# Conformal equivalence of square and upper-plane, including part of the boundary

I was reading the book "Conformal Invariants" of L. Ahlfors, and seen (P.74) he says that "It is in fact obvious that the part of the teichmuller annulus (for R=1) in the upper plane is conformally equivalent to a square". Where the teichmuller annulus (for R=1) is $\mathbb{C}/(-1,0)\cup(1,\infty)$.

I don't even sure what he means, and in particular can't see why it is an obvious fact.

As I understand it, he means that there exists a conformal map from $\mathbb{H^+}/(-1,0)\cup(1,\infty)$ onto a square, including two of it's (opposite) edges. without the need for the edges (only the interior) it is trivial (due to Riemann mapping theorem) and the map is well known (Schwarz–Christoffel in this case is just the elliptic integral of first kind). But the interior is not enough since I (and Ahlfors too) need to calculate the extremal length (\conformal modulus) of the region using this fact [the interior is also equivalent to any other rectangle, while they have diffrent ext. length]. Can someone help me here please?

In fact, i am trying to find an elementary argument for the more general fact (also Page 74) that the modulus of teichmuller annulus with some specific R is (up to a multiplicative constant) the inverse of the modulus of teichmuller annulus with 1/R ($\Lambda(R)\Lambda(R^{-1})={1\over4}$). Any idea how to prove this will be usefull.

• The standard conformal map from the upper half plane onto the unit disk $(z-i)/(z+i)$ sends your intervals to the arcs from $1$ to $i$ and from $-1$ to $-i$, respectively. It seems "obvious" that this can be mapped to the square in the desired fashion, and I don't think a rigorous argument will be very hard to find, though I wouldn't exactly know how to do it off the top of my head. Sep 20, 2016 at 19:12

A simply connected region with 4 marked boundary points is called a quadrilateral. A conformal map of quadrilaterals is a conformal map of the regions which sends the marked boundary points to the marked boundary points. Each quadrilateral is conformally equivalent to a rectangle with vertices marked (this is a simple exercise with Schwarz and Christoffel).

A square is characterized (among all rectangles) by the property that it has a conformal automorphism which acts on the vertices cyclically (1 to 2 to 3 to 4 to 1, with natural enumeration). This automorphism is just a rotation by $90$ degrees.

The upper half of the Teichmuller ring in question is the upper half-plane with $-1,0,1,\infty$ marked. It has a conformal automorphism $f(z)=(1+z)/(1-z)$ which maps $-1\to 0\to 1\to\infty\to -1$. Therefore this quadrilateral is conformally equivalent to a square.

• thanks alot. It makes sense now with simple reasoning. On the other hand, i can't see how i can use a similar argument for the general relation $\Lambda(R)\Lambda(R^{-1})={1\over4}$ which Ahlfors cite with "the proof of which is left as an exercise." at the same paragraph. Also - is there any proof that not relay on Schwarz and Christoffel directly? Sep 21, 2016 at 19:07
• Hint. The product of the moduli of quadrilaterals (a,b,c,d) and (b,c,d,a) is $1$. Sep 21, 2016 at 19:40
• Awesome, thanks! can't understand how I didn't see it before. Sep 27, 2016 at 18:52
• Great answer. I was watching a video on string theory by Suskind where he says that a square is a conformally equivalent to a rectangle, but not in a way that preserves the vertices. The audience (and me) was surprised, but this answer clearly explains why.
– Joël
Jan 14, 2022 at 21:58
• Every two simply connected regions (other than the whole pane) are conformally equivalent. This is Riemann's theorem. So why people were surprised? Jan 14, 2022 at 23:29