4 ms·
A question for people who know more topology(?) than I do: what determines this? Is it possible to construct a surface that requires an arbitrary number of dime
by wiml 3y ago
A question for people who know more topology(?) than I do: what determines this? Is it possible to construct a surface that requires an arbitrary number of dimensions for its embedding?
- dullcrisp 3y agoGood question. Based on googling [0] it looks like any surface can be embedded in four dimensions (and any n-manifold can be embedded in n+2 space). But I don’t know enough to explain the construction. Might have something to do with the tangent bundle, it sounds like? [0] https://math.stackexchange.com/questions/2960093/why-do-we-need-4-dimensions-to-embed-a-two-dimensional-shape-on-a-surface https://math.stackexchange.com/questions/2960093/why-do-we-n...
- xelxebar 3y agoIf by surface you mean a 2-dimensional manifold, then no. The Nash embedding threorms give upper bounds for how big the embedding target dimension needs to be for any n-dimensional manifold. In the case of 2-dimensional surfaces, at worst you'll need 51 = 2×(2+1)×(3×2+11)/2 dimensions. Mind this is for isometric embeddings (ones that don't squish things around in a sense), which is kind of the most stringent constraint possible. If you allow squishing things around, then the upper bounds get even smaller. https://en.wikipedia.org/wiki/Nash_embedding_theorems#Ck_embedding_theorem https://en.wikipedia.org/wiki/Nash_embedding_theorems#Ck_emb...