Talk:Exterior algebra/Archive 1

Archive 1

Archive 2

Archive 3

Wedge product

I might want to revive the "wedge product" page, with an eye to producing something more concrete. Would that be a problem? Gene Ward Smith 01:26, 30 Jan 2004 (UTC)

The way things are now, the main 'titles' are exterior power, tensor product and tensor algebra, and symmetric power, for one good reason. In defining the various products, from a mathematical point of view, one should explain the space in which the product has its range. This is how multilinear algebra is done, in current mathematics. It is indeed not how it is done in applications; there you define some operation, and if it is 'external' there is no immediate answer about where the product actually lies. The wedge of two vectors is just an 'expression'.

Obviously the ideal treatment gives the heuristic basis a good space. also. But as far as I'm concerned splitting up the discussion over many pages is a way of ducking the issue.

Charles Matthews 08:25, 30 Jan 2004 (UTC)

It's not clear to me if this is a yes, a no or a maybe. The point is, should there not be a place where the wedge product is defined in a way which makes sense to nonmathematicians, with an eye towards computation and applications? If so, where would this be? Gene Ward Smith 08:36, 30 Jan 2004 (UTC)

For the record, I more or less agree with Gene Ward Smith. The exterior algebra is fundamental enough to warrant some examples of the wedge product early in the article. Non-mathematicians use differential forms as well, but they aren't necessarily comfortable with terms like "unital" or "associative algebra." Wikimathematicians: please try to say what it is before delving into details. We are all guilty of this to some extent, but I've decided to tag the article. Thanks, 151.204.6.171 00:28, 5 November 2005 (UTC)

The minor (matrix) page would be a good location for that, considering that in algorithmic terms what you do is form a vector of minors.

As was found on the tensor page, there is really too much 'hanging off' multilinear algebra concepts in various directions for there to be any agreement on the correct encyclopedic approach. In this case my own inclination is to centralise material on this page.

Charles Matthews 15:39, 30 Jan 2004 (UTC)

I am a non-mathematician, with essentially no exposure to the theory of algebras. I arrived here looking for an operating definition (a formula for calculating) a Wedge Product. A couple of comments:

1) It's confusing to me that the definition of an operation (wedge product) is wound into an article about a form of algebra. I take it from context that the reason is that the Wedge Product *generates* the exterior algebra. For mathematicians, that seems to be definition enough, but it's not very meaningful to me. We have separate articles for (vector) "dot product", "cross product", etc., so why is it wrong to separate the discussion of a particular operation from the general mathematical landscape in which it is used?

2) I know it's really hard to balance informal exposition/examples with formally correct statements, and I have no problem with needing to read down through the article to find the more accessible expositions. But I was pretty disappointed at what appears to be a circular definition: All of the formulas defining how to calculate a wedge product have wedge products (of basis vectors) on the right hand side of the equals sign.

3) A final small point, maybe also an artifact of confusion over whether this is an article about an algebra or about an algebraic operation: Nowhere does it say that the (Inverted V symbol, whatever it's called) is the symbol for "wedge product". If you're writing an article about an operation, you should explicitly say that.

I haven't edited the article because I'm unqualified to do so. Please take this as constructive, and thanks to all the mathematicians who've worked on this!Ma-Ma-Max Headroom (talk) 23:19, 6 February 2010 (UTC)

Exterior algebra/Exterior power

I think we should rename this page exterior algebra (currently a redirect) instead of exterior power. It seems to make more sense to define the space in which the product lives before defining the product (See Charles' comment above). At any rate, this would be more consistent with the symmetric algebra and symmetric power pages. Normally I would just do this, but as there has been some discussion regarding the name, I thought I would ask first.

The pages currently redirecting here are:

• exterior algebra
• exterior product
• wedge product
• Grassmann algebra

-- Fropuff 00:38, 2004 May 22 (UTC)

One day ... we'll have all this linear algebra stuff sorted out. Yes, exterior algebra is the consistent top-level name. Charles Matthews 08:38, 22 May 2004 (UTC)

Ok, I moved the article over to exterior algebra, and rewrote most of it, starting with the algebra itself but emphasizing the rules governing the wedge product. I also redirected outer product to here. I suppose a section about applications and example calculations would be a good addition. What other applications besides differential n-forms are there? AxelBoldt 18:48, 4 Oct 2004 (UTC)

Plücker coordinates. I see my arguments above; but my reaction to having the algebra introduced, before the graded pieces, isn't so positive. Oh well, perhaps there is no ideal treatment. Charles Matthews 18:56, 4 Oct 2004 (UTC)

logical conjuction

Is this related to Logical conjunction? They use the same symbol

Not so you'd notice. Charles Matthews 20:39, 16 Nov 2004 (UTC)

---

I've seen the "wedge" notation $\wedge ^{k}V$, is it the same as the lambda thing $\Lambda ^{k}V$?. Also, wedge product is supposed to cancel out when k > n (n being the dimension or V), so we wouldn't need to go up to infinity in the direct sum, just up to n, right?.--Xavier 08:32, 2005 Mar 31 (UTC)

Yes, both notations are used. And yes, if V happens to be finite dimensional then there are only finitely many terms in the direct sum. -- Fropuff 15:03, 31 March 2005 (UTC)

Algebraic properties

Overall, this looks like a nice article. But I would like to see a special section devoted to the algebraic properties of the exterior algebra. Notably missing is the interior product of the dual space of V and the exterior algebra over V. Apart from the wedge product (exterior multiplication), this is the second most important algebraic feature of the exterior algebra (consider, for instance, the Lie derivative and its formulation using the deRham operator and the interior product).

Furthermore, I'm really not eager to see this topic exported to the bialgebra stub. Rather it should be listed as an example once bialgebra has matured.151.204.6.171 18:27, 4 November 2005 (UTC)

I've added a section on the interior product, giving three different definitions of it. Please do whatever needs to be done with this material in order to integrate it better into the intended style of the article. 151.204.6.171 19:49, 4 November 2005 (UTC)

Defining properties

Previously this article stated that "alternating" meant that $v\wedge v=0$, and that this entailed the other two properties ($u\wedge v=-v\wedge u$, $v_{1}\wedge v_{2}\wedge \cdots \wedge v_{k}=0$ for linearly dependent $v_{k}$). This is clearly false, as the zero operator ($u\bullet v=0$) is associative, bilinear, and ... er, self-zeroing, I suppose. I've assumed that "alternating" actually means "anticommutative", and have edited accordingly (since the other two properties are easily derivable from this).  –Aponar Kestrel (talk) 05:38, 12 May 2006 (UTC)

Alternating does in fact mean $v\wedge v=0$ whereas antisymmetric means $u\wedge v=-v\wedge u$. Over most fields these are equivalent since one implies the other. But over fields of characteristic 2, alternating is a stronger property than antisymmetry (every alternating form is antisymmetric but not vice-versa). This is why the definition is the way it is. See discussion at bilinear form. -- Fropuff 20:19, 12 May 2006 (UTC)

Wedge product contradiction

Consider a ^ b ^ c. This = a ^ (b ^ c) = a ^ (-c ^ b) = -(a ^ c) ^ b = -(-c ^ a) ^ b = c ^ a ^ b. This also = (a ^ b) ^ c = -c ^ (a ^ b) = -c ^ a ^ b. Therefore c ^ a ^ b = -c ^ a ^ b and so any triple wedge product is zero. But I do not think that this is true! Please point out the error. AeroSpace 15:59, 29 May 2006 (UTC)

I think the wedge product of two vectors is not itself a vector, so the step (a ^ b) ^ c = -c ^ (a ^ b) is incorrect. In fact, (a ^ b) ^ c = c ^ (a ^ b). Someone please correct me if I'm wrong. —Keenan Pepper 19:47, 29 May 2006 (UTC)

Yes that is right, if a is a p form and b is a q form then a ^ b = ((-1)^pq) b ^ a. AeroSpace 06:36, 30 May 2006 (UTC)

Universal property and construction

Under this heading, we find, in passing: ... We thus take the two-sided ideal I in T(V) generated by all elements of the form v⊗v for v in V, ...

Now the character I see for the product in V appears as an "unknown character" empty square when I view the page in UTF-8. Could we please replace this by whatever character is customary for Wikimathematicians to use for the product (presumably the dot product) in a vector space? yoyo 14:16, 21 August 2006 (UTC)

Example: the exterior algebra of Euclidean 3-space

This example seems to jump from the abstract definition of a wedge product to an algorithm for computing it given two vectors. Could someone fill in that leap? —Ben FrantzDale 20:35, 11 September 2006 (UTC)

not quite sure what you mean there, what's the leap? the wedge product distributes over addition so one simply multiplies according to the definition. Mct mht 07:16, 13 September 2006 (UTC)

Parallelogram

The wedge product of two vectors does not represent a parallelogram, but just the whole plane containing them, and the area of that parallelogram. That's because the wedge product of any pair of independent vectors (yielding the same area) is the same algebraic object. So what is said at the beginning is misleading. Ylebru 16:28, 31 December 2006 (UTC)

Exterior to what? Ausdehnungslehre means "extension theory", not "exterior theory"

Translation from German: This talk section is mainly about the terminology used by Grassmann himself in German. Grassmann used two different terms:

• "äußere" (literally "outer", or exterior), for the "produkt" he defined, known today as the exterior product, and
• "Ausdehnung" (extension, expansion, dilation; see German Dictionary), for his algebraic theory. Namely, in the title of his book (see References), he used the word "Ausdehnungslehre", where "Ausdehnung" means extension, and "slehre" means theory.

