Talk:Riemannian manifold/Archive 1

Archive 1

Riemann for Anti-Dummies

Deleted external link to http://wlym.com/tiki/tiki-index.php?page=Pedagogicals Riemann for Anti-Dummies. The site was of questionable authority and had a very non-neutral point of view.

You were right to delete the link, thanks. I found two more articles with the same link and also deleted it there. However, for future reference, you accidentally deleted a bit more than just the link: categories like [[Category:Riemannian geometry| ]] and interwiki links like [[de:Riemannsche Mannigfaltigkeit]] . Please leave them next time. Anyway, I fixed it now. -- Jitse Niesen (talk) 03:14, 19 February 2007 (UTC)

For The Lay(er) Man

Is there any chance that one could have a paragraph (or two) about What this means with less of the math? I realize that one would have to make assupmtions etc. in order to explain this to others. E.Lefraw 23:19, 23 February 2007 (UTC)

Untitled question

I've come to terms with the fact that I will never understand this.

You know, the article is much too short. It could stand to have many examples and explanatory sentences added. Perhaps you could help us decide how best to improve it, with focused questions. at what point does the exposition lose you? if you ask for more, you might receive it... - Lethe

I added a paragraph to the article. I should probably look it over again tomorrow to make sure it's coherent, but I want to make the basic idea of a Riemannian manifold transparent; an inner product gives you a notion of length on a vector space, and a Riemannian metric gives you a notion of length on a smooth manifold by defining an inner product on each tangent space, pointwise, in a smooth way. to pass from a notion of length of a tangent vector to length of a curve on the manifold, all you have to do is integrate. There is a lot more you can do with Riemannian geometry, but if you can grasp that starting notion, then the rest should fall into place. I guess what I'm saying is, if you think the article is incomprehensible, you may be right. Asking for clarification might be a good idea.... Lethe

-- I think my new paragraph here kinda sticks out like a sore thumb. I don't like it, and perhaps it belongs in a different article (differential geometry, perhaps?) I'm looking for feedback, i guess. - Lethe 00:17, Jul 17, 2004 (UTC)

Hi Lethe, relatively new contributor here. What I'm thinking would be useful, is some initial mention for motivation -- why did Riemann introduce this concept, what could it effectively model? Including a few sentences outlining such information, optimally as a continuance of the first paragraph, would substantially improve the overall quality of this article. Geo.per 19:21, 26 June 2006 (UTC)

I went back and added a few section headers. Lumping everything under Introduction seems inappropriate. Actually, a more thorough list of properties should probably be added. Geo.per 19:26, 26 June 2006 (UTC)

I agree with the idea that this article was too small and not enough documented. That's why I have added some further things regarding the metric and the metric space ... Hope this is coherent for everybody --Zadigus (talk) 10:14, 29 January 2008 (UTC)

Question about the first sentence

I have not yet given up the hope to understand this article. Let me start with the beginning: what confuses me in the first sentence is that g is used for two different purposes: once it is the metric tensor and second time it is the inner product.

I know from classical differential geometry that one needs the vector product in order to calculate the metric tensor. But these two things are surely not one and the same. TomyDuby (talk) 02:48, 7 September 2008 (UTC)

Yes, g is the inner product and the metric tensor. They are one and the same thing, by the canonical isomorphism

$L^{2}(TM)\to T^{*}M\otimes T^{*}M$

from the space of bilinear mappings on the tangent bundle to the tensor product of the cotangent bundle with itself. More details can be found in the metric tensor article. siℓℓy rabbit (talk) 03:02, 7 September 2008 (UTC)

Thanks. TomyDuby (talk) 10:15, 7 September 2008 (UTC)

Riemannian metrics: Question

What is ν(M) (see main article)? TomyDuby (talk) 02:56, 7 September 2008 (UTC)

I think you mean ${\mathcal {V}}(M)$. This is a somewhat idiosyncratic notation for the sheaf of vector fields on M. I have removed it, since the article already suffers from too much unexplained notation. siℓℓy rabbit (talk) 03:07, 7 September 2008 (UTC)

Yes I meant ${\mathcal {V}}(M)$. Thanks for clearing this. TomyDuby (talk) 10:15, 7 September 2008 (UTC)

Another two questions from the same section:

Let $\{\left({\frac {\partial }{\partial x_{i}}}\right)_{p}\}_{i}$ be a basis of tangent vectors over $p\in M$. What are the x_i-s? Are these the coordinates of point p in Rⁿ?

