Talk:Hodge dual

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1 Which Grad?
2 Definition of *
- 2.1 Alternative approach?
- 2.2 Clarifying existing definitions
3 Hodge dual of an arbitrary tensor
4 Hodge dual generalizes the cross product
5 underpinning de Rham
6 Yet another definiton
7 signature and Hodge squared
8 pseudo riemannian case

[edit] Which Grad?

In the Hodge dual article, the following statement appears:

"The combination of * and the exterior derivative d generates the classical operators div, grad and curl, in three dimensions."

Grad is linked, but it points to a disambig page.

Which of the following two possible definitions of "grad" is the correct one to use?

Help in figuring out where to point this link would be appreciated. Kevyn 13:58, 29 Jun 2004 (UTC)

It's the first of those. Thinking about it, the * operator is not really involved in defining grad. Charles Matthews 14:19, 29 Jun 2004 (UTC)

No, grad a is just da. You need to combine d and * if you want to do any more interesting things with d because d^2=0. Francis Davey 07:53, 11 Aug 2004 (UTC)

This is not correct. The gradient of a scalar field f is obtained by forming the 1-form df (which is a section of the cotangent bundle) and then using duality (Riemannian metric!) to obtain the associated vector field v = grad f. This is a subtle but important point. ASlateff 128.131.37.74 05:47, 11 February 2007 (UTC)

[edit] Definition of *

[edit] Alternative approach?

I have done some reading and found some much more direct and appealing (to me) definitions of duality operators, one of which I believe is the operator we are calling the hodge star here. I suspect that both should be mentioned. I have this treatment from Schutz's book on mathematical methods for theoretical physics.

First one relies on the existence of a volume form (not a vector -- and I think we should be careful to distinguish them at this stage). Suppose we have $ω$ as a volume form, then the dual of v is $ω(v)$ in the obvious sense. This provides a map from p-vectors to (n-p) forms. If we have a volume n-vector as well (which may be needed as well?) this dual is really easy to understand since it can be visualised directly.

As a result of the above insight, I was able to immediately understand Maxwell's equations in 3D exterior algebra form. Cool. I am afraid it knocks the inner produce definition below into a cocked hat.

"cocked hat"? is that some kind of English slang? I'm not sure what it means. I assume something like "makes it irrelevant"? But it's not irrelevant. They are just two different ways of writing the same thing. -Lethe | Talk

Second, by using an inner product G:V->(V->F) we can then make a map from p-vectors to (n-p) vectors like this: $G (ω(v))$ .

Hmm... So actually, you don't really need to use a volume form, you just want to use the volume element of the exterior algebra of the dual of our vector space? If this is all you need it for, then my comments above whining about not doing differential topology too early are not relevant.

We were talking at cross purposes. In the books I have read a form is a member of the exterior algebra of the dual of a vector space. Given any vector space I can form (1) an exterior algebra of vectors and (2) an exterior algebra of forms. If I chose the vector space to be in the tangent space of a differentiable manifold, then the forms will be differential forms, but vector spaces can arise in other ways. In particular I may not need a differential structure to impose them on a manifold. Indeed I may not need the machinery of manifolds to study whatever vectors I want.

OK, so in my mind the word "form" is short-hand for "differential form", so everytime you say "form", i thought you were trying to sneak in some differential stuff. But it's just a matter of terminology, and perhaps I'm confused about which usage is standard. I will check some textbooks on monday. So anyway, forget all my complaints above. You seem to be on the right track then. I am going to delete that list above, since it is irrelevant and distracting from the thread of this discussion. -Lethe | Talk

So I do mean a volume form, that is an object which maps an n-vector to a scalar. The definition I gave above has two advantages (1) its basis independent; (2) it is a closed form definition of the dual, rather than defining a dual as the unique object which satisfies an equation. I think that is a very satisfying way of defining a mathematical operation.

OK, so you are comparing the following two definitions:

$\langle\zeta\wedge *\alpha,\omega\rangle = \langle\zeta, \alpha \rangle$

and

G (ω(v))

and you feel the second one is easier... Let me think about that for a bit. I find both definitions pretty cumbersome, so while I agree with your complaints about the definition I gave, I don't really like your definition so much either. For one thing, while yours ends up with a more straightforward expression, it gets there in a more roundabout way, relying on the construction of two exterior algebras, one on the vector space and one of the dual space. I see that as not so expedient. But let's see what others think as well. And I wouldn't object to having more than one definition (or more than two, since we already have two). -Lethe | Talk

I also suspect that the p-vector (n-p)-form equivalence is "more natural" than the p-vector (n-p)-vector equivalence, which requires a metric tensor, but I cannot quite see if that is right.