Paolo.dL

Please consider the idea of changing the article title and/or using a redirection. Grassmann algebra should be called "extended (linear or vector) algebra", or “extensive algebra”, or "extension algebra", rather than "exterior algebra". The authors who called it "exterior algebra" seem to have had little respect for its creator. In 1844, Hermann Grassmann published his work entitled "The Linear Extension Theory, A New Branch of Mathematics". The famous matematician W.K. Clifford published in 1878 his masterpiece entitled "Applications of Grassmann’s extensive algebra". This is probably the most authoritative reference to Grassmann's work (see other references below). Obviously, Grassmann intended to propose an extension, that is a generalization, of the concept of vector (the n-vector), and operations involving those objects. Those objects are not “exterior” (i.e. “external”) to some other objects, because they include the 1-vectors. An algebra defines a collection of objects and all of the operations which can be performed on them. The expression "exterior algebra" is totally misleading. The Grassmann's theory (algebra), in the author's intention and terminology, is not exterior or external to another theory (algebra). It is not a dress changing only exteriorly the look of a previous existing algebra. Moreover, the exterior product is just one of the operations defined in extended algebra, including for instance an interior product as well. So, why “exterior” algebra? Exterior to what? Please notice that I have no objection on the use of the expression “exterior product”. This comment only deals with the use of the expression “exterior algebra”. Paolo.dL

Article change suggestion: For the first rows of the article, I suggest the following: In mathematics, the exterior (or extended) algebra (also known as Grassmann algebra.... Somewhere in the article, I would also explain that the expression "exterior algebra", although used by many people, is terminologically improper (see above). Wikipedia should sometimes give also an educative input and correct misuse of terminology, which is possible even in mathematics. Paolo.dL

Related aside (homogeneous transformation). Another example of widely misused terminology is the expression "homogeneous transformation matrix", widely and rather conventionally used (unfortunately) by many experts in computer graphics, robotics, biomechanics, to indicate 4x4 transformation matrices containing 4-D homogeneous coordinates. These matrices are used to perform non-linear (affine and projective) transformations of vectors in 3-D space. This is neither an homogeneous matrix, nor it performs homogeneous transformations. It performs transformations that are non-homogeneous in 3-D. In 4-D, the transformation becomes linear and homogeneous. But all matrices, even 2x2 or 3x3 matrices, perform linear and homogeneous transformations (see discussion on this topic on the BIOMCH-L mailing list). Paolo.dL

What Grassmann did was to give an intrinsic geometrical notion of n-dimensional extended quantities (i.e., Ausdehnungen), which roughly coincides with the present day basis-independent definition of a vector space. He is also concerned with the general theory of Strecken, or stretches, comprised of different elements of the vector space: thinking of these as homotheties (stretches) along multiple line segments in the space. The Aussere Mulitplikation der Strecken literally translates as the exterior (or outer) product of stretches (cf. with http://de.wikipedia.org/wiki/Graßmann-Algebra) I believe the translation extérieur (or exterior instead of outer) was settled on by people like Gaston Darboux and Elie Cartan of the fin de siècle French geometry school. (Produit externe, it seems likely, already had other connotations at the time.)

So, there is absolutely no historical revisionism in calling these exterior algebras. This is totally consistent with Grassmann's own writings. More importantly, they weren't truly revived until 50 years later by the French geometers who quite clearly labelled them "exterior".

Unrelated aside (homogeneous transformation). I agree with you about the improper term homogeneous transformation for a projective general linear transformation. The classical term for such things is homographic transformation, and I fully support a revival of this more precise terminology. Silly rabbit 16:46, 6 May 2007 (UTC)

Clarification 1. I have no objection on the use of the expression “exterior product”. I repeat that my comment only deals with the use of the expression “exterior algebra”. In short, I agree with Silly Rabbit that "Aussere Multiplikation" means "exterior multiplication", however "Ausdehnungslehre" means "extended theory" (see above Translation from German).

References. Nothwithstanding the undeniable authority of those (Gaston Darboux, Elie Cartan and others) who proposed and used it, the expression "exterior algebra" is both terminologically and historically questionable. Let me add five references about Grassmann's theory:

• Lloyd C Kannenberg, 1862 "A New Branch of Mathematics: The Extension Theory". The first english translation (second edition)
• Whitehead, 1898. "A Treatise on Universal Algebra". Probably the best and most complete exposition on the Ausdehnungslehre in English ("universal" is more similar in meaning to extended or extensive than to exterior)
• Forder, 1941. "The Calculus of Extension". The University Press. Another complete exposition on the Ausdehnungslehre
• W.K. Clifford, 1878. "Applications of Grassmann’s Extensive Algebra". This is by far the most authoritative reference to Grassmann's work. And it is a masterpiece of a prestigious matematician (see Clifford algebra).
• J.M. Browne, 2007, "Grassmann algebra - Exploring Applications of Extended Vector Algebra with Mathematica". Published on line - click here. This is the most recent contribution, and it includes an interesting hystorical review.

Paolo.dL

I see what you're saying (especially with the Whitehead reference) that Grassmann's work on the subject of algebras represented much more than merely the Grassmann algebras (as mathematicians know and love today). Grassmann's unique approach to the subject was the first approach to algebraic structures from a "modern" symbolic point of view. This probably deserves some mention, but I think it goes too far to say that Grassmann's algebras are eo ipso any different than the article says they are. I'm surprised that you didn't bring up Peano's 1888 book Calcolo geometrico secondo l'Ausdehnungslehre di H. Grassmann preceduto dalle operazioni della logica deduttiva which (prior to Whitehead) situates Grassmann's work into the modern axiomatic tradition (or, indeed, starts the modern axiomatic tradition). Silly rabbit 18:44, 6 May 2007 (UTC)

I've added a History section (at the bottom of the page, for now). You seem qualified to give it a look and make additions/subtractions/suggestions. I'm still keeping the alternative term extended algebra (and extensive algebra) as a footnote. This isn't in common parlance today (and has probably never been so, even in the 19th century outside of a very small group of mathematicians). Moreover, for reasons I've already indicated, it is still not clear to me that this is a more accurate terminology than exterior. Still, you raise interesting issues about the philosophical threads originating with Grassmann and culminating with Whitehead, that deserve some treatment. Silly rabbit 19:34, 6 May 2007 (UTC)

Thanks to Silly Rabbit for adding the History section. However, there's still something in which we don't completely agree. Silly Rabbit wrote (first reply): "...calling these exterior algebras [...] is totally consistent with Grassmann's own writings." I showed that the contrary is (in my opinion undeniably) true, but Silly Rabbit seems still not convinced. He refers to my Whitehead reference, but my main point has got little to do with it.

Main point. My main point is about the terminology used by Grassmann himself in German, and about its evident meaning and incontestable translation (see above Translation from German). Let's have some respect for the author. Moreover, as I showed in my original comment, the modern incorrect translation of Grassmann's terminology makes no sense. Paolo.dL

Google's contribution. Moreover, it is not even true that the correct translation ("extended algebra" or "extension algebra") is not used "in common parlance today", as Silly Rabbit maintains. Try and search for "extended algebra" on Google; the number of results is 34.000, about 1/4 of the results of "exterior algebra". And there is even a book - the title of which contains that expression - written in 2001-2007 (see bibliographic list above). Thus:

• There's nothing wrong in saying that Grassmann's original theory is also called "extended algebra"
• If you like, there's even a mission that an encyclopedia should have: try and give, when possible, an educative input about misuse of terminology, which is possible even in mathematics.

But if you don't agree on the second point, the first is enough for me. If not educative, it cannot be denied that Wikipedia should be at least complete. Paolo.dL

Except that a Grassmann algebra isn't called an extended algebra (with the possible exception of maybe three or four people). I googled extended algebra too, and none of the top page hits were relevant: an extended algebra appears to have other meanings in the theory of quantum groups and Hopf algebras. (So that's actually an argument against putting it here.) I'm not even convinced Grassmann himself called it an extended algebra (in the sense of a count noun). Ausdehnungzahlen? Also your link to the mathematica book seems to refer to a class of algebras including the exterior algebras and things like Clifford algebras. Anyway, the fact that we're even having this discussion suggests strongly against including the term extended algebra: Wikipedia has a neutral point of view policy (See Wikipedia:NPOV). Silly rabbit 10:30, 7 May 2007 (UTC)

Among the "three or four people" that you mention, there are five mathematicians who wrote widely quoted books or papers about Grassmann (see references), including a paper by W.K. Clifford. Does neutral point of view mean cancelling history? In my opinion this discussion means that Wikipedia is not yet neutral (or complete) w.r. to this topic. Is it neutral policy to deny the existence of the above-listed five references?

An Aristotelic syllogism - Silly Rabbit wrote (in the article's section about Hystory): "Ausdehnungslehre, or extension theory, which referred more generally to an algebraic (or axiomatic) theory of extended quantities. Let me use a syllogism, based on Silly Rabbit's text:

• Thesis - If Grassmann's theory is algebraic, and
• Antithesis - if Grassmann calls it theory of extension,
• Synthesis - we can conclude that Grassmann theory is the "algebraic theory of extension" (in short, "extension algebra").

I explained before the reason why I believe this terminology is the only one that makes sense.
However, of course I agree that the most commonly used expression nowadays is "exterior algebra". Paolo.dL 11:53, 7 May 2007 (UTC)

You still can't provide a single reference where the author unambiguously defines an extended algebra to be an alternating algebra generated in degree 1 by a vector space. Not one! Grassmann doesn't do it, Clifford doesn't do it, Whitehead doesn't do it. Not even your favorite mathematica book does it. And don't straw-man me. I read the Clifford paper, and it's a bloody 8 page review article which also doesn't do it. Go find a real reference, Page number, Line number. Then we'll talk. Until then, it's you whose trying to rewrite history. Not me. Silly rabbit 11:55, 7 May 2007 (UTC)

Do we agree on the definition of "algebra"? Is it different from an algebraic theory, in your opinion (see my syllogism above)? We need at least a third opinion, don't you agree? So, I ask everybody: Shouldn't there be a reference to the expression "external algebra" in Wikipedia, associated with Grassmann's algebraic theory? Paolo.dL

Meaning of the word "algebra". I hope you don't mind me interrupting. The word "algebra" has two meanings in English. It can mean both a field of study, as in algebra, and a mathematical structure consisting of a set with certain operations (addition, scalar multiplication, multiplication), as in algebra over a field. Silly rabbit's reference to a countable noun means that he has the second meaning in mind, and that is the meaning being used in the first sentence of this article. In contrast, the title "Applications of Grassmann's extensive algebra" uses the first meaning (if it used the second meaning, it would have been "Applications of Grassmann's extensive algebras"). The question is: do you have any evidence of somebody calling Λ(V) an or the "extended algebra"?

