Talk:Exterior algebra

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A request for clarification

In the early part of this article, with the example headed: Cross and Triple products, at the point where there occurs the text

"Bringing in a third vector...."

when looking at the form of the product of three vectors

many readers will surely be very tempted to draw the conclusion that a wedge product of three vectors is equated with the result of performing the binary wedge operation twice in succession. In other words, that uvw may be identified with (uv) ∧ w and in particular that

But from what I understand looking down further to the remainder of the article, I suspect such a supposition would be incorrect. Might a sentence or two be inserted at this early point, within the example, to clarify this situation one way of the other?

This relates to a more general point that it might perhaps be usefully clarified very briefly, even within these early examples, what kind of object the result of a wedge product is. For example, that the wedge product of two 3-vectors is not itself (as most readers will already know that a cross product, by contrast, is) another 3-vector.

Thanks. — Preceding unsigned comment added by (talkcontribs)

In answer to your question, the wedge product is associative, so uvw = (uv) ∧ w. But I think for the purposes of this elementary discussion it is more helpful to think of the wedge product of three vectors on its own, rather than as an iterated wedge product. For your second question, when you multiply elements of different degrees, the degrees add like polynomials. This is already mentioned in the lead. Sławomir Biały (talk) 13:18, 8 May 2012 (UTC)
Sławomir, thank you very much for your fast response. I would like to clarify briefly what is - and was when I asked the question - in my mind. Firstly, to me it does not follow from associativity that uvw = (uv) ∧ w. I understand associativity as a property of binary operations (in the sense of mappings from AxA to A, for some set A). Associativity says that, for such an operation, (uv) ∧ w = u ∧ (vw), for all triples (u,v,w) in A3. But my question was rather: Would I be correct, in this case, to identify the presented ternary operation, written uvw, and with the value as supplied in the text, as being equivalent to an iterated binary operation at all; or, instead in this case, is the ternary version of the ∧ operation rather to be interpreted as something separately defined in its own right, and not constructible out of the building block of the binary version? I did not find the answer obvious from the existing text of this example - And I in fact guessed (wrongly it now seems, for your reply - although I am still a bit unclear) that the answer was the latter: i.e. I guessed that uvw was not to be interpreted as an entity that I should assume could be built as a combination of binary wedge applications. (Of course, I understand that any associative binary operation in a simple way unambiguously defines a meaning for arbitrarily long finite expressions: a ternary, 4-ary, etc extension of itself. But this knowledge did not answer my question, since I was not sure whether for example e1 ∧ e2 even belonged to a set of objects to which the binary wedge operation could be applied.)
I think this illustrates a common difficulty of explaining mathematics to someone who is seeing it for the first time. Typically the reader (I in this case) will be asking themselves questions that are difficult for the writer to anticipate. (So for this reason, the reader will not always be finding the discussion as elementary as was intended.)
An important question is whether many of the readers who are not familiar with wedge product will - on reading this text - be puzzling in a way similar to the way that I was. I am afraid I don't know the answer to that. (I am a reader with probably a few thousand hours of previous exposure to different areas of abstract algebra; and also - some decades ago - a 1st class maths degree.)
Thanks again for your response. — Preceding unsigned comment added by (talkcontribs) 17:51:59,‎ 8 May 2012 (UTC)
Note that in this context, when we refer to a k-vector in this context, the k refers to the grade of the entity, not the dimensionality. Thus 0-vector is a scalar (1-dimensional), a 1-vector is a standard n-dimensional vector (where n is the dimensionality of the geometric space), a 2-vector is what we call a bivector, and so on. I note you referred to a 3-vector, meaning what we'd call a 1-vector here.
The wedge product is a true binary operation, with true associativity in the ordinary sense, and the algebra as a whole is closed under the wedge product. What may be confusing is that the wedge product maps specific grades (distinct subspaces of the the algebra, sharing only the zero element) in a way that is not closed. In particular, it maps p-vector and a q-vector onto a p+q-vector. Thus, a 1-vector and a 1-vector are mapped onto a 2-vector (bivector). A 1-vector and a 2-vector (and also a 2-vector and 1-vector, as per the associativity) are mapped onto a 3-vector.
This all should be reasonably clear from the lead, but it is easy to have some misconception due to quirky terminology, such as your interpretation of a "3-vector". (We'd say the cross product of two 1-vectors in an algebra over a three-dimensional vector space is another 1-vector. The algebra is a larger space than the original vectors space. Try re-reading it with this in mind.) — Quondum 19:39, 8 May 2012 (UTC)

Rank of a ... multivector?

