# An example against the periodic tiling assumption

Rachel Greenfeld and I simply uploaded our article on arXiv”An example against the periodic tiling assumption“. That is the total model of the consequence I introduced on this weblog just a few months in the past. periodic tiling assumption associated to Grunbaum-Shephard And Lagarias-Wang. The article took a bit longer than anticipated as a consequence of a technical problem that we weren’t conscious of on the time of the announcement and wanted a workaround.

In additional element: The unique technique, as described within the announcement, was to create a “tiling language” that might encode a specific “sequence”.-adic Sudoku puzzle ”after which present that the second kind of puzzle has non-periodic options provided that: ${P}$ was a big sufficient prime quantity. It appears that evidently the second half of this technique labored, however there was an issue with the primary half: Our tile language (utilizing this) ${2}$-group valued capabilities) to encode arbitrary boolean relationships between boolean capabilities and in addition (utilizing) ${{bf Z}/p{bf Z}}$-value capabilities) to encode “clock” capabilities akin to ${n mapsto n hbox{ mod } p}$ it was part of us ${P}$-adic Sudoku puzzle, however we have not been in a position to get these two sorts of capabilities to “discuss” to one another as wanted to code. ${P}$-adic Sudoku puzzle (primary downside, if ${H}$ a finite commutative ${2}$-group then there aren’t any non-trivial subgroups. ${H times {bf Z}/p{bf Z}}$ not included in ${H}$ or insignificant ${{bf Z}/p{bf Z}}$ course). Consequently, we needed to change our “”.${P}$-adic Sudoku puzzle” by “${2}$-adic Sudoku puzzle” mainly means changing the prime ${P}$ by a sufficiently giant power ${2}$ (we imagine ${2^{10}}$ will suffice). This fastened the encoding problem, however ${2}$-adic Sudoku puzzles had been a bit extra complicated than others. ${P}$-adic situation, because of the following motive. Beneath is a pleasant evaluation train:

Theorem 1 (Linearity in three instructions means full linearity) To offer permission ${F: {bf R}^2 rightarrow {bf R}}$ be an everyday perform that’s affine-linear on each horizontal line, diagonal (slope line) ${one}$) and anti-diagonal (slope line ${-one}$). In different phrases, for any ${c in {bf R}}$capabilities ${x mapsto F(x,c)}$, ${x mapsto F(x,c+x)}$And ${x mapsto F(x,cx)}$ each affine perform ${{bf R}}$. Later ${F}$ is an affine perform on ${{bf R}^2}$.

Certainly, the property of being affine in three instructions signifies that there’s the second-order type related to Hessian. ${nabla^2 F(x,y)}$ disappears at any level ${(1,0)}$, ${(1,1)}$And ${(1,-1)}$and so it ought to disappear all over the place. In actual fact, the smoothness speculation isn’t essential; We go away this as an train to the reader. If one is substituted, the identical expression seems to be true. ${{bf R}}$ with cyclic group ${{bf Z}/p{bf Z}}$ if ${P}$ unusual; that is the important thing for us to indicate ${P}$-adic Sudoku puzzles have a (roughly) two-dimensional affinity construction, which can be utilized later in additional evaluation to indicate that they’re truly non-periodic. Nonetheless, the corresponding declare for cyclic teams ${{bf Z}/q{bf Z}}$ when can it fail ${Q}$ is powerful sufficient ${2}$! Really the overall type of capabilities ${F: ({bf Z}/q{bf Z})^2 rightarrow {bf Z}/q{bf Z}}$ takes the affine type on every horizontal line, diagonal, and diagonal

$displaystyle F(x,y) = Ax + By + C + D frac{q}{4} y(xy)$

for some integer coefficients ${A B C D}$. This extra time period “pseudo-affine” ${D frac{q}{4} y(xy)}$ causes some extra technical issues, however finally seems to be manageable.

Throughout the writing course of, we additionally found that the coding a part of the proof turns into extra modular and conceptual when two new definitions, “expressible property” and “poorly expressible property”, are added. These ideas are considerably just like: ${Pi^0_0}$ sentences and ${Sigma^0_1}$ sentences in arithmetic hierarchyor algebraic sets And semi-algebraic sets in actual algebraic geometry. Roughly talking, an expressible property is a property of a set of capabilities. ${f_w: G rightarrow H_w}$, ${w in {mathcal W}}$ from a commutative group ${G}$ finite abelian teams ${h_w}$such that the characteristic may be expressed on the graph by way of a number of tiling equations

$displaystyle A := { (x, (f_w(x))_{w in {mathcal W}} subset G times prod_{w in {mathcal W}} H_w.$

For instance, the property that two capabilities have ${f,g: {bf Z} rightarrow H}$ the distinction with respect to a continuing may be expressed by way of the tiling equation

$displaystyle A oplus ({0} times H^2) = {bf Z} times H^2$

(vertical line check) and in addition

$displaystyle A oplus ({0} times Delta cup {1} times (H^2 backslash Delta)) = G times H^2,$

The place ${Delta = {(h,h): h in H }}$ is the diagonal subgroup ${H^2}$. A poorly expressible characteristic ${P}$ is an existential measurement of some expressible trait ${P^*}$so a set of capabilities ${(f_w)_{w in W}}$ suits property ${P}$ if and provided that there’s an extension of this tuple by some extra performance that matches the characteristic ${P^*}$. It seems that weak expressible options are turned off below a collection of helpful operations, permitting us to simply create fairly complicated weakly expressible options from a “library” of easy weakly expressible options, akin to a fancy pc program may be created. easy library routines. Specifically, we can “program” our Sudoku puzzle as a weakly expressive characteristic.

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