# Is there a map from a segment to a triangle?

Is there a conformal map from a sector in the circle with 60 degrees angle to a equilateral triange that maps the points 0, 1 and $exp (\frac{\pi i}{3})$ to the vertices of a triangle?

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Yes, as long is the interiors are concerned. The Riemann mapping theorem guarantees this. –  Hagen von Eitzen Oct 13 '12 at 13:24

I realized that my question is quiet easy, we can just map a segment to a circle as well as a triangle to a circle by Riman theorem. Then the property we do need that the vertices of triangle map to the same points as "vertices" of a segment. That can be done by adding an automorphism of a circle, that maps three first images on the boundary to the three second images on the boundary. Then we just have to take a composition.

So, the form of the domains in this problem is not important as far as it concerns only three points on the boundary.

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Yes, assuming you mean the interiors of these regions. See the Riemann Mapping Theorem. However, you should know that the proof of this theorem is not effective. That is, it won't give you an explicit description of the map. It is often the case with these sort of questions that we know, by the theorem, that a conformal bijection exists, yet nobody has any idea how to explicitly construct such a function.

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Many thanks for your answer but I've forgotten the most important point - the "vertices" should map to vertices. I've changed the question. –  Olga Oct 13 '12 at 13:37
That is, you want to include the boundary? Then no such mapping exists because there are angles of 90° that cannot correspond to 60° angles. However, there is a Riemann map of the open disc to the open hexagon that keeps the origin fixed. By symmetry, this map can be sliced into six sectors mapped to six triangles. Have a look at thiy, maybe. –  Hagen von Eitzen Oct 13 '12 at 13:43
Yes, I do want a conformal mapping of interiors, without boundary. –  Olga Oct 13 '12 at 17:52

For something explicit, consider the Schwarz-Christoffel mapping. Together with $z\mapsto \frac{i-iz}{z+1}$ this produces a map $f$ of the open unit disc via upper halfplane to a regular hexagon. With suitable adjustments you can achieve $f(0)=0$ and $f'(0)>0$. Then the restriction of $f$ to $\{z\mid 0< \arg z<\frac\pi3, 0<|z|<1\}$ will be the mapping you want (including that $f(z)$ tends to the vertices if $z$ tends to a vertex; this follows by symmetry). On the other hand, $f$ cannot be extended conformally to the vertices themselves.

To elaborate on the "symmetry handwaving" above: Let $D$ be the open unit disc and $X$ the open hexagon. We know that the automorphisms of $D$ have the form $\phi_{u,a}\colon z\mapsto u\frac{z-a}{1-\bar a z}$ with $|u|=1$ and $|a|<1$. If we have an arbitrary biholomorphic $f_1\colon D\to X$, let $b=f_1^{-1}(0)$. Then $f_2:=f_1\circ \phi_{1,-b}$ is also biholomorphic $D\to X$, but this time we know $0\mapsto 0$. Let $c=f_2'(0)$. Then $c\ne0$ and $f_3:=f_2\circ \phi_{\frac{|c|}c,0}$ is also biholomorphic $D\to X$ and as promised we have $f_3(0)=0$, $f_3'(0)>0$. Assume $f_4\colon D\to X$ is biholomorphic with $f_4(0)=0$ and $f_4'(0)>0$. Then $f_3^{-1}\circ f_4\colon D\to D$ is an automorphism, i.e. it is some $\phi_{u,a}$. From $\phi(0)=0$ we conclude $a=0$ and then from $\phi'_{u,a}(0)>0$ we conclude $u=1$. It follows that the automorphism is the identity, i.e. $f_4=f_3$. There is exactly one biholomorphic map $D\to X$ that maps $0\mapsto 0$ and has positive real derivative at $0$. Now observe that $f_5\colon z\mapsto \overline{f_3(\bar z)}$ is also a biholomorphic map $D\to X$ with $f_5(0)=0$ and $f_5'(0)>0$. Hence $f_5=f_3$, which implies that $f_3(z)$ is real if $z$ is real. In fact, injectivity and positive derivative at $0$ impliy that $f_3$ maps $[0,1)$ to itself monotonuously. Also observe that $f_6\colon z\mapsto e^{-\frac\pi3} f_3(e^{\frac\pi3} z)$ has $f_6(0)=0$, $f_6'(0)>1$, hence $f_6=f_3$ and we conclude that $f_3$ maps the line $e^{\frac\pi3}\cdot [0,1)$ to itself. In summary, $f_3$ maps the region bounded by the straight line from $0$ to $1$, the straight line from $0$ to $e^{\frac\pi3}$ and the circular arc from $0$ to $e^{\frac\pi3}$ biholomorphically to the triangle with vertices $0$, $1$ and $e^{\frac\pi3}$.

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Do I understand correctly that I actually can choose $Arg f'(0)$ to be any angle I want (by Riemann theorem construction) and that determines my map? And as far as I understand, for any choice of $Arg f'(0)$ the map will still map vertices to vertices, isn't it? –  Olga Oct 13 '12 at 17:57
I would like to precise the argument via symmetry principle: so, I do have a map from the unit circle to a hexagon. Let me now choose a point that maps to a vertice, and a corresponding diameter of a unit circle. Why does this diameter map to a line connecting edges of hexagon? Because only in this case we can apply symmetry principle. –  Olga Oct 13 '12 at 18:03
Thank you very much for help! By the way, doest it follow, that diameters of the circle go to straight lines in hexagon from your argument? –  Olga Oct 14 '12 at 9:00