The lead says

The rank of any element of the exterior algebra is defined to be the smallest number of simple elements of which it is a sum.

While I can make sense of this statement, it seems to have no value to me, whereas a similar statement ties up with the more normal and useful concept of rank:

The rank of any k-vector is defined to be the smallest number of simple k-vectors (i.e. k-blades) of which it is a sum.

Considering that I've just removed a whole lot of the curious term k-multivector, I wonder whether this might be part of a similar confusion. Can anyone enlighten me about the intended meaning here (or even just correct the lead)? — Quondum 14:02, 2 September 2012 (UTC)

Yes, this is what was meant. Feel free to implement the change if you think it's clearer. Sławomir Biały (talk) 00:04, 6 September 2012 (UTC)
Will do. What might not have been clear is that "any element of the exterior algebra" can be the sum of nonzero elements of differing grades, whereas "any k-vector" cannot be. This makes the two statements nontrivially different in meaning, and hence it is a matter of correctness, not merely a matter of clarity. — Quondum 06:28, 6 September 2012 (UTC)
Yes, right. It should have read "homogeneous element". Sławomir Biały (talk) 12:40, 6 September 2012 (UTC)

I'm not exactly sure what the issue is with the term "k-multivector". This term cerntainly appears in the literature outside of Wikipedia. I don't object to the change though. Sławomir Biały (talk) 13:06, 6 September 2012 (UTC)

I hadn't encountered it before this article, and the article used both terms apparently without highlighting their equivalence (though I do see that it was defined in the article). I've not encountered it in any other WP article, though I see it occurs in Multivector, which hapazardly switches between the terms without indicating their equivalence, or even defining a k-multivector. If its use is notable, then we should make this clear as an alternative term for k-vector. In some ways, using "k-multivector" (as in "a homogeneous grade-k multivector") is preferable, as it potentially removes the confusion with "k-dimensional vector". The term k-multivector (or its equivalents, p-multivector etc.) seems to occur in only a handful of books, whereas k-vector or p-vector seems to be about three orders of magnitude more prevalent. — Quondum 12:19, 8 September 2012 (UTC)
My thoughts as well, that "k-multivector" is marginally preferrable, because of the potential for confusion over the term "k-vector". But I don't want to insist on it. Sławomir Biały (talk) 12:28, 8 September 2012 (UTC)
It is in the lead where the potential for confusion is the greatest, so it probably makes sense to draw attention to not-to-be-confused-with terms such as 4-vector (I've had to help someone out of this confusion before, I forget where). In association with this, the alternative and unambiguous term k-multivector could be mentioned (in the lead or later), provided that we feel that it is adequately notable. — Quondum 12:57, 8 September 2012 (UTC)
I've dealt with this by adding a footnote in the lead. Since it is only a footnote, I felt the burden of notability to be lessened, and included the mention of a k-multivector.

Excess implied algebraic structure in the lead

I am parking this as a note for a necessary correction to the lead (anyone feel free, I may get around to it in due course). The statement in the lead

The magnitude of u ∧ v can be interpreted as the area of the parallelogram...

suggests that the magnitude of a bivector is defined as part of the definition of an exterior algebra. This implication is incorrect: the definition of an exterior algebra (with the minimal structure that qualifies it as an exterior algebra) is without any concept of magnitude or interior product, though this doesn't prevent such structure from being added. It might even make sense to point out in the lead that no concept of magnitude is needed in the definition. Penrose, for one, stresses this in The Road to Reality. — Quondum 06:53, 6 September 2012 (UTC)

This very issue is discussed in a footnote. The point of saying this at all, though, is so that lay persons get an idea of what the wedge product is. Sławomir Biały (talk) 12:37, 6 September 2012 (UTC)
Good point about giving an idea. It was slack of me not to read the footnote, though it'd be good to make this more blatant. What is always true is that the scalar multiplier of a parallel reference bivector (without using the structure of an interior product) will be the same as for the area of the parallelograms with any interior product whatsoever (except for null vectors, when the areas will be zero). It would be nice to find concise wording that can draw on this proportionality concept without resorting to the structure of an interior product. — Quondum 16:01, 6 September 2012 (UTC)
This probably takes too many words I think to do conpellingly in the lead. The proportionality idea is discussed somewhat in the motivation section. Sławomir Biały (talk) 12:30, 8 September 2012 (UTC)

