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From the answer https://math.stackexchange.com/a/3743490/ the "impossibility" of the Penrose triangle and the Escher cube in Euclidean space arises because they violate the condition that a closed loop of orthogonal edges must return to the same $z$-coordinate. In Nil geometry, the $z$-coordinate is coupled to the area swept out in the horizontal ($x,y$) plane. A closed loop in the $x,y$ plane (like a square) creates a vertical shift proportional to its area.

I wonder if the impossible cube can be embedded isometrically into a curved three-dimensional space such that its edges are also geodesics.

impossible cube

▲ In a standard Euclidean cube, the faces are merely adjacent; in the "impossible" embedding, the boundary curves of the faces can become topologically linked.

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  • The highly upvoted linked answer is nonsensical. In order to make a valid question you first has to define what you mean by an "impossible cube": You just have a picture but not a definition. Is your "impossible cube" a metric space? Then what is the metric that you are using? Is it just a topological space? Then "isometric" makes no sense, but you can talk about a topological embedding, which is, of course, possible. – Moishe Kohan Feb 03 '25 at 02:19

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