4 ms·
No, a Klein bottle is an object that can not exist in 3d. The one you probably know is an "immersion" into 3d (making the object intersect itself, which it woul
by master-lincoln 3y ago
No, a Klein bottle is an object that can not exist in 3d.
The one you probably know is an "immersion" into 3d (making the object intersect itself, which it wouldn't do in it's original shape in 4d)
- notso411 3y ago[dead]
- dullcrisp 3y agoI don’t suppose it’ll help at all if I say that a Klein bottle is a two-dimensional surface and four is the smallest dimensional Cartesian space into which it can be embedded.
- wiml 3y agoA 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...