Diagrams for n-vectors and n-forms (n = 1, 2, 3)

If it's ok I added a diagram to the Motivating examples section, so that readers get the geometric interp first thing. It's disappointing that there seem to be no diagrams for the interpretation of an n-form on WP (unless I have not looked enough). Maschen (talk) 02:02, 8 September 2012 (UTC)

I think that similar diagrams would be valuable at Dual space or Linear functional for geometrically interpreting linear functionals (covectors) on a vector space in terms of the annihilator of the functional and its parallel planes. The planes should be flat, semi-transparent, blatantly deviate from perpendicular, and the plane that the functional maps to 1 should stand out. References for this interpretation (e.g. Penrose's The Road to Reality) should be available. Interested? — Quondum 07:16, 8 September 2012 (UTC)
Definitely. Do you mean to include patterns of stacked surfaces and to run vector/s through them (hence the inner product = number of surfaces intersected)? Thanks, I'll definitely get round to this. Maschen (talk) 07:25, 8 September 2012 (UTC)
Also are you referring to sections 12.3-12.4 in Road to Reality? Before going any furher you might see the recent edit history of linear functional, we could make modifications from there? Maschen (talk) 07:41, 8 September 2012 (UTC)
Yes, and yes. I'm not too sure why 1-form and Linear functional have not been merged; AFAIK there is no distinction (other than the disciplines/contexts in which they are used). The diagram that you temporarily added gives the idea, but has detail that does not apply. It would have to be illustrated in a pure vector space without any inferred differential structure, i.e. with a clear origin from which all the vectors originate, and no "d" notation. And preferably don't use the term inner product in this context; strictly speaking it does not apply (scalar product might be permissible, and Penrose denotes it with a (rather big) dot). Nevertheless, the action of the linear functional on a vector is (in a continuous sense) equal to the number of planes intersected as you say. — Quondum 10:31, 8 September 2012 (UTC)
There is a discussion at Talk:one-form about this. I think the reason for the lack of merger is that a 1-form is usually thought of as a linear functional that varies from point to point. Sławomir Biały (talk) 12:55, 8 September 2012 (UTC)
PS on merging: the terms 2-form and bilinear functional are generally not equivalent, which might be a partial argument for not merging, but I find that unpersuasive. — Quondum 12:29, 8 September 2012 (UTC)


Small gripe is that the article hasn't talked much about forms at the point of the graphic. If possible, I would split the left and right hand sides of the graphic, and include the right-hand side in the section on alternating forms. Also, the vectors do not need to be basis vectors; they can be any vectors and still this interpretation is valid. It would be better to replace the symbol ei with some other symbol, like vi or u,v,w as in the text. And, as Quondum says, the planes in all graphics should be flat, not curved (for this article). Sławomir Biały (talk) 12:57, 8 September 2012 (UTC)

A wealth of valuable responses. The reason for:
  • drawing curved surfaces was that for other curvilinear coordinates they are not planes (e.x. a 3-form in spherical coordinates is a mesh of concentric spherical surfaces, circular planes, and conical surfaces, coord. curves of r, θ, φ, as you both know of course). I presume that my own drawing doesn't emphasize that well enough for a reader to understand though anyway (given that the curves are nearly flat), and yes for this article they should be planes.
  • using basis n-vectors/n-forms was to emphasize the correspondence between the index notation and the geometric interpretation, and since each direction can be scalar multiplied by numbers and added to form a n-vector/n-form.
  • including basis n-vectors and n-forms side by side in the same picture was for immediate comparison.
It's not a problem to split and correct into new images, and will be done. Also I agree with the merge of Linear functional and 1-form. Again thanks for feedback, Maschen (talk) 17:57, 8 September 2012 (UTC)
Linear functionals (1-forms) α, β and their sum σ and vectors u, v, w, in 3d Euclidean space. The number of (1-form) hyperplanes intersected by a vector equals the inner product.[1]
Here is the diagram suggested by Quondum.
Any problems please say.
Unfortunately, although Penrose is an extremely good writer and illustrations are clear, those sections mentioned are a little confusing and I didn't see them as clearer/more direct than MTW depicts n-forms as linear functionals. This diagram is reproduced from MTW, even the letters are identical so that readers of MTW can recognize it and that readers gain a feel for common notation in the image as well as text.
(Right now still in the process of splitting/fixing the other image). Maschen (talk) 18:23, 8 September 2012 (UTC)

