Talk:Hodge star operator

message: 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?
* a mathematical operator in vector calculus, a form of the derivative is taken (gradient) 
*a unit of angular measurement, synonymous with the gon 
Help in figuring out where to point this link would be appreciated. Kevyn 13:58, 29 Jun 2004 (UTC)
thread_title: Which Grad?

message: 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)
:The tensor formula (summation with a fully anti-symmetric tensor) automatically discards any part of the tensor that is not fully antisymmetric, but I don't think that this should be regarded as taking the Hodge dual, which should IMO only be defined on a k-form or k-vector. I think it should be rephrased to make this clear. —Quondum 06:38, 25 January 2016 (UTC)
replies: text: I modified the relevant statement so that it does not imply that the tensor formula for the Hodge dual is actually taking the Hodge dual when applied to an arbitrary tensor. —Quondum 05:00, 27 January 2016 (UTC)
thread_title:  Hodge dual of an arbitrary tensor 

message: Exterior algebra is a subalgebra for Geometric algebra. In GA the closest thin to hodge is a pseudoscalar I
I = e_1e_2 ... e_n = * 
in GA geometric product is defined
ab = a \cdot b + a \land b
 
so 
Ib = I \cdot b + I \land 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)
thread_title:  Yet another definiton 

message: I once added the equation
:**=(-1)^{k(n-k)+s}id
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)
thread_title:  signature and Hodge squared 

message: I believe that the correct version of the \delta is given by, when acts on k+1 forms, 
:\delta = (-1)^{(n-k)(k+1)} s\; *d*
Especially, with this convention, the Lapalacian operator on 1-form has the correct expression 1, see the Lemma 27 of Chapter 7 in 2.
--Vanabel (talk) 14:25, 29 December 2018 (UTC)
thread_title:  The signature of the codifferential operator 

message: 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)
replies: text: I corrected the presentation a bit. However, I think that the specific relation between d and δ is motivated by the desire to have adjoint operators and this is not a miraculous consequence of the definition given at the start of this section. Therefore, I think that the order of the presentation should also be changed. Bas Michielsen (talk) 16:33, 26 August 2009 (UTC)
thread_title:  codifferential is adjoint of d 

message: 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)
replies: text: Sorry, I was going to reply.  Your edit looks reasonable to me.  Something more can obviously be said about how the star operator is generally density-valued in the absence of a metric.  siℓℓy rabbit (talk) 19:00, 4 August 2008 (UTC)
thread_title:  Tensor notation 

message: 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)
thread_title:  Inner product of ''k''-vectors 

message: 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)
replies: text: It's not especially clear, but the exterior product of zero vectors is usually taken to be a scalar. In this case with unit vectors the value should be 1. See e.g. Exterior algebra#Basis and dimension, where Λ0(V) is the space of exterior products of zero vectors.--JohnBlackburnewordsdeeds 17:53, 12 February 2010 (UTC)
thread_title:  Formal definition of the Hodge star of ''k''-vectors 

message: 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)
thread_title:  Sign incompatibility 

message: 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)
thread_title:  Explanation Section 

message: 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)
thread_title:  Revised formal definition 

message: 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)
replies: text: My guess would be that special cases of it were used by others before Hodge developed the general definition. JRSpriggs (talk) 05:09, 16 July 2012 (UTC)
thread_title:  "introduced in general by" in lead? 

message: 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)
thread_title:  Explanation Section 

message: I have been editing this article for consistency on the assumption that what is meant by "inner product" \langle\cdot,\cdot\rangle 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:
* a general metric tensor (symmetric bilinear form)
* a hermitian form (symmetric sesquilinear form).
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)
thread_title:  Generality of "symmetric bilinear form" 

message: 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)
replies: text: You are right, mappings are directed and relations are not. The text was a mix-up of the two. I made the correction. Bas Michielsen (talk) 08:46, 25 February 2014 (UTC)
thread_title: A mapping "from.....and...."?