-- Jitse Niesen (talk) 12:34, 7 May 2007 (UTC)

I greatly appreciate Jitse Niesen's clarifying contribution. No, I have no evidence about somebody calling Λ(V) an "extended algebra". Thus, I am not supporting anymore the suggestion I gave above regarding changes on the first rows of the article. This solves some misuderstandings in the previous discussion, but leaves two points still valid:

• P1) Re: "algebra over a field". The algebra Λ(V) should be called "extended" rather than "exterior" (even though most likely it has never been called that way) because it includes V as a subspace. Therefore, it is neither exterior (=external, outer) to V nor to anything else. You should agree that a set cannot include another and be exterior to it at the same time. The two concepts are just incompatible, aren't they? (Grassmann, if he were alive, would most likely call it that way). About that, I would like Silly Rabbit and Jitsie Niesen to agree with me (this doesn't imply an agreement about the need and content of possible article changes). I know that it is not pleasant to admit that everybody (including our masters) are using an improper terminology, but please indulge me. You can still safely maintain that we had better not even to think about changing a deeply rooted conventional terminology, even when it is improper.
• P2) Re: "algebra". We all agree that the expression "extended (or extensive) algebra" has been and is being used with reference to Grassmann's algebraic theory (and not just marginally in some web pages, but in book and scientific paper titles). But where's the Wikipedia article (or redirection, or article section) about it? This may require a revision of this article, perhaps more extensive than that I suggested above.

Paolo.dL

Re: P1) I always thought that it's called "exterior algebra" because it is an algebra with multiplication given by the exterior product, just like a "concatenation algebra" is an algebra where the multiplication is concatenation of words. I'd agree that Λ(V) is not exterior to V or anything else.

Re: P2) I wouldn't have a problem with having "extensive algebra" redirect to this article if this article is a fair reflection of Grassmann's Ausdehnungslehre (I don't know the full extent of the Ausdehnungslehre).

-- Jitse Niesen (talk) 01:59, 8 May 2007 (UTC)

Re: P1) I see Jitse Niesen's point and I thank him for the note. But "concatenation" is a name, whilst "exterior" is an adjective, and when written before "algebra" it qualifies the algebra, not the product, unfortunately. It means that the algebra is exterior to something else. If it were called "exterior product algebra", the name would be partial, but it would be understandable and wouldn't create a contradiction with its meaning (esterior to V but including V... see above). On purpose, I inserted above the "related aside" about the improperly-so-called "homogeneous transformation matrices". Notice that "Homogeneous" is an adjective. Of course I know these are just exceptions: I like mathematics because of its intrinsic coherence, which in most cases is reflected in its terminology.

Article change suggestion 2. What if we add a redirection from "extended algebra" to "exterior algebra" and, at the end of the introduction, a simple phrase of this kind: The expression extended algebra would be linguistically more suitable to indicate Λ(V) than the widely accepted conventional name "exterior algebra", because Λ(V) includes V and is therefore not exterior to it. It would be also more coherent with Grassmann's terminology (he called his algebraic theory "Lineal Ausdehnungslehre", which literally means "Linear extension theory") and with that used in the first historical references in English translating and discussing Grassmann's work (see section on History below). However, the expression "extended algebra" has been and is being currently used only to indicate Grassmann's original theory (see section on History below), and not its modern implementation in abstract algebra, which is discussed in this article. I am not an expert in Grassmann's theory, I can't do much more than that. But I think this note would be useful to the world and would show due respect for Grassmann, who didn't receive much recognition when he was alive, but is probably the grandfather of modern algebra. Paolo.dL

To add something like this to the article would be expressing a clear preference, or point of view, in the matter of nomenclature. It neither is, nor should it be, the stated mission of Wikipedia to override the opinions of Wise Men (and Women) in the field of algebra and geometry who have long ago settled on the use of the term exterior algebra. As for the etymological significance of the term exterior for the product defined on this algebra, who knows. I suspect that Whitehead may provide the most illuminating clues. He views the Calculus of Extension as a system of what he terms manifolds consisting of reference elements in a space, along with a law of composition or aggregation of reference elements contained in manifolds of orders a and b to the manifold of order a + b. As far as I can tell, these manifolds are completely separate entities from one another, and it is only through the law of aggregation that one can go from one manifold to another (hence exterior). Furthermore, there are also clues to indicate that the algebra Λ(V) is a logically different beast altogether than the extensive calculus:

(4) This symbolism [of aggregation] and its interpretations can have no applications unless a subregion is more than a mere aggregate of its contained elements. It is essentially assumed that a subregion can be treated as a whole and that it possesses certain properties which are symbolized by relations between the elements of the derived manifold of the appropriate order. Thus a subregion of the manifold of the first order, conceived as an element of a positional manifold of a higher order, is the seat of an intensity and the term which symbolizes it always symbolizes it as a definite intensity.

(5) A positional manifold whose subregions possess this property will be called an extensive manifold.

(Whitehead 1898, III.I.(4)-(5) p. 177.)

One may not, as seems to be indicated, add elements of different derived manifolds because such an operation would not symbolize an intensity as conceived of by Whitehead. Hence, strictly speaking, the exterior algebra has more structure to it than Whitehead's conceptualization of Grassmann's calculus. Whether or not this was Grassmann's idea is anybody's guess I suppose, but this is a strong indication that it was. Silly rabbit 12:09, 8 May 2007 (UTC)

Thank you very much for this informative contribution.

• 1) I have read again the Neutral point of view article, and I now agree with Silly Rabbit that my preposition expresses preference and therefore a non-neutral point of view. Thus, it violates Wikipedia's policy.
• 1b) Hence, I (or someone else) should find a way to inform Wikipedia's readers about the bare fact: the existance of the expression "external algebra" and its use related to Grassmann's theory. Jitse Niesen, would you mind to give a try? You seem to be the most neutral, in this discussion.
• 1c) I also have an additional concern, possibly less legitimate than 1b. This phrase possibly violates Wikipedia's policy: "Λ(V) includes V and is therefore not exterior to it". But it is also perfectly correct! Really, I can't imagine how somebody could deny such an elementary deduction, which is, however, not published anywhere (except for here, in this talk). Therefore, it is not a fact. I.e., we cannot write: "although Wise Men of the fin de siècle French geometry school decided to call it "exterior algebra", Albert Einstein maintained that...". I am asking advice: wouldn't it be possible to somehow justify the presence of the above-mentioned phrase in a Wikipedia article?
• 2) Silly Rabbit writes about the use of the term "exterior" to indicate the product defined by Grassmann. I am surprised. That's something on which we agreed from the beginning, and there's no need to quote Whitehead about this. Grassmann himself called his product Aussere Mulitplikation, which means exterior multiplication, and this discussion is about the name of the algebra, not the name of the product. Notice that, according to "my own logic" (let's call it that way) the product is indeed exterior to its domain (i.e. the result of the exterior multiplication is outside the subset of its operands), and exterior is therefore an adjective coherent with the definition of the product.
• 3) I am a biomechanist, not a mathematician. Although I appreciate Silly Rabbit's expertise and highly value his effort to contribute to this discussion, I do not know what the "law of aggregation", the "intensity" and the "seat of an intensity" are. Is there somebody (perhaps Silly Rabbit himself) who would like to translate or try an easier approach to support Silly Rabbit's conclusion?
• 4) Do not recur to Wise Men, please. There are many examples of wise men who found useful algorithms, wrote successfull formulas, but used terribly inadequate terminology. One example: D'Alembert and his imaginary, fictitious, appearant "inertial force" (e.g., the "centrifugal force"). Newton, if he were alive, would explain that this terminology is against nothing less than the logical foundations of mechanics. He spent much of his lifetime trying to convince others that there does not exist a centrifugal force, to stress the distinction between inertia and force (Newton's first law), and to fight against the "appearance" (inertial forces are appearant). Newton's outstanding theory stemmed from the study of the elliptical motion of planets (inspired and initiated by Kepler). Newton realized that they were acted upon by a centripetal force, while the absence of this force, and not a centrifugal force, produced rectilinear uniform motion. D'Alembert naively ignored Newton's effort, and brought back prejudice in science, by just using terminology incompatible with Newton's insightful teaching, unfortunately associated with a perfectly legitimate and useful rearrangement of Newton-Eulero's equations of motion. However, scientists and engineers have been using extensively D'Alembert's naive terminology... There have been dramatic examples in history about Wise Men being just men and being wrong, with the world blindly believing in them. And not only in the field of terminology! Before Galileo and Newton there was Aristoteles. He was for several centuries acclaimed as "The Wisest Man", but unfortunately everybody trusted him too much, and whenever somebody tried to say that the Earth was not the heart of the universe, somebody else could answer: "you are crazy, don't you know about Aristoteles and all of these Wise Men who for centuries embraced geocentrism? How dare you say they were wrong? Do you think you are smarter than all of them?". Now we know that was the case, indeed. But before our "awakening", this even became a religious issue, with the Catholic Church contrasting Galileo's teaching with forced exile.

