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From the answer https://math.stackexchange.com/a/3743490/ it is possible to embed isometrically (bent but not stretched) Penrose triangle into curved three dimensional space, something called "nil geometry".

I wonder if the impossible cube also can be embedded isometrically into a curved three-dimensional space.

impossible cube

▲ An impossible cube, with adjacent squares having linking number $1$ or $-1$.

hbghlyj
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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
  • @MoisheKohan Yes, it is a metric space – hbghlyj Feb 03 '25 at 08:04

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