I can't see anyway that the Hodge dual would be more natural on the dual exterior algebra. To be sure, you need the metric and the orientation in both cases. -Lethe | Talk

Also, my understanding was that * required that a space be oriented, i.e. that there are two *'s, and the choice of orientation selects one (I guess that might come from your defintion of determinant). Francis Davey 19:33, 14 Aug 2004 (UTC)

I have never heard anyone say that there are two * maps, although I suppose in a sense, that is true, because there are two possible orientations for the vector space. But you might just as well say that because there are infinitely many possible inner products on the vector space, there are infinitely many possible Hodge dual maps. It's true, but not very useful, I think. In practice, the inner product and the orientation are usually given to you beforehand, leaving you with a single Hodge dual map. -Lethe | Talk

Also easy to visualise (although inner products are harder to visualise in exterior algebra, but that's hardly surprising because they require some more geometry). -

Can we try approaching the Hodge dual in this way? Start with an abstract definition if you like (which will help the mathematicians of this world who can't bear to understand anything (Hey!! I resent that!) but need only a formal definition and all else will follow -- scary people). Then some motivation using a volume element, plus possible some pictures, followed by some more alternative derivitions (perhaps a basis dependent one as well).

Apologies for the above. I am stuggling to understand things from textbooks which combine two irritating features I have found with mathematics textbooks: (1) far too little chat or expansion for me to gain any kind of intutition as to what is being said -- i.e. of the definition/theorem/proof kind; and (2) not enough rigour for me to have a clue what is being said. I am trying to understand a remark at the moment which talks about the closure of a set of lie brackets, but never tells me what that means. I have had particular difficulty with the hodge star, many definitions being like this article, leaving just enough unsaid for me not to be sure what the definition actually is, but not enough for me to really understand it. I am getting there, but think that something really clear could be written here.

I don't know how you're currently doing with the Lie bracket stuff, but I can think of a really simple explanation of what that remark probably means, so let me give it to you, and perhaps you will find it useful. One Lie algebra that we all know and love is sl(n), the set of nxn traceless matrices. One can easily check that the product of two traceless matrices need not be traceless. However, the Lie bracket (commutator) of two traceless matrices is necessarily traceless (by the cyclicity of trace), thus one says that the Lie bracket closes on the set of traceless matrices. Of course, under more general definitions, axioms of closure become sort of trivial in a sense... -Lethe | Talk

Actually, I am surprised there isn't a page that motivates exterior algebra in the same way. Once one understands it (and I think I now do) one can immediately "see" all of electromagnetism. Francis Davey 23:29, 13 Aug 2004 (UTC)

certainly there is a lot of work do be done around here -Lethe | Talk

So I guess my point is, this article should not be meant as a fastest approach to understanding electromagnetism (that would belong in electromagnetism), but rather a complete and general definition and description of the purely mathematical topic that is the Hodge dual. Certainly the article needs some examples, though, and in an examples section, it would be nice to explain how this applies to electromagnetism. -Lethe | Talk 19:01, Aug 14, 2004 (UTC)

Agreed. I think most of the discussion is how to structure the definition and explanation section. Examples are probably much easier and the reader can be pointed elsewhere for some of them. Francis Davey 19:33, 14 Aug 2004 (UTC)

This article would benefit greatly from some examples, I think. -Lethe | Talk

[edit] Clarifying existing definitions

I wondered if someone more mathematical than me could give me a better geometric understanding of *. Can it be defined without reference to an (orthonormal basis)?

