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I posted this on the new math.SE website but didn't get much of a response, so I am reposting it here.

Suppose $f$ is a continuous $\mathbb{R}$-valued function on $\mathbb{R}^n$. What type of conditions on $f$ guarantee it is a polynomial up to homeomorphism. That is, when can I find a homeomorphism $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $\phi^* f = f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$?

Some related questions:

  • A necessary condition in the case of $n = 1$ is that point inverse images of $f$ must be finite (since a polynomial has only finitely many roots). Is this sufficient?
  • What if we replace $\mathbb{R}$ by $\mathbb{C}$?
  • What if we look at smooth functions and diffeomorphism instead? (I tried playing around with the implicit function theorem but didn't get anywhere).
  • What about the complex analytic case?

I'm not quite sure how to tag this, so feel free to edit them.

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In the case $n=1$ a necessary (and I would guess sufficient) condition for $f$ to be expressible as a nonconstant polynomial composed with a homeomorphism is: (1) there are only finite many values of $x$ at which $f(x)$ has a local maximum or minimum and (2) $f$ is proper (i.e. tends to plus or minus infinity at each end of the line). A function like $x+2 sin x$ has finite point preimages but infinitely many local maxima. The case $n=2$ seems much more interesting/harder. –  Tom Goodwillie Aug 3 '10 at 18:34
    
Topological classification of continuous real functions ($n=1$) may be obtained by considering their intervals of monotonicity. In general, differentiable and analytic classifications are harder than topological. A keyword is "singularity theory". –  Victor Protsak Aug 3 '10 at 18:42
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A theorem of Whitney says that every continuous function is homotopic to a smooth one, perhaps one could use this to reduce the question to the case where the original function is smooth? Though I'm not at all sure how you would get a homeomorphism of R^n from the homotopy... –  Dylan Wilson Aug 3 '10 at 18:46
    
In the real valued case, $n=1$, a necessary and sufficient condition might be that $f$ makes a finite number of oscillations and is unbounded as $x\to\pm\infty$. –  Piero D'Ancona Aug 3 '10 at 20:35
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It might be good to begin with a local question: Which germs of real-valued functions are topologically conjugate to polynomials. –  Tom Goodwillie Aug 3 '10 at 22:09
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1 Answer

I'm going to go out on a limb and make a partial conjecture based on Tom Goodwillie's comment. A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p(x)$ only if it is topologically conjugate to a continuous function $q(x0$ of finite type which is nowhere constant. By "finite type" I mean that there is a tiling of $\mathbb{R}^n$ by finitely many convex regions, such that the restriction of $q(x)$ to each region is non-constant and either linear or of the form $1/\phi(x)+c$, where $\phi$ is linear and $c$ is constant.

In this version of the answer, I'm going out on a limb for a second time. I first conjectured that $q$ should simply be linear on each convex piece, and Richard Borcherds quickly found a counterexample to that in two variables.

I don't mean this to be a sufficient condition, since clearly it is not sufficient when $n=1$. A "finite type" function in the above sense can be bounded, while a polynomial cannot be bounded. Maybe it is a sufficient condition as a topological characterization of rational functions with no poles. For polynomials specifically, there are strong restrictions on the behavior at infinity, but per Richard's example, they are somewhat looser than I first thought.

There is a relevant pair of results due to Whitney and Goresky. Whitney proved that every analytic variety in $\mathbb{R}^n$ is a Whitney stratified space. Goresky proved that every Whitney stratified space in $\mathbb{R}^n$ is supported on a piecewise smooth triangulation (but not necessarily one which is finite type). It is easy to ride roughshod over subtleties as I already did, but these results seem like a good way to get started with the problem.

The smooth and complex cases of the problem seem more complicated for various reasons.

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I suppose you meant to add that the piece-wise linear function is nowhere constant. –  Arend Bayer Aug 3 '10 at 20:14
    
Oh yes, you are right. –  Greg Kuperberg Aug 3 '10 at 20:17
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Here is a possible counterexample to the conjecture above: The real polynomial x^2 + (xy-1)^2 is bounded below but does not achieve its lower bound 0, which I think implies that it is not conjugate to a piecewise linear function of finite type. –  Richard Borcherds Aug 3 '10 at 20:30
    
Whoops, clearly my conjecture needs to be adjusted. I didn't properly think through projectivization of the question. –  Greg Kuperberg Aug 3 '10 at 20:51
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