About the previous File:Gradient 1-form.svg (now redrawn) -------> (#moved down below)

Agreed with above (no "inner" product, flat level surfaces etc.), except that if Φ (now f, since this is more likely to be used for a function) is a scalar function (0-form) then surely the exterior derivative of it dΦ = α is a 1-form (surface stack) at least locally (Poincare's lemma)? To that end I fixed the image, but still include the statement in the caption.

(Another reason for reverting was my once-confusion between "circular" notation: dΦ for exterior derivative of Φ, and dΦ for gradient (MTW terminology) of Φ in place of ∇Φ, hence my incorrect statement "dΦ = <dΦ, dr> = <∇Φ, dr>", completely wrong... since the diff form dΦ is the total differential (in a more general way, as the directional derivative, not an entry in the scalar product)...). Maschen (talk) 20:44, 8 September 2012 (UTC)

I agree that these diagrams are much better than Penrose's version. I would remove any arrow associated with the functional (1-form); rather use some technique to associate varying values (e.g. annotation and/or a colour variation from one plane to the next). Why three subdiagrams for σ? Only one is needed, and removal of the arrow makes them identical. It'd also be nice to have the spacing for β visibly different from that of α (and, being really picky, changing angles so that none can be interpreted to be perpendicular: arrows to planes, arrows to arrows). The captions need some tweaking, but that can wait. The annotation of f is quite not correct: f is varying from plane to plane according to the caption. Despite my quibbles, this is looking good. — Quondum 22:31, 8 September 2012 (UTC)
File:Gradient 1-form.svg: See what you mean now... How to generally annotate the planes of constant values (not just using "f = 12, 13, 14...")? Using f = c + 1, c + 2, c + 3... where c is any initial constant? Or just use "f = a" for the first and "f = b" for the last plane, where a < b are constants? Or just remove them?
File:1-form linear functional.svg: The reason for the arrow normal to/on the surface stacks is because
  • IMO it is clumsy to have a separate arrow to one side (like MTW actually draw),
  • in this way the geometric significance of normal (to coord plane) is emphasized,
  • if the arrows are removed, how else to more directly indicate the positive sense? As you suggest, I think a colour variation (lightest from first, darkest to last) sounds like the best solution, but then the other split images need updating... also extra words are needed in the captions just to explain that... (the arrow has an automatic implication)
There isn't much of a problem with the arrow to the surfaces, but it is less clutter/redundancy in some important respects. I would think the reader can gain the idea of a 1-form as a linear functional far easier using the current orientations. Changing all the angles would mean changing all the inner product values... which is going to take time to make such changes (but I'll still do it)...
I will prepare the changes but not implement them till further comments are proposed. Maschen (talk) 23:02, 8 September 2012 (UTC)
In order of your bullets:
  • There is no purpose to the arrow aside from indicating which side of the stack is increasing. This could be indicated by a "+" and a "−" on the other side in the absence of annotation; with annotation it is entirely unnecessary.
  • There is no concept of perpendicularity in this diagram; we must de-emphasize any suggestion of it as much as possible.
  • Answered by my first bullet.
The annotation does not need f= at all. We only need -δ, 0, δ, 2δ etc. Including 0 is important. I think ot would be better to use real values (no delta): -0.5, 0, 1, 2.5 or suchlike. The reast is easily handled in the caption. — Quondum 23:52, 8 September 2012 (UTC)
PS: There is not much gained by changing the angles – give that a miss. — Quondum 23:54, 8 September 2012 (UTC)
Ok that makes much more sense (thinking too much about "tangent and normal...").
Although again (and I'm sorry), the +/- is exactly equivalent and actually more ambiguous than an arrowhead, which again has the automatic implication for directed increase (to a reader: "+/-" what? Explain by "The sense of the 1-form is + to -"?). Also in books like Penrose's, covectors/1-forms are actually drawn as full arrows identically to vectors. No, nothing is more simple or more intuitive than the arrowhead: we need a diagram that averages between books like MTW and Penrose. Also, it is graphically far quicker to draw an arrow than anything else (typing +/-), and colour gradients can take time... Of course, arrows are definitely redundant using numerical annotation, or using a colour gradient.
Sorry to continue arguing like this... I'd be happy to change to numerical annotation/a colour gradient while waiting a day or so for more comments before reloading them anyway (admittedly a bit self-centred on my own preferences...) Thanks, Maschen (talk) 00:36, 9 September 2012 (UTC)
My (rather emphatic) objection to the embedded arrows, notwithstanding Penrose's use thereof (note: putting it to the side as you say MTW does turns it into an annotation, which would be equivalent to my + and −; this is an important difference between Penrose and MTW), is precisely that it would immediately be assumed to be a vector, or to have some true content related to a specific vector, which it doesn't (the only content of such a vector would be to indicate which plane maps to 1). We should not "average" the two diagrams, we should select the most correct features. IMO, we should ignore Penrose, since his diagram is clearly deficient. Think about it this way: a covector is nothing other than a map from vectors to scalars. The depiction we are making is a contour map of the vector tips that map to each scalar value. This "mapping of vector tips" rather than "number of surfaces intersected" may be far easier to put across. — Quondum 07:47, 9 September 2012 (UTC)

