x_3,x_3^2+x_2-x_4,5x_4)$$ has polynomial components, but is not surjective as the restriction $x_i>0$ implies we cannot get a $1$ in the first coordinate.

Any help or advice is greatly appreciated. I think the answer is no but am not sure how to prove it.

share|improve this question
4  
Sometimes the answer is yes. In two dimensions, we can take the real and the imaginary part of $[z-(1+i)]^4$ where $z=x_1+ix_2$, which immediately takes care of all even $n\ge 4$ (the degree 4 is to high to work for $n=2$ itself. –  fedja Sep 15 '10 at 21:09
2  
This seems to be a duplicate of mathoverflow.net/questions/11904/… –  j.c. Sep 15 '10 at 21:51
    
Well, except for the degree restriction. –  j.c. Sep 15 '10 at 21:54

2 Answers 2

up vote 14 down vote accepted

Complete Version

This is motivated by the complex cubing map described below, which generalizes to the cubing map on quaternions:

For $n > 1$, the map analogous to cubing a complex number or a quaternion is f_3: $ (x_1, x_2, \dots, x_n) \rightarrow (x_1^3 -3 x_1( x_2^2 + x_3^2 + \dots + x_n^2), -x_2^3 + 3 x_2 x_1^2, \dots, - x_n^3 + 3 x_n x_1^2)$. This map takes any plane through the $x_1$ axis to itself, and acts like the complex cubing map in such a plane.

Therefore, the image of the cone $C_{60}$ within a 60 degree angle of the $x_1$ axis is all of $\mathbb R^n$.

There are linear transformations that take the positive orthant to nearly all of a half-space, in particular, there are linear images that contain $C_{60}$.

Composing, we get a surjective homogeneous degree 3 polynomial map from the positive orthant to all of $\mathbb R^n$.

The dimension 1 case is impossible with any degree polynomial, as discussed in Richard Palais's response, and the degree 2 cases cannot be done with degree < 3, as discussed below.

NEW VERSION

earlier version at end.

For n = 2, if we consider a homogeneous quadratic polynomial, it acts on lines through the origin as a rational map of degree 2. This is either a double cover map of $S^1$ to itself, equivalent to the complex map $z \rightarrow z^2$ , or it folds the circle in half to an interval. The only chance it has to be surjective is the double cover case. The image of any interval under the complex squaring map may be surjective on the circle of lines, but when lifted to its double cover, the circle of rays, its image is an interval, since there are lines which are only hit once. Therefore, no homogeneous degree 2 polynomial is surjective.

Adding a linear or constant term doesn't change the limiting action on rays unless the homogeneous part is degenerate and is 0 on some line, as for instance $(x,y) \rightarrow (0, y^2)$ (up to linear transformations in the domain and range). Adding lower degree terms obviously can't make this surjective. So, for $n=1$ or $2$, the answer is NO.

There are degree 3 polynomial maps of $\mathbb R^2$ to itself that take the positive quadrant to the whole plane: first, compress the plane to fatten the quadrant to make an angle more than 120 degrees, then cube it as a complex number.

In even dimension $2m \ge 4$, a solution in complex coordinates, as noted by fedja in comments, is $(z_1, \dots, z_m) \rightarrow (z_1^4, \dots, z_m^4)$, and a solution of degree 3 is obtained by performing the degree 3 map above coordinatewise.

For odd dimensions $2m+1 \ge 7$, a solution of degree 6 can be obtained as the degree 3 solution on $2m$ coordinates followed by the complex squaring map on the last coordinate with any of the first $2m$. The degree of the composition is 6.

This leaves dimensions 3 and 5 unanswered. Maybe with a little more cleverness?...

Earlier Version

Note, added: I was hasty reading the question, and didn't pay attention to degree restrictions. The comment of fedja to the question took this into account, and answered it as well as this. The missing cases at the moment seems to be $n = 3, 5, 7$.

In the complex plane, the map $z \rightarrow z^4$ maps the positive quadrant surjectively to $\mathbb C$. Expressed in terms of real and imaginary parts, this satisfies your question for dimension $n = 2$.

You can get a positive answer for any $n \ge 2$ by composing copies of this function applied to pairs of coordinates (the pairs can overlap). For dimension $n = 1$, it is impossible

share|improve this answer
1  
The degree is bounded by the dimension! Also, how do you show that you have full image when the pairs overlap? –  fedja Sep 15 '10 at 21:45
    
Very nice argument ! I'll add a few words below on why it is false for $n = 1$. –  Dick Palais Sep 15 '10 at 21:47
1  
If you apply the $z \rightarrow z^4$ map to the first two coordinates, the image is an $\mathbb R^2 \times $ an orthant in the other coordinates. Now leave the first coordinate alone: we have at least an orthant in any other pair of coordinates. The most efficient way I see to use this technique would be to keep using disjoint pairs of coordinates until at most one is left over. So far it's degree 4. Combine the last coordinate with any one of the earlier ones, and you just have to square. This seems to make it degree 8 for $n$ odd, degree 4 for $n$ even. –  Bill Thurston Sep 15 '10 at 22:15
2  
@Spiked Math. Would coordinate k < degree k be good for something? If it can be disproved, would that imply the 2n-conjecture? What is your field, and what is the 2n-conjecture? It was a fun question to think about. –  Bill Thurston Sep 16 '10 at 17:38
1  
@Bill. If it can be disproved for coord $k$ $\leq$ degree $k$, then this would imply that the 2n-conjecture is true. The original problem is an inverse eigenvalue problem which deals with the construction of a matrix with prescribed eigenvalues (where the matrix has a specified property). The polynomial and degree constraint on $F$ comes from looking at the characteristic polynomial and viewing the coefficients as polynomials in the matrix entries. More information is available at www.uwyo.edu/bshader/notes.pdf –  Spiked Math Sep 17 '10 at 3:16

For n = 1, it is well-known and obvious that a polynomial $P(x)$ approaches either $+ \infty$ or else $-\infty$ as $x \to \infty$, so in particular it has all its values in either the positive half-axis or else in the negative half-axis for say $x > M$. But then it cannot cover the whole other half-axis on the compact interval $[0,M].

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.