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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)
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 as a volume form, then the dual of v is 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.
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: .
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).
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)
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)?
Suppose I have a metric G:V -> V -> F
(where F is the underlying field and V a vector space over F), what is * defined as?
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.
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;
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 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.
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)
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 , the dual isn't unique from that definition because 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)
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)
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 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)
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)
Exterior algebra is a subalgebra for Geometric algebra. In GA the closest thin to hodge is a pseudoscalar I
in GA geometric product is defined
so
Exterior algebra is ancommutative so the dot product term is zero.
Inverse of k-vector A is
So Hodge is a pseudoscalar Q.E.D.
eg.
-- 213.169.2.205 18:03, 14 November 2005 (UTC) aps
I once added the equation
to this article. At some point in its history, someone (I'm looking at you here, linas) changed it to
I've fixed the error, but I notice that a similar weirdness is in the equation
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)
I believe that the correct version of the is given by, when acts on forms,
Especially, with this convention, the Lapalacian operator on 1-form has the correct expression 1 [1], see the Lemma 27 of Chapter 7 in 2 [2].
-- Vanabel ( talk) 14:25, 29 December 2018 (UTC)
Can someone file a reference for the pseudo riemannian case?
I would like to point out a correction which should be made to the section proving that the codifferential is the adjoint of d. It is true that it is the adjoint, but the proof offered is not correct. The article currently states that we can get the required identity from integrating d of the volume form since that is zero. But to integrate forms, the degree of the form integrated must match the dimension of the manifold over which integration takes place, and integrating d of the volume form would violate this. In fact, the proof uses Stokes' Theorem (see for example p. 318ff of Differential Forms in Mathematical Physics, C. von Westenholz, for the details of the proof). Thank you. Idempotent 09:27, 13 September 2007 (UTC)
I fear that the expression given in the 'Tensor notation for the star operator' section is incorrect. I noticed that while I was trying to prove that it is covariant under coordinate transformations, but I think the easiest way to see there is something wrong is to note what happens if η is a 0-form. In this case, there are no indices on it, and we just get *η = ε, which can only be a well-defined tensor equation if the right-hand-side is also a tensor, i.e., ε should mean the Levi-Civita tensor (the volume form), not the symbol (which is a tensor density).
I would go ahead and make the change, but I just wanted to make sure I'm not missing anything here.
Edit: I guess I did go on and changed the article since I don't know if anybody is ever going to read this talk page.
Legendre17 ( talk) 18:55, 4 August 2008 (UTC)
The paragraph on the inner product of k-vectors contained a reference to volume forms which I think should not be here. I edited the paragraph such that there is a clearer separation between the general k-vector case and the specific application to exterior differential forms on manifolds. However, I think that this should not appear here at all. Bas Michielsen ( talk) 12:18, 28 August 2009 (UTC)
In fact there are a few more issues with this page in its actual state. The most important one, I think, is that there is a mix between the general definitions in terms of k-vectors and the special case of exterior differential forms. I would prefer to group everything related to the special case into a dedicated section somewhat like "Hodge duality on manifolds." The present mixing up is most disturbing when there are references to "the metric". In the general case, this should be the metric of the underlying vector space V introduced at the start. But in the the context of exterior differential forms on manifolds, the g used is apparently the metric on the tangent bundle and not the one on the k-forms in the cotangent space, of which the coefficient matrix is the inverse of the one conventionally used.
A less important issue, perhaps, is the use of the name "dual." I think it would be much clearer if dual would only be used when there is actually a metric pairing between two spaces. I think that the whole idea of the Hodge dual is motivated by this idea. The fact that there is an isomorphism between spaces of complementary dimensions does not justify the name dual in itself. Bas Michielsen ( talk) 09:18, 31 August 2009 (UTC)
In the equation for the formal definition, it does not seem clear what happens in the case k=0 - LHS seems ill-defined - or the case k=n - RHS seems ill-defined.
As I understand it 'wedge-ing' together all of the vectors of the orthonormal basis gives a unit volume, so I guess the answer should be 1 for those cases, but the definition doesn't seem to say that.
