Area preserving bijection between square and circle that is conformal on the perimeter?

Question

Is there a bijection between a square and circle that is (1) area-preserving (2) conformal (angle-preserving) on the perimeter (except at the corners)?

This map satisfies (1) but not (2). This theorem constructively proves the existence of map satisfying (2) but not (1).

You might have a look at https://en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping — Jack D'Aurizio, Jan 10 '18 at 17:43
Having got straight what you mean, I don't see why there could not be such a map, not that I have one handy... — David C. Ullrich, Jan 10 '18 at 18:02
I am not sure if this solves the problem, but take a look at this version of Moser's theorem: Diffeomorphisms and volume-preserving embeddings of noncompact manifolds, by R. E. Greene and K. Shiohama, Trans. Amer. Math. Soc. 255 (1979), 403-414. — Moishe Kohan, Jan 14 '18 at 11:26

score 4 · Accepted Answer · edited Jan 18 '18 at 21:41

Suppose $f$ is the desired map, and let $\mathbf J$ be its Jacobian matrix. On the boundary (except the corners), $f$ is conformal, so $\mathbf J$ is a multiple of a rotation matrix. But $f$ is also area-preserving, so $\det\mathbf J = 1$, so $\mathbf J$ actually is a rotation matrix. Therefore $f$ preserves lengths along the boundary. But this is impossible, because the boundaries of a square and a circle of equal area do not have the same length.

Area preserving bijection between square and circle that is conformal on the perimeter?

1 Answers1