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I am starting to read "General Relativity" by Robert Wald. A little bit of my physics background: I am pursuing a mathematics major and I have not taken any physics courses, just Analysis and Differential Geometry.

My doubt is mainly about the definition of manifold that the author states. His definition is similar to the one I saw in my mathematics courses in the sense that a manifold $M$ locally looks like $R^k$ (where $k$ is the dimension of the manifold), but it differs in the existence of ambient space. Generally, in my class we defined a manifold as a $k$-dimensional object embedded in $R^n$ (for $n > k$), but Wald states that " in many situations - most importantly in general relativity - one is given a manifold without embedding of it in $R^{n}$" (page 14). What does it meand that a manifold, in general relativity, may not be embedded in $R^{n}$? Then, we have to treat a manifold as an object itself without taking care of the space it lives? As long as I understand is that the definition given in the book doesn't take care about the space where the manifold lives, but I am still a little bit confused as to why is this?

As you see I got a little bit confused with this slightly different definition. I hope someone can clarify this point.

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    $\begingroup$ Notice that Wald's wording does not imply that the manifold cannot be embedded. Only that it is presented without an embedding attached. $\endgroup$ – jacob1729 May 12 '20 at 17:43
  • $\begingroup$ Possible a duplicate of physics.stackexchange.com/questions/547140/… $\endgroup$ – Metropolis May 12 '20 at 19:47
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There are two ways of defining a manifold, the extrinsic and intrinsic points of view.

One is the one you saw in your courses: you define a manifold as a $k$ dimensional surface embedded in $\mathbb R^n$ with all the conditions that make it a manifold (locally euclidian, etc...). This is the "extrinsic" definition, it is a good way of defining a manifold, but if we want to start from the principle that spacetime is a manifold, it would be really weird to assume that there is a big $n$ dimensional Euclidean space, with $n>4$, in which our spacetime is embedded.

Instead, we can use the "intrinsic" definition and define a manifold simply as a locally Euclidean topological space that is Hausdorff and second countable. This removes any mention of a mysterious space in which spacetime is embedded. Fortunately, these two definitions are equivalent, thanks to the Whitney embedding theorem, which says that any manifold defined in this way may be embedded in $\mathbb R^n$ for $n$ suitably large.

There are mathematical advantages in using the intrinsic definition (it's more flexible and general, in a sense), but the physical motivation for using it is that it stresses that it is not necessary to see curved spacetime as a curved surface in a flat ambient $n$ dimensional space, even if mathematically such a space can be found, it is just a mathematical trick and the actual structure of spacetime is independent of it. And of course, whenever a spacetime is "given" to you, it is never given in the form of an embedded surface, because as I said there is no such space in which it is naturally embedded, hence why one has to work without an ambient space in general.

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  • $\begingroup$ Thanks a lot !! $\endgroup$ – Luis Carlos May 13 '20 at 0:49
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    $\begingroup$ Here I will note that the Whitney embedding theorem does not guarantee isometric embeddings in case of (pseudo-)Riemannian manifolds. The stronger Nash embedding theorem states that there are isometric embeddings as well for the positive definite csse, and also gives an upper bound on the dimension for the ambient space. As far as I am aware, embedding theorems exist also for Lorentzian/pseudo-Riemannian geometries, but there are no upper bounds on the dimension of the embedding space. $\endgroup$ – Bence Racskó May 13 '20 at 15:32

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