On Mon, Oct 25, 2004 at 01:50:32PM -0500, David Gale wrote:
1. Believe me, I am very familiar with the story of "Hilbert's third". ...
My apologies; I didn't understand what you were asking about.
As an aside, I had the intention of sending these links to the group anyway. They are certainly math and they're supposed to be fun. I'd welcome reaction from any of you.
http://mathsite.math.berkeley.edu/intro.html http://mathsite.math.berkeley.edu/main.html
Unfortunately, none of the exhibit itself works with free software, so I cannot view it. It sounds quite interesting, though.
2.Dylan, your remarks about the problem are not correct. You wrote " Then P and P' are equidecomposable iff they have the same area and same Dehn invariant." "It was solved early last century".
Not true. Dehn only proved the necessity. The sufficiency was proved (?) seventy years later by a Danish mathematician named Sydler and a clear proof was given by Borge Jessen shortly afterwards. The proof is difficult and uses homological algebra (!). The book "Hilbert's Third Problem" by Vladimir Boltyanskii 1978 is the definitive reference.
Thanks for the correction!
3. Bill, Do your remarks imply that the Dehn-Hilbert problem can be reduced to our rectangle problem??? This would be astonishing because our method for finding the equi-decompositions is quite elementary. By the way, when I say "our", the main work was done not by me but by Clifford Gardner. Following your suggestion I'm sending a copy of this to Walter Neumann.
I think the import of Bill's message was that the same homological techniques that can be used to prove the converse of the Dehn-Hilbert problem can also be used to solve your question. Your proof is doubtless much easier than that.
4. Just for emphasis let me say again, what we have is an algorithm. Given two finite sets of rectangles with the same tensor area it finds a common dissection. Such a natural problem, you'd think it would be known, but is it?
Can you say a bit more about the proof? I would guess that essentially from the definition of tensor product you can get an algorithm that possibly uses negative area rectangles, and the difficulties are avoiding those? In any case, I would ask Walter Neumann if he knew about previous work. Peace, Dylan