Yes. and in fact you find this definition in the article:

$\langle\zeta\wedge *\alpha,\omega\rangle = \langle\zeta, \alpha \rangle,$

you can prove that such a *α exists for each α and is unique. Note that there is no choice of basis and no reference to components in this definition. On the other hand, I think it is harder to see what the Hodge star actually does with this equation. -Lethe | Talk

That's not enough! Indeed this whole article is tantalising and frustrating because it leaves so much unsaid (see later). In this case, I don't know what <,> means when applied to elements of the exterior algebra. The article just says the inner product induced on the exterior algebra, is there such a one, is it unique, what is it? Its not in the article on exterior algebra as far as I can see. Sorry to complain, but I have spent all day trying to puzzle this out. Francis Davey 17:27, 11 Aug 2004 (UTC)

Indeed, you are correct. Without saying what the induced inner product is, this definition has no meaning. That needs to be added to some article (exterior algebra, i suppose). Let me give the definition here, and if you like, we'll put it in the appropriate place:

$\langle e_1\wedge\dotsb\wedge e_k,f_1\wedge\dotsb\wedge f_k\rangle=\det(\langle e_i,f_j\rangle)$

Well at first I thought this might help, but I am afraid its just given me more to puzzle about. I am not sure what you mean by $\det(\langle e_i,f_j\rangle)$ . For example, I understood your equation to mean:

$\langle e_1\wedge e_2,e_3\rangle=\det(e_1,e_2,e_3)$

This is not what I meant with that equation (nor do I see any meaningful way to interpret this equation (what is the determinant of an ordered list of vectors?), but anyway).

but that can't be right because that would mean $\langle e_2,e_1\rangle = -1$

The inner product is only defined between k-vectors and k-vectors, in other words, the formula above doesn't work between a 2-vector and a 1-vector, as you are trying to do. This results in attempting to take the determinant of a 2x1 matrix (illegal operation). Although I suppose it might make sense (and perhaps people do) define all k-vectors to be orthogonal to all j-vectors, for k≠j.

In other words your definition simply pushes the mystery somewhere else. I will ponder on what you have said. Francis Davey 17:35, 12 Aug 2004 (UTC)

Let's make the explain the notation of my definition, and then see an example, shall we?

$\det(\langle e_i,f_j\rangle)$

means the determinant of a matrix with

$\langle e_i,f_j\rangle$

in the i^th column and j^th row (or maybe I mean j^th column, i^th row)

therefore

$\det(\langle e_i,f_j\rangle)=\det\begin{pmatrix}\langle e_1,f_1\rangle&\cdots&\langle e_1,f_j\rangle&\cdots&\langle e_1,f_k\rangle\\\vdots&\ddots&\vdots&\ddots&\vdots\\\langle e_i,f_1\rangle&\cdots&\langle e_i,f_j\rangle&\cdots&\langle e_i,f_k\rangle\\\vdots&\ddots&\vdots&\ddots&\vdots\\\langle e_k,f_1\rangle&\cdots&\langle e_k,f_j\rangle&\cdots&\langle e_k,f_k\rangle\\\end{pmatrix}$

that last is a pain in the arse to write, which explains the much shorter, if harder to recognize, shorthand that I used above. Notice that the entries of the matrix are the inner product of 1-vectors. This is how the induced inner product of the exterior algebra is built out of the inner product of the underlying vector space (member of which I call 1-vectors, when considered as being included in the exterior algebra). This inner product was given to us, and we have built an inner product for the whole algebra out of it. Now for an example. Assume e₁, ..., e_n to be orthonormal (i.e. 〈e_i,e_j〉=δ_ij)

$\langle e_1\wedge e_2,e_2\wedge e_1\rangle=\det\begin{pmatrix}\langle e_1,e_2\rangle& \langle e_1,e_1\rangle\\\langle e_2,e_2\rangle& \langle e_2,e_1\rangle\end{pmatrix}=\det\begin{pmatrix}0&1\\1&0\end{pmatrix}=-1$

and

$\langle e_1\wedge e_2,e_3\wedge e_1\rangle=\det\begin{pmatrix}\langle e_1,e_3\rangle& \langle e_1,e_1\rangle\\\langle e_2,e_3\rangle& \langle e_2,e_1\rangle\end{pmatrix}=\det\begin{pmatrix}0&1\\0&0\end{pmatrix}=0$

which demonstrates the point that I mentioned above that the way we interpret this definition is that it is the inner product that takes the orthonormal basis {e_i} and basically promotes the corresponding basis of the entire exterior algebra to be orthonormal. In other words, if {e₁,e₂,e₃} is an orthonormal basis of 1-vectors, then {e₁∧e₂,e₂∧e₃,e₃∧e₁} is an orthonormal basis for the 2-vectors. I believe Charles says the exact same thing below. OK, so if you followed this stuff about determinants, then look again at what I said about the inner product of k-vectors and how the properties of the determinant match up with the properties of the wedge product, etc. (currently that's in the paragraph directly below this one) - Lethe | Talk

