5 ms·
Both of those items are 3D - while the dress transforms over a 4th time dimension.
by kahunalu 3y ago
Both of those items are 3D - while the dress transforms over a 4th time dimension.
- MichaelZuo 3y agoDon't all dresses transform over time? Eventually 100% of all possible dresses will transform into dust.
- dambi0 3y agoWhat about dresses repurposed into non-dresses prior to their dusty demise.
- MichaelZuo 3y agoEventually those will too.
- Razengan 3y agoI mean everything transforms over time.
- omoikane 3y agoReminds me of https://xkcd.com/209/ https://xkcd.com/209/ "So the kayak travels through time?" "Sure! Just like everything else!"
- dclowd9901 3y agoThis applies to everything so it’s moot.
- master-lincoln 3y agoNo, 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...
- lavela 3y agoDoes it really though? The heat setting process doesn't seem reversible so it's just one step in the initial manufacturing and after that it doesn't seem to be expected to change.