Template:Outdent Ok, agreed about Penrose (but please still keep the citation in the captions, for reader's interest). As an alternative or addition to the colour gradient, we could put a small arrow over/under/to one side of the symbol for the 1-form, not the surface stack itself, but the symbol, in the sense of increase, something like this: or ? When applying the colour gradient to the alternating multiforms, the colour gradients can merge in places making it less clear which is the increasing sense (still visible). Just a small arrow is needed. This is a (far more) compact equivalent to MTW, so their originality is preserved. I wish I thought of this before... I'm not still "fighting for the arrow", just a possible extra fine little which may make all the difference? Thanks, Maschen (talk) 09:33, 9 September 2012 (UTC)

The colour gradient might introduce more problems than it's worth, as you point out. What the only thing that is necessary is a way of indicating the plane that maps to zero (which might be obvious if the origin is clear), and a sense of the scale factor as you move between planes, which could be indicated by annotation of at least one other plane with a scalar value. Where the scale factor is not of interest but the sign is, the small arrow may work (and I agree that it is preferable to + and −); it could also be included in diagrams with annotated plane(s) if desired for uniformity. I think that it's essential to get the concept of a covector (1-form) as simple and clear as possible.
As one goes to the higher-degree forms, the diagrams become inaccurate, a sort of short-hand for "the wedge product of these 1-forms", and so the finer detail becomes unimportant; the point then is largely to keep an echo of the diagram of the 1-form for each of the factors. — Quondum 12:08, 9 September 2012 (UTC)
Ok - happy that we are now settled on simply keeping it as simple as MTW. No faff; just surfaces with either:
  • numerical annotations (for scale and illustrate constant slices, I plan to use −1, −0.5, 0, +0.5, +1, which is a very simple sequence and has positive/negative fractions/integers included so readers are not restricted to think they are only nat numbers) exactly as you suggest, or
  • small arrows as I suggest,
that's it. 0_^ Just a couple of questions:
  • Are the senses of σ exactly as originally depicted in File:1-form linear functional.svg:? If so how to include? I think just annotations for that one would be better, no arrows.
  • It worries me that you say the multiforms are inaccurate depictions when Sławomir said the interpretation is valid (after his suggested corrections). Why the inaccuracy?
Thank you! Maschen (talk) 13:29, 9 September 2012 (UTC)



Geometric interpretation of a 1-form α as a stack of hyperplanes of constant value, each corresponding to those vectors that α maps to a given scalar value shown next to it along with the "sense" of increase. The zero plane (purple) is through the origin.