Paolo.dL

Re: point 4. I think perhaps in the heat of debate, you missed my point entirely. You're entitled to your opinion. But insofar as that opinion differs from views widely held by experts in an area you yourself admit that you are unacquainted with, it is best to keep it out of an encyclopedia article. You're free, of course, to convince others that you are right. To publish papers. To argue with your colleagues. But Wikipedia isn't the place for a paradigm-shift. Silly rabbit 11:49, 9 May 2007 (UTC)

Clarification 2. Let me clarify. I believe Silly Rabbit missed that I am not anymore proposing a "paradigm-shift", as he/she calls it above (see my point 1). I am mainly proposing to inform Wikipedia readers about facts (see my point 1b). A great part of what I wrote above is a bare objective fact, and not just a logic deduction (see 1b and 1c). These facts we can and should say in the article. About the "views" widely held by experts, I am not sure they are "views". Sometimes experts just use terminology they find in the literature, not because they like it, but for reasons of convenience (see my comments on D'Alembert's principle in point 4). Also, I couldn't find anything directly against my simple logic, in what you wrote (see my point 1c). And simple logic is something I am entitled to use, even though I am not a mathematician, isn't it? Doesn't that deserve a direct answer? Paolo.dL

I've already given you my direct answer. If you can find a single source which says something to the effect of:

An extended algebra is a collection of objects equipped with a pair of operations + and ∧, satisfying... (etc.)

then we can talk. So far, I don't see that in any of the references you have provided. Perhaps the closest is the Whitehead book. But as I've already indicated, an exterior algebra (as conceived in this article) is a different logical beast altogether. Whitehead, and Grassmann it would seem, describe a sort of calculus in the sense of propositional calculus (which, indeed, seems to be Whitehead's chief intent in the aforementioned book), not what we would call an algebra in modern terms. In mathematics, there are also things called Boolean algebras (and Heyting algebras) which model the propositional calculus. Would you demand that each of these be renamed to propositional algebras? This would in fact be an easier case to make, I suspect, since Boolean algebras are semantically equivalent to the first order propositional calculus. Exterior algebras contain more structure than Grassmann's extensive calculus, though. Silly rabbit 12:40, 9 May 2007 (UTC)

Clarification 3. I am not maintaining and never maintained that "extended algebra" is or was used to indicate an "algebra over a field" (see above: Niesen's contribution "meaning of the word algebra" and my answer). I explained above that I am not suggesting anymore to rename anything (see points 1, 1b, 1c, and how much they differ from my tentative suggestion n.2). Paolo.dL

Can you (a) sign properly, and (b) not personalise your remarks in that way? The page title is sound. Any historical remarks can be added in the section on history, Or under the Grassmann article. Charles Matthews 08:27, 10 May 2007 (UTC)

Completing History section. Anybody knows who where the authors who first introduced the expression "exterior algebra"? As I showed, Grassmann's theory was referred to by Grassmann himself and other authors as Extension Theory or Extensive algebra... Who proposed a different name for the "modern version" of that algebra? (we will never know why, I guess). That's an interesting piece of information for the history section. I thought they were Gaston Darboux and Elie Cartan of the fin de siècle French geometry school, but I read again what Silly Rabbit wrote above and I am not so sure anymore. Silly Rabbit, do you happen to know? Paolo.dL 09:53, 11 May 2007 (UTC)

Bourbaki suggests that it was Cartan, but they seem a bit diffident on the matter. Silly rabbit 12:44, 11 May 2007 (UTC)

Request for redirections. Someone added notes into the article, according to this talk (thanks), and I edited some of them. Since this talk started because I was looking for "Extended Vector Algebra" in Wikipedia and couldn't find it, I believe we should also insert a redirection to "Exterior algebra" from (1) "Extended vector algebra", (2) "Extensive algebra", (3) "Extension Theory" (4) "Calculus of Extension" (see notes). However, I can't do it. If you agree, please do it. The readers will find in the notes and History section that these expressions do not refer to the modern algebra over a field concept, as we learned in this talk (see also Re: P2 above). Paolo.dL 13:00, 11 May 2007 (UTC)

Extended vector algebra is a neologism, if you ask me, but I don't see any harm in adding a redirect. Extension theory is dangerously close to Extension (mathematics) which is a disambiguation page. So redirecting here could potentially cause confusion. I can add exterior algebra to its list of (many) disambiguations, if that will satisfy you. Silly rabbit 12:40, 11 May 2007 (UTC)

I agree, Silly Rabbit. Please do it, thanks. What about "Calculus of Extension"? Paolo.dL 12:44, 11 May 2007 (UTC)

I've made the redirects you request. Silly rabbit 12:55, 11 May 2007 (UTC)

Neologisms. Thank you. It is interesting to notice that also "exterior algebra" was a neologism, when it was created, and that the latter was not coherent with previously used terminology and has a literal meaning inconsistent with its modern formal definition, whilst the neologisms "extended vector algebra" (suggested by Browne) and "extensive algebra" (suggested by Clifford) are both definition-consistent and more faithful to Grassmann's terminology. I keep thinking that the success of the expression "exterior algebra" is a (legitimate) matter of convenience, not a matter of wise choice by those who just quoted the first author (Cartan?) who used that expression to indicate modern developements on the subject. This is probably my last contribution to this exciting talk. Thanks to everybody. Paolo.dL 10:58, 14 May 2007 (UTC)

Order of components in exterior product formula

Relevance. I believe it is extremely important to be clear in Wikipedia about this point, because lack of knowledge in this case means high likelyhood of mistakes when data describing k-vectors or multivectors (in the form of their "scalar components") are exchanged between different scientists or processed by different computer programs. Let's start with a simple example:

Example A) This formula is published in the exterior algebra article, and refers to vectors in R³:

$\mathbf {A} =\mathbf {u} \wedge \mathbf {v} =(u_{1}v_{2}-u_{2}v_{1})(\mathbf {i} \wedge \mathbf {j} )+(u_{1}v_{3}-u_{3}v_{1})(\mathbf {i} \wedge \mathbf {k} )+(u_{2}v_{3}-u_{3}v_{2})(\mathbf {j} \wedge \mathbf {k} )$

and since it is also valid as a simple example of geometric product, namely

$\mathbf {u} \mathbf {v} =\mathbf {u} \wedge \mathbf {v}$, if $\mathbf {u} \perp \mathbf {v}$

it is sufficient to point out a more general and complex problem, regarding the convention for representing any kind of multivector constructed over R^N as a vector with 2^N elements.

1) "Internal" ordering. Is it advisable to use $\mathbf {i} \wedge \mathbf {k}$? I have always seen $\mathbf {k} \wedge \mathbf {i}$ in books and other web pages (mainly concerning geometric algebra, which has the same historical roots as exterior algebra). Of course, $\mathbf {k} \wedge \mathbf {i}$ = - $\mathbf {i} \wedge \mathbf {k}$.

2) Scalar component ordering. The three components appear to be ordered according to an unusual criterion. I am not sure about the correct (or conventional, or most frequently used) order. I guess there are other two possibilities:

Example B) This is my best guess about how Grassmann would order it, and it is also the order adopted by several contemporary authors who wrote about geometric algebra:

$\mathbf {A} =\mathbf {u} \wedge \mathbf {v} =(u_{1}v_{2}-u_{2}v_{1})(\mathbf {i} \wedge \mathbf {j} )+(u_{2}v_{3}-u_{3}v_{2})(\mathbf {j} \wedge \mathbf {k} )+(u_{3}v_{1}-u_{1}v_{3})(\mathbf {k} \wedge \mathbf {i} )$

Example C) And this is also a nice criterion, which is also compatible with the numerically equivalent cross product:

$\mathbf {A} =\mathbf {u} \wedge \mathbf {v} =(u_{2}v_{3}-u_{3}v_{2})(\mathbf {j} \wedge \mathbf {k} )+(u_{3}v_{1}-u_{1}v_{3})(\mathbf {k} \wedge \mathbf {i} )+(u_{1}v_{2}-u_{2}v_{1})(\mathbf {i} \wedge \mathbf {j} )$

Example D) Cross product:

$\mathbf {a} =\mathbf {u} \times \mathbf {v} =(u_{2}v_{3}-u_{3}v_{2})\mathbf {i} +(u_{3}v_{1}-u_{1}v_{3})\mathbf {j} +(u_{1}v_{2}-u_{2}v_{1})\mathbf {k}$

Since this point can be generalized to multivectors in N-dimensional spaces, I moved the relevant discussion in Talk:Geometric algebra#Is there a convention about the order of multivector components?, but if you prefer to post your comment here, you are free to do so, and I'll insert there a link to this section. Paolo.dL 17:20, 31 May 2007 (UTC)

Being a novice, I prefer the form 2c) above because it is displayed in the same form as the vector product. Also the ring order is maintained in the i, j & k product's, as it would - had they been considered as the quaternionic elements. Scot.parker 21:26, 8 June 2007 (UTC)

Hopf algebra structure of exterior algebras

I would like to have a comment on the Hopf algebra structure of an exterior algebra. As it is pointed out in the article, an exterior algebra may be equipped also with a coproduct and counit map, and it is stated that together with these, an exterior algebra becomes bialgebra. It is also stated that if one takes the Hodge duality map (defined by a prechosen volume form), it becomes an antipode map of this bialgebra, thus if becomes a Hopf algebra. However, to my understanding, this is not true. The Hodge dual (defined by any prechosen volume form) does not preserve the identity element, by definition (it maps it into the prechosen volume form). However, it is clearly seen from the definition of the coproduct and counit on the exterior algebra (and from the definition of an antipode), that it should preserve the identity element! Do I see something wrong?

A further question. How can be seen that the definition of the coproduct on an exterior algebra is independent of the choice of the generators? In the text it is defined via a prechosen set of generators. To verify that the definition is "good", one has to see, that this definition is independent of the choice of the generators. How can this be seen?

AndrasLaszlo 11:11, 1 September 2007 (UTC)

This same difficulty with the antipode was recently discussed on sci.math, [1], and I believe everyone agreed that Hodge duality could not possibly be the antipode, all for reasons like you state. Can someone check if this is a small mistake, or whether those two sentences should just be struck from the article? JackSchmidt (talk) 14:42, 21 November 2007 (UTC)

Wedges of Vectors

The page has that $\wedge$ is alternating on V. But that doesn't make any sense.

Wedge is defined on $\Lambda (V)\times \Lambda (V)$. My question is: "What to change this to?" Should it be replaced by $v,u\in \Lambda (V),\;\omega ,\eta \in \Lambda (V),v,u{\text{( or }}\omega ,\,\eta )\in V^{*}$, or something else altogether?"