This is the only possible definition one can make that reduces to the given inner product on 1-vectors, that respects the antisymmetry of the wedge product (due to the antisymmetry of the determinant under exchange of columns), that respects the symmetry of the inner product (due to the invariance under transposition of the determinant), and has the nice property that multivectors made of orthonormal vectors are orthonormal. It is defined without respect to any basis. It extends to p-vectors which are not a series of 1-vectors wedged together by linearity. With this definition in available, at least all the pieces should be in place to begin trying to understand the definition of the Hodge dual. Also note that this sentence from the article: "the all wedge products of elements of orthonormal basis in V form an orthonormal basis of exterior algebra" is another attempt to define the inner product on the exterior algebra, though this definition is incomplete and basis dependent. Use the definition I give above instead, until we can fix the basis dependent definition. -Lethe | Talk

Suppose I have a metric G:V -> V -> F

surely you mean g:V⊗V -> F?-Lethe | Talk

sorry, I find it clearer to use curried functions, they are equivalent. Francis Davey 17:27, 11 Aug 2004 (UTC)

Can you explain what you mean? I don't know what a "curried function" is, and i can't quite understand how what you wrote can be equivalent to what i wrote. -Lethe | Talk

I am sorry, its my computer science background. In a lot of algebras there is an isomorphism between A⊗B -> C and A -> (B -> C). This is essentially what Cartesian Closed Categories are all about. If you think about it, using my definition G u is a member of V -> F, in other words a 1-form. You can see that (G u) v would be the same as g(u,v) in your definition. I remember a lot of my fellow students in maths having problems with dual vector spaces, but this intuition makes it easier.

If you are happy with functions being first class objects (which computer scientists are) you don't always have to tuple things up first. Francis Davey 19:49, 11 Aug 2004 (UTC)

Ahh.. OK, I am with you. Without the parentheses, I didn't really understand what you meant (to me that looks like a sequence of maps, not a map from a space to a space of maps). That and the use of the unfamiliar term "curried" threw me off, but now I see. Mathematicians do similar constructions in category theory all the time, like to give examples of spaces that are canonically isomorphic. OK, so I agree that the inner product can be thought of as a map V -> (V -> F). Cool, I'm happy with that (although I have very little CS, and don't know what "first class objects" are, but nevermind). By the way, you are using the word metric here where I would use the word inner product. Lots of people consider them interchangeable (especially physicists). Do you use them interchangeably? -Lethe | Talk

Surely an inner product will define you a metric tensor and vice versa? Or have I missed something? Francis Davey 19:33, 14 Aug 2004 (UTC)

(where F is the underlying field and V a vector space over F), what is * defined as?

*:Ω^k(M) -> Ω^n-k(M)-Lethe | Talk

Hmmmm, that tells me what the domain and codomain are, but its not quite a definition. Francis Davey 17:27, 11 Aug 2004 (UTC)

No, of course it is just specifying the domain and codomain. the definition is supplied above.-Lethe | Talk

One of the nice things about exterior calculus, as far as I understand it, is that one can go quite a long way in understanding certain equations without having to use components, which I find more satisfying. Can I do the same with *?

Francis Davey 07:53, 11 Aug 2004 (UTC)

The alternative definition of the hodge star using basis vectors is hopeless (and incorrect) since it doesn't state that the hodge star is a linear map, which wasn't something that was clear to me at all. Only if its linear can the definition be adequate.

Let's not say incorrect, but rather just incomplete (incompleteness is admittedly a form of incorrectness, but a weaker one than is implied by the unqualified use of the word incorrect)-Lethe | Talk

The article opens with something that might help a reader to "intuit" what the hodge star is all about. But it suffers from several problems: (1) the alphas are vectors but the omega is a form, so the equation cannot be right;

A differential form is a member of the exterior algebra of the vector space of cotangent vectors, right? well you can make an exterior algebra over any vector space. In this article, which vector space we're talking about is left unspecified, since the Hodge dual can be defined on any exterior algebra with inner product. With me so far? so don't think of that first equation as one with forms and vectors. that equation has a p-vector wedged with an (n-p)-vector on the left-hand side (together they make an n-vector), and the volume element (which is an n-vector) on the right-hand side. So the equation works out just as it should.-Lethe | Talk

2. because the hodge star is linear, if you double a vector, you double its star, thus quadrupling the volume of the wedge product of the two. The obvious understanding of the first equation is of something which is not linear, so that doubling a vector, halves its dual.