Is the derivative ∂ / ∂ x_i the derivative of coordinates of point p = (x₁, ..., x_n) giving (0, ..., 1, ..., 0)?

TomyDuby (talk) 10:15, 7 September 2008 (UTC)

I have made some changes to the offending section. Let me know if it is clearer. Thanks, siℓℓy rabbit (talk) 12:49, 7 September 2008 (UTC)

I think that this is much better. Thanks! TomyDuby (talk) 18:06, 7 September 2008 (UTC)

Riemannian metrics – a counter example

I think that it would be useful to have a simple example in which the inner product is not continuous. TomyDuby (talk) 18:12, 7 September 2008 (UTC)

Would this be better delt with here or at metric tensor? It seems more relevant to understanding the tensor itself rather than understanding what a Riemannian manifold is. An example is, at any rate, easy to give. On Euclidean space Rⁿ, define

$g_{p}(v,w)={\begin{cases}v_{p}\cdot w_{p}&p\not =0\\2v_{p}\cdot w_{p}&p=0\end{cases}}$

-siℓℓy rabbit (talk) 20:15, 7 September 2008 (UTC)

Thanks for the count erexample. Again, I learnt something!

Where to put it? I think that it might be better at metric tensor. TomyDuby (talk) 20:37, 11 September 2008 (UTC)

Hilbert space?

I am new to topology. It appears that a Riemannian manifold is a manifold in which the tangent space is a Hilbert space. Is that all there is to it? —Ben FrantzDale 01:15, 5 May 2007 (UTC)

A finite-dimensional one over the reals, but yes. Also the inner product must vary smoothly from one tangent space to the next. -- Fropuff 16:41, 5 May 2007 (UTC)

Thanks. Why does it have to be finite-dimensional? —Ben FrantzDale 17:14, 5 May 2007 (UTC)

It has to be the same dimension as the manifold and manifolds are usually finite dimensional. —Preceding unsigned comment added by 82.31.208.151 (talk) 18:52, 18 March 2010 (UTC)

Notation S^2

I'm in a grad-level geometry course right now, and neither in the course nor any of the texts I've used have I seen the notation that occurs here "More formally, a Riemannian metric $g$ is a section of the vector bundle $S^{2}T^{*}M\,$." What is meant by $S^{2}$ here? Is this like a subbundle operator on $T^{*}M$? Can someone who knows either describe the notation (in the article). I think the intention is that $S^{2}T^{*}M$ is the subbundle of the tensor bundle that is the disjoint union of 2-tensor fields $TM\times TM\rightarrow \mathbb {R}$, where the subbundle selects those tensor fields that are symmetric and positive definite.

The notation I have seen for this would be 'a Riemannian metric is a smooth section of $\Lambda ^{2}M$' (using Lee's notation, or maybe you could say it's a smooth section of $(T^{*})^{2}M$, though I don't know if that notation is used anywhere), such that the tensor field is symmetric and positive definite. --Tekhnofiend (talk) 00:26, 28 February 2011 (UTC)

I interpret $S^{2}(T^{*}\!M)$ here as being the symmetric square of the cotangent bundle. I may try to add this clarification. --Dylan Moreland (talk) 20:12, 7 May 2011 (UTC)

Possible plagiarism?

Hi, I've noticed that in the section "Riemannian metrics" the "Examples" subsection is taken word for word from Do Carmo's book, is this a problem? 66.215.164.91 (talk) 03:54, 18 January 2013 (UTC)

Yes, it's definitely a problem. The content in question was added in this pair of edits that substantially expanded the article. This is going to need to be examined carefully to determine if there is more infringing content. Sławomir Biały (talk) 23:01, 19 January 2013 (UTC)

This is bad... (just saying). -- Taku (talk) 04:21, 20 January 2013 (UTC)

Is that all it is?

A Riemann space is defined in the lede as "a real smooth manifold M equipped with an inner product on the tangent space at each point that varies smoothly".

