Bonjour. I’m looking for a conformal mapping that transform a square into a circle, a cube into a sphere, eventually a rectangle into an ellipse.
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1What is a "complex isometry"? – Eric Wofsey Mar 23 '19 at 19:48
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A complex transformation which preserves distance – DINEDINE Mar 23 '19 at 19:49
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2Assuming you're using the usual Euclidean distance, no such isometries exist. – Eric Wofsey Mar 23 '19 at 19:51
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Separately, what do you mean by "complex transformation"? – Eric Wofsey Mar 23 '19 at 19:52
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A complex transformation is any complex plane mapping such that to a point M assign a unique point M’. – DINEDINE Mar 23 '19 at 20:02
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Well, how on earth could a complex transformation turn a cube into a sphere? – Eric Wofsey Mar 23 '19 at 20:04
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Should I understand that there’s no transformation that isometricly transform a cube to a sphere? – DINEDINE Mar 23 '19 at 20:14
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That’s it @Somos – DINEDINE Mar 23 '19 at 20:31
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You may find MSE question 1587209 "Conformal mapping from square to disk as inverse of hypergeometric function" of interest. Also MSE question 1015205 "Confromal mapping circle onto square (and back)". In general elliptic functions are used in such a case. – Somos Mar 23 '19 at 20:35
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1Maybe of interest https://arxiv.org/ftp/arxiv/papers/1509/1509.06344.pdf – cgiovanardi Mar 24 '19 at 01:24
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@EricWofsey : You might be interested in the Autagraph projection which converted a sphere to a tetrahedron, which could be unfolded to a triangle with which you could pave the plane. – Camion Mar 01 '23 at 09:39
1 Answers
If you have some knowledge about complex integration, here is a function that maps the open unit disk onto the open square with vertices $K(\pm 1 \pm i)$.
$$z_0 \ \to \ Z=\varphi(z_0)=\int_{[0,z_0]}\frac{dz}{\sqrt{1+z^4}}\tag{1}$$
($[0,z_0]$ denotes the line segment from $0$ to $z_0$).
and $K=0.927037...=\int_0^1 \frac{dx}{\sqrt{1+x^4}}$ (real integral).
See on Fig. 1 a representation of the action of $\varphi$ on a polar grid.
It is a conformal mapping (with the meaning of angles preservation) : look at red and blue curves ; after the transformation, their right angles are preserved). BUT a conformal transformation isn't an isometry : it doesn't preserve distances.
Function $\varphi$ defined in (1) is an elliptic function (or more exactly the inverse of an elliptic function).
Fig. 1.
Different remarks :
The poles of $\dfrac{1}{\sqrt{1+z^4}}$ are the four 4th roots of $-1$, i.e., the vertices of a square at $K'(\pm 1 \pm i)$ where $K'=\tfrac{1}{\sqrt{2}}$. We could have used integrand $\dfrac{1}{\sqrt{1-z^4}}$ instead ; the resulting transformation differs from $\varphi$ by a rotation with angle $\pi/4$.
A nice application of this transformation : mapping in the geographical meaning of the word : it is the "Peirce projection" map (https://en.wikipedia.org/wiki/Peirce_quincuncial_projection) ; see as well this question (Conformal mapping circle onto square (and back))
One can more generally map the upper half plane $H$ onto the interior of any polygon $P$ by the so-called Schwarz-Christoffel transformations (https://en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping). Then by combining such a transformation with a transformation like $Z=\dfrac{z-i}{z+i}$, one can map the unit disk D onto the interior of any polygon.
The previous reference is connected to this very nice document : (http://archive.bridgesmathart.org/2016/bridges2016-179.pdf).
Interest in elliptic functions arose with Euler, Legendre and Gauss (years 1750-1820) for real arguments ; they were extended to complex arguments genially in years 1830-1850, mainly by Jacobi. They play a large rôle in many parts of mathematics ; for example they are instrumental in the rather recent proof of Fermat's theorem. For the very interesting history of elliptic functions, with for example the connection with the lemniscate curve, see (http://users.mai.liu.se/vlatk48/teaching/lect2-agm.pdf) and document (9781461457244-c1.pdf).
Fig. 1 has been obtained using the following Matlab program (complex integration is rather easy with this language).
clear all;close all;hold on;axis equal off;
NR=50;NT=100;
t=zeros(NR,NT+1);s=t;
NRmax=NR+1;
scatter(1,0,30,'filled');text(1.05,0.1,'1');
for p=1:NR;
for q=1:NT+1;
z0=exp(3*(p/NRmax-1))*exp(i*2*pi*q/NT);
s(p,q)=z0;
z=quadl(@(z)(1./sqrt(1+z.^4)),0,z0);%integration
t(p,q)=z;
end;
end;
K=1.1;
plot([0,0],[-K,K]);plot([-K,K],[0,0]);
for p=30:NR;
plot(s(p,:),'r');
end;
for q=1:NT;
plot(s(30:NR,q),'b');
end;
d=3;
K=1.1;
plot([0,0]+d,[-K,K]);plot([-K,K]+d,[0,0]);
plot(exp(i*(0:pi/100:2*pi)),'k')
plot(0.927038*[1-i,1+i,-1+i,-1-i,1-i]+d,'k');
scatter(1+d,0,30,'filled');text(1+d+0.05,0.1,'1');
for p=30:NR;
plot(t(p,:)+d,'r');
end;
for q=1:NT;
plot(t(30:NR,q)+d,'b');
end;
More precisely, function $\varphi$ is defined by $$ \varphi(z) = -\sqrt{i} \, F\bigl(i \sinh^{-1}(\sqrt{i} \, z) \,|\, -1 \bigr). $$ where $F(z | m)$ is the incomplete elliptic function of first kind.
Wolfram:

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1@JeanMarie Looks like my second edit has not been approved. I fixed an error: square root of $i$, not fourth root. And thanks to you, I didn't know this transformation. – Stéphane Laurent Jul 11 '22 at 21:12
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Similarly, as documented in Ahlfors' "Complex Analysis", there are elliptic-integral expressions for the conformal maps of (reasonable) polygons to circles. Of course, these do not extend conformally to the boundary. – paul garrett Jul 11 '22 at 22:31
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Jean-Marie> Please see https://laustep.github.io/stlahblog/posts/square2circle.html :-) Thanks to your answer. – Stéphane Laurent Feb 25 '23 at 13:10
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