Consider a smooth finite curve $\gamma$ without intersections in $\mathbb{R}^2$. Consider the family of smooth maps $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $T(\gamma)$ is a straight line.

Which ways are there to distinguish one of these transformations as the one (straightening map of $\gamma$), e.g. the one closest to the identity transformation?

How could this transformation be constructed from $\gamma$?


Consider a straight line $PQ$ and all other curves $\gamma$ connecting $P$ and $Q$, each of them approximating the straight line better or worse according to a given distance function $\delta$ between curves.

Suppose there is one distinguished straightening map $T$ for one of those curves $\gamma_0$ with given distance $\delta(PQ,\gamma_0)$. Then the question arises how the images of $PQ$ and $\gamma_0$ under $T$ behave, i.e. what can be said about $\delta(PQ,\gamma_0)$ vs. $\delta(T(PQ),T(\gamma_0))$?

  • 1
    $\begingroup$ When you write "is a straight line", do you mean the entire line or a line segment? // If $T$ is conformal on the entire plane, it's a similarity. // There is no isometry or similarity unless $\gamma$ is already a straight line, so these conditions look excessively strict. $\endgroup$
    – user53153
    Jan 7, 2013 at 18:48
  • 1
    $\begingroup$ I mean a line segment connecting the points $T(P),T(Q) \in \mathbb{R}^2$. $\endgroup$ Jan 7, 2013 at 18:52
  • $\begingroup$ So I will dismiss them... Thanks for the hint. $\endgroup$ Jan 7, 2013 at 18:53

2 Answers 2


Up to composition with a similarity, there is a unique conformal map of $\overline{\mathbb C}\setminus \gamma$ onto the complement of a line segment; by the Riemann mapping theorem. However, this map will not have a continuous extension to $\mathbb R^2$ unless $\gamma$ is already a straight line. This is because smooth curves are removable for continuous holomorphic functions: if a function is globally continuous and is holomorphic on the complement of a curve, then it's globally holomorphic. On the plus side, there are numerical methods for constructing conformal maps, it is not just an abstract existence assertion.

I do not know of an easy way to create a distinguished global diffeomorphism that straightens out a given smooth curve. The first thing that comes to mind is to minimize $F(T)=\int_{\mathbb R^2} \|DT-I\|^2$ among all such diffeomorphisms: i.e., look for the least squares approximation to the identity. The functional is convex but the set of admissible maps isn't, hence there is no guarantee that the minimum will be attained at a unique point, or at all. We can always take a sequence $(T_n)$ such that $F(T_n)\to \min F$, but the limit may fail to be smooth on $\gamma$, or to be injective.

Yet another approach is to begin with a canonical map between $\gamma$ and a line segment (i.e., a length-preserving one), and try to extend it to a diffeomorphism or at least to a homeomorphism. There are extension methods, but they are neither very explicit nor canonical. This paper comes to mind; unfortunately behind a paywall and in any case it deals with curves and maps of low smoothness.

[Edit] If you can give up injectivity, I can suggest a canonical way to create a (mostly) smooth surjective map $T$ that maps $\gamma$ onto a line segment $I$, and also a similar map $S$ that maps $I$ onto $\gamma$. To construct $T$, begin with a length-preserving map $f:\gamma\to I$. Using vector notation, write $f(x,y) = (x,y) + (u_1(x,y),u_2(x,y))$ where $u_1$ and $u_2$ are scalar functions defined (so far) only on $\gamma$. For $i=1,2$ there exists a unique bounded function $v_i$ such that

  • $v_i$ is continuous and bounded on $\mathbb R^2$
  • $v_i=u_i$ on $\gamma$
  • $v_i$ is harmonic in $\mathbb R^2\setminus \gamma$; that is, $\Delta v_i=0$.

In other words, $v_i$ is the harmonic extension of $u_i$. In yet other words, it's the solution of the Dirichlet problem with boundary data $u_i$.

Now let $T(x,y)=(x,y)+(v_1(x,y),v_2(x,y))$. This map is continuous on $\mathbb R^2$, infinitely differentiable outside of $\gamma$, and agrees with $f$ on the curve $\gamma$. Its surjectivity can be proved as follows: since $v_1,v_2$ are bounded, the image of a sufficiently large circle under $T$ will be roughly a circle of about the same size (it will loop around the origin and neighboring points, in particular). The key words here are winding number and degree theory. Nothing guarantees that $T$ will be injective, but if $\gamma$ is not too wiggly, it might be.

