17
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Geva Yashfe Jun 25 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 $d>1$. $\endgroup$ – Robert Bryant Jun 25 at 11:26
  • $\begingroup$ Thank you @GevaYashfe. (Not quite what I was looking for, but good to know either way.) $\endgroup$ – qp10 Jun 25 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 at 11:31
  • $\begingroup$ @qp10: In that case, you should have specified $d>1$ in your question. $\endgroup$ – Robert Bryant Jun 25 at 11:34
24
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That's convincing. Thank you! $\endgroup$ – qp10 Jun 25 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. $\endgroup$ – Robert Bryant Jun 25 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. $\endgroup$ – Iosif Pinelis Jun 25 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 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. $\endgroup$ – Robert Bryant Jun 25 at 16:14
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.