Combining the ideas into one would lead to this -------->

Maschen (talk) 13:49, 9 September 2012 (UTC)
This is a very nice picture. Do you think anyone might read anything into them being depicted horizontally? I've edited the caption; see what you think.
As you did in your other diagram, it may make sense to include say two arbitrary vectors from an origin on the "0"-surface, touching two surfaces, so that the caption can deal with them as example mappings.
Perhaps I should have said "non-isomorphic", not "inaccurate". The diagram of the 1-form is complete in the sense that it contains all information necessary to calculate the 1-form (assuming you know how the space is calibrated in terms of vectors), and vice versa, with no excess information. The diagram of a higher-grade form contains excess information: it depicts three 1-forms and their information. You can construct a 3-form from three 1-forms, but you cannot determine the original 1-forms from a 3-form. So that is far tricker to make an intuitive picture, and for now I cannot think of a better way than what you've done. So in summary: vectors, and 1-forms we can depict geometrically very well ("faithfully"); the rest are not "faithful" depictions but are as good as anything I can think of. — Quondum 16:13, 9 September 2012 (UTC)
Thanks for your clarifying response, the isomorphic concept is a good point, but those split images are to show the elements of the graded algebra in steps (what would really be fascinating is n ≠ integer - fractal !? Ok off a tangent for this thread but still). You mentioned to include vectors though the surfaces, which is the point of File:1-form linear functional.svg: (see above for the redraw). This one here is a stand-alone 1-form diagram.
About the captions, by all means do rewrite them! For this one it is more precise and nice but what is n? The dimension of the vector space? We know but the reader may not. Writing "(n − 1)-surfaces" literally sounds like "the number of surfaces is n − 1", perhaps just write "n-hypersurface"? Maschen (talk) 17:40, 9 September 2012 (UTC)
The "surfaces" in question are actually just (hyper-)planes. The caption should indicate this. In general, any function, not just a linear one, can be visualized in terms of its level surfaces. But, of course, in that case they will not be planes. Sławomir Biały (talk) 18:44, 9 September 2012 (UTC)
Caption tweaked, thanks. (For those that rewrite captions just do it, no need to cross out/underline things as previous versions can always be recovered and it's easier to read). Maschen (talk) 19:18, 9 September 2012 (UTC)
A quibble about "level", which here I know means "of constant value", but might be misinterpreted by most readers to mean "horizontal". I changed the caption accordingly, plus removed excess detail of dimension (hyperplane says it all). — Quondum 21:31, 9 September 2012 (UTC)
Phew... what a long thread I caused. Aside from fixing captions, is everything now fine? Maschen (talk) 22:10, 9 September 2012 (UTC)
It would be nice if it was clear that the 0-plane always contained the origin (via axes?), a fact I'd ordinarily put into the caption, but that threatens to overload it. That the planes are uniformly spaced is also a requirement, but I think that the diagram suggests that adequately. — Quondum 07:27, 10 September 2012 (UTC)
For this one it may be ok, for the others it may clutter. How do you want the axes oriented? I.e. if we use x, y, z then x, y in the horizontal planes and z in vertical direction cutting through the planes? Maschen (talk) 07:35, 10 September 2012 (UTC)
Yes, axes would be unnecessary elsewhere, especially if they have other details such as vectors that will achieve the same as a side effect. They can also be very thin, no markings: just mutually perpendicular crosshairs (obviously mutually perpendicular in 3-space, projected onto the plane, accuracy unimportant). They can be oriented any way (and yes, visibly through planes) as long as they are not parallel or perpendicular to the planes, as that might make an incorrect implicit suggestion. The 1-form representation can be rotated in space if need be. If not rotated, z could come up through the nearer corner of the top plane, and x and y can go through near the left and right near edges of the −0.5 plane, or x down towards us through the near −0.5 corner fractionally to the left and y can go horizontally right. — Quondum 13:48, 10 September 2012 (UTC)
I'll get to it, just busy cleaning up Polynomial greatest common divisor right now. Thank you. Maschen (talk) 13:58, 10 September 2012 (UTC)
Template:Done Maschen (talk) 14:46, 10 September 2012 (UTC)
Very, very nice. — Quondum 14:59, 10 September 2012 (UTC)
I absolutely apologize for crossing out "thank you" just then, indeed - thanks ...... *blush* Maschen (talk) 15:19, 10 September 2012 (UTC)

Going back to where we started, shall we add the 1-form diagrams to 1-form/dual vector/linear functional (possibly differential form)? Maschen (talk) 22:15, 10 September 2012 (UTC)