From the computational viewpoint, solving the Dirichlet problem isn't hard (Matlab is really good at it), but an explicit analytic solution is unlikely to be found. The chances of getting an explicit formula are greater if one exchanges the roles of $\gamma$ and $I$ in the preceding argument: that is, begin with a length-preserving map $g:I\to\gamma$, extend $g-\operatorname{id}$ harmonically to $\mathbb R^2\setminus I$, then add the identity map $\operatorname{id}$ back. The advantage is that solving the Dirichlet problem in $\mathbb R^2\setminus I$ is easier. This is because $\mathbb R^2\setminus I$ can be conformally mapped onto the unit disk by an elementary function (inverse of $z+z^{-1}$, up to scaling), and on the disk the Poisson kernel is known explicitly.

[Edit 2] I worked out the case of mapping the line segment $L=[-\pi/2,\pi/2]$ onto the upper semicircle $\gamma$; as noted earlier, this is easier than going the other way. The length-preserving map $g:L\to\gamma$ is $g(z)=ie^{iz}$. We want to extend the function $g(z)-z=ie^{iz}-z$ harmonically to $\mathbb C\setminus L$; and then add $z$ to the extension. To this end, introduce a conformal map of the unit disk $\mathbb D=\{z:|z|<1\}$ onto $\mathbb C\setminus L$; the formula is $\varphi(\zeta)=\frac{\pi}{4}(\zeta+\zeta^{-1})$. Now we have $g\circ \varphi - \varphi$ defined on $\partial \mathbb D$ and must extend it harmonically inside of $\mathbb D$. Rather than deal with the Poisson integral, I prefer to expand into a trigonometric series and use the fact that the harmonic extension of $e^{i k t}$ is $r^{|k|}e^{i k t}$.

So, in terms of $\zeta=e^{it}$ we have $\varphi(e^{it})=\frac{\pi}{4}(e^{it}+e^{-it})$, which extends to $\frac{\pi}{4} r(e^{it}+e^{-it})$. That was easy.

Next, $g\circ \varphi$ takes the form $$g\circ \varphi(e^{it})= i\exp \left( \frac{\pi i}{4}(e^{it}+e^{-it})\right)$$ where I used the series for $\exp$ and the binomial formula: $$g\circ \varphi(e^{it})= i \sum_{n=0}^\infty \frac{1}{n!}\left(\frac{\pi i}{4}\right)^n \sum_{k=0}^n \binom{n}{k} e^{(2k-n)it}$$ The harmonic extension of $g\circ \varphi(e^{it})$ is $$i \sum_{n=0}^\infty \frac{1}{n!}\left(\frac{\pi i}{4}\right)^n \sum_{k=0}^n \binom{n}{k} r^{|2n-k|} e^{(2k-n)it}$$

Putting everything together: the map $S:\mathbb C\to\mathbb C$ which extends $g$ in the way described above can be written as $S=\widetilde S\circ \varphi^{-1}$ where $$\widetilde S(re^{it}) = i \sum_{n=0}^\infty \frac{1}{n!}\left(\frac{\pi i}{4}\right)^n \sum_{k=0}^n \binom{n}{k} r^{|2n-k|} e^{(2k-n)it} \, - \frac{\pi}{4} r(e^{it}+e^{-it}) + \varphi(re^{it})$$ I don't expect the process of plugging $\varphi^{-1}$ into $\widetilde S$ to be enjoyable. Therefore, I visualize the map $S$ by drawing

  1. images of concentric circles of radius $<1$ under $\varphi$; these are simply ellipses with foci $\pm 1$;
  2. images of the same circles under $\widetilde S$; these are precisely the images of the above ellipses under $S$.

    These are the ellipses around $L$ (one of which degenerates into $L$)


and these are their images under $S$ (one of which coincides with $\gamma$)


Clearly, the map is not injective and not even open: there is a lot of folding going on.