Albear-And ( talk) 17:17, 12 February 2010 (UTC)
It seems that in the section on the codifferential on manifolds, the article uses the positive Laplacian (in the sense of functional analysis) while in the very last section, the opposite sign convention is used for Δ. This is likely to confuse people. Should the two different uses be left in, with a note to warn people, or should one of them be changed? They are both natural in their two separate contexts, so I would vote for the former. Maybe use \nabla^2 for the three-dimensional one, and note that it is the negative of the general Laplacian. -- Spireguy ( talk) 02:43, 8 June 2010 (UTC)
I may be wrong, but I wanted to point out what I believe is a sign error in the 'Examples' section, in the subsection 'Four dimensions', I believe the wrong sign has been applied to all the examples of 2-forms given. Consider the first one first instance:
. But the earlier definition
where is an even permutation of ,
implies that
.
Also, the expressions under 'Four dimensions' contradict the assertion in the section 'Duality' that
,
for if and , then this becomes ; but
and
implies
.
Look at http://www.dr-gert-hillebrandt.de/mathe_uni.htm http://www.dr-gert-hillebrandt.de/pdf/Universitaet/Mathe/Der%20Stern-Operator.pdf — Preceding unsigned comment added by 77.8.136.118 ( talk) 11:16, 26 March 2015 (UTC)
Clarification? Thanks - Erik, 00.55 UTC 8th April 2012 — Preceding unsigned comment added by 93.96.22.178 ( talk) 00:54, 8 April 2012 (UTC)
Regarding my recent addition: I've posted a message on the Maths WikiProject talk page asking for feedback. — Fly by Night ( talk) 18:37, 1 July 2012 (UTC)
"Let W be a vector space, with an inner product \langle\cdot, \cdot\rangle_W. For every linear function f \in W^* there exists a unique vector v in W such that f(w) = \langle w, v \rangle_W for all w in W. The map W^* \to W given by f \mapsto v is an isomorphism. This holds for all vector spaces, and can be used to explain the Hodge dual."
Consider the vector space V over Q consisting of all finite sequences of rational numbers. With the inner product given by summation over the multiplication of corresponding pairs, and point-wise addition and scalar multiplication. This is a countable set. V* consists of the set of all finite and infinite sequences of rational numbers. This set is uncountable. According to this section, the two spaces are isomorphic. Clearly this does not hold for all spaces.
Also note: it is claimed to hold for all vector spaces, which is necessarily false, since W must be an inner product space. — Preceding unsigned comment added by 149.89.17.110 ( talk) 18:16, 12 September 2013 (UTC)
I have revised the section Formal definition of the Hodge star of k-vectors, since it was incomplete and had a few problems. I found the Tevian Dray reference particularly suited for drawing on for this article, and would encourage others to use it. I have removed the need for any reference to a basis in the definition of ω, drawn attention to that a choice of orientation must be made and importantly have removed reliance on another article for the crucial definition of the "inner product" (better named a bilinear form throughout?) as extended to k-vectors. I haven't indicated that s=⟨ω,ω⟩=±1 is determined by the signature of the bilinear form. I invite others to review and edit for correctness/clarity etc. My intention was to achieve an understandable and complete definition, but others may feel that I've not achieved this and I'd value their input. — Quondum ☏ 13:57, 4 July 2012 (UTC)
I am confused by the phrase "introduced in general by" in the lead. I am unsure of what was intended and do not know the actual history, and am thus hesitant to simply remove the words "in general". Anyone know what's best here? — Quondum ☏ 19:12, 15 July 2012 (UTC)
The explanation section is a nice attempt at getting a coordinate free construction of the Hodge star operator, but it is not immediately clear why the formula in the explanation section gives the desired formula in the formal definition section.
Jfdavis ( talk) 19:18, 26 September 2012 (UTC)
I have been editing this article for consistency on the assumption that what is meant by "inner product" is a general symmetric bilinear form, but some of the wording suggests to me that in some contexts a symmetric sesquilinear form (i.e. a Hermitian form) is used. This would suggest that two distinct types generalization of "inner product" are used in the definition of the Hodge dual according to context:
Can someone confirm whether both of these generalizations are normally understood to be permitted in the definition of the Hodge dual? It will not be too difficult to include both cases in the definition. — Quondum 17:11, 14 October 2012 (UTC)
Shouldn't the sentence "the Hodge star operator establishes a one-to-one mapping from the space of k-vectors and the space of (n−k)-vectors." instead say "between the space of k-vectors and the space of (n−k)-vectors" or "from the space of k-vectors to the space of (n−k)-vectors." ?
