A polynomial map from $\Bbb R^n$ to $\Bbb R^n$ mapping the positive orthant onto $\Bbb R^n$?

Question

Question: Is there a polynomial map from $\Bbb R^n$ to $\Bbb R^n$ under which the image of the positive orthant (the set of points with all coordinates positive) is all of $\Bbb R^n$ ?

Some observations:

My intuition is that the answer must be 'no'... but I confess my intuition for this sort of geometric problem is not very well-developed.

Of course it is relatively easy to show that the answer is 'no' when $n=1$. (In fact it seems like a nice homework problem for some calculus students.) But I can't seem to get any traction for $n>1$.

This feels like the sort of thing that should have an easy proof, but then I remember feeling that way the first time I saw the Jacobian conjecture... now I'm wary of statements about polynomial maps of $\Bbb R^n$ !

See http://mathoverflow.net/questions/38868/vector-valued-functions for some apparently minimal degree maps when n>1. — j.c., Sep 16 '10 at 02:24

score 16 · Accepted Answer · edited Apr 13 '17 at 12:58

16

The map $z\in\mathbb C\mapsto z^4\in\mathbb C$, when written out in coordinates, is a polynomial map which sends the closed first quadrant to the whole of $\mathbb R^2$---and by considering cartesian products you get the same for $\mathbb R^{2n}=\mathbb C^n$.

Later: as observed in a comment by Charles, this can be turned into a solution for the open quadrant by composing with a translation, as in $z\in\mathbb C\mapsto (z-z_0)^4\in\mathbb C$ with $z_0$ in the open first quadrant.

edited Apr 13 '17 at 12:58

Community

1

answered Jan 15 '10 at 19:41

Mariano Suárez-Álvarez

46,795

This yields maps for all even dimensions. Is there a simple answer to the odd-dimensional case? – j.c. Jan 15 '10 at 19:44
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Yes, because you can use this trick for any pair of coordinates in $\mathbb{R}^n$, regardless of the parity of $n$, and then compose until positivity has been eliminated for all coordinates. – Greg Kuperberg Jan 15 '10 at 19:45
The question asks about the open orthant, not the closed one! – Alicia Garcia-Raboso Jan 15 '10 at 19:48
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Then take $(z-z_0)^4$ and it should be patched, when $z_0$ is in the open first orthant. – Charles Siegel Jan 15 '10 at 19:50
As Charles suggests, the open orthant is equivalent to the closed orthant for this question, because you can compose with translation. – Greg Kuperberg Jan 15 '10 at 19:52
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Thanks very much indeed, this is a great answer! (Now I wonder what can be said about the form of such polynomial maps.) – Louis Deaett Jan 15 '10 at 20:38

A polynomial map from $\Bbb R^n$ to $\Bbb R^n$ mapping the positive orthant onto $\Bbb R^n$?

1 Answers1