Images were made in Maple 16, the code is below. (j/9)^(1/2) is the radius of the jth circle to be drawn; chosen to produce reasonable spacing of the curves. The series in the formula for $\widetilde S$ was terminated at $n=6$, where the terms got very small.

display(seq(complexplot(g((j/9)^(1/2),t), t=0..2*Pi, color=ColorTools:-Color([j/9,0,1-j/9])), j=1..9));
h:=(r,t)->I*sum((Pi*I/4)^n/factorial(n)*sum(binomial(n,k)*exp((2*k-n)*I*t)*r^abs(2*k-n),k=0..n),n=0..6) - Pi/2*r*cos(t) + Pi/4*(r*exp(I*t)+1/r*exp(-I*t));
display(seq(complexplot(h((j/9)^(1/2),t), t=0..2*Pi, color=ColorTools:-Color([j/9,0,1-j/9])), j=1..9));
  • $\begingroup$ This is marvelous! $\endgroup$ Jan 7, 2013 at 23:05
  • $\begingroup$ This is a fantastic answer! I think the folding is a result of mapping $−π/2$ to $1$ and $π/2$ to $−1$, rather than the other way around. But this indicates another difficulty: if $γ$ were already a straight line, say between $-i\pi/2$ and $i\pi/2$, one would probably want $S$ to be a rigid motion of the plane (in this case $S(z)=iz$), but then the $v_i$ would have to be affine and not bounded. $\endgroup$
    – user856
    Jan 9, 2013 at 5:05
  • $\begingroup$ @RahulNarain That's a very good point. Perhaps the words "closest to the identity transformation" in the question should be interpreted as "closest to an orthogonal transformation". That is, a canonical map should minimize $\int_\mathbb{R^2} \|DT-O\|^2$ where minimum is taken over all diffeomorphisms and all orthogonal matrices $O$. Of course, the attainment of minimum is not guaranteed, for the same reasons as in the 2nd paraghraph of my answer. $\endgroup$
    – user53153
    Jan 9, 2013 at 5:47
  • $\begingroup$ "This is because smooth curves are removable for continuous holomorphic functions": Can you recommend a reference for this result? I want to learn more about this sort of thing. $\endgroup$
    – user856
    Jan 25, 2013 at 11:35
  • $\begingroup$ @RahulNarain The proof of this particular result (for smooth curves) is based on Morera's theorem: since continuity is given, we just need to show that integrals over triangles are zero. It should go like this with additional complications. (Moving and/or subdividing triangles to avoid the curve). // Some references: EOM article, L. Zalcman, "Analytic capacity and rational approximation" Springer-Verlag Lecture Notes 50, 1968; and Chapter 8 of "Uniform algebras" by Gamelin. $\endgroup$
    – user53153
    Jan 27, 2013 at 2:30

Consider the family $\mathcal{F}$ of rectifiable functions $f:[0,1]\rightarrow \mathbb{R}$ with $f(0) = f(1) = 0$. Consider these functions as approximations of the straight line segment $PQ$ with $P = (0,0), Q = (1,0)$. (In $\mathbb{R}^2$ there are many more curves connecting $P,Q$ but they would just add extra detours and can be ignored in the context of approximations.)

Consider for a given function $f_0$ the family of maps $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $T((x,y)) = (L(xy),0)$ for all $y = f_0(x)$ and $L(xy)$ the arclength of $f_0$ from $(0,0)$ to $(x,y)$.

Maybe one should restrict the question to this family of curves and maps that isometrically straighten the "curve" $f_0$. (Ideally there is just one such map?)

  • 1
    $\begingroup$ This is addressed by the 3rd paragraph of my answer. There are infinitely many ways to extend a length-preserving map between curves to a global smooth map. E.g., imagine composition with a diffeomorphism that is identity in a neighborhood of the curve. $\endgroup$
    – user53153
    Jan 7, 2013 at 20:25
  • $\begingroup$ @Pavel. Sorry, I have been carried away, and I should have read your answer more carefully. Nevertheless: Isn't there any chance to find a distinguised map in this rather restricted family of maps? $\endgroup$ Jan 7, 2013 at 20:34
  • $\begingroup$ I don't know. Maybe a concrete example would be illustrative: what do you think would be the best way to straighten out a semicircle? $\endgroup$
    – user53153
    Jan 7, 2013 at 21:42
  • $\begingroup$ Consider $f:[-1,1]\rightarrow[0,1]$ with $f(x) = \sqrt{1-x^2}$. As "the" best (because simplest) way to straighten this semicircle I would consider the map $T$ with $T(x,y) = (\arcsin(x),0)$. $\endgroup$ Jan 7, 2013 at 22:09
  • $\begingroup$ You don't want $T$ to be surjective? That was somehow my impression, since you had $T:\mathbb R^2\to\mathbb R^2$ in the original question. The formula $T(x,y)=(\dots, 0)$ maps everything to a line: this is more of squashing than straightening. $\endgroup$
    – user53153
    Jan 7, 2013 at 22:14

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.