Tashiro ( talk) 07:54, 25 February 2014 (UTC)
Inclusion of the following ideas would be clarifying (from Vaz; da Rocha (2016), An Introduction to Clifford Algebras and Spinors):
This would fit neatly into this article. It is also an approach that avoids the non-obvious extension of the bilinear form through the use of a determinant. — Quondum 18:27, 8 August 2017 (UTC)
I notice that the operations usually used in tensor analysis under names such as dual tensor and Hodge dual are
and its inverse, which are really the quasi-Hodge isomorphisms, though in physics indices are generally raised and lowered so freely that the related tensors with raised and lowered indices are considered equivalent. In this article, it would probably be best to clarify the distinction. — Quondum 17:09, 10 August 2017 (UTC)
The article recently changed title from Hodge dual to Hodge isomorphism without prior consultation of the community which is standard when changing titles. By WP:COMMONNAME, any of the alternatives below is better.
Hit count using Google Scholar:
The present title gets this:
The terminology to be used in WP isn't what the terminology ought to be (according to some), but it should reflect what the terminology is. If there are logical arguments saying that the article refers to the map (which Hodge dual certainly can), as opposed so mapped elements, then Hodge star is a superior choice because it matches the notation, and is less likely to be confused with a single mapped element than Hodge dual. Besides, people apparently use it.
An article title's primary task is to be found by those searching for it. As for the current title, it should be reduced to "the Hodge star is an isomorphism". YohanN7 ( talk) 09:53, 11 September 2017 (UTC)
See also WP:SET. —DIV ( 120.17.18.193 ( talk) 10:16, 13 August 2018 (UTC))
Is it really necessary to mix up notions of the Hodge star operator on differential forms with notions of a Clifford algebra (in the subsection Three dimensions of the section Examples)? I believe this creates much more confusion than clarity on the subject.
Also, use of the Einstein summation convention (in Expression in index notation) is not at all well-explained here. The mere link to the article Ricci calculus certainly does not help, since that article is about much more than the Einstein summation convention.
Go ahead and use a summation sign so that readers have a chance of understanding what is written! 50.205.142.35 ( talk) 22:08, 15 February 2020 (UTC)
Magyar25, I am a bit concerned about this edit, for a few reasons, including that
— Quondum 21:58, 11 March 2020 (UTC)
Notation: I think the { {math|| } } environment really sux, and should be deprecated. Also, \bigwedge is too tall and obnoxious for inline formulas; \wedge is understated and elegant.
Magyar25, I am going to revert the section content of Hodge star operator § Three dimensions to before these edits. It introduces undefined notation (det of a list of vectors, equating differential forms to matrices, hatted vectors, "matrix multiplication" of vectors), and introduces concepts that are not relevant and downright confusing in the context (Lie algebras, infinitesimal rotations, orthogonal groups, lie brackets, badly defined tensor products, etc.). — Quondum 00:38, 4 May 2020 (UTC)
Uh...the articles states that in two dimensions:
But that means that
Now the square of a pseudoscalar isnt always -1 but in 2 and 3 dimensions it certainly is. So shouldnt the answer be -1?
Just granpa ( talk) 17:12, 19 June 2020 (UTC)
The five-pointed star notation seems to be that fabled thing called WP:OR; at least I've never seen it used anywhere but here. It wouldn't be a big deal except that it conflicts with the Moyal product and more generally with star product on the algebra of symbols, which always use the five-pointed star. What's more, this all being differential geom, there are texts that use both symbols at the same time ... so using the five-poited star here seems like a bad idea... are there any texts, anywhere at all, besides WP, that use the five-pointed star for the Hodge star?