Guardian of Light 05:29, 10 March 2008 (UTC)

V is contained in $\Lambda (V)$. To say that the wedge product is alternating on V means just what the article says: $v\wedge v=0$ for all v ∈ V. Silly rabbit (talk) 05:35, 10 March 2008 (UTC)

I agree with that, but only as V*, which seems and abuse of notation. Thoughts? Guardian of Light (talk) 05:55, 10 March 2008 (UTC)

I think maybe there is something not being communicated here. Forget the dual space. The product $\wedge$ is defined on Λ(V). Since V is contained inside Λ(V), the product can be applied to two elements of V, producing another element of Λ(V). The dual space does not come up. Silly rabbit (talk) 05:58, 10 March 2008 (UTC)

I agree, so far as I know, $V\not \subseteq \Lambda (V)$ Guardian of Light (talk) 07:04, 10 March 2008 (UTC)

No, no, no. As I've said repeatedly, V is contained in its exterior algebra. If you want to split hairs, then there is a canonical linear monomorphism of V onto a subspace of the exterior algebra. But that's what people usually mean when they say "is contained in." Silly rabbit (talk) 12:06, 10 March 2008 (UTC)

E(V) is defined as a quotient of T(V) by I, and T(V) is a direct sum of V and some other things. The intersection of I and the direct summand V is 0, so the subspace (V+I)/I of E(V) corresponding to V in T(V) is naturally isomorphic to V/(V n I) = V by one of those isomorphism theorems. Another way of seeing this, is that elements of E(V) are labeled by sums of tuples of elements of V, and those that are labeled by sums with just one term, and where that one term is just a 1-tuple, form a subspace, and the map v -> (v) is an isomorphism of V into E(V). If you write things with little wedgies, then the map looks even simpler: v -> v, since we want the elements without any wedgies. JackSchmidt (talk) 13:58, 10 March 2008 (UTC)

(Comment. By E(V), JackSchmidt means what I and the article call Λ(V).) Yes, that's the full construction (or one of them) of the exterior algebra. The bottom line is that V ⊂ E(V), and that $w\wedge w=0$ for all w ∈ V. Silly rabbit (talk) 14:15, 10 March 2008 (UTC)

I don't mean to 'split hairs' as it were, and if it comes off that way, I apologize. What I mean is that--so far as I have been taught--$\Lambda (V)=\bigoplus _{i=0}^{\infty }\Lambda ^{i}(V)$ with $\Lambda ^{0}(V)=\mathbb {R}$ and each $\Lambda ^{j}$ is a set of alternating tensors (which are functionals, not vector spaces). When you put it in the language of the monomorphism, that was what I was querying about earlier, since I know V and V* are isomorphic, but you had also told me to forget the dual space, so I assumed you meant I was missing something else. So now I'm just a bit confused as to what you are trying to help me understand. If what you mean is that we do the wedging (technically only) in V* and then transferring back to V, then I understand what you mean by the wedge product on V so to speak. If that's it, I have a further follow up question (relating to the article) and if not, I ask you further explain as I'm not sure I understand what you mean. Guardian of Light (talk) 21:11, 10 March 2008 (UTC)

(unindent) I'm not exploiting any unspecified isomorphism of V* and V. Literally, the dual space is not required for the definition of the alternating algebra of V. The confusion is because you are using a non-standard definition which is dual to the one given in the article. In the terminology of the article (and most standard mathematical treatments — see the references listed at the end of the article), $\Lambda ^{k}(V)$ is not the set of alternating forms of degree k on V, but rather is given as a certain quotient of the tensor algebra of V (see JackSchmidt's remark above). The space of all alternating forms of degree k is canonically identified with the dual vector space:

$\Lambda ^{k}(V)^{*}.$

In finite dimensions, this can be identified with $\Lambda ^{k}(V^{*})$ (see the section on Alternating multilinear forms in the article).

If you are used to seeing this on a manifold, then you may have noticed that some authors denote the space of k-forms by

$\Lambda ^{k}T^{*}M$

This is the exterior algebra of the cotangent bundle. It does not consist of alternating multilinear forms on T*M, but rather on the dual vector bundle TM, the tangent bundle. Silly rabbit (talk) 21:24, 10 March 2008 (UTC)

That I can concur with then. Thank you for explaining! Guardian of Light (talk) 21:54, 10 March 2008 (UTC)

Note on Harvard citation convention

In my latest edit, I applied Harvard citation convention throughout the article. I would like to point out that there's a circumstance in which the name of the author should not be within parentheses. For instance, please consider the difference between these two sentences:

1. Kannenberg (2000) published a translation of Grassmann's work in English.
2. (Kannenberg 2000) published a translation of Grassmann's work in English.

In the first sentence, the subject is Kannenberg (and he published a translation). In the second, in my opinion the subject appears to be Kannenberg's book, rather than Kannenberg. Since a book cannot publish a translation, the second sentence is not correct. In order to use the convention shown in sentence 1, you need to use the template {{harvcoltxt|Kannenberg|2000}} Paolo.dL (talk) 18:03, 5 April 2008 (UTC)

Recent edits

I suggest that User:ProperFraction should float edits here for discussion before implementing them. For instance, I have no idea why the word "Motivating" in "Motivating examples" is confusing. It doesn't seem confusing to me. Perhaps ProperFraction could explain why he/she feels the need to change things in subtle inessential ways? silly rabbit (talk) 10:52, 1 May 2008 (UTC)

What does the term 'motivating' mean? It means nothing to me.--ProperFraction (talk) 23:52, 3 May 2008 (UTC)

• Another example: Why does ProperFraction feel it is better to split the (already quite short) paragraph on the exterior product into two paragraphs? Paragraph breaks aren't just to make text "digestible", they should also be made in logical places. I think having one paragraph devoted to the exterior product is enough. Originally, there was one large paragraph serving as a lay introduction to the exterior product and the exterior algebra. Although I see some logic in splitting this into two, we should still try to keep the number of paragraphs in the lead short. Per WP:LEAD, the recommended maximum number of paragraphs is three, and we are already running afoul of that. Also, moving 90% of the text out of the lead and into a separate Explanation section is unacceptable. As it is, the lead currently summarizes, explains, and sets the context for the article. This is exactly what a lead is supposed to do. silly rabbit (talk) 11:01, 1 May 2008 (UTC)

Im just trying to make this article more accessible to non mathematicians (like me) shorter paragraphs are easier to digest than a large mass of text IMO. In the lede, the paragraphs themselves should be short to avoid the readers eyes glazing over!

Silly rabbit, would you care to explain why a section attempting to explain the subject ( called Explanation for want of a better term) is unacceptable?--ProperFraction (talk) 23:58, 3 May 2008 (UTC)

Couple of points: First, I spent a lot of time getting the lead to conform to the Wikipedia stylistic guidelines. Currently, the article has been submitted for peer review, and I am trying to get it up to GA status. As the article now stands, the lead (1) provides context for the subject of the article, and (2) summarizes the main points of the article. This is what the lead is supposed to do. Moving most of the lead into a separate section is counterproductive, unless you have an alternative lead in mind. It certainly goes against the manual of style.

Secondly, I'm not married to the idea of having three and only three lead paragraphs. Currently there are four (I've let one of your changes stand). However, I don't see why every sentence should have a paragraph break of its own, nor is it clear what this has to do with whether a reader is a mathematician. Any reader willing to tackle the article should also be willing to read paragraphs. These aren't page-long essays or anything.

Third, and finally, the word "Motivating" in the "Motivating examples" section is important since, from the mathematical point of view, the algebra has not actually been defined at this point of the article. It is thus a clue to the reader that these two examples come before the actual definition. This is quite common in many mathematics articles on Wikipedia, and it is done specifically to make the articles more accessible to a lay audience. silly rabbit (talk) 00:50, 4 May 2008 (UTC)

OK. I think its a very unhelpful lede but well see what the reviews have to say.--ProperFraction (talk) 05:28, 4 May 2008 (UTC)

I'm sorry that you feel the lead is unhelpful. I don't really understand how giving each sentence its own paragraph or removing most of the lead would make it more helpful. Perhaps you would like to see the article look more like the Encyclopedia of Mathematics article:

• "Exterior algebra/Archive 1", Encyclopedia of Mathematics, EMS Press, 2001 [1994]

This is a bit more concise, but doesn't give the exterior algebra much context. Would you prefer a lead that doesn't mention the role of the algebra in linear algebra? Perhaps the article should just start with the universal construction, and leave out all the motivation and applications. silly rabbit (talk) 17:53, 4 May 2008 (UTC)

Saint-Venant?

I have removed the following line from the text. This needs additional verification before it can be included here, since "similar ideas" covers a lot of terrain. Joseph-Louis Lagrange was publishing "similar ideas" at least a century beforehand. Anyway, if there was a priority dispute for the invention of the exterior algebra (as the sentence suggests), it should be extensively documented, not just in tertiary source like a dictionary.

Saint-Venant also published similar ideas of exterior calculus for which he claimed priority over Grassmann.

Cheers, silly rabbit (talk) 21:31, 3 May 2008 (UTC)

I think that you were rather premature to remove. It is a pretty well known and documented priority dispute, and that is why it appears in a reliable tertiary source. Have a look also at MacTutor [2] where its says

Saint-Venant developed a vector calculus similar to that of Grassmann which he published in 1845. He then entered into a dispute with Grassmann about which of the two had thought of the ideas first. Grassmann had published his results in 1844, but Saint-Venant claimed (and there is little reason to doubt him) that he had first developed these ideas in 1832. Again it would appear that Saint-Venant was unlucky. Itard writes in [1]:-

Saint-Venant used this vector calculus in his lectures at the Institut Agronomique, which were published in 1851 as "Principes de mécanique fondés sur la cinématique". In this book Saint-Venant, a convinced atomist, presented forces as divorced from the metaphysical concept of cause and from the physiological concept of muscular effort, both of which, in his opinion, obscured force as a kinematic concept accessible to the calculus. Although his atomistic conceptions did not prevail, his use of the vector calculus was adopted in the French school system.

