# Approximation of smooth diffeomorphisms by polynomial diffeomorphisms?

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

• 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 Jun 25, 2020 at 11:23
• 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 $d>1$. Jun 25, 2020 at 11:26
• Thank you @GevaYashfe. (Not quite what I was looking for, but good to know either way.)
– qp10
Jun 25, 2020 at 11:28
• 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?
– qp10
Jun 25, 2020 at 11:31
• @qp10: In that case, you should have specified $d>1$ in your question. Jun 25, 2020 at 11:34

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.

• That's convincing. Thank you!
– qp10
Jun 25, 2020 at 12:05
• @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. Jun 25, 2020 at 13:06
• 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. Jun 25, 2020 at 13:57
• @IosifPinelis : Such non-linear 'bi-polynomial' diffeomorphisms exist; e.g. for $d=2$, the Hénon map is an explicit example.
– qp10
Jun 25, 2020 at 14:03
• @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. 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 $$(x,y)\mapsto(x,y+x^2)$$).