BEGIN:VCALENDAR VERSION:2.0 PRODID:researchseminars.org CALSCALE:GREGORIAN X-WR-CALNAME:researchseminars.org BEGIN:VEVENT SUMMARY:Gabor Somlai (E\\" otv\\" os Lor\\' and University and R\\' enyi Institute) DTSTART:20250521T143000Z DTEND:20250521T145500Z DTSTAMP:20260423T041653Z UID:CANT2025/28 DESCRIPTION:Title: Pushing the gap between tiles and spectral sets even further\nby Gab or Somlai (E\\" otv\\" os Lor\\' and University and R\\' enyi Institute) as part of Combinatorial and additive number theory (CANT 2025)\n\nLecture held in CUNY Graduate Center - Science Center (4th floor).\n\nAbstract\nF uglede conjectured that a bounded measurable set in a locally compact topo logical space endowed with Haar measure is spectral if and only if it is a tile and Fuglede also confirmed the conjecture for sets whose tiling comp lement is a lattice and for spectral sets one of whose spectrums is a latt ice.\n\nThe conjecture was disproved by Tao in the case of finite abelian groups where the counting measure plays the role of the Haar measure. \nTa o constructed a spectral set in $\\mathbb{Z}_3^5$ of size 6\, that is not a tile. This construction was lifted to the $5$ dimensional Euclidean spac e\, where the original conjecture was mostly studied. \n\nLev and Matolcsi verified Fuglede's conjecture for convex sets in $\\mathbb{R}^n$ for ever y positive integer $n$. The key of proving the harder direction of the con jecture is to introduce the weak tiling property and prove that all spectr al sets are weak tilings.\n\nOne of the goals of our work was to answer a question of Kolountzakis\, Lev and Matolcsi\, whether there is a weak tile that is neither a tile nor spectral. There is such a set which apparentl y makes it harder to prove the spectral-tile direction of the conjecture i n the remaining open cases. \n\nThe other result towards structurally dist inguishing spectral sets and tiles was a disproof of a conjecture of Green feld and Lev. They conjectured that the product of two sets is spectral if and only if both of them are spectral. A similar property holds for tiles \, but the product of a non-spectral set with a spectral set can be spectr al. \nFinally\, we obtain an easy characterization of tiles using the spec tral property.\n LOCATION:https://researchseminars.org/talk/CANT2025/28/ END:VEVENT END:VCALENDAR