Talk:Metric signature

Negative eigenvalues not mentioned in the definition

The article states "The signature (v, p, r) of a metric tensor g (or equivalently, a real quadratic form thought of as a real symmetric bilinear form on a finite-dimensional vector space) is the number (counted with multiplicity) of positive and zero eigenvalues..." It seems that there was earlier a mention of negative eigenvalues, but the current version doesn't mention it. Is there any particular reason for the removal? --Kishore G (talk) 06:37, 10 September 2018 (UTC)[reply]

I found that it was removed earlier this year, but I'm not sure if the reasoning is correct. Right now, the sentence mentions three numbers (v,p,r), but mentions only the values of two of them. Kishore G (talk) 06:39, 10 September 2018 (UTC)[reply]

null subspace

I think there is a point of confusion here. The article says the number of zero eigen-values of the metric matrix can also be interpreted as the maximal dimension of null subspace. The definition of nullspace is not mentioned though. If interpreted as the space over which g(v,v) turns out to be zero, then it is obviously wrong. On the other hand, I think it refers to {v s.t g(v,w)=0 for all w}. This is indeed called the nullspace, while the former is called the null cone (I think). I'm not sure how to make this a bit more transparent in the article, or possibly create a new article for nullspace. — Preceding unsigned comment added by 139.18.242.203 (talk) 16:49, 21 December 2017 (UTC)[reply]

q

What's q? Did I miss something? --Paul 01:04, 10 November 2005 (UTC)[reply]

q is any positive integer, I guess. -lethe ^{talk +} 20:01, 27 February 2006 (UTC)[reply]

Concerning a merge with Metric signature, I'd like to comment that signature is a linear phenomenon. What makes most sense to me would be bilinear form explaining the signature phenomenon (linear case) and possibly linking to an article about Sylvester's rigidity theorem (dunno if this is the canonical name in english) -- this theorem is the basis for the whole thing. Metric signature should explain that at every point of the space, the metric tensor can be assigned a signature which is independent on the point. -- unreg 07:35, January 4 2006 (UTC)

In English it's called Sylvester's law of inertia. -lethe ^{talk +} 20:09, 27 February 2006 (UTC)[reply]

In the discussion of matrices, shouldn't the signature of the second matrix be (0,0,2), by the definition of signature? Jswetnam 16:29, 2 August 2007 (UTC)[reply]

non-sentence

HEY: in the "definition" section, the sentence: "If φ is a scalar product on a finite dimension vector space V, the signature of the matrix which represents φ for a basis" is not a sentence. I don;t know how to correct it since I came to this article trying to understand the damn thing myself. TechnoFaye Kane 04:38, 6 September 2008 (UTC)[reply]

describing relevance

It would be nice to have a section on relevance, i.e. Where is the concept of Metric signature useful. —Preceding unsigned comment added by Dnordon (talk • contribs) 09:21, 4 March 2009 (UTC)[reply]

Eigenvalues: not

A metric tensor does not have eigenvalues, so to say that "The signature of a metric tensor (or more generally a symmetric bilinear form, thought of as a quadratic form) is the number of positive, negative and zero eigenvalues of the metric" is, simply put, outright incorrect. Only a linear transform (a type (1,1) tensor), which maps vectors into the same space, has eigenvectors; a metric tensor (a type (0,2) tensor), in contrast, maps vectors into the dual space. This article, consequently, will have to be rewritten without reference to eigenvectors. — Quondum 13:33, 23 October 2012 (UTC)[reply]

To be fair the article only mentions eigenvalues, which can be interpreted as solutions λ of det(g-λI)=0. The determinant of a metric makes no sense covariantly speaking, but it does provide a scalar tensor density. It is common in physics because multiplication bub the square root of this density makes d⁴x covariant. That being said, reformulation would be nice.T R 15:08, 23 October 2012 (UTC)[reply]

This confuses the metric tensor with the matrix of its components with respect to a given basis. Anyhow, a formulation that respects the context (tensors) and hence does not rely on matrix representations (or tensor densities, for that matter) is appropriate here IMO, notwithstanding the matrix approach being used by many authors. I have rewritten Pseudo-Riemannian manifold#Metric signatures in terms of an arbitrary orthogonal basis and may get around to doing the same here eventually (probably without adequate references, though). I also like the more abstract approach in Metric signature#Geometrical interpretation of the indices of the dimensionalities of a decomposition into (I presume mutually orthogonal) positive definite, negative definite and null subspaces, though the presentation there is not rigorous. — Quondum 16:09, 23 October 2012 (UTC)[reply]

I suggest that if you rewrite this, you actually do so based on proper sources. Carroll, for example, gives a rather thorough treatment of the subject in his book "spacetime and geometry". His approach is based on going to local inertial coordinates. This has the advantage over the approach you took in Pseudo-Riemannian manifold that it does not confuse tensors and tensor fields. (A metric is the latter not the former).T R 22:27, 23 October 2012 (UTC)[reply]

Every tensor field is also a tensor in the tangent space at each point of the manifold, and any statement that does not relate to differential structure applies to this tensor, so I disagree with your final parenthesized statement. This is however not particularly important here; I am far more concerned with the blatant mathematical incorrectness of the use of eigenvectors in the statement about the signature, even though the result in terms of number of +1, −1 and 0 values comes out the correctly. It would definitely be preferable, however, as you say, to work from a good reference (I'll look at what I have, which is rather limited). — Quondum 16:56, 24 October 2012 (UTC)[reply]

There is no use of eigenvectors.T R 18:50, 24 October 2012 (UTC)[reply]

The article repeatedly refers to eigenvalues, which imply the existence of eigenvectors and vice-versa. Sorry if I confused the terms, though this is immaterial here. My paltry set of books does not seem to deal adequately with the metric signature, which means I will not be able to adequately make the necessary change. — Quondum 21:30, 24 October 2012 (UTC)[reply]

Quondum, your definition doesn't work. Among other problems, if you happen to include a null-vector in the orthogonal "test set", it will most likely yield the wrong answer.

One and the same mathematical object can have several equally valid interpretations. The object in question is

A type (0, 2) tensor
A (nondegenerate) symmetric bilinear form
A (full-rank) quadratic form

It is in the capacity of being a quadratic form that the objects signature is defined (see L & L). Tensors do not, as you seem to think, have a signature in general, quadratic forms do, and in the language of quadratic forms, the matrix representing the tensor/quadratic form/bilinear form is a linear map ℝⁿ → ℝⁿ, and has as such both eigenvalues and eigenvectors, whether you like it or not.

Ironically, when asked how to find the signature, you replied "diagonalize" (your talk page). That amounts exactly to finding eigenvalues (and to be complete, eigenvectors).

Even if you manage to salvage your definition and get it sourced, the definition is practically useless because it doesn't provide a clue to how to find the signature efficiently. The proper context to define it is in the context of the quadratic form. This way, there is an efficient algorithm (not in article, built on completing squares of the explicit quadratic form). YohanN7 (talk) 12:09, 15 September 2015 (UTC)[reply]

So I repaired the damage. But the whole article needs an overhaul. YohanN7 (talk) 09:50, 2 October 2015 (UTC)[reply]