On Sun, Jun 21, 2015 at 12:59 PM, Warren D Smith <warren.wds@gmail.com> wrote:
--Dear delinquents: if you are going to find problems with what I say, I'd appreciate hearing it directly.
My bad. They're (mostly) not on math fun. But generally we share our problems here, don't we?
RWG: Kids. You can't tell them anything these days. Here's Julian trash-talking to some of his fellow young delinquents: Julian Ziegler Hunts wrote: Sure, but first I'd like to point out that he has not proved that any of these tile the plane, and I particularly do not believe that #1 does. He shows that each is the union of scaled copies of itself, but this does not suffice to prove that it tiles the plane—you would also need to prove that the tile has nonempty interior (the negligible-intersection part of being a tiling would follow). It is true that the scaling involved in the constructions would guarantee the tile has dimension 2, but only if the intersection between the scaled copies is negligible (being of smaller dimension would suffice, as would having measure zero if the whole has positive measure (which is not guaranteed by having dimension 2!)). And having dimension 2 does not guarantee that it has non-empty interior or will tile the plane. A simple counterexample is to remove all points with rational coordinates from the unit square; it is then the union of four non-intersecting half-scale copies of itself but does not tile the plane. Julian
--In the meantime, I also was developing skepticism for related reasons. Things are not so easy. To reach nirvana, it seems to me we want to prove the tile T has an interior which is both nonempty and arc-connected. If so, then the claim that T is tiled by some suitably-scaled copies, should recursively provide a tiling of the whole plane, as I said. But sadly I did not provide such proof. (All I proved was: T is bounded, and T is tiled by some suitable scaled copies.) Nor did I even compute good pictures.
However, Roth's theorem suggests that the points generated by my SUM (if truncated at some finite #digits) are never too close together. http://mathworld.wolfram.com/RothsTheorem.html https://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem I mean, in plain 1 dimensional binary, T = SUM(j>=0) 2^(-j) b_j, the summands if the SUM is truncated to 0<=j<n are 2^(-n) apart. In certain 2 dimensional binary systems like T = SUM(j>=0) sqrt(-2)^(-j) b_j they are about 2^(-n/2) apart. With an algebraic number radix, those things are again going to happen to within a factor of 2^o(n), note the LITTLE o... is that right? (Also, mine are not merely algebraic, they also arise from Pisot or Salem numbers, which probably provide extra juice for arguments of this kind.)
This muttering about Roth seems to me to be inadequate to prove tiling of the plane, but likely *is* adequate to see that the tile I have created, is not a Cantor set or something like that which would obviously have been a failure.
If you are skeptical about my tile #1 (which actually was the one I was least skeptical about) then I'd like to know: why is the Rauzy fractal an ok tile, but my tile #1 is not? https://en.wikipedia.org/wiki/Rauzy_fractal
AHA!! So that's what they're called! Soon after I sent Martin Gardner the Flowsnake (before or after 1981?), Martin mailed me a "letter" from the now famous but still mysterious Robert Ammann. It was a bunch of pencil sketches on graph paper of such frac-tiles, almost certainly including that one, with chords drawn on them, cryptically labeled with polynomials. Martin was anxious for details, but Ammann's mailing address kept changing. When a letter asking his occupation finally caught up to him, Ammann responded that he was a parking lot attendant! --rwg (Or maybe Rauzy is not actually ok either. But pictures have been drawn
for it and it sure looks ok.)
So, bottom line: I do not know if my tiles really tile, but it seems to me that if they have nonempty arc-connected interior, then they must. That seems most likely to me in cases #1 and #3. I'd like to see pictures. Final question: Is it possible to create an "einstein" tile T (which tiles the plane by itself, but only in an aperiodic way) as
T = SUM(j>=0) R^(-j) d_j
where (1) each d_j is selected from a fixed set of K complex numbers in all possible ways, (2) R is a fixed algebraic complex number with |R|=sqrt(K) got by multiplying the unit-norm root of a "Salem polynomial" by sqrt(K).
It has been a long-open question whether an einstein tile exists. This construction might provide one with (say) K=5. But you'd need to understand more than I do about when this construction method works.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)