I'm not sure what you mean by "doubling a vector". do you mean wedging a 1-vector with itself? this will always get you zero...-Lethe | Talk

I meant multiplying by a scalar of value 2. That is *(2a)=2*a

Ahh... I see. So I'm not sure off-hand, but I think that first equation is only meant to apply to orthonormal vectors. So you can't go doubling things and still expect it to apply. But make no mistake, the Hodge star is indeed a linear operator. And the defining equation will apply to any vectors, so you may double to your heart's content. -Lethe | Talk

I am not sure I understand what a standard basis is either, but the whole first part of this article is so muddled that it makes no difference.

I suppose standard basis means "orthonormal basis", though it would be nice if we made that explicit.-Lethe | Talk

HELP! I want to understand what this is, but at the moment I have no complete definition. Francis Davey 17:27, 11 Aug 2004 (UTC)

Well, I have thought for some time that this article needs to be rewritten a bit. Maybe we can work out some better explanations on this talk page, which will lead to a clearer article. So if you have more questions, please be my guest!! -Lethe | Talk

Yes, my attempt to provide an introductory idea doesn't seem to have worked out, at least without more detail. For a standard orthonormal basis, * of a wedge of a subset of the basis is the wedge of the complementary subset, up to sign. Eg for dimension 3, you exchange the wedge of e₁ and e₂ with e₃, and so on by cyclic permutation. That's what one is trying to generalise, really. Charles Matthews 09:53, 12 Aug 2004 (UTC)

That is quite helpful. The problem is that a lot of books give definitions like ones on this page, which aren't quite complete. I didn't know * was supposed to be linear, and the intuition that I got was that given a volume element u, *v is given by $*v \wedge v = u$ , the dual isn't unique from that definition because $*e_1=(e_2\wedge e_3 + e_1)$ also holds (why is that not a problem with your definition, or are you only saying that * operates on basis vectors in the way described). More seriously the "larger" a vector, the "smaller" the dual. However it has a nice geometric intuition attached. Francis Davey 17:35, 12 Aug 2004 (UTC)

Firstly, you may have missed it, but the linearity of the Hodge star operator is mentioned in the first sentence. Secondly, as I said above, I think that first equation is only meant to apply to orthonormal vectors, so you can't do this construction that leads to the ambiguity here. While the first equation is a quick way to calculate a Hodge dual, it is not a satisfactory definition, for the reasons you point out. -Lethe | Talk

[edit] Hodge dual of an arbitrary tensor

the new texts reads that it is possible to take the Hodge dual of an arbitrary tensor, not just an antisymmetric one. Is that true? It's news to me! -Lethe | Talk 18:27, Apr 29, 2005 (UTC)

[edit] Hodge dual generalizes the cross product

I don't agree with the assertion that the Hodge dual generalizes the cross product in three dimensions. I previously removed that assertion from the article here with edit summary "the Hodge dual does not generalize the cross product to higher dimensions, although it is part of the def in 3 dimensions; rewrite introduction (this intro is redundant".

Here are some differences between the vector cross product and the Hodge dual:

The vector cross product is a binary operation while the Hodge dual is unary (it's not even a vector product).
The vector cross product is closed on the underlying vector space while the Hodge dual is not.

The only relationship between the Hodge dual and the vector cross product is that the Hodge dual applied to the exterior product gives you the vector cross product, in the three dimensional case. One might make the argument that the exterior product (or the Lie bracket) generalize the vector cross product, but it would be very hard to make that case for the Hodge dual. In any event, it isn't the motivation behind the definition, and therefore does not deserve to be the first sentence.

