Area-preserving map between rectangles and fat polygons - MathOverflow

Area-preserving map between rectangles and fat polygons

2011-06-01T19:53:13Z

Are there any well-known continuous area-preserving maps between fat convex polygons and fat rectangles? Specifically, given a fat convex polygon $C$, is there a natural way to choose a fat rectangle $R$ and a continuous mapping $f:C\rightarrow R$ such that area of sub-regions is preserved? I've avoided giving a precise definition of "fatness", but two common definitions are:

*The aspect ratio of the minimum bounding box of $C$ is bounded by some constant (see page 5 of http://portal.acm.org/citation.cfm?id=1137901).

*The ratio between the diameters of the smallest circle containing $C$ and the largest circle contained in $C$ is bounded by some constant (see http://books.google.com/books?id=QS6vnl8WlnQC&pg=PA588).

Answer by André Henriques for Area-preserving map between rectangles and fat polygons

2011-06-01T20:32:33Z

This problem is treated in section 18 ("Monge problem for polytopes") of Igor Pak's book: http://www.math.ucla.edu/~pak/geompol8.pdf (without any reference to fatness).

Answer by Dylan Thurston for Area-preserving map between rectangles and fat polygons

2011-06-01T20:34:54Z

There are lots of ways to do this; without more constraints (or indication of what is desired), it's hard to pick a best one. But here's one. We'll exploit the fact that for any two triangles, there is a unique affine linear map that takes one to the other (with vertices in a specified order); if the two triangles have the same area, this will be area-preserving.

Start by chopping up $C$ into triangles $T_1,\dots,T_k$ by drawing lines connecting the vertices to the center.
Now take $R$ to be a square with the same area as $C$. Divide up the boundary of $R$ into $k$ intervals $I_1,\dots,I_k$ whose lengths are proportional to the area of the $T_i$. (Some of the intervals may make a turn at the corner.) Divide the square into regions $S_i$ by connecting the ends of $I_i$ to the center of the square. Most of the $S_i$ will be triangles, except for the four at the corners, which are quadrilaterals.

Now, the area of $S_i$ is equal to the area of $T_i$, since the area of $S_i$ is proportional to the length of its base (as all $S_i$ have the same height), which is proportional to the area of $T_i$. Now just map $T_i$ to $S_i$ in an area preserving way. For the $S_i$ that are not at the corners, just take the affine linear map from above. For the $S_i$ that do go around the corner, cut both $S_i$ and $T_i$ into two triangles in the same proportion of area, and map the corresponding triangles to each other.

Answer by Anton Petrunin for Area-preserving map between rectangles and fat polygons

2011-06-02T07:26:57Z

It is easy to construct a volume preserving map between any two shapes of equal volume with triangular Jacobian matrix. (Gromov used such map from the given domain to ball to prove isoperemetric inequality.)