Itard is the same ref as ours J Itard, Biography in Dictionary of Scientific Biography (New York 1970-1990).

I have a copy of "A history of vector analysis". Fortunately it is also on google books. See p84. [3] I have not yet got hold of other original references but the title of this one says alot "F D Kramar, The geometrical algebra of Saint-Venant, Grassmann, Hamilton (Russian), in 1968 Proc. Second Kazakhstan Interuniv. Sci. Conf. Math. Mech. (1965), 236-237." Billlion (talk) 07:58, 4 May 2008 (UTC)

See also Desmond Fearnley-Sander, Hermann Grassmann and the Creation of Linear Algebra, The American Mathematical Monthly, Vol. 86, No. 10 (Dec., 1979), pp. 809-881. See page 811."In 1847 Grassmann had wanted to send a copy of the Ausdehnungslehre to Saint-Venant (to show that he had anticipated some of Saint-Venant's ideas on vector addition and multiplcation". This claim comes from Engls biography of Grassman. It is still not clear from this that Saint Venant was claiming priority on the exterior product. But I think what is really needed is a good look at StV's paper "De l'interpretation geometrique des clefs et des determinants algebraiques" Billlion (talk) 09:33, 4 May 2008 (UTC)

I have self-reverted, since I found it suspicious that none of my sources even mention Saint-Venant. Apparently there are multiple sources to substantiate this, so I apologize. silly rabbit (talk) 12:02, 4 May 2008 (UTC)

Too technical?

User:ProperFraction seems to feel that the article is too technical. I am curious to know why this is, since the article clearly explains its subject, has motivating examples, and (see the peer review) is even less technical than some would like to see. In my opinion, some generality has been sacrificed for clarity already. If ProperFraction is willing to discuss constructive directions for the article, then he/she is free to discuss them here on the talk page. But tagging the article with administrative templates is completely pointless, and will ultimately serve no purpose without some accompanying discussion. silly rabbit (talk) 02:33, 12 May 2008 (UTC)

I would say it is certainly not too technical. I have a lot of experience trying to explain exterior algebra to undergraduate maths students and to engineers and the style of this article with a brief summary of definition followed by some illustrative examples is about as simple as it gets.Billlion (talk) 06:24, 12 May 2008 (UTC)

Try to explain it to me then. (I am not a degree level mathematician as I sure most readers aren't either)--ProperFraction (talk) 23:44, 12 May 2008 (UTC)

This isn't Dummipedia. There is a certain required degree of mathematical maturity that is expected before someone should attempt to tackle this article. Minimum prerequisites include some familiarity with abstract algebra and vector spaces. There are links in the text for readers unfamiliar with the terms being used and, furthermore, there are books listed at the end of the article aimed at a broad variety of different backgrounds. This encyclopedia article needs to be, well, encyclopedic rather than pedagogical.

Now referring to your most recent edit, the proposed sectioning was completely wrong, since the so-called "Formal definition" does not actually contain a definition (formal or otherwise) and the paragraph about category theory is also not part of the putative "definition". As I have already explained above, the definition comes after the "Motivating examples". (Hence the modifier "Motivating".) Finally, I should ask why you are intent on making edits to an article on a subject that you admittedly and quite obviously do not understand. silly rabbit (talk) 00:43, 13 May 2008 (UTC)

The fact that I, an engineering graduate, do not understand the article means I feel it needs to be simplified for all to understand. I could write lots of stuff that you dont understand, but what good is that to an encyclopedia?--ProperFraction (talk) 00:59, 23 May 2008 (UTC)

It is important to preview an edit before saving it. The last few edits by PF created self-links, links to disambiguation pages, and improper section headings. The placement of the technical tag definitely violates community consensus. Normally I don't like merely reverting good faith edits, but these definitely degraded the article. JackSchmidt (talk) 01:56, 13 May 2008 (UTC)

OK you make the article comprehensible than!--ProperFraction (talk) 00:59, 23 May 2008 (UTC)

link to Wirtinger inequality

I see there is extensive discussion here of decomposable vectors (sometimes known as simple vectors). There is a closely related result called Wirtinger's inequality for the comass. If it is appropriate it may be helpful to add a link to that page. 132.70.50.117 (talk) 15:45, 19 May 2008 (UTC)

exterior derivative, anyone?

I find it a little odd that the lengthy piece on the exterior algebra does not contain any mention of the exterior derivative, namely the differential that makes the exterior algebra into a differential algebra in addition to being a graded algebra. This is of course of central importance in de Rham theory as well as just about anywhere else the exterior algebra is actually used. Is this treated elsewhere perhaps and so not covered here so as to avoid duplication? If so, a link should be provided. Katzmik (talk) 04:41, 2 June 2008 (UTC)

There is a separate article on exterior derivative, which should be linked to in this article. However, that treats only the differential-geometric case, and does not make it explicit that the construction of the exterior derivative is essentially an algebraic construction. Indeed, the algebraic de Rham complex used in algebraic geometry relies on this purely algebraic definition (which works in differential geometry as well, once one defines cotangent spaces / bundles / sheaves as appropriate universal (bundles / sheaves of) modules of R-differentials of smooth functions). Once one has the universal derivation d from smooth functions (regular functions in algebraic geometry,...) to the module of differentials, the definition of the exterior differentials gets dictated by the desire to obtain the de Rham complex; this is due to the extension properties of differentials applied to this situation (e.g., Bourbaki: Algebra III §10.9).

The natural place to discuss all of this would be the article on derivation (abstract algebra), which unfortunately redirects currently to the more specialised differential algebra article. Stca74

(talk) 06:38, 2 June 2008 (UTC)

Well, I added a comment on de Rham cohomology here, see if you like it. Katzmik (talk) 15:19, 4 June 2008 (UTC)

Exterior product and wedge product really the same? o.O

As the first sentence in this article implies, the exterior product and the wedge product is the same thing. But according to the wedge product article, these to things are not to be confused with the other. So, how is it really with this thing? Is it the same or not? --Kri (talk) 00:29, 15 December 2008 (UTC)

The wedge product article is talking about something else that is sometimes called the "wedge product". The exterior product is also called the wedge product, but it is different from what the other article calls the wedge product. That is, there are two different things called the wedge product, and you shouldn't confuse them with one another. siℓℓy rabbit (talk) 00:51, 15 December 2008 (UTC)

Insufficient condition for the definition

In the following, I'll assume that $\mathrm {ch} K\neq 2$. So only condition (1) is, logically, independent. My question is, condition (1) is not enought to define an exterior algebra. For example, for a given linear space ${V}$, define $\wedge$ so that $u\wedge v=0$ for all $u,v\in V$. With such a "wedge product", $\Lambda (V)\cong V\oplus K$ is associative with unit 1, satisfying condition (1), of course. But, it is not an exterior algebra when $\dim V>1$.

I'm afraid another condition should be added: whenever $v_{1},\ldots ,v_{k}\in V$ are linearly independent, $v_{1}\wedge \ldots \wedge v_{k}\neq 0$.

I'm a newbie. Please show me if there is anything I've done improperly. Thanks.

--Bin Zhou (talk) 06:05, 4 May 2009 (UTC)

Revision of the definition, and reason

I'm sorry that, although there is a notice in the source code not to change the definition, I still made some revision of the definition directly in the page. The points of revision and corresponing reason are as follows.

(1) In the former definition, it is said that

Formally, the exterior algebra is a certain unital associative algebra over the field K, containing V as a subspace.

In my revision, it becomes

Formally, the exterior algebra is certain a unital associative algebra over the field K, generated by a K-linear space V as well as the unit.

For example, if $W$ is a K-linear space containing a proper linear subspace $V$, $\Lambda (W)$ also satisfies the former definition, while only $\Lambda (V)$ could be called the exterior algebra of V.

(2) In the former definition, the essential property of $\wedge$ is $v\wedge v=0$ for arbitrary $v\in V$. However, it does not specify whether $v_{1}\wedge \ldots \wedge v_{k}$ is nonzero when envolved vectors (in $V$ only) are linearly independent. Then I could define an operator $\wedge$ by $u\wedge v=0$ for all $u,v\in V$. Obviously, $V\oplus K$ with such an operator satisfies the former definition. But, when $\dim V>1$, it is not the exterior algebra we are referring to. According to the revised definition, it is no longer an exterior algebra.

• The implied feature (3) has been included in the definition, parallelly to feature (1). On the other hand, it is still implied by feature (1). From the point of view that a definition should be based on a minimum logical foundation, a better revision should be like this: the operator $\wedge$ satisfies (i) $v\wedge v=0$ for any $v\in V$, and (ii) $v_{1}\wedge \ldots \wedge v_{k}\neq 0$ whenever $v_{1},\ldots ,v_{k}\in V$ are linearly independent. So I keep the original implied feature (3) for further revision by someone else.

--Bin Zhou (talk) 09:20, 6 May 2009 (UTC)

The part you changed is not a definition (though I agree there is a very large notice in the source code saying that it is), just a vague description. Unfortunately, even your extra condition "(ii)" above is not enough to define the exterior algebra since it does not disallow v ∧ w = u ∧ v ∧ w even when u,v,w are linearly independent. The definition is given much lower in the article, in the section "formal definition".

To fix the lead to conform to WP:LEAD I suggest deleting everything from "Formally," until the first section, and replacing it with a summary of the article. JackSchmidt (talk) 14:41, 6 May 2009 (UTC)

I totally agree with you. Once I thought it could be accepted as a definition. As you've shown, it cannot. Within the summary as you suggested, I think that all envolved formulae (such as the associative law, the distributive law, and so on) should be given explicitely. This makes more sense than pointing out "$\Lambda (V)$ is an associative unital algebra over $K$ with the property...", especially for those not familiar with abstract algebra.

