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I'm going to go out on a limb a little bit and make a specific conjecture. A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p$ if and only if it is topologically conjugate to a continuous, piecewise linear function $q$ 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 $q$ to each region is linear.

As Tom Goodwillie points out, this conjecture is clearly a necessary condition when $n=1$ and it should be a sufficient condition as well. It seems very likely to be sufficient using the Lagrange interpolation theorem and some version of the Weierstrass approximation theorem.

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). Also, every polynomial $p$ on $\mathbb{R}^n$ has a projectivization, which is then a polynomial section of a line bundle on $\mathbb{R}P^n$. In other words, if $p$ has even degree it comes from a symmetric function on a sphere $S^n$, while if it has odd degree it comes from an antisymmetric function. Since $p$ has an extension to a compact manifold, this is already pretty good evidence that the conjecture is at least a necessary condition.

The smooth case seems more complicated in high dimensions because the polynomial $p$ could have a high-degree singularity somewhere in $\mathbb{R}^n$, and a diffeomorphism can only make limited changes to the singularity. When $n=1$, there should be an answer similar to the one before, but in general a polynomial in $n$ dimensions up to diffeomorphism is as complicated as a polynomial hypersurface in $n-1$ dimensions up to a projective transformation.

Likewise the complex case (up to homeomorphism, say) seems as complicated as the topological classification of complex algebraic varieties.

Well, the above doesn't work per a simple counterexample from Richard Borchards. There should be some natural notion of finite type to answer the question, but it is not as simple as what I suggested.

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.

I'm going to go out on a limb a little bit and make a specific conjecture. A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p$ if and only if it is topologically conjugate to a continuous, piecewise linear function $q$ 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 $q$ to each region is linear.

As Tom Goodwillie points out, this conjecture is clearly a necessary condition when $n=1$ and it should be a sufficient condition as well. It seems very likely to be sufficient using the Lagrange interpolation theorem and some version of the Weierstrass approximation theorem.

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). Also, every polynomial $p$ on $\mathbb{R}^n$ has a projectivization, which is then a polynomial section of a line bundle on $\mathbb{R}P^n$. In other words, if $p$ has even degree it comes from a symmetric function on a sphere $S^n$, while if it has odd degree it comes from an antisymmetric function. Since $p$ has an extension to a compact manifold, this is already pretty good evidence that the conjecture is at least a necessary condition.

The smooth case seems more complicated in high dimensions because the polynomial $p$ could have a high-degree singularity somewhere in $\mathbb{R}^n$, and a diffeomorphism can only make limited changes to the singularity. When $n=1$, there should be an answer similar to the one before, but in general a polynomial in $n$ dimensions up to diffeomorphism is as complicated as a polynomial hypersurface in $n-1$ dimensions up to a projective transformation.

Likewise the complex case (up to homeomorphism, say) seems as complicated as the topological classification of complex algebraic varieties.

I'm going to go out on a limb a little bit and make a specific conjecture. A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p$ if and only if it is topologically conjugate to a continuous, piecewise linear function $q$ 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 $q$ to each region is linear.

As Tom Goodwillie points out, this conjecture is clearly a necessary condition when $n=1$ and it should be a sufficient condition as well. It seems very likely to be sufficient using the Lagrange interpolation theorem and some version of the Weierstrass approximation theorem.

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). Also, every polynomial $p$ on $\mathbb{R}^n$ has a projectivization, which is then a polynomial section of a line bundle on $\mathbb{R}P^n$. In other words, if $p$ has even degree it comes from a symmetric function on a sphere $S^n$, while if it has odd degree it comes from an antisymmetric function. Since $p$ has an extension to a compact manifold, this is already pretty good evidence that the conjecture is at least a necessary condition.

The smooth case seems more complicated in high dimensions because the polynomial $p$ could have a high-degree singularity somewhere in $\mathbb{R}^n$, and a diffeomorphism can only make limited changes to the singularity. When $n=1$, there should be an answer similar to the one before, but in general a polynomial in $n$ dimensions up to diffeomorphism is as complicated as a polynomial hypersurface in $n-1$ dimensions up to a projective transformation.

Likewise the complex case (up to homeomorphism, say) seems as complicated as the topological classification of complex algebraic varieties.

Well, the above doesn't work per a simple counterexample from Richard Borchards. There should be some natural notion of finite type to answer the question, but it is not as simple as what I suggested.

I'm going to go out on a limb a little bit and make a specific conjecture. A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p$ if and only if it is topologically conjugate to a continuous, piecewise linear function $q$ of finite type. By "finite type" I mean that there is a tiling of $\mathbb{R}^n$ by finitely many convex regions such that the restriction of $q$ to each region is linear.

As Tom Goodwillie points out, this conjecture is clearly a necessary condition when $n=1$ and it should be a sufficient condition as well. It seems very likely to be sufficient using the Lagrange interpolation theorem and some version of the Weierstrass approximation theorem.

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). Also, every polynomial $p$ on $\mathbb{R}^n$ has a projectivization, which is then a polynomial section of a line bundle on $\mathbb{R}P^n$. In other words, if $p$ has even degree it comes from a symmetric function on a sphere $S^n$, while if it has odd degree it comes from an antisymmetric function. Since $p$ has an extension to a compact manifold, this is already pretty good evidence that the conjecture is at least a necessary condition.

The smooth case seems more complicated in high dimensions because the polynomial $p$ could have a high-degree singularity somewhere in $\mathbb{R}^n$, and a diffeomorphism can only make limited changes to the singularity. When $n=1$, there should be an answer similar to the one before, but in general a polynomial in $n$ dimensions up to diffeomorphism is as complicated as a polynomial hypersurface in $n-1$ dimensions up to a projective transformation.

Likewise the complex case (up to homeomorphism, say) seems as complicated as the topological classification of complex algebraic varieties.

I'm going to go out on a limb a little bit and make a specific conjecture. A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p$ if and only if it is topologically conjugate to a continuous, piecewise linear function $q$ 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 $q$ to each region is linear.

As Tom Goodwillie points out, this conjecture is clearly a necessary condition when $n=1$ and it should be a sufficient condition as well. It seems very likely to be sufficient using the Lagrange interpolation theorem and some version of the Weierstrass approximation theorem.

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). Also, every polynomial $p$ on $\mathbb{R}^n$ has a projectivization, which is then a polynomial section of a line bundle on $\mathbb{R}P^n$. In other words, if $p$ has even degree it comes from a symmetric function on a sphere $S^n$, while if it has odd degree it comes from an antisymmetric function. Since $p$ has an extension to a compact manifold, this is already pretty good evidence that the conjecture is at least a necessary condition.

The smooth case seems more complicated in high dimensions because the polynomial $p$ could have a high-degree singularity somewhere in $\mathbb{R}^n$, and a diffeomorphism can only make limited changes to the singularity. When $n=1$, there should be an answer similar to the one before, but in general a polynomial in $n$ dimensions up to diffeomorphism is as complicated as a polynomial hypersurface in $n-1$ dimensions up to a projective transformation.

Likewise the complex case (up to homeomorphism, say) seems as complicated as the topological classification of complex algebraic varieties.