Is it possible to (locally) approximate an arbitrary smooth diffeomorphism by a polynomial diffeomorphism?

More precisely: Let $f:\mathbb{R}^d\rightarrow\mathbb{R}^d$ be a smooth diffeomorphism for $d>1$. For $U\subset\mathbb{R}^d$ bounded and open and $\varepsilon>0$, is there a diffeomorphism $p=(p_1, \cdots, p_d) : U\rightarrow\mathbb{R}^d$ (with inverse $q:=p^{-1} : p(U)\rightarrow U$) such that both

  • $\|f - p\|_{\infty;\,U}:=\sup_{x\in U}|f(x) - p(x)| < \varepsilon$, $\ \textbf{and}$
  • each component of $p$ and of $q=(q_1,\cdots,q_d)$is a polynomial, i.e. $p_i, q_i\in\mathbb{R}[x_1, \ldots, x_d]$ for each $i=1, \ldots, d$?

Clearly, by Stone-Weierstrass there is a polynomial map $p : \mathbb{R}^d\rightarrow\mathbb{R}^d$ with $\|f - p\|_{\infty;\,U} < \varepsilon$ and such that $q:=(\left.p\right|_U)^{-1}$ exists; in general, however, this $q$ will not be a polynomial map.

Do you have any ideas/references under which conditions on $f$ an approximation of the above kind can be guaranteed nonetheless?

$\textbf{Note:}$ This is a crosspost from https://math.stackexchange.com/questions/3689873/approximation-of-smooth-diffeomorphisms-by-polynomial-diffeomorphisms

  • $\begingroup$ I know very little about this, but people have invested some time in finding some form of inversion procedure for a polynomial diffeomorphism: en.wikipedia.org/wiki/… Maybe this is already useful for you $\endgroup$ Jun 25, 2020 at 11:23
  • $\begingroup$ Note that this is impossible when $d=1$, since, in that case, a polynomial map with polynomial inverse must be linear. I think it is very unlikely to be true for any $d>1$. $\endgroup$ Jun 25, 2020 at 11:26
  • $\begingroup$ Thank you @GevaYashfe. (Not quite what I was looking for, but good to know either way.) $\endgroup$
    – qp10
    Jun 25, 2020 at 11:28
  • $\begingroup$ Thanks for your comment, @RobertBryant. The impossibility of the case $d=1$ was already pointed out in the stack-exchange version of the question (link above); I share your scepticism for $d>1$, but maybe there's some definite counterexample in the literature? $\endgroup$
    – qp10
    Jun 25, 2020 at 11:31
  • $\begingroup$ @qp10: In that case, you should have specified $d>1$ in your question. $\endgroup$ Jun 25, 2020 at 11:34

2 Answers 2


The answer is 'no', because polynomial mappings with polynomial inverses preserve volumes up to a constant multiple.

To see why this property holds, suppose that $p:\mathbb{R}^d\to\mathbb{R}^d$ is a polynomial mapping with polynomial inverse $q:\mathbb{R}^d\to\mathbb{R}^d$. Then $p$ and $q$ extend to $\mathbb{C}^d$ as polynomial maps with polynomial inverses. This means that the Jacobian determinant of $p$ on $\mathbb{C}^d$ is a complex polynomial with no zeros and hence must be a (nonzero) constant.

Now, consider a diffeomorphism $f:\mathbb{R}^d\to\mathbb{R}^d$ that is radial, i.e., $f(x) = m(|x|^2)x$ for some smooth function $m>0$. One can easily choose $m$ in such a way that $m(4)=1/2$ and $m(9)=4/3$, so that $f$ maps the ball of radius $2$ about the origin diffeomorphically onto the ball of radius $1$ about the origin while it maps the ball of radius $3$ about the origin diffeomorphically onto the ball of radius $4$ about the origin.

Let $\epsilon>0$ be very small and suppose that $\|f-p\|_{\infty;U} <\epsilon$ for $U$ chosen to be some very large ball centered on the origin. Then $p$ maps the sphere of radius $2$ about the origin to within an $\epsilon$-neighborhood of the sphere of radius $1$, while it maps the sphere of radius $3$ about the origin to within an $\epsilon$-neighborhood of the sphere of radius $4$. It's easy to see from this that $p$ cannot have constant Jacobian determinant.

