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I'm self-studying smooth manifolds and differential geometry. I had a hard time understanding that a manifold can be described without a parametric or implicit equation (or not embedded in $\mathbb{R}^n$). A lot of progress was made by reading other questions, in particular this one. However, I can't come up with an example of a metric and a manifold with that metric that do not come from a parametric/intrinsic function. In other words, I can't imagine a manifold and its metric without embedding it. Could you provide an example, please?

Other questions:

  1. When describing a manifold: does parametric/implicit equation mean embedded manifold, and vice versa?

  2. Once one uses coordinates, does it mean that it's already using an embedding?

  3. Can one define a metric on an intrinsically defined manifold without using coordinates?

  4. In order to "measure things" (like distances and curvature) does one necessarily need coordinates and an embedding?

Ale
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  • The Nash embedding lemma says basically that every Riemannian metric on a manifold can be thought of as coming from embedding said manifold in Euclidean space with the standard metric. With that in mind, could you clarify what exactly it is you're asking about here? – Arthur Sep 30 '22 at 12:24
  • I'm sorry for my lack of precision, I'm very new to all this. After reading about the embedding lemma, I (think) what I mean in question 1 is: even if one can think of the Riemannian manifold as coming from an embedding, can one still describe it without using said embedding? Because the lemma first states that there's a Riemannian manifold $(M,g)$, so that means that a metric is already in place. Does this make any sense? – Ale Sep 30 '22 at 12:45
  • Maybe this example will help: consider the circle as a manifold. The usual metric is symmetric with respect to rotating the circle. But you can also imagine the circle as a racetrack where the material of the track changes across the course, so it takes longer to cover some stretches of the track than others. You can formalize this by saying that the track has a new metric that is not the usual one. This metric can be realized in a Euclidean embedding by stretching the circle out in the right places so as to compensate for the varying surface materials, but that's not essential. – Elchanan Solomon Sep 30 '22 at 12:57
  • Anything that can be done on a differential manifold can be regarded as coming from an embedding or as a purely intrinsic notion: this is what Whitney's embedding Theorem says. The Nash embedding Theorem (I'm surprised some people call it a lemma!) pretty much tells the same thing in the Riemannian case: anything that can be done on a Riemannian manifold can be seen as induced by an isometric embedding or as a purely intrinsic notion. In some context, it is easier to work with one setting rather the other – Didier Sep 30 '22 at 15:46
  • Using coordinates doesn’t mean that you are using an embedding. Coordinates mean that you identify a subset of the manifold with an open subset of $\mathbb{R}^n$, where $n$ is the dimension of the manifold. An embedding means that you put your whole manifold to some $\mathbb{R}^N$ and typically $n<N$. Also you can give a Riemann metric on a smooth manifold without embedding: locally using coordinates you can pull back the standard Riemann metric on $\mathbb{R}^n$ and you can smear these together with a partition of unity. In practice this is unusable, this is just an existence statement. – Laci Sep 30 '22 at 16:13

1 Answers1

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Here is a nice example: complex projective space $\mathbb{CP}^n$ is a space parameterizing $1$-dimensional complex subspaces of $\mathbb{C}^{n+1}$. It turns out to be a manifold, and it can be described as the quotient of the sphere $S^{2n+1}$ by a particular action of $S^1$. This description as a quotient means it doesn't come with any kind of embedding into a Euclidean space, and actually it's not clear a priori that $\mathbb{CP}^n$ embeds into Euclidean space at all!

(This is exactly analogous to the way working with abstract groups rather than permutation groups - groups together with an embedding into the symmetric group - frees us to consider quotient groups, which don't come with any such embedding. In the very beginnings of group theory life was difficult because only the concept of a permutation group was available which made it awkward to talk about quotients.)

$\mathbb{CP}^n$ also has a natural metric called the Fubini-Study metric which can be defined using a convenient choice of local coordinates given by homogeneous coordinates. This does not require choosing an embedding to define; local coordinates are local, an embedding is global.

So, addressing your specific questions:

  1. No, one can define an embedding without choosing a parameterization or giving an implicit equation, although I don't have a nice example at hand.

  2. No, coordinates are local and an embedding is global. Again, as in the above example, homogeneous coordinates can be defined even without knowing that $\mathbb{CP}^n$ admits any embedding into Euclidean space whatsoever.

  3. Yes, but I don't know a nice example. A complicated example is given by the Weil-Petersson metric on Teichmuller space.

  4. In theory all you need is a metric, but in practice it's common to work in a convenient choice of local coordinates. You do not need an embedding.

Qiaochu Yuan
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  • The Fubini-Study metric can also be defined by requiring the quotient map $S^{2n+1}\to \Bbb{CP}^n$ to be a Riemannian submersion: it is thus an example of a natural metric defined without referring to any embedding or local coordinates (well, one could argue that some embedding is hidden behind that construction, since we use the Hopf fibration which is intrinsically linked to the embedding of the sphere into $\Bbb C^{n+1}$) – Didier Sep 30 '22 at 17:35
  • Indeed, I was going to comment that for 3. you should just take any homogeneous space $G/H$ with $G$-invariant metric. Now a little bit of algebra replaces coordinates. – Ted Shifrin Sep 30 '22 at 17:53