The perfect spot for [1] is Linear functional#Visualizing linear functionals. We haven't clarified the status of 1-form (when does it imply a differential structure?), and at this point I'm hesitant to use these diagrams where a differential structure is implicit. Candidate places for [2] include one-form and dual vector. — Quondum 07:00, 11 September 2012 (UTC)
Ok forget the diff forms article, that was just a suggestion. We have spent much time on this so I will leave them here for just one more day for others to comment before we show them in those articles you link to (unless you or someone beats me to it)... ^_^ Maschen (talk) 07:37, 11 September 2012 (UTC)
Suppose we've waited long enough; let's add them. Maschen (talk) 23:08, 11 September 2012 (UTC)

New images for scalar multiplication

{{#invoke:see also|seealso}}

Maybe they'll help, especially for clarifying pictorially how scalars can "slide" out of an exterior product. M∧Ŝc2ħεИτlk 12:11, 18 August 2013 (UTC)

I don't think the first illustration is really relevant for this article, and the last illustration (while interesting) might be a distraction, since most of the article is not about the exterior product of forms. One thing I would like to see is an illustration of the wedge product of three vectors that includes the orientation. The orientation on a solid ball can be thought of as giving the ball a counterclockwise spin around any given axis. I think something on the cube should be drawable. Sławomir Biały (talk) 13:25, 18 August 2013 (UTC)
You seem to be trying to illustrate multilinearity in the exterior product. This is not too obvious initially from the illustrations (aside from the unfortunate coincidence of multipliers chosen). I see little value in illustrating this; an algebraic treatment of multilinearity is more direct and less demanding on the reader. If this was what you were trying to illustrate, I don't think that they'll help. — Quondum 18:13, 18 August 2013 (UTC)
I'm just showing what scalar multiplication is like in this context nothing more. Let's forget the images if they are not helpful, but I'm not sure how "obviously" most readers will interpret the scalar multiplication of forms (vectors are easy, IMO not so much forms).
Agreed the first is not relevant, and the third overkill, but the third was for comparison with vectors.
About the exterior product of three vectors, there already is the interpretation in the first diagram in the article lead, but agreed that the shape need not be a parallelepiped, and could be anything. A diagram for a 3-vector as different shapes is in preparation and will be posted shortly hopefully. (Would a parallelepiped, sphere, and an irregular splodge, all with orientations, be ok?) M∧Ŝc2ħεИτlk 19:34, 18 August 2013 (UTC)
The exterior product of three vectors u, v, w has the geometric interpretation as an oriented volume. Three possible shapes are shown: a parallelepiped, a sphere, and an irregular lemon shape. The actual shape is irrelevant to the exterior product.
Note: I changed the above images to links to save space.
For now here is a PNG, since SVG and PDF didn't work for some reason (the shading shouldn't be a problem...). An SVG/PDF version can be produced later, but for now... is this what Sławomir Biały meant? By all means criticize. M∧Ŝc2ħεИτlk 19:55, 18 August 2013 (UTC)
The axis should be an oriented axis (i.e., an axis with a north pole designated). Sławomir Biały (talk) 20:23, 18 August 2013 (UTC)
Ach, yes missed that (else the orientation is ambiguous), it is fixed now with an arrow at the tip of the axis for the north pole. M∧Ŝc2ħεИτlk 21:32, 18 August 2013 (UTC)
The leftmost (cube) should have the v arrow blued a bit, as though it is being seen through the cube, to give the right visual cues. The axis an equatorial arrow don't make sense to me; I would have thought that the orientation of a bounded volume is best depicted by a local circulating direction on the boundary, as with the cube; I've seen you do this on other irregular shapes. — Quondum 01:11, 19 August 2013‎ (UTC)
The leftmost parallelepiped does have a hint of blue, it's just too pale to see (now fixed). The "global" orientation on the sphere and irregular shape are still the same as for the local circulating arrows you're thinking of. Another version to show this is below. M∧Ŝc2ħεИτlk 09:56, 20 August 2013 (UTC)
The exterior product of three vectors u, v, w has the geometric interpretation as an oriented volume. Three possible shapes are shown: a parallelepiped, a sphere, and an irregular lemon shape. The actual shape is irrelevant to the exterior product.
Yup, better colouring. I'm getting the hang of the "global orientation", as a spin on an axis in 3d as suggested by Sławomir. Every part of an oriented volume is oriented in this sense, so would it not make sense to shrink the axis-and-spin figures to represent a local property? This would also allow a few of of them in a diagram, which makes it clearer that the choice of axis for the spin figure is completely arbitrary. I quite like the idea, since it is independent of any boundary. — Quondum 21:51, 20 August 2013 (UTC)
About localizing the spin... Do you mean to add an axis at each circulation? I thought by convention the orientation was just left or right on the surface, nothing more. If there is any axis it would be the outward normal to the surface, but I don't see why that's necessary. Or have I missed your point? M∧Ŝc2ħεИτlk 08:13, 24 August 2013 (UTC)
Yes, I do mean adding an oriented axis at each local circulation, but more to the point, that this circulation-around-a-vector applies at each point throughout the volume, like a bag of spinning tops (each with only one sharp point). Where a spin axis happens to penetrate a boundary (the tops whose points touch the cloth of the bag), aligned to match an outward surface normal, this picture corresponds to the local circulation on a surface. This picture is thus independent of but consistent with the local-circulation-on-a-boundary picture. No boundary needs to be defined at all, making it a more powerful picture. It works on an unbounded manifold, whereas the surface circulation needs the construct of subdividing an unbounded volume into bounded regions, and a rule for transferring the outward circulation from one artificial region to the opposite circulation on the boundary of a neighbouring region. The spinning-top analogy simply requires continuous transfer of orientation throughout the volume along every path. The surface circulation also does not as obviously work for a disjoint boundary ("outwards" remains unambiguous, but transferring circulation direction from one portion of a boundary to another disconnected portion is less obvious). — Quondum 11:18, 24 August 2013 (UTC)
I can follow most of what you're saying, but still thought the circulations on the surfaces were enough, and still think the axes are superfluous. Anyway, shall get round to modifying this in time. M∧Ŝc2ħεИτlk 07:40, 25 August 2013 (UTC)
On the surfaces, we are in agreement: the axis is superfluous. I was not suggesting a change to the diagrams with the surface circulations (those are already plenty neat and complete), only to the ones already with the "spin around an axis" picture suggested by Sławomir. It is in the alternate picture where spins depicted locally at any point throughout a volume as circulation around a vector where the spinning top picture applies. One could also simply shrink the "spin on an axis" symbol (circulation around a vector) into the volume in the diagram, though I still feel a single such symbol might create the incorrect impression that the axis chosen has some significance. — Quondum 11:51, 25 August 2013 (UTC)
OK. Trying to orientation throughout the volumes one way or another would be useful, but it would become messy... M∧Ŝc2ħεИτlk 11:33, 27 August 2013 (UTC)