To my favour, the following points should contain in the summary:

• $\Lambda (V)$ is a unital associative graded algebra over a field $K$, $\Lambda (V)=\bigoplus _{k=0}^{\infty }\Lambda ^{k}(V)$, with $\Lambda ^{0}(V)=K$ and $\Lambda ^{1}(V)=V$ for certain a linear space $V$ over $K$. (In fact, $=$ should be $\cong$.) (This excludes the possibility of $v\wedge w=u\wedge v\wedge w$ for some $u,v,w\in V$ or even in $\Lambda (V)\setminus K$.)
• Its multiplication is called the wedge product or the exterior product and denoted by $\wedge$. Moreover, for any $k,l\in \mathbb {Z}$ and $k,l\geqslant 0$, the restriction of $\wedge$ from $\Lambda ^{k}(V)\times \Lambda ^{l}(V)$ to $\Lambda ^{k+l}(V)$ is surjective. (This excludes the possibility to identify a larger algebra (which contains $\Lambda (V)$ as a subalgebra) to be the exterior algebra of $V$.)
• For any $v_{1},\ldots ,v_{k}\in V$, $v_{1}\wedge \ldots \wedge v_{k}=0$ if and only if they are linearly dependent. Especially, $v\wedge v=0$ for each $v\in V$, implying that $u\wedge v=-v\wedge u$. (Even when the characteristic of $K$ is 2, this is valid because of the formal identity $-1=1$.)
• Useful formulae derived from the above features:

$\alpha \wedge \beta =(-1)^{kl}\beta \wedge \alpha$ for any $\alpha \in \Lambda ^{k}(V)$, $\beta \in \Lambda ^{l}(V)$;

$a\wedge \alpha =a\alpha$ for any $a\in K,\alpha \in \Lambda (V)$ (because $1\in K$ is identified with the unit of $\Lambda (V)$);

$(\alpha \wedge \beta )\wedge \gamma =\alpha \wedge (\beta \wedge \gamma )$;

$(a\beta )\wedge \gamma =\beta \wedge (a\gamma )=a(\beta \wedge \gamma )$ for any $a\in K$;

$(\alpha +\beta )\wedge \gamma =\alpha \wedge \gamma +\beta \wedge \gamma$, $\alpha \wedge (\beta +\gamma )=\alpha \wedge \beta +\alpha \wedge \gamma$;

$\Lambda (V)=\bigoplus _{k=0}^{n}\Lambda ^{k}(V)$ when $n=\dim V$ is finite dimensional, and $\dim \Lambda ^{k}(V)={n \choose k}$;

$\Lambda ^{k}(V^{*})\cong \Lambda ^{k}(V)^{*}$ when $V$ finite dimensional. (To my favour, I prefer the notation $\Lambda _{k}(V)$.)

Maybe the above formulae can be listed in a section. The best place will be the section Formal definitions and algebraic properties. The quotient algebra $T(V)/I$ can be explicitely labelled as the definition of $\Lambda (V)$. I mean, give it a title "Definition".

---Bin Zhou (talk) 09:15, 7 May 2009 (UTC)

These should definitly be there.--Gustav Ulsh Iler (talk) 07:16, 6 March 2010 (UTC)

Which of these exactly are not already in the article? Sławomir Biały (talk) 23:26, 13 March 2010 (UTC)

Lead change

I have attempted to implement Jack's suggestion from the previous section, to make the lead comply with WP:LEAD. I have removed the middle paragraph, and attempted to preserve what aspects of it are obviously important for giving context to the exterior product (e.g., we must say somewhere in the lead that the product is alternating). Let me know if this is an improvement. Sławomir Biały (talk) 13:02, 22 May 2009 (UTC)

I like it. I think it introduces and summarizes the article fairly well, while leaving it to the article itself to introduce and summarize humanity's vast knowledge of the exterior algebra. JackSchmidt (talk) 15:00, 22 May 2009 (UTC)

Lacking Examples, should be a few more.

Please, I implore that there should be more examples of different kinds worked out, especially in the Index Notation section, such as giving tensors and then finding their wedge product and also writing them in elementry form.

For example, Let x,y,z be vectors in R^5 and F(x,y,z) = 2x₂y₂z₁ + x₁y₅z₄, G(x,y,z) = x₁y₃ + x₃y₁, h(x)=w₁ - 2w₃

Then find Alt(F) and Alt(G) and Alt(F)Λ h and write them in terms of elemenary alternating tensors.

Also, more examples should be done in particular sections as there is too much theory and not that many examples which is a shame as the examples which are present I find really make you understand the theory.

I would say there should be at least 2 examples for each section or subsection.--Gustav Ulsh Iler (talk) 07:05, 6 March 2010 (UTC)

"Alternating"

Since the word "alternating" is used without definition in the intro, and alternating form redirects here, would it be better to use the word "anticommutative" here, instead? -- Karada (talk) 04:07, 19 March 2010 (UTC)

From the very first paragraph:

"Also like the cross product, the exterior product is alternating, meaning that u ∧ u = 0 for all vectors u, or equivalently u ∧ v = -v ∧ u for all vectors u and v."

Emphasis my own. "Anticommutative" is also correct under usual circumstances, but it is wrong in characteristic two, which is probably why "alternating" is used in most discussions of the exterior algebra. This distinction is emphasized, for instance, in the Bourbaki text referenced in the article. So I think we should leave it the way it is. Sławomir Biały (talk) 10:59, 16 April 2010 (UTC)

Hamilton quaternions

It would be helpful to mention the connection to Hamilton quaternions H. The exterior algebra on R^2 gives the Hamilton quaternions H, if I am not mistaken. Let's see, 0-th degree terms give the scalars R in H. If we set i=e_1 in R^2 and j=e_2 in R^2 then the product e_1 wedge w_2 corresponds to k, and the usual relations are satisfied. Of course, the graded structure is not respected, but at any rate they are isomorphic as algebras. It may be worth pointing this out. At least one recent editor at WPM was not aware of the fact that the term "algebra" is not used in "exterior algebra" in the same sense as in "linear algebra", so the quaternion algebra example may help clarify this. Tkuvho (talk) 18:56, 19 January 2011 (UTC)

The problem is though the product isn't the same, as e.g. the product of i=e_1 with itself is zero in the exterior algebra but -1 in the quaternion algebra. They are better related through geometric algebra. GA uses the same structure as the exterior algebra, a graded structure, but with a different product the geometric product which generalises the exterior product. The quaternions then arise over R^2 for a particular signature as you describe, though they are more usually derived from the even sub-algebra over R^3, as they can then be identified with the rotors in R^3.--JohnBlackburne^words_deeds 19:08, 19 January 2011 (UTC)

Good point, sorry. Perhaps some dumbed-down version of this can go into the article. Tkuvho (talk) 19:10, 19 January 2011 (UTC)

Geometric algebra seems to be a different name for Clifford algebra. Is this another difference between physicists and mathematicians? Anyway it would be helpful to describe the exterior algebra on R^2. I am still confused whether one should get Hamilton quaternions or the matrix algebra. Did you imply above that you do get the quaternions? What's the identification? I just tried a diagonal basis but it seems to give the wrong result. In principle one should describe the exterior algebra for R^1 as well but it is not a very revealing example. Tkuvho (talk) 19:25, 19 January 2011 (UTC)

See Geometric algebra and Clifford algebra for the differences between them, or any good reference. As noted you can't identify quaternions with the exterior algebra, or at least if you do you don't get anything useful. At best you get the part of the product which can be identified with the cross product, but the cross product is already covered in the article. In GA the even sub-algebra over Euclidian R^3 is the quaternions: not only is it isomorphic but it can be used to do rotations in R^3 the same way, with the product being the geometric product so far more useful than the quaternion product.--JohnBlackburne^words_deeds 19:50, 19 January 2011 (UTC)

My point is that we should calculate out the example of the exterior algebra on R^2. I don't remember what it is and don't have any books in front of me right now. The fact that this is not calculated out in the page is a definite shortcoming. perhaps you get C+C? At any rate this should be clarified. Tkuvho (talk) 19:55, 19 January 2011 (UTC)

It couldn't be C+C because it's not commutative, so there is a good chance it's 2x2 matrices. If so, this should be one of the first examples discussed here, the matrix algebra being a familiar and concrete object. Tkuvho (talk) 20:07, 19 January 2011 (UTC)

The 2D algebra is worked out, with it's most obvious application the area of a parallelogram. I can't think of anything else that should be added: there's not a lot to the algebra in 2D so most examples will relate to the one given. There's not much we can do if you don't remember it: if you can find a reference for it then maybe someone could use that to add something.--JohnBlackburne^words_deeds 20:22, 19 January 2011 (UTC)

If there is a simple matrix algebra presentation it should be included, but I can't figure it out right now. It doesn't seem to be the 2x2 matrices since there doesn't seem to be an element that squares to -1 (analogous to 90 degree rotation). Tkuvho (talk) 20:31, 19 January 2011 (UTC)

Antisymmetric plus scalar?