Doesn't that also describe an ordinary sphere in R³ (Euclidian 3-Space)? --67.162.165.126 (talk) 04:24, 20 May 2013 (UTC)

Well, indeed, the "ordinary" sphere (seen as a submanifold of R³, inheriting by restriction the usual inner-product) is an example of a Riemannian manifold. See the last paragraph (starting with "Every smooth submanifold") of section "Overview". In general, however, there is no need to immerse a manifold in some Euclidean space in order for it to acquire a metric; normally, you would define a manifold regardless of an immersion (i.e., without reference to an ambient Euclidean space), and you would simply add the metric (as some extra structure), as required by the context; that is basically what this whole article is about. Pierdeux (talk) 17:37, 12 June 2013 (UTC)

Incorrect classification of article?

I don't know if this is the right place to state these concerns, but I think that there are three significant issues about the way this article is classified and written (feel free to disagree with me):

Firstly, shouldn't this article be characterized under WikiProject Geometry?

Secondly, Riemannian manifolds are one of the most principal and fundamental objects of study in geometry (particularly in differential geometry), so shouldn't this page be rated as 'High Importance' at the least? Moreover, the article about the metric tensor, which is fundamentally related to this article, is rated high-importance, so should this not be as well? (Perhaps the two should be merged?)

Thirdly, because of the importance of Riemannian manifolds in geometry, I think that the intro needs a touch-up (in particular it should contain a non-technical piece for the layman; I may put this in when I get time). SillyBunnies (talk) 06:07, 22 June 2014 (UTC)

Differential properties enabled by a metric tensor

I agree in with this edit of the lead. On a differentiable manifold without a metric tensor, a form of gradient of a scalar field (the index-lowered version, which is a co-vector field) is still well-defined, and the index-raised version is, in itself, not adding enough to merit a mention here, especially considering that it is not always clear whether "gradient" refers to the index-lowered or -raised version and this would need clarification. A similar (but trickier) argument applies to the divergence of a vector field: the n-form derivative of (n – 1)-form (e.g. the current density 3-form J_αβγ) is well-defined and is similar to a scalar; this is related to the divergence of a vector field using considerably less additional structure than a metric tensor. The subtleties and lack of evident added value brought by the metric tensor here make their inclusion in the lead unsuitable.

However, there is differential structure that is enabled by the metric tensor. The above two operations cannot be combined without it, and in particular the full structure of a metric tensor is necessary (AFAICT: this is WP:OR) for any form of the Laplace operator to be well-defined. Wave equations (or harmonic equations in the Riemannian context) are consequently only meaningful in the presence of a metric tensor. This seems like a worthwhile addition to the article, if it can be sourced. —Quondum 13:31, 14 August 2018 (UTC)

Yes, defining the Laplace operator would require the metric. JRSpriggs (talk) 23:54, 14 August 2018 (UTC)

The induced metric

I think that all the bullet points starting from the second one in the Examples section are actually about different ways that a metric can be induced and how to carry that process mathematically... I believe they would rather belong in a Immersed manifold/induced metric section, each in different subsections. — Preceding unsigned comment added by Vyrkk (talk • contribs) 21:53, 27 October 2019 (UTC)

Introduction: mention differentiability of vector fields?

Should the introduction mention that the vector fields X and Y have to be differentiable (otherwise there probably are choices for X and Y that would make fulfilling the smoothness condition of g impossible), or is this clear enough from context? 128.130.48.152 (talk) 11:32, 24 May 2013 (UTC)

I have just independently noticed (the differentiability of the fields X and Y is stated now) that the constraint of differentiability of X and Y is insufficient. For every nondegenerate g, one can always choose differentiable X and Y such that g(X, Y) is non-smooth. This implies that no manifolds meet this definition. Some repair is evidently needed. —Quondum 14:01, 29 January 2020 (UTC)

Smoothness is totally irrelevant to the actual definition of a Riemannian metric, especially when it comes to much modern work. Shouldn't the introduction just say that one assigns a positive-definite inner product to each tangent space?

Gumshoe2 (talk) 03:38, 3 May 2020 (UTC)

If one wants to define smoothness of the metric, the best way to do it (since it trivially extends to defining any desired regularity of the metric) is to use local coordinates. I'll make the edits

Gumshoe2 (talk) 03:41, 3 May 2020 (UTC)