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 $\Bbb R^n$ !


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.

  • $\begingroup$ This yields maps for all even dimensions. Is there a simple answer to the odd-dimensional case? $\endgroup$ – j.c. Jan 15 '10 at 19:44
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    $\begingroup$ 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. $\endgroup$ – Greg Kuperberg Jan 15 '10 at 19:45
  • $\begingroup$ The question asks about the open orthant, not the closed one! $\endgroup$ – Alberto García-Raboso Jan 15 '10 at 19:48
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    $\begingroup$ Then take $(z-z_0)^4$ and it should be patched, when $z_0$ is in the open first orthant. $\endgroup$ – Charles Siegel Jan 15 '10 at 19:50
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    $\begingroup$ Thanks very much indeed, this is a great answer! (Now I wonder what can be said about the form of such polynomial maps.) $\endgroup$ – Louis Deaett Jan 15 '10 at 20:38

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