I think a concrete representation of the exterior algebra on R^2 in terms of a matrix algebra may be helpful. I have the impression that 3x3 matrices which are sums of scalar matrices with antisymmetric matrices may be isomorphic to the exterior algebra over R^2; this needs to be checked. One problem with the page is that the wedge product appears out of nowhere. It is not explained how one constructs it. The discussion of the wedge product in the 2-dimensional case looks concrete, but only to those who already understand it. Eventually the page gets to constructing the exterior algebra in terms of the tensor algebra, by a quotient by an ideal. This cannot be said to be a concrete construction. Motivating exterior algebras in terms of what a reader is presumably already familiar with from a basic course in linear algebra may help break the barrier that users are complaining about at WPM. Tkuvho (talk) 21:58, 19 January 2011 (UTC)

No, this doesn't work, either. To build a matrix algebra model for the exterior algebra it seems one needs to work with upper triangular matrices, to account for strong nilpotency in the exterior algebra. This works right away for the exterior algebra over R^1. This can be represented by upper triangular 2x2 matrices A with a_{11}=a_{22}. In other words, we take the scalar matrices plus the nilpotent upper-triangular one [ [0 1] [0 0] ]. To do this for the exterior algebra over R^2 it might be necessary to go to 4x4 matrices. Still, this may be helpful for someone familiar with the matrix algebra but not yet with the exterior algebra, as a first example. At the very least we should include the rank one example. Tkuvho (talk) 05:40, 20 January 2011 (UTC)

While I think it's interesting to construct the exterior algebra as a matrix algebra, it's not clear to me that this will help readers with a background in just linear algebra. The idea of constructing a matrix representation of an algebra is fairly abstract, and in the typical science curriculum is only encountered in courses on quantum mechanics and physical chemistry (and then in a way that is probably not easily understood as a general concept). Unfortunately, taking the topic of the previous thread into account as well, I also can't really think of any good way to introduce the "algebra" idea. Sławomir Biały (talk) 14:09, 20 January 2011 (UTC)

What's wrong with saying that as an algebra (i.e. set with a pair of operations satisfying the usual rules), the exterior algebra in the rank-1 case is isomorphic to a concrete subalgebra of the 2x2 matrices? This might take some of the mystery out of it. After all, even in a basic algebra course, one typically "takes an element of M{n,n}(R), etc.", so the matrix algebra should be far less mysterious to a beginner than the exterior algebra, whose construction as currently stated involves the tensor algebra, which means that in order to get anywhere with exterior algebras, he will have to learn about tensor powers first! Tkuvho (talk) 14:34, 20 January 2011 (UTC)

Sorry, I should have been clearer. Nothing is wrong with saying this, but I don't think it would solve what I think is your original issue, that people generally do not understand the term "algebra" as referring to individual objects. (Thus, "matrix algebra"="linear algebra", not "an algebra consisting of matrices"). If you think adding the case of R^1 to the end of the motivation section will help, then by all means go for it. However, I don't really think that anyone will be able to get anywhere with the exterior algebra using matrix representations. In practice, the wedge product is characterized by "what it does" not "what it is"; that's the essence of the universal construction. But where one needs to say "what it is" to beginners, usually (in intermediate linear algebra courses for instance), the exterior algebra of the dual space is the first thing we define via the isomorphism to the algebra of alternating forms on the space. That is to some extent an abuse, but I can at least envision an article carefully arranged to put that definition first. It seems likely to me that focusing on this sort of approach might ultimately be more satisfactory in the sense of reaching about the same audience as your matrix algebra suggestion. Sławomir Biały (talk) 15:02, 20 January 2011 (UTC)

Thoughts on the new lead?

I have made some changes to the lead, following all of the discussion at WT:WPM. I now feel that the article somewhat flouts WP:LEAD, in that the lead is actually supposed to summarize the article but now only the last paragraph really does that. But, there it is. I welcome any feedback. Sławomir Biały (talk) 00:39, 20 January 2011 (UTC)

The lead has a nice elementary explanation of the wedge product, but not of the exterior algebra. The other additions to the lead are helpful but are not really related to the discussion at WPM which was how to make this more accessible to a novice with, say, one year of linear algebra background. Tkuvho (talk) 05:45, 20 January 2011 (UTC)

In drafting the new lead, I was reminded of a saying that a (now departed) physicist friend of mine, who was used to writing research grants, often used: "A mathematician will use two words where three would suffice." The intent of this is to point out that mathematicians tend to be unnecessarily economical in their writing. It seemed that there was an element of this in the discussion at WT:WPM, and I felt that, to a certain extent, simply adding more words might help to improve the readability for someone not used to digesting mathematical text. In the process, of course, I also added more content to the lead that I felt would be illuminating.

But since there wasn't a clear identification of what the barrier to understanding was (beyond "I have a background in college mathematics, and I don't understand the first paragraph"), it's hard to know how to address it. But I am accepting as given that readers of the article have a good familiarity with cross products and determinants, and that they know a little about matrices. At most US universities, we teach these things to first and second year undergraduate math, physics, chemistry, economics, and engineering students. (At my university, multivariable calculus is required for these majors; linear algebra is for all but chemistry I think.) I think they are not unreasonable demands to expect from a reader of this article. In any event, we are not a textbook, and cannot be expected to break things down further than this. To paraphrase Eintein, "Simplify as far as possible, but no further." Sławomir Biały (talk) 13:57, 20 January 2011 (UTC)

My impression is that the beginner who expressed himself at WPM did not register the fact that the term "algebra" in "exterior algebra" is not used in the sense of some generalisation of "linear algebra". We are defining a new algebraic structure akin to the matrix algebra. Limiting ourselves in the introduction to saying that it is "an abstract algebraic structure" just does not do anything for him. He is more likely to be comfortable with the term "matrix algebra" than "abstract algebraic structure". In this sense, perhaps an explicit description of the rank-1 case in terms of the upper-triangular 2x2 matrices (with equal eigenvalues) may be helpful. Also, the page assumes the existence "ex nihilo" of the wedge product, and only gets to constructing things using tensor algebra several sections later. The explanation of the rank-2 case in terms of areas is wonderful, but it still does not explain how you construct such a thing. Exhibiting it as occurring in the familiar framework of the matrix algebra may be one way of doing it. The vector product example in R^3 is a wonderful example but it is somewhat limited as far as the wedge product is concerned. Note that the latter is associative but not the former. Tkuvho (talk) 14:21, 20 January 2011 (UTC)

Good point, I hadn't detected this issue. However, I don't think we should call it a "matrix algebra". At any rate this doesn't fix the issue, as one is probably even more likely to think that "matrix algebra" refers to "linear algebra". The best we can do is make it grammatically clear that algebra is used here as a count noun: hence "the exterior algebra is an algebraic structure..." etc. Sławomir Biały (talk) 14:25, 20 January 2011 (UTC)

The viewpoint in the lead is a bit confused right now. On the one hand, we want to have everything abstract, invariant, and coordinate-free, which of course works well with tensor powers. On the other hand, we want to talk about "lengths" of "blades", which of course pre-supposes a background metric, which defeats the whole purpose of the invariant approach. The introduction is way too long. Tkuvho (talk) 14:48, 20 January 2011 (UTC)

I am considering rearranging the lead a bit to put the linear algebra paragraph at the end, to segue in with the functoriality of the exterior algebra. I think that there should then be enough transition between the geometric intuition of the first paragraph, and the concern over invariants of the part on linear algebra. Sławomir Biały (talk) 15:23, 20 January 2011 (UTC)

Another option is to kill the second paragraph altogether. Sławomir Biały (talk) 15:32, 20 January 2011 (UTC)

With the title being exterior algebra, I think we have to mention it in some detail :) — Kallikanzarid^talk 19:52, 20 January 2011 (UTC)

Are blades an often-used term? I think we can safely throw them out. — Kallikanzarid^talk 19:50, 20 January 2011 (UTC)

Two undid revisions

http://en.wikipedia.org/w/index.php?title=Exterior_algebra&oldid=409032113 'Algebraic structure' is too vague, and it doesn't achieve much at this cost. 'graded algebra' is concrete while still making it clear that this is an algebraic topic. IMO Anyone curious will find better explanation by clicking on the latter link, too.

http://en.wikipedia.org/w/index.php?title=Exterior_algebra&oldid=409030292 'Something you can integrate' is confusing to beginners, because for them something they can integrate is a function (elementary calculus on curves and surfaces does not explain integrands as forms so they are perceived as but a notation). 'Something like a differential' is more helpful for two reasons:

1. Well, that's what it is :)
2. College graduates actually know what differentials are,
3. There is a theorem in calculus on curves about whether Pdx + Qdy + Rdz is a differential. While the fact that it is a differential form is not usually explained, this should hopefully ring the bell when differentials are invoked. — Kallikanzarid^talk 21:08, 20 January 2011 (UTC)

It would be better to mention graded algebra only after the graded structure is introduced. As for differentials, while I am sympathetic to this point of view, bringing in new notation dx^i seems excessively cryptic. Maybe it's better to say nothing, but leave the bare link to differential form. Sławomir Biały (talk) 21:15, 20 January 2011 (UTC)

It's your call — Kallikanzarid^talk 21:16, 20 January 2011 (UTC)

Maybe you can just leave the part about differentials, and cut out the part about k-forms. Just saying that differential forms serve as a vector space for the exterior algebra would be helpful, since it's exactly the topic being discussed. — Kallikanzarid^talk 21:21, 20 January 2011 (UTC)

It can hardly be helpful because it is an error. The exterior algebra is not applied to the vector space of differential forms. It is applied pointwise to each cotangent space. The entire Bourbakist infestation in the lede should be deleted. Tkuvho (talk) 21:31, 20 January 2011 (UTC)

It depends on your point of view. In practice it's the exterior algebra over the sheaf of 1-forms, or it's the bundle if exterior algebras. Either way it's a more general construction. Also, see my post above regarding Bourbakist infestation. We have to mention this because most of the article is about it. Refer to WP:LEAD, that we already stretch to the breaking point. Eliminating the one thing that the lead is supposed to do seems counterproductive. Sławomir Biały (talk) 21:39, 20 January 2011 (UTC)

Thank you for the correction, the point remains, however :) — Kallikanzarid^talk 21:45, 20 January 2011 (UTC)

BTW, what is Bourbakism? — Kallikanzarid^talk 22:00, 20 January 2011 (UTC)

Just to clarify, my last post was directed at Tkuvho. I think I have implemented a solution that should be mutually satisfactory? I have included a link to differentials, while keeping the more intuitive language about things that can be integrated over higher-dimensional manifolds. Sławomir Biały (talk) 12:53, 21 January 2011 (UTC)

I'm rather satisfied with the situation now. Unfortunately, this surfaces the problem with line integral: on that page it's (generally incorrectly) defined in terms of a dot product, which will impede the understanding of its connection to differential forms (and the ability to actually see a differential form there, too). — Kallikanzarid^talk 13:28, 21 January 2011 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Exterior algebra/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Top importance may be too high, but this construction is so fundamental in mathematics! I have not thought about how such an important article could be improved: please replace this comment by suggestions. Geometry guy 21:46, 20 May 2007 (UTC)

Last edited at 21:46, 20 May 2007 (UTC). Substituted at 21:15, 4 May 2016 (UTC)