Added remark: The group $\mathrm{SDiff}(\mathbb{R}^d)$ consisting of volume-preserving diffeomorphisms of $\mathbb{R}^d$ is a 'Lie group' in Sophus Lie's original sense (i.e., a group of diffeomorphisms defined by the satisfaction of a system of differential equations; in this case, that the Jacobian determinant be equal to $1$).

The subgroup $\mathcal{SP}(\mathbb{R}^d)\subset \mathrm{SDiff}(\mathbb{R}^d)$ consisting of volume-preserving polynomial diffeomorphisms with polynomial inverses however, is not a 'Lie subgroup' in Lie's original sense when $d>1$, as it cannot be defined by the satisfaction of a system of differential equations: It contains all of the mappings of the form $p(x) = x + a\,(b{\cdot}x)^m$ where $a,b\in\mathbb{R}^d$ satisfy $a\cdot b = 0$ and $m>1$ is an integer (indeed, $p^{-1}(y) = y - a\,(b{\cdot}y)^m$), plus, it contains $\mathrm{SL}(d,\mathbb{R})$ and the subgroup consisting of the translations. Using this, it is easy to show that, for any $f\in\mathrm{SDiff}(\mathbb{R}^d)$ and for any integer $k$, there exists a $p\in \mathcal{SP}(\mathbb{R}^d)$ such that $f$ and $p$ have the same Taylor series at the origin up to and including order $k$. Thus, $\mathcal{SP}(\mathbb{R}^d)$ cannot be defined by a system of differential equations (in Lie's sense).

Using this Taylor approximation property, one can prove that $\mathcal{SP}(\mathbb{R}^d)$, like $\mathrm{SDiff}(\mathbb{R}^d)$, acts transitively on $n$-tuples of distinct points in $\mathbb{R}^d$ for any integer $n$. Whether one can prove that $\mathcal{SP}(\mathbb{R}^d)$ can 'uniformly approximate' $\mathrm{SDiff}(\mathbb{R}^d)$ on compact sets is an interesting question.

  • $\begingroup$ That's convincing. Thank you! $\endgroup$
    – qp10
    Jun 25, 2020 at 12:05
  • 2
    $\begingroup$ @qp10: It's still an interesting question if one assumes that $f$ is volume preserving. For volume preserving $f$, the statement is correct when $d=1$, because $f$ itself must be polynomial. However, for $d>1$, there are non-polynomial volume preserving diffeomorphisms, so the question becomes interesting again. I don't see any obvious reason that the volume-preserving diffeomorphisms that are polynomial with polynomial inverse (which is obviously a subgroup of the volume-preserving diffeomorphisms) does not approximate all volume-preserving diffeomorphisms uniformly on compact subsets. $\endgroup$ Jun 25, 2020 at 13:06
  • $\begingroup$ Do you think there is any polynomial diffeomorphism of degree $\ge2$ with polynomial inverse? To me, this already seems implausible, let alone the approximation property. $\endgroup$ Jun 25, 2020 at 13:57
  • 1
    $\begingroup$ @IosifPinelis : Such non-linear 'bi-polynomial' diffeomorphisms exist; e.g. for $d=2$, the Hénon map is an explicit example. $\endgroup$
    – qp10
    Jun 25, 2020 at 14:03
  • 1
    $\begingroup$ @IosifPinelis: In fact, there are tons of such volume-preserving polynomial mappings with polynomial inverses as soon as $d>1$. Any mapping of the form $p(x,y) = \bigl(x,y+h(x)\bigr)$ where $h$ is a polynomial in one variable is such, and once you start composing these with maps of the form $p(x,y) = \bigl(x+g(y),y\bigr)$ where $g$ is a polynomial in one variable and toss in the invertible linear maps, you'll rapidly generate a huge number of such maps that do not fix any variable and are not obviously invertible. $\endgroup$ Jun 25, 2020 at 16:14

An illustration for one of the examples in the answer by Robert Bryant. It is supposed to convey the feeling of something extremely rigid, unyielding and inflexible.

Image of the square $[-1,1]\times[-1,1]$ under the map $(x,y)\mapsto(x-y^2-2x^2y-x^4,y+x^2)$ (composite of $(x,y)\mapsto(x-y^2,y)$ with $(x,y)\mapsto(x,y+x^2)$).

enter image description here


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