I am removing the sentence again. -Lethe | Talk 03:00, August 10, 2005 (UTC)

The goal of that sentence was to provide a simple, informal introductory comment that allowed a typical college sophomore to grasp the general idea. Technically, you are right; its not a binary op, etc. but pseudovectors do show up all over the place in college physics, and thus I felt the article should include some simple touchstone. How about The Hodge star is commonly used to define the cross product of two vectors in three dimensions. Would that work? linas 00:06, 16 August 2005 (UTC)

Except that the hodge dual is not commonly used for the vector cross product at all. Most books that deal with the subject simply define the vector cross product by means of its components. A less geometric definition, but surely you can't argue that the hodge dual definition is common. So I still can't get on board having that sentence in the first paragraph. As far as something to indicate the fact that the Hodge dual depends on the orientation, and how it changes vectors in a vector rep to vectors in a pseudovector rep, I agree, that's worth mentioning. The corollary that the vector cross product of two vectors is a pseudovector should follow, sure. -Lethe | Talk 01:00, August 16, 2005 (UTC)

Right. And so the article on the cross product does not begin by stating that its the "Hodge dual of the exterior product". What I like to see, in any math that I read, are those sentences that make my brain go "ah ha!" and light up that lightbulb, as soon as possible. I like to see the ah-ha-inducing language early, even if this means that its somewhat inaccurate or poorly defined; the correct, detailed definition can be supplied once the reader is hooked. Anyone who is interested in the Hodge dual and wants to read about it will already know all about the vagaries of the cross product, and so here, the "ah ha" moment comes with the realization "its like this other thing that I've seen before". linas 00:42, 17 August 2005 (UTC)

[edit] underpinning de Rham

furthermore, in what sense does the Hodge decomposition "underpin" de Rham cohomology? The most we can say is that the Hodge decomposition provides us with a privileged representative of the cohomology class, namely, the harmonic form. This of course depends on the choice of metric, while the cohomology groups are topological invariants, so it is a stretch to say that it is an underpinning. -Lethe | Talk 03:09, August 10, 2005 (UTC)

It was just casual, informal, language so that I could wikilink to the article that covers cohomology. It was meant only to be a more literate way of saying "see also cohomology", nothing more. linas 00:06, 16 August 2005 (UTC)

I guess it would be nice to have a link to de Rham cohomology. Note that pure de Rham cohomology itself is a topological invariant, and doesn't depend on a Hodge dual, so this isn't relevant there, strictly speaking. I guess what we could do is say something like ..., which in the case of compact Riemannian manifolds leads to the Hodge decomposition of differential forms, which uses concepts from de Rham cohomology. But it's still weirdly nebulous, and without an explicit explanation, adds nothing to the article. Maybe just a "See also" link at the bottom would be better. And I guess we should write the Hodge decomposition article, it will all be explained in there. -Lethe | Talk 01:20, August 16, 2005 (UTC)

"See also" works for me. I thought I'd added a short section on Hodge decomposition to some article, but I can't remember where now, and appearently didn't create a re-direct page. Oh ... I added it to, well, lookee there ... Mind if I just #redirect until someone splits it out into a full-fledged article? linas 00:42, 17 August 2005 (UTC)

[edit] Yet another definiton

Exterior algebra is a subalgebra for Geometric algebra. In GA the closest thin to hodge is a pseudoscalar I

$I = e 1 e 2 ... e n = *$

in GA geometric product is defined

$ab = a \cdot b + a \and b$

$Ib = I \cdot b + I \and b$

Exterior algebra is ancommutative so the dot product term is zero.

Inverse of k-vector A is

$A^{-1} = A^{\dagger} / |A|^2$

So Hodge is a pseudoscalar Q.E.D.

eg.

$\vec J = *_\sigma \vec E = \sigma I E$

$\vec E = *_\sigma^{-1} \vec J = \frac{I}{\sigma} \vec J$

--213.169.2.205 18:03, 14 November 2005 (UTC) aps

[edit] signature and Hodge squared

I once added the equation

* * = ( - 1) k (n - k) + s i d

to this article. At some point in its history, someone (I'm looking at you here, linas) changed it to

$**\eta=(-1)^{k(n-k)}s\;\eta.$

I've fixed the error, but I notice that a similar weirdness is in the equation

$\delta = (-1)^{n(k+1)+1} s\; *d*$

for the codifferential. I haven't fixed it, since I don't know the correct convention off the top of my head. Furthermore, I'm wondering if the fact that the same weirdness appears twice means that perhaps it's not simply a mistake, perhaps someone thinks that signature does more than flip the sign. Does anyone want to stand up for this equation? -lethe ^{talk +} 17:41, 8 April 2006 (UTC)

[edit] pseudo riemannian case

Can someone file a reference for the pseudo riemannian case?

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