The conversion to the five-pointed star happened in this edit, by a one-time contributor. I guess no one else has minded, so far. 67.198.37.16 ( talk) 16:34, 29 October 2020 (UTC)
In the section Codifferential, this passage appears:
"The most important application of the Hodge star on manifolds is to define the codifferential on k-forms. Let
where is the exterior derivative or differential, and for Riemannian manifolds."
But there is no mention of any application other than Riemannian manifolds.
I suggest either explaining something about the other cases, or else just setting s = 1 (modifying formulas appropriately) and deleting all references to this mysterious s. 2601:200:C000:1A0:F1D0:4B34:8877:BEE3 ( talk) 18:08, 7 October 2021 (UTC)
In an encyclopedia article, it is never a good idea to use the Einstein summation convention (repeated indices get summed) when the use of the symbol is much clearer, especially to non-professionals.
Another thing that ought to be avoided is the use of the numeral 1 to indicate an identity map from some mathematical object X to itself. Much better would be either to use the symbol 1X or the symbol idX, since that avoids ambiguity.
The section titled Geometric explanation begins as follows:
"The Hodge star is motivated by the correspondence between a subspace W of V and its orthogonal subspace (with respect to the inner product), where each space is endowed with an orientation and a numerical scaling factor. Specifically, a non-zero decomposable k-vector corresponds by the Plücker embedding to the subspace with oriented basis , endowed with a scaling factor equal to the k-dimensional volume of the parallelepiped spanned by this basis (equal to the Gramian, the determinant of the matrix of inner products )."
But suppose that an orthonormal basis {e1, ..., ek} of W is replaced by {2e1, ..., 2ek}. Then the Gramian determinant will be of a k×k matrix with 4's on the diagonal and zeroes elsewhere. Its determinant is 4k, which is the square of the actual volume of 2k spanned by the corresponding parallelotope (a k-dimensional cube of side 2).
Since this section is the only place in the article where the subject of the article (Hodge star) is actually defined, I hope someone knowledgeable about the subject can fix this mistake.
PS I also believe that referring to the Plücker embedding is entirely out of place here and contributes nothing of value to this article, where it serves mainly as a distraction from the already technical content of the Hodge star.
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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)
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 as a volume form, then the dual of v is 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.
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: .
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).
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)
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)?
Suppose I have a metric G:V -> V -> F
(where F is the underlying field and V a vector space over F), what is * defined as?
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.
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;
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 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.
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)
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 , the dual isn't unique from that definition because 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)
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)
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 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)
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)
Exterior algebra is a subalgebra for Geometric algebra. In GA the closest thin to hodge is a pseudoscalar I
in GA geometric product is defined
so
Exterior algebra is ancommutative so the dot product term is zero.
Inverse of k-vector A is
So Hodge is a pseudoscalar Q.E.D.
eg.
-- 213.169.2.205 18:03, 14 November 2005 (UTC) aps
I once added the equation
to this article. At some point in its history, someone (I'm looking at you here, linas) changed it to
I've fixed the error, but I notice that a similar weirdness is in the equation
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)
I believe that the correct version of the is given by, when acts on forms,
Especially, with this convention, the Lapalacian operator on 1-form has the correct expression 1 [1], see the Lemma 27 of Chapter 7 in 2 [2].
-- Vanabel ( talk) 14:25, 29 December 2018 (UTC)
Can someone file a reference for the pseudo riemannian case?
I would like to point out a correction which should be made to the section proving that the codifferential is the adjoint of d. It is true that it is the adjoint, but the proof offered is not correct. The article currently states that we can get the required identity from integrating d of the volume form since that is zero. But to integrate forms, the degree of the form integrated must match the dimension of the manifold over which integration takes place, and integrating d of the volume form would violate this. In fact, the proof uses Stokes' Theorem (see for example p. 318ff of Differential Forms in Mathematical Physics, C. von Westenholz, for the details of the proof). Thank you. Idempotent 09:27, 13 September 2007 (UTC)
I fear that the expression given in the 'Tensor notation for the star operator' section is incorrect. I noticed that while I was trying to prove that it is covariant under coordinate transformations, but I think the easiest way to see there is something wrong is to note what happens if η is a 0-form. In this case, there are no indices on it, and we just get *η = ε, which can only be a well-defined tensor equation if the right-hand-side is also a tensor, i.e., ε should mean the Levi-Civita tensor (the volume form), not the symbol (which is a tensor density).
