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The question was edited several times. Most recent version, suggested by Fedja:

Does there exist an open set $U\subset \mathbb R^n$ (n>1) that contains balls of arbitrarily large radius and such that no polynomial mapping $p\colon \mathbb R^n\to\mathbb R^n$ takes $U$ onto $\mathbb R^n$? (Take $n=2$ if you prefer.)


Older version (with bounty) was answered by Fedja in the affirmative:

Does there exist a topological ball $U\subset \mathbb R^n$ of infinite volume such that no polynomial mapping $p\colon \mathbb R^n\to\mathbb R^n$ takes $U$ onto $\mathbb R^n$?


Discussion:

Motivated by a recent question, I wonder if there is a geometric characterization of open sets $U\subset \mathbb R^n$ that can be mapped onto $\mathbb R^n$ by a polynomial $p$. Let $U$ be a topological ball to simplify matters. The following tail volume condition is necessary for the existence of such $p$.

(TVC) $\int_1^{\infty} r^{m} |U\setminus B(r)|=\infty$ for some $m>0$.

Indeed, the absolute value of the Jacobian of $p$ must have infinite integral over $U$. Since the Jacobian is a polynomial, $\int_U |x|^N dx=\infty$ for some $N$. The latter can be rephrased in terms of tail volume: $\int_1^{\infty} r^{N-1} |U\setminus B(r)| =\infty$. The example of $U=\{(x,y)\in \mathbb R^2\colon x>1, 0<y<x^{-M}\}$ shows that (TVC) is somewhat sharp. This particular $U$ can be mapped onto $\mathbb R^2$ by $(x,y)\mapsto (x,x^{M+1}y)$ followed by translation and the power map $(x+iy)\mapsto (x+iy)^8$.

I am interested in other obstructions besides small tail volume, as well as in reasonably general sufficient conditions. Initially I hoped to get a stronger necessary condition from an affirmative answer to the question below, but Bjorn Poonen answered it in the negative.

"If $f\colon U\to\mathbb R^n$ is a polynomial surjection, does there exist $\epsilon>0$ such that $p(U\cap B(r))$ contains $B(r^\epsilon)$ for large $r$?"

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Any motivation? –  Anton Petrunin Jan 16 '10 at 5:35
    
You should repost the new version, or let fedja post it. –  Harry Gindi Jan 18 '10 at 5:19

2 Answers 2

up vote 10 down vote accepted

It is enough to construct a sequence of pairwise disjoint disks $D_j$ of infinite total volume so that the image $\bigcup_j f(D_j)$ has $0$ density for any polynomial mapping $f:\mathbb R^2\to\mathbb R^2$. Then you can connect them by very thin passageways to get a topological disk and add just a finite area to the image.

The idea is that near each point $x$ with $|x|=r$, the mapping $f$, either $|\mbox{det}\phantom{,}Df| < r^{-5}$, or $f$ is essentially linear in the square $Q$ with side length $r^{-M}$ centered at $x$ where $M$ depends on $f$ only and $f(Q)$ has diameter less than $1$. Now, we can create a locally finite covering of the entire plane with squares $Q$ such that the side length of each $Q$ is less than $e^{-|c(Q)|}$ where $c(Q)$ is the center of $Q$ and put into each square $Q$ a small disk $D$ whose area is o-small of the area of $Q$ as $c(Q)\to\infty$ (which doesn't contradict the infinite volume restriction). If we forget about the image of a large disk centered at the origin depending on $f$, and the image of those squares on which the determinant is small (both have finite area), then the remaining image is the union of small approximate ellipsoids inside much larger but still small approximate parallelograms.

The final observation is that each generic point can be covered by only $N=N(f)$ parallelograms (Bezout's theorem), so the total area of parallelograms intersecting a large disk is just a constant multiple of the area of that disk and the area of the corresponding ellipsoids is much smaller.

This formally answers Leonid's question about TVC but I'm not very happy with this bubble bath construction. So, let us ask whether there is a set containing sequence of disks of radii tending to $\infty$ that cannot be mapped onto $\mathbb R^2$ by a polynomial mapping.

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Nice answer, fedja! –  Bjorn Poonen Jan 18 '10 at 5:33

The answer to the precise version of the question (the one asked at the end) is no.

The map $(x,y) \mapsto (x,xy)$ maps $\mathbb{R}^2$ onto all of $\mathbb{R}^2$ except the positive and negative parts of the $y$-axis. Following this with $z \mapsto z^3$ (where $z=x+iy \in \mathbb{C}$) yields a polynomial surjection $(x,y) \mapsto (f_1(x,y),f_2(x,y))$ such that the preimage of $(0,0)$ is the $y$-axis. Define $f(x,y,z) = (f_1(x,y),f_2(x,y),z)$, so $f$ defines a surjection $\mathbb{R}^3 \to \mathbb{R}^3$. Let $U := \mathbb{R}^3 - \lbrace (0,y,z) : y \le e^z \rbrace$, so $U$ is a topological ball. The value of $f$ on each deleted point $(0,y,z)$ is the same as its value at $(0,e^z+1,z) \in U$, so $f|_U$ is still surjective.

For $r>0$, if $u \in U$ and $f(u) = (0,0,\log r)$, then $u=(0,y,\log r)$ with $y>r$, so $|u|>r$. So $f(U \cap B(r))$ does not even contain $(0,0,\log r)$, let alone $B(r^\epsilon)$ for large $r$ (for fixed $\epsilon>0$).

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It doesn't show this; sorry about that. All my example does is answer your last question. –  Bjorn Poonen Jan 17 '10 at 3:25

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