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By the Nash embedding theorem, it seems the definition of riemannian smooth manifold is equivalent to the zero locus of some functions $f_1,\dots,f_r : \mathbb{R}^n\to\mathbb{R} $ which are $C^\infty$ and which differentials are zero at no point.

This alternate definition is simpler, because it does not introduce partial charts, the metric tensor is just the restriction of the euclidean inner product and the Levi-Civita connection is just the orthogonal projection of the usual derivative of vector fields. And as a bonus, the zero locus approach is closer to the definition of affine schemes in algebraic geometry. Pseudo-riemannian manifolds can also be defined this way, under the additional assumption of stable causality.

So what is the advantage of the standard definition of riemannian manifold through charts and custom metric tensors? Is it just that the dimension increase of the Nash embedding $n(n+1)(3n+11)/2$ makes computations intractable? The story of an internal observer's viewpoint is not very convincing, because the only riemannian manifolds with rigid body motions (as would be a true internal observer) are the ones with constant curvature, so a tiny part of riemannian manifolds.

V. Semeria
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    It's just a lot easier to write down manifolds and metrics on them than it is to write down Nash embeddings. For example if $M$ is a Riemannian manifold and $G$ a group acting freely and properly by isometries then so is the quotient $M/G$. Do you know how to write down a Nash embedding of such a quotient in general, even given such an embedding for $M$? I certainly don't. – Qiaochu Yuan Oct 04 '23 at 21:03
  • @QiaochuYuan so the reason really is just easier computations? – V. Semeria Oct 04 '23 at 21:03
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    Well, I don't know about "just." How would you even prove that statement about quotients I just made without using an intrinsic definition of a Riemannian manifold? This situation is closely analogous to early group theory where all groups were understood to be groups of permutations. You can do some stuff with this definition but it's difficult to understand what a quotient group is this way. The abstract definition makes taking quotients easy and that's one of its main advantages. – Qiaochu Yuan Oct 04 '23 at 21:05
  • I don't understand the objection in the last sentence. You can be an observer in a manifold with non-constant curvature. For example, we humans are observers in the 4-manifold of the universe, which has non-constant curvature according to general relativity. – Nick Alger Oct 04 '23 at 21:24
  • Right, but I don't see how that is relevant to the issue of whether or not observers require constant curvature. – Nick Alger Oct 04 '23 at 21:28
  • This approach immediately runs into problem if you want to use this to study Hermitian manifolds, for example. – user10354138 Oct 04 '23 at 21:35
  • @NickAlger In the riemannian case (positive definite metric), the only observer's rigid body motions I know is local isometries. So that's a different and approximated treatment of observers compared to general relativity. And along such isometry paths the curvature is constant. – V. Semeria Oct 04 '23 at 21:36
  • I guess I still don't understand why you think that observers require rigid body motion. Observers are idealized as being very small, perhaps a point; they don't have any internal structure on length scales for which curvature would matter. – Nick Alger Oct 04 '23 at 21:40
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    Actually, the definition in the 1st paragraph is wrong. Besides what others above said, I would add that Nash's theorem breaks down in the closely related case of Finsler manifolds. – Moishe Kohan Oct 04 '23 at 23:00
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    I want to echo @MoisheKohan’s complaint about the misdefinition. But forget about metrics for a moment. Suppose you want to work only with embedded smooth submanifolds (which is reasonable for undergraduates first encountering them). How do you begin to deal with projective spaces and Grassmannians? Do they even exist as smooth manifolds? Then add the layer of invariant metrics, as @Qiaochu mentioned in the first comment. – Ted Shifrin Oct 04 '23 at 23:19
  • Adding to good points already made, the proposed formulation makes intrinsic geometry dependent on extrinsic geometry. <> The Nash embedding gives existence but not uniqueness. Locally, many surfaces (such as those of constant Gaussian curvature) in three-space are flexible, with an infinite-dimensional space of isometric deformations. For one thing, that throws a wrench into the Levi-Civita connection is just the orthogonal projection of the usual derivative of vector fields: This "definition" depends on an embedding, which is Far From Unique. – Andrew D. Hwang Oct 06 '23 at 12:56

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The comments note multiple obstacles to basing Riemannian geometry on manifolds embedded in Euclidean space. In the hope of distilling major objections for posterity:

  1. Intrinsic geometry is a thing. (Before Gauss's Theorema Egregium this was not apparent. The fact tends to get hidden because nowadays we teach Riemannian geometry starting from intrinsic geometry, e.g., Kobayashi-Nomizu Volume I.)
  2. Defining Riemannian manifolds in terms of embeddings casts intrinsic geometry in terms of extrinsic geometry.
  3. In a particularly strong sense, intrinsic geometry of a Riemannian manifold (traditional Riemannian geometry) does not uniquely determine extrinsic geometry (isometrically embedded submanifolds): Every Riemannian manifold has an infinite-dimensional space of isometric embeddings in a sufficiently high-dimensional Euclidean space. (Twice the Nash dimension is enough to bend the ambient space into an arbitrary Cartesian product of plane curves.)
  4. As multiple comments note, extrinsic geometry is Difficult To Work With for specific, interesting Riemannian manifolds, does not play nicely with Riemannian quotients and submersions, and flatly does not accommodate other geometric structures.

<>

To reiterate my comment: The Nash embedding gives existence but not uniqueness. Locally, many surfaces (such as those of constant Gaussian curvature) in three-space are flexible, with an infinite-dimensional space of isometric deformations. For one thing, that throws a wrench into "the Levi-Civita connection is just the orthogonal projection of the usual derivative of vector fields": This proposed definition depends on an embedding, which is Far From Unique.

Andrew D. Hwang
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