3 ms·

No, but Nash was the first (AFAIK) to prove that every Riemannian manifold can be isometrically embedded in an Euclidean space of sufficiently high dimension.

by kkylin 5y ago

No, but Nash was the first (AFAIK) to prove that every Riemannian manifold can be isometrically embedded in an Euclidean space of sufficiently high dimension. See, e.g., [1,2]. Previous embedding theorems (see Whitney embedding [3]) preserved topology but not necessarily geometry. [1] https://en.wikipedia.org/wiki/Embedding#Riemannian_and_pseudo-Riemannian_geometry https://en.wikipedia.org/wiki/Embedding#Riemannian_and_pseud... [2] https://en.wikipedia.org/wiki/Nash_embedding_theorem https://en.wikipedia.org/wiki/Nash_embedding_theorem [3] https://en.wikipedia.org/wiki/Whitney_embedding_theorem https://en.wikipedia.org/wiki/Whitney_embedding_theorem

rsj_hn 5y ago
> Previous embedding theorems (see Whitney embedding [3]) preserved topology but not necessarily geometry. The definition of Riemmannian manifold fairly straightforwardly means that every compact smooth (I think you only need C^2) manifold can be isometrically embedded in a high enough dimensional euclidean space (with the euclidean metric) just by using the fact that every riemannian manifold is locally euclidean, which means we can cover it with balls on which there are local isomorphisms, and from these we can pick a finite subcover and partition of unity that glues together for a global isometry. Basically you are gluing together local solutions to a PDE. This is something often covered in a first year course on geometry. For the details of this approach worked out, see this expository paper: https://www.math.mcgill.ca/gantumur/math580f12/siyuan.lu.pdf https://www.math.mcgill.ca/gantumur/math580f12/siyuan.lu.pdf To the best of my knowledge, this result doesn't even have a name attached to it (I could be wrong, but it was always presented as just a series of unnamed propositions whenever I saw this). That's not the hard part, the hard part is putting an explicit bound on the size of the target space other than "finite". So the genius of Nash was to put a strong bound on the dimension of the target space as well as the size of the target space. Rather than using general compactness and existence arguments, this requires careful arguments and innovative geometric ideas and is really a suprising result.
- kkylin 5y ago
  I have to admit my memory about these things is quite hazy. About the mathematics, I do not have anything useful to add. About the history, you may be right, but my recollection seems mostly compatible with the brief history recounted in [0], which is also cited by the note you linked. The author of the note also repeats the assertion about Nash's role in the subject (of global isometric embeddings), and explains that the simpler proof presented is due to Gunther (which came later I think). I tried to find a more detailed history, but in a few minutes' searching on the web (and flipping through the textbooks I can access from home) did not find one. Somewhat surprising given interest in the subject! Possibly this is covered in, e.g., Spivak. [0] https://bookstore.ams.org/surv-130/12 https://bookstore.ams.org/surv-130/12
  - rsj_hn 5y ago
    Perhaps I wasn't clear in my answer, but I was pointing out that just for the existence of embeddings, this follows simply from compactness arguments -- it is Proposition 3.4 in the paper I cited (not even a theorem). However this dimension depends on the finite subcover which in turn depends on (M, g). That is a completely different question that trying to find a dimension that is independent of (M, g) and only depends on the dimension of M rather that it's geometry. The AMS article you cited starts off only discussing this second question. So again, if you just want to know "can I embed a manifold", the answer is "of course you can". But if you ask, "Can I embed every manifold of dimension n into a space of dimension f(n)", then that is a much harder question, deserving of a person's name and the status of "theorem". The first example of the paper is a conjecture that f(n) = n(n+1)/2. But if you are just asking, "who showed you can embed a given manifold into a given dimension isometrically", then I maintain this follows rather simply from partitions of unity and finite subcovers, and was known pretty much at the same time as the definition of Riemannian manifold was formalized.

by kkylin 5y ago

rsj_hn 5y ago

> Previous embedding theorems (see Whitney embedding [3]) preserved topology but not necessarily geometry. The definition of Riemmannian manifold fairly straightforwardly means that every compact smooth (I think you only need C^2) manifold can be isometrically embedded in a high enough dimensional euclidean space (with the euclidean metric) just by using the fact that every riemannian manifold is locally euclidean, which means we can cover it with balls on which there are local isomorphisms, and from these we can pick a finite subcover and partition of unity that glues together for a global isometry. Basically you are gluing together local solutions to a PDE. This is something often covered in a first year course on geometry. For the details of this approach worked out, see this expository paper: https://www.math.mcgill.ca/gantumur/math580f12/siyuan.lu.pdf https://www.math.mcgill.ca/gantumur/math580f12/siyuan.lu.pdf To the best of my knowledge, this result doesn't even have a name attached to it (I could be wrong, but it was always presented as just a series of unnamed propositions whenever I saw this). That's not the hard part, the hard part is putting an explicit bound on the size of the target space other than "finite". So the genius of Nash was to put a strong bound on the dimension of the target space as well as the size of the target space. Rather than using general compactness and existence arguments, this requires careful arguments and innovative geometric ideas and is really a suprising result.

kkylin 5y ago
I have to admit my memory about these things is quite hazy. About the mathematics, I do not have anything useful to add. About the history, you may be right, but my recollection seems mostly compatible with the brief history recounted in [0], which is also cited by the note you linked. The author of the note also repeats the assertion about Nash's role in the subject (of global isometric embeddings), and explains that the simpler proof presented is due to Gunther (which came later I think). I tried to find a more detailed history, but in a few minutes' searching on the web (and flipping through the textbooks I can access from home) did not find one. Somewhat surprising given interest in the subject! Possibly this is covered in, e.g., Spivak. [0] https://bookstore.ams.org/surv-130/12 https://bookstore.ams.org/surv-130/12
- rsj_hn 5y ago
  Perhaps I wasn't clear in my answer, but I was pointing out that just for the existence of embeddings, this follows simply from compactness arguments -- it is Proposition 3.4 in the paper I cited (not even a theorem). However this dimension depends on the finite subcover which in turn depends on (M, g). That is a completely different question that trying to find a dimension that is independent of (M, g) and only depends on the dimension of M rather that it's geometry. The AMS article you cited starts off only discussing this second question. So again, if you just want to know "can I embed a manifold", the answer is "of course you can". But if you ask, "Can I embed every manifold of dimension n into a space of dimension f(n)", then that is a much harder question, deserving of a person's name and the status of "theorem". The first example of the paper is a conjecture that f(n) = n(n+1)/2. But if you are just asking, "who showed you can embed a given manifold into a given dimension isometrically", then I maintain this follows rather simply from partitions of unity and finite subcovers, and was known pretty much at the same time as the definition of Riemannian manifold was formalized.