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I'm not nearly smart enough to state this confidently, but doesn't [1] imply that exotic embeddings can always be replaced by larger Euclidean embeddings? [1]
by vslira 2y ago
I'm not nearly smart enough to state this confidently, but doesn't [1] imply that exotic embeddings can always be replaced by larger Euclidean embeddings?
[1] https://en.wikipedia.org/wiki/Nash_embedding_theorems https://en.wikipedia.org/wiki/Nash_embedding_theorems
- hansvm 2y agoNo. Consider a square (with graph shortest path distances) for example. All euclidean embeddings have a minimum error of 20-40% or so. If you try to embed any graph containing that subgraph (embedding friends, advertisers, words, ...), you'll similarly have guaranteed error. Relaxing the square to a squircle, you'll see that error even for a really simple manifold. Riemannian manifolds are a bit more special.