I would go ahead and make the change, but I just wanted to make sure I'm not missing anything here.
Edit: I guess I did go on and changed the article since I don't know if anybody is ever going to read this talk page.
Legendre17 ( talk) 18:55, 4 August 2008 (UTC)
The paragraph on the inner product of k-vectors contained a reference to volume forms which I think should not be here. I edited the paragraph such that there is a clearer separation between the general k-vector case and the specific application to exterior differential forms on manifolds. However, I think that this should not appear here at all. Bas Michielsen ( talk) 12:18, 28 August 2009 (UTC)
In fact there are a few more issues with this page in its actual state. The most important one, I think, is that there is a mix between the general definitions in terms of k-vectors and the special case of exterior differential forms. I would prefer to group everything related to the special case into a dedicated section somewhat like "Hodge duality on manifolds." The present mixing up is most disturbing when there are references to "the metric". In the general case, this should be the metric of the underlying vector space V introduced at the start. But in the the context of exterior differential forms on manifolds, the g used is apparently the metric on the tangent bundle and not the one on the k-forms in the cotangent space, of which the coefficient matrix is the inverse of the one conventionally used.
A less important issue, perhaps, is the use of the name "dual." I think it would be much clearer if dual would only be used when there is actually a metric pairing between two spaces. I think that the whole idea of the Hodge dual is motivated by this idea. The fact that there is an isomorphism between spaces of complementary dimensions does not justify the name dual in itself. Bas Michielsen ( talk) 09:18, 31 August 2009 (UTC)
In the equation for the formal definition, it does not seem clear what happens in the case k=0 - LHS seems ill-defined - or the case k=n - RHS seems ill-defined.
As I understand it 'wedge-ing' together all of the vectors of the orthonormal basis gives a unit volume, so I guess the answer should be 1 for those cases, but the definition doesn't seem to say that.
Albear-And ( talk) 17:17, 12 February 2010 (UTC)
It seems that in the section on the codifferential on manifolds, the article uses the positive Laplacian (in the sense of functional analysis) while in the very last section, the opposite sign convention is used for Δ. This is likely to confuse people. Should the two different uses be left in, with a note to warn people, or should one of them be changed? They are both natural in their two separate contexts, so I would vote for the former. Maybe use \nabla^2 for the three-dimensional one, and note that it is the negative of the general Laplacian. -- Spireguy ( talk) 02:43, 8 June 2010 (UTC)
I may be wrong, but I wanted to point out what I believe is a sign error in the 'Examples' section, in the subsection 'Four dimensions', I believe the wrong sign has been applied to all the examples of 2-forms given. Consider the first one first instance:
. But the earlier definition
where is an even permutation of ,
implies that
.
Also, the expressions under 'Four dimensions' contradict the assertion in the section 'Duality' that
,
for if and , then this becomes ; but
and
implies
.
Look at http://www.dr-gert-hillebrandt.de/mathe_uni.htm http://www.dr-gert-hillebrandt.de/pdf/Universitaet/Mathe/Der%20Stern-Operator.pdf — Preceding unsigned comment added by 77.8.136.118 ( talk) 11:16, 26 March 2015 (UTC)
Clarification? Thanks - Erik, 00.55 UTC 8th April 2012 — Preceding unsigned comment added by 93.96.22.178 ( talk) 00:54, 8 April 2012 (UTC)
Regarding my recent addition: I've posted a message on the Maths WikiProject talk page asking for feedback. — Fly by Night ( talk) 18:37, 1 July 2012 (UTC)
"Let W be a vector space, with an inner product \langle\cdot, \cdot\rangle_W. For every linear function f \in W^* there exists a unique vector v in W such that f(w) = \langle w, v \rangle_W for all w in W. The map W^* \to W given by f \mapsto v is an isomorphism. This holds for all vector spaces, and can be used to explain the Hodge dual."