message: 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:
*"Hodge star" - 8300
*"Hode dual" - 6200
*"Hodge star operator" - 5500
*"Hodge operator" - 4600
*"Hodge dual operator" - 460
*"Hodge map" - 360
The present title gets this:
*"Hodge isomorphism" - 280
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)
replies: text: Google scholar needs some care of interpretation, not least of which is occurrence of a term with unrelated meaning.  Google n-grams are more reliable for excluding terms, but is also not suitable for determining frequency of a term with a given meaning.  This Google Ngram chart is probably adequate for ruling out "Hodge dual operator", "Hodge map", "Hodge mapping" and "Hodge isomorphism".  Considering that one is a subphrase of the other, the frequency of "Hodge star" should be subtracted from that of "Hodge star operator", giving four phrases with roughly equal frequency of use in recent times.  "Hodge dual" has multiple meanings, suggesting that its use to refer to the mapping rather than to its output element might even have a lower frequency than the others do, an as noted it has ambiguity issues.  So I would think that the title should be "Hodge star operator", "Hodge star" or "Hodge operator".  —Quondum 03:19, 17 September 2017 (UTC)

text: Anyone of the four frequent ones would do. I like the ones with "star" in them for it matches the notation. As for ambiguity, I'd expect this article to answer both questions about the map and what is the dual of my single element. YohanN7 (talk) 13:59, 18 September 2017 (UTC)
replies: text: I have a very mild preference for Hodge star operator, but would like opinions from others too.  —Quondum 02:12, 19 September 2017 (UTC)
replies: text: When it comes to nitpicking, I prefer an article's name that best denotes the map, even when this is defined by the effect on single elements. So map, operator, and isomorhism are preferable to me, because dual and star alone, appear to me as leaning to the image. Isomorhism might be too revealing already, and map too genral. Dual in Dual operator is too wishy-washy a word in my perception; i.e., -tadaa, here it comes- "Hodge star operator" is my favourite. :) Purgy (talk) 06:07, 19 September 2017 (UTC)
replies: text: This closely echoes my feelings.  Unless we hear from someone else in the near future, I'll move it to "Hodge star operator" as apparently being the consensus, or at least acceptable to all (unless, of course, someone beats me to it).  —Quondum 01:38, 22 September 2017 (UTC)
thread_title:  Hodge xxxx 

message: Uh...the articles states that in two dimensions:
: {\star} \, 1 = dx \wedge dy 
: {\star} ( dx \wedge dy ) = 1 .
But that means that 
: {\star} {\star} \, 1 =  1 .
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)
replies: text: The section on duality states that, and I quote: for \eta\in {\textstyle\bigwedge}^k V in an -dimensional space , one has :{\star} {\star} \eta = (-1)^{k(n-k)} s \eta , where  is the parity of the signature of the inner product on , that is, the sign of the determinant of the matrix of the inner product with respect to any basis.  So for n=2 and K=0 or 2 and s=1, the article is correct. If you are doing Minkowski-anything or string theory, then s=-1 and that would explain the pseudoscalar comment which I don't otherwise understand. 67.198.37.16 (talk) 17:08, 29 October 2020 (UTC)
thread_title:  Dual of the dual in 2 dimensions should be negative 

message: 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)
replies: text: Ugh. See, for example, phase space quantum mechanics which uses \star for the Moyal product. In order to do phase space QM with differential forms, one needs the Hodge star as a plain-old asterisk. I'm sorely tempted to bulk-change the star back to an asterisk in this article.  If you are reading this, please express an opinion. 67.198.37.16 (talk) 04:41, 3 November 2020 (UTC)
thread_title:  Five-pointed-star notation !? 

message: In the section Codifferential, this passage appears:
"The most important application of the Hodge star on manifolds is to define the codifferential  \delta  on k-forms.  Let
:\delta = (-1)^{n(k-1) + 1} s\ {\star} d {\star} = (-1)^{k}\, {\star}^{-1} d {\star} 
where d is the exterior derivative or differential, and s = 1 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)
thread_title:  Why the parameter "s" ???