My reaction was similar to Dylan's. However, I don't remember hearing this problem formulated in this way before---it's a nice phenomenon. Just to be more explicit about what Dylan was alluding to, you can think of each edge of a polyhedron as determining a rectangle, in cylindrical coordinates, whose length is the length of the edge and whose width is the angle. An equidecomposition of polyhedra leads to an equidecomposition of these rectangles in Gale's sense, except that any rectangle of width 2 pi can be created or thrown away "for free". One person who has worked on related questions more recently is Walter Neumann; if you need references, he probably has a pretty good grip. This is all connected to the homology of classifying spaces for Lie Groups given the discrete topology. E.g. the fancy terminology for equidecomposition classes of rectangle case as described is the 2nd homology group of R with the discrete topology (which is well-known since as an abstract group R is just a free abelian group on lots of generators, if you accept the axiom of choice.) The only point not captured directly by homology is the issue of positive vs. negative rectangles. Bill Thurston On Oct 25, 2004, at 8:50 AM, Dylan Thurston wrote:
On Sun, Oct 24, 2004 at 03:54:57PM -0500, David Gale wrote:
I would like to know whether the following stuff is known.
Well known: There does not exist a set of rectangles which will tile both a 1 by 2 rectangle and a square of side sqrt(2) (although they both have the same area). We say these sets are not EQUI-TILABLE.
GENERAL PROBLEM: Let S be a union of a finite set of rectangles and let S' be another such set. Under what conditions are S and S' equi-tilable.
ANSWER: Think of reals as a vector space over Q. For a rectangle of width a and height b we can form their TENSOR PRODUCT a(x)b. The TENSOR AREA of the set S is summation(a_i(x)b_i) summed over the rectangles of S.
THEOREM. The sets S and S' are equi-tilable if and only if they have the same tensor area.
This is surely known. There's a harder problem, one of Hilbert's problem, solved by Dehn:
Question. When are two polyhedra in R^3 equidecomposable into polyhedra? [Allow arbitrary polyhedra, not just rectangles]
Answer. For each edge, consider l(x)theta, where l is the length of the edge and theta is the angle of the edge, as an element of R(x)(R/2pi), tensor product over Q. The sum of this quantity over all edges of a polyhedron P is called the Dehn invariant of P. Then P and P' are equidecomposable iff they have the same area and same Dehn invariant.
I don't know the references, sorry. It was solved early last century.
Peace, Dylan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun