Skip to main content
21 events
when toggle format what by license comment
May 5, 2023 at 14:55 history edited Wolfgang
edited tags
S Apr 3, 2023 at 1:06 history bounty ended CommunityBot
S Apr 3, 2023 at 1:06 history notice removed CommunityBot
Mar 31, 2023 at 17:24 comment added Nick S Also, I think that one version of the Socolar-Taylor tiling is a topological disk(all you need is to connect the tile to the isolated pieces by the right line segments), but not the closure of its interior. Finally, while the SCD tile is aperiodic by our definition, it does have skew symmetry.
Mar 31, 2023 at 17:19 comment added Nick S The SCD tile can probably be extended to any dimension $d \geq 3$. The problem lies in what do you mean by aperiodic tile... In the field interested in this question, Aperiodic Order, an aperiodic tile/set of tiles is any set of tiles which can tile $\mathbb R^d$ via translates and rotates, but any tiling of $\mathbb R^d$ has no translational symmetry. By this definition, the new einstein tile is actually not an einstein tile (you need the tile and its reflection) but the SCD tile is aperiodic. The 2-dim Socolar-Taylor tile may or may not be aperiodic depending on what a tile is.....
Mar 31, 2023 at 16:35 history edited Nicholas James CC BY-SA 4.0
added 9 characters in body
Mar 29, 2023 at 13:23 history edited Nicholas James CC BY-SA 4.0
added 1 character in body
Mar 28, 2023 at 13:30 history edited Nicholas James CC BY-SA 4.0
deleted 11 characters in body
Mar 26, 2023 at 3:48 comment added Seewoo Lee @WillSawin Thank you for pointing it out. I just realized that we can make aperiodic tiling just with squares in a similar way, but that's not what we want. In case of hat prism, we can fix a single tiling of $\mathbb{R}^2$ by hats and take a product with $[0, 1]^{(n-2)}$ (so just stack it into the other directions without any perturbations) - this gives periodic tiling, right?
Mar 25, 2023 at 23:47 comment added Will Sawin @SeewooLee The point is not just to find a tile that admits an aperiodic tiling, but also to ensure it doesn't admit any periodic tilings. It is not at all obvious that the hat prisms don't admit such tilings.
S Mar 25, 2023 at 23:39 history bounty started Nicholas James
S Mar 25, 2023 at 23:39 history notice added Nicholas James Draw attention
Mar 24, 2023 at 20:28 comment added Seewoo Lee Adding to Tao’s comment, I think this would give a construction. Consider the “hat prisms” which are just hat x [0,1]^(n-2). The paper says that there are uncountably many different tilings. Now tile each “layers” of R^n with different tilings of hat prisms (just consider a product of tiling in R^2 and $[0,1]^{(n-2)}$. If you randomly shift all the tiles in each layer by random vectors in $\mathbb{R}^2$, I believe this would give aperiodic tiling. Maybe finding an aperiodic monotile in a essentially different way would be more interesting.
Mar 24, 2023 at 14:37 history edited Nicholas James CC BY-SA 4.0
deleted 2 characters in body
Mar 24, 2023 at 0:36 history edited Nicholas James CC BY-SA 4.0
deleted 3 characters in body
Mar 23, 2023 at 23:57 comment added Nicholas James @TerryTao: Thank you for your response! Out of curiosity, what would the “annoyingly technical arguments” that you mentioned involve exactly? Would they be unrelated to the arguments made in the paper in question?
Mar 23, 2023 at 21:51 comment added Terry Tao I would imagine that taking the Cartesian product of the new aperiodic monotille in ${\bf R}^2$ with a generic fundamental domain of ${\bf R}^{n-2}/{\bf Z}^{n-2}$ should give aperiodic monotile in ${\bf R}^n$, though proving this claim rigorously may require some annoyingly technical arguments.
Mar 23, 2023 at 20:52 history edited Nicholas James CC BY-SA 4.0
Fixed wording and listed authors explicitly so proper credit is given.
Mar 23, 2023 at 18:58 history edited Nicholas James CC BY-SA 4.0
Fixed formatting.
S Mar 23, 2023 at 18:45 review First questions
Mar 23, 2023 at 18:46
S Mar 23, 2023 at 18:45 history asked Nicholas James CC BY-SA 4.0