Consider the vector space V over Q consisting of all finite sequences of rational numbers. With the inner product given by summation over the multiplication of corresponding pairs, and point-wise addition and scalar multiplication. This is a countable set. V* consists of the set of all finite and infinite sequences of rational numbers. This set is uncountable. According to this section, the two spaces are isomorphic. Clearly this does not hold for all spaces.
Also note: it is claimed to hold for all vector spaces, which is necessarily false, since W must be an inner product space. — Preceding unsigned comment added by 149.89.17.110 ( talk) 18:16, 12 September 2013 (UTC)
I have revised the section Formal definition of the Hodge star of k-vectors, since it was incomplete and had a few problems. I found the Tevian Dray reference particularly suited for drawing on for this article, and would encourage others to use it. I have removed the need for any reference to a basis in the definition of ω, drawn attention to that a choice of orientation must be made and importantly have removed reliance on another article for the crucial definition of the "inner product" (better named a bilinear form throughout?) as extended to k-vectors. I haven't indicated that s=⟨ω,ω⟩=±1 is determined by the signature of the bilinear form. I invite others to review and edit for correctness/clarity etc. My intention was to achieve an understandable and complete definition, but others may feel that I've not achieved this and I'd value their input. — Quondum ☏ 13:57, 4 July 2012 (UTC)
I am confused by the phrase "introduced in general by" in the lead. I am unsure of what was intended and do not know the actual history, and am thus hesitant to simply remove the words "in general". Anyone know what's best here? — Quondum ☏ 19:12, 15 July 2012 (UTC)
The explanation section is a nice attempt at getting a coordinate free construction of the Hodge star operator, but it is not immediately clear why the formula in the explanation section gives the desired formula in the formal definition section.
Jfdavis ( talk) 19:18, 26 September 2012 (UTC)
I have been editing this article for consistency on the assumption that what is meant by "inner product" is a general symmetric bilinear form, but some of the wording suggests to me that in some contexts a symmetric sesquilinear form (i.e. a Hermitian form) is used. This would suggest that two distinct types generalization of "inner product" are used in the definition of the Hodge dual according to context:
Can someone confirm whether both of these generalizations are normally understood to be permitted in the definition of the Hodge dual? It will not be too difficult to include both cases in the definition. — Quondum 17:11, 14 October 2012 (UTC)
Shouldn't the sentence "the Hodge star operator establishes a one-to-one mapping from the space of k-vectors and the space of (n−k)-vectors." instead say "between the space of k-vectors and the space of (n−k)-vectors" or "from the space of k-vectors to the space of (n−k)-vectors." ?
Tashiro ( talk) 07:54, 25 February 2014 (UTC)
Inclusion of the following ideas would be clarifying (from Vaz; da Rocha (2016), An Introduction to Clifford Algebras and Spinors):
This would fit neatly into this article. It is also an approach that avoids the non-obvious extension of the bilinear form through the use of a determinant. — Quondum 18:27, 8 August 2017 (UTC)
I notice that the operations usually used in tensor analysis under names such as dual tensor and Hodge dual are
and its inverse, which are really the quasi-Hodge isomorphisms, though in physics indices are generally raised and lowered so freely that the related tensors with raised and lowered indices are considered equivalent. In this article, it would probably be best to clarify the distinction. — Quondum 17:09, 10 August 2017 (UTC)
The article recently changed title from Hodge dual to Hodge isomorphism without prior consultation of the community which is standard when changing titles. By WP:COMMONNAME, any of the alternatives below is better.
Hit count using Google Scholar:
The present title gets this:
The terminology to be used in WP isn't what the terminology ought to be (according to some), but it should reflect what the terminology is. If there are logical arguments saying that the article refers to the map (which Hodge dual certainly can), as opposed so mapped elements, then Hodge star is a superior choice because it matches the notation, and is less likely to be confused with a single mapped element than Hodge dual. Besides, people apparently use it.
An article title's primary task is to be found by those searching for it. As for the current title, it should be reduced to "the Hodge star is an isomorphism". YohanN7 ( talk) 09:53, 11 September 2017 (UTC)
See also WP:SET. —DIV ( 120.17.18.193 ( talk) 10:16, 13 August 2018 (UTC))
Is it really necessary to mix up notions of the Hodge star operator on differential forms with notions of a Clifford algebra (in the subsection Three dimensions of the section Examples)? I believe this creates much more confusion than clarity on the subject.
