# uniformization theorem - squares and circles

I am trying to understand the uniformization theorem and get some intuition about it.

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to one of the three domains: the open unit disk, the complex plane, or the Riemann sphere.

Specifically, the Riemann mapping theorem states that every simply connected open subset of the complex plane that is different from the complex plane itself admits a conformal and bijective map to the open unit disk.

So, an open square is conformally equivalent to the open circle. But how do I find the bijection?

What about a closed square - is it conformally equivalent to the open circle, and how?

• flickr.com/photos/sbprzd/362529354 May 23, 2013 at 17:45
• The closed square is not a Riemann surface – it isn't even a manifold without boundary! May 23, 2013 at 18:03
• In general it is hard to write closed forms for the conformal maps. But sometimes there are integral expressions for them and good numerical ways of finding them. Try en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping for a start, which says how to conformally map polygons onto the disc. (The wiki article has your question about the square as an example, and brainjam's picture is also very nice for developing intuition.)
– GCD
May 24, 2013 at 3:32
• Thank you all, this is very helpful. May 24, 2013 at 5:07

(1) In practical terms, it is slightly easier to work with upper half-plane instead of the open unit disk. The composition with $(z-i)/(z+i)$ then gives a map onto the disk. The Schwarz–Christoffel method gives a practical way to find a conformal map of upper half-plane to a polygon.