Talk:Smooth Manifold admits Riemannian Metric

Meaning of "admits"
I agree that "admits" for a firstcomer sounds vague. At the same time there are plenty of source showing this exact formulation of the theorem. Improvement is needed, but I worry that this may require reading several proofs (which are not present in my books) and then finding the common denominator. Already my comment about sufficient structure provided by smooth manifolds was an observation of types of proofs, not a copy of any text.--Julius (talk) 23:35, 9 November 2022 (UTC)


 * Is the "admits" not defined in your book? Then I think we need to take the definition from another source. It is really vague. To understand the theorem, I first need to think about what it should be. Does it means that there exists some diffeomorphism between the given smooth manifold and some Riemann manifold? I am not sure how I can formulate this theorem correctly. --Usagiop (talk) 00:30, 10 November 2022 (UTC)


 * Vaguely, however, I know what you mean. --Usagiop (talk) 00:40, 10 November 2022 (UTC)


 * I am not sure if we can get rid of "admits". There are plenty of lecture notes, books, sites like math.stackexchange, mathoverflow, ncatlab which use this word. However, if we could clarifiy what it means, that would be great. From what I understand, this states that there exist a subset of bilinear maps which satisfy the requirements for it to be a tensor. But still this is a distilation of various dicsussions.--Julius (talk) 08:41, 10 November 2022 (UTC)


 * It should be part of the definition of the Riemann metric. See how "admits an infimum" and "admits a supremum" are defined. I'll leave it all up to you guys. --prime mover (talk) 08:55, 10 November 2022 (UTC)


 * Of course, if a metric is admitted then surely it must satisfy the definition of a metric. The problem is that I was taught that metric is an additional structure you attach to a smooth manifold. Now it sounds as if there is already a metric on smooth manfolds and it is just a question whether you use it or not. I would like to formulate this in a set-builder notation, i.e. there exists a non-empty set of metrics constructed solely from the topological and differentiable structures of a given smooth manifold. But then I see examples of a smooth manifold embedded in higher-dimensional Euclidean spaces (which looks like an additional structure to the original smooth manifold) inducing a metric, and I don't know how to fit them consistently in my picture. Replacing "admits" with "has" or "exists" does not sound like a valid solution either.--Julius (talk) 10:21, 10 November 2022 (UTC)

The point I'm hoping we can reach is that a "Riemannian manifold" may be definable on a topological object which may not actually be a smooth manifold. We then use the word "admit(s)" as and when the topological objst is such that a Riemannian manifold *can* be defined on it. It is then the task of this page here to prove that a smooth manifold *does* admit a Riemannian metric.

It would be the same as (hypothetically) defining a "metrizable topology" as being "a topology that admits a metric".

The problem of course is that the implementation on is (probably) not using the exact same words on this page as on that page. This is because the original implementation of Definition:Riemannian Metric was from the point of view of a popular work for entertainment rather than from that of a rigorous text. That original text was all about discussing Riemann's achievements at the most basic level of abstraction, and should be made as rigorous as the source works available. In this case Lee has been used for both pages, so there should be at least a modicum of consistency. I don't have the Lee work. I do have the Auslander and MacKenzie work but havent cracked it open past the first paragraph yet. --prime mover (talk) 10:58, 10 November 2022 (UTC)