Also, use of the Einstein summation convention (in Expression in index notation) is not at all well-explained here. The mere link to the article Ricci calculus certainly does not help, since that article is about much more than the Einstein summation convention.
Go ahead and use a summation sign so that readers have a chance of understanding what is written! 50.205.142.35 ( talk) 22:08, 15 February 2020 (UTC)
Magyar25, I am a bit concerned about this edit, for a few reasons, including that
— Quondum 21:58, 11 March 2020 (UTC)
Notation: I think the { {math|| } } environment really sux, and should be deprecated. Also, \bigwedge is too tall and obnoxious for inline formulas; \wedge is understated and elegant.
Magyar25, I am going to revert the section content of Hodge star operator § Three dimensions to before these edits. It introduces undefined notation (det of a list of vectors, equating differential forms to matrices, hatted vectors, "matrix multiplication" of vectors), and introduces concepts that are not relevant and downright confusing in the context (Lie algebras, infinitesimal rotations, orthogonal groups, lie brackets, badly defined tensor products, etc.). — Quondum 00:38, 4 May 2020 (UTC)
Uh...the articles states that in two dimensions:
But that means that
Now the square of a pseudoscalar isnt always -1 but in 2 and 3 dimensions it certainly is. So shouldnt the answer be -1?
Just granpa ( talk) 17:12, 19 June 2020 (UTC)
The five-pointed star notation seems to be that fabled thing called WP:OR; at least I've never seen it used anywhere but here. It wouldn't be a big deal except that it conflicts with the Moyal product and more generally with star product on the algebra of symbols, which always use the five-pointed star. What's more, this all being differential geom, there are texts that use both symbols at the same time ... so using the five-poited star here seems like a bad idea... are there any texts, anywhere at all, besides WP, that use the five-pointed star for the Hodge star?
The conversion to the five-pointed star happened in this edit, by a one-time contributor. I guess no one else has minded, so far. 67.198.37.16 ( talk) 16:34, 29 October 2020 (UTC)
In the section Codifferential, this passage appears:
"The most important application of the Hodge star on manifolds is to define the codifferential on k-forms. Let
where is the exterior derivative or differential, and for Riemannian manifolds."
But there is no mention of any application other than Riemannian manifolds.
I suggest either explaining something about the other cases, or else just setting s = 1 (modifying formulas appropriately) and deleting all references to this mysterious s. 2601:200:C000:1A0:F1D0:4B34:8877:BEE3 ( talk) 18:08, 7 October 2021 (UTC)
In an encyclopedia article, it is never a good idea to use the Einstein summation convention (repeated indices get summed) when the use of the symbol is much clearer, especially to non-professionals.
Another thing that ought to be avoided is the use of the numeral 1 to indicate an identity map from some mathematical object X to itself. Much better would be either to use the symbol 1X or the symbol idX, since that avoids ambiguity.
The section titled Geometric explanation begins as follows:
"The Hodge star is motivated by the correspondence between a subspace W of V and its orthogonal subspace (with respect to the inner product), where each space is endowed with an orientation and a numerical scaling factor. Specifically, a non-zero decomposable k-vector corresponds by the Plücker embedding to the subspace with oriented basis , endowed with a scaling factor equal to the k-dimensional volume of the parallelepiped spanned by this basis (equal to the Gramian, the determinant of the matrix of inner products )."
But suppose that an orthonormal basis {e1, ..., ek} of W is replaced by {2e1, ..., 2ek}. Then the Gramian determinant will be of a k×k matrix with 4's on the diagonal and zeroes elsewhere. Its determinant is 4k, which is the square of the actual volume of 2k spanned by the corresponding parallelotope (a k-dimensional cube of side 2).
Since this section is the only place in the article where the subject of the article (Hodge star) is actually defined, I hope someone knowledgeable about the subject can fix this mistake.
PS I also believe that referring to the Plücker embedding is entirely out of place here and contributes nothing of value to this article, where it serves mainly as a distraction from the already technical content of the Hodge star.