Template:Od I don't see how it would become messy. Replacing the large yellow circular arrow and axis vector with say three of the same, only say 10%–15% of the size, distributed through the volume with random orientation and located to avoid collision with existing features (i.e. the vectors being wedged) should do it. Also, the leftmost figure (the parallelopiped) is inconsistent with the others, because it uses the surface circulation picture. — Quondum 12:36, 27 August 2013 (UTC)

Now I see what you mean clearer, thanks for clarifying that. Yes the parallelepiped will be changed in the axis figure. M∧Ŝc2ħεИτlk 12:39, 27 August 2013 (UTC)
Done, better? M∧Ŝc2ħεИτlk 13:03, 27 August 2013 (UTC)
You've captured the idea nicely. A minor suggestion: the direction of spin of the small figures is unclear, but this could be easily fixed by colour cues: make the nearer half of the circulating arrow visibly brighter (less blued). Then you can decide whether retaining the "global" spin figures is useful or not. — Quondum 02:56, 28 August 2013 (UTC)

Determinant module

The treatment of determinant seems rather cursory. I would suggest a section Determinant module treating the top exterior power of a vector space and its relationship to the determinant of a linear map: this might be a good target for a redirect Determinant module, currently requested at WP:RA/M. It might mention that a module is free iff the determinant module is free, see {{#invoke:citation/CS1|citation |CitationClass=book }}. Deltahedron (talk) 19:39, 2 February 2014 (UTC)

More on the relationship with the determinant would certainly be a welcome addition. There is already a little in the section on functoriality, but it's fairly minimal at present. Sławomir Biały (talk) 22:51, 2 February 2014 (UTC)
  1. {{#invoke:citation/CS1|citation |CitationClass=book }}