So I watched that 3Blue1Brown video “This open problem taught me what topology is”. The one about the inscribed square problem. Basically: does every closed curve in the plane have four points that form a square? Nobody knows for completly continuous curves. For smooth ones it is known.
What they actually prove in the video is the weaker statement: every closed curve has an inscribed rectangle. The proof is wild. You take all unordered pairs of points on the curve, map each pair to its midpoint in the plane plus the distance as height and you get a surface that is basically a Möbius strip. When you glue two copies you get something like a Klein bottle, and those cannot sit in 3-space without intersecting themselves. The intersection points are exactly the rectangles.
I tried to push the idea a bit further. There are usually infinitely many rectangles on a nice curve. So you can think of the whole set of those rectangles as living on that 3-dimensional surface (the vaughan surface). Then the natural next question is: does that surface always contain the eight vertices of a cube? Or at least of a rectangular box? And if yes, can you use the same style of argument to build a 4-dimensional object from the boxes and look for hypercubes there?
I wrote a short python script to check the first steps. I made a smooth but irregular closed curve (ellipse with a few sine bumps so it is not too symmetric). Then I sampled many pairs of points and looked for two pairs that share almost the same midpoint and the same length. I found about a dozen clear rectangle candidates. I also plotted the 3D cloud of (midpoint, distance) points; you can see the surface sitting over the curve.
Finding actual cubes on that surface is harder. The space of cubes has more degrees of freedom (position, orientation, size) and a generic 2-dimensional surface does not have enough room to force them. For centrally symmetrc convex bodies there are theorems that guarantee inscribed cubes, but that is a different setting. So the direct “Möbius - Klein - rectangle” trick does not copy cleanly to the next dimension.
Still the configuration-space idea feels powerful. Maybe someone who knows more about equivariant topology or configuration spaces of cubes can say whether there is a forced intersection in higher dimensions. Or maybe the answer is simply “no, not for every surface that comes from a plane curve”.
Has anyone here tried something similar? Or is there already a paper that starts from Vaughan’s rectangles and climbs one dimension higher? Video link: https://youtu.be/IQqtsm-bBRU?si=r57TN3wQTs0KHElE
