Talk:Geometric algebra/Archive 3

The lead (again)
Just so changes to the lead don't come as a surprise: I would like to de-mystify the discussion of multivectors currently in the lead by using (parts of) the lead we came up with at Geometric_algebra/Sandbox. I just find what is currently there too cryptic. I'm currently in the process of reevaluating what is written at the sandbox and meshing it with the current lead. Rschwieb (talk) 14:36, 30 May 2013 (UTC)
 * In the end, what is it we're doing when we use GA to solve a problem? We attach some geometric significance to elements of the algebra, and then algebraic operations on those elements correspond to geometric operations. The objects with clear geometric significance are blades, not general multivectors. Even rotors, despite their usefulness, can be extremely difficult to imagine geometrically, but blades (in the "vector space interpretation" of Dorst, Fontijne, and Mann) always have simple geometric meaning.
 * With that in mind, it strikes me that we could, instead of saying GA is an algebra whose elements are interpreted in such-and-such a way, say that GA is the algebra of lines, planes, volumes, and so on, and worry about the terminology (blades, multivectors, and such) later on. Muphrid15 (talk) 16:09, 30 May 2013 (UTC)
 * @Rschwieb: Demystification is good, so go ahead. The target audience of this article should be anyone who has a moderate grasp of vector algebra, and potentially no experience of abstract algebra. The sandbox version is in many ways an improvement, but take care not to lose interesting and especially motivating information.  The 2nd paragraph of the current lead has some of interesting applications/fields (e.g. electromagnetism) that should be mentioned in the lead.  The mention of STA, CGA and GC could be replaced by more familiar terms such as relativity, geometric manipulations and the extension of calculus to GA.  In terms of motivation, we can keep an echo of Hestenes: that it is "better" in many ways than anything else.
 * @Muphrid15: Your idea is good, but your words can be made even more concrete by dropping the explicit mention of algebra (it is in the name, after all) and adding familiar actions to the description, e.g. rotation, intersection etc. — Quondum 20:09, 30 May 2013 (UTC)
 * To be honest, I'm kind of at a loss for what to call GA, then, even though I agree that it's redundant to say it's an algebra. Muphrid15 (talk) 22:01, 30 May 2013 (UTC)
 * Sorry, I didn't intend to be so cryptic. What I was suggesting was to change your statement to something like: A GA manipulates lines, planes, volumes etc. using rotation, intersection and the like in a direct way in any number of dimensions.  The wording may need work, but this is what I meant be dropping "algebra".  It leaves no abstract words in the outline description, and hopefully is effectively what you were aiming at.  — Quondum 02:06, 31 May 2013 (UTC)

Grading? What?
Did I miss something? Everything I've seen regarding graded rings/vector spaces/whatever regards the product as only summing the individual grades to get the new grade. The geometric product doesn't do this (it does a sum and a difference), so while I think GA's grading is a natural extension to grading as the term is typically understood, can we really just say it is a graded vector space? Muphrid15 (talk) 18:22, 31 May 2013 (UTC)
 * There seem to be two different definitions of "graded algebra" in use here. I was going with the one that was consistent with the article graded algebra that had already been linked, which states explicitly that Clifford algebras are Z2-graded algebras.  The footnote I removed said, "For example, when vv=v⋅v≠0, the product is grade-0 and not grade-2, as a graded algebra would require."  But for a Z2-graded algebra, this is fine because $$2 \equiv 0 \pmod 2$$.  The idea is that an even multivector times an even multivector gives an even multivector, an even multivector times an odd multivector gives an odd multivector, and an odd multivector times an odd multivector gives an even multivector. Teply (talk) 19:03, 31 May 2013 (UTC)
 * Yeah Q and I ran into this once before. The algebra is a graded vector space (but not a graded algebra with the geometric product) under the usual grading. I totally forgot about the Z_2 grading, because I've never read about it outside of wikipedia. That would probably be good to have... Rschwieb (talk) 20:52, 31 May 2013 (UTC)
 * There's some discussion of the distinction at Clifford algebra. You could link to there without getting too bogged down with the details here. Teply (talk) 21:15, 31 May 2013 (UTC)

Starting it off...
The initial discussion is backwards. Clifford algebras emerged as a generalization of Clifford's original formulation of an algebra to describe real geometric objects and transformations. It gradually dawned on mathematicians that the framework could be expanded to include basis elements that square to 0, and to -1. From that has emerged the rich set of extensions that have informed much of physics and which are now doing the same to computer science, but with a difference. Geometry has once again been re-introduced into the picture, just as Clifford and Grassmann intended.

The article needs to start by telling this story. The abstraction, with GA being cast into that wider framework can follow. That's the proper way to involve the audience. There is also a need for a citation to Clifford's original AMS paper. That's available online and there should be a link. ColdSun (talk) 17:58, 3 March 2014 (UTC)


 * Hi there! By the "initial discussion," do you mean the lead? Or everything before the history section? Please consider that math articles are not mandated to carry a chronological form in their leads or in their bodies. Usually a quick modern development is preferred. However, we recognize history is certainly important in its own right, and if you think your idea can improve the history section of the article, then that would certainly be welcome.
 * Finally, which "original AMS paper" are you thinking of? I guess it's not the American Mathematical Society since that was formed after Clifford's death, but I gather you might be thinking of Applications of Grassman's Extensive Algebra in the American Journal of Mathematics, or something similar maybe. Regards Rschwieb (talk) 14:01, 4 March 2014 (UTC)


 * Thanks for the reply. I do mean the lead. While I understand the nature of mathematical exposition, a good hook is not a bad idea. The historical framework does that. A simple re-working of the existing lead might do just that. The information is there, but I believe it can be re-cast.
 * Yes, the paper I was referring to is the one you cite. Also, thank you for catching that. It wasn't the AMS but the American Journal of Mathematics which published that work. ColdSun (talk) 18:07, 6 March 2014 (UTC)


 * Two comments and one suggestion before you begin. The lead is already a little too involved at the moment, so I hope the length and complexity does not increase much. Secondly (you may already be prepared for this) is that sometimes even "simple re-works" by one editor can cause conflict with other interested editors. That said, there is probably room for improvement on that historical bit at the end of the lead. Take a shot at improving the historical bit, but be conservative :) Citing that paper is a good addition I don't think anybody will oppose. Regards Rschwieb (talk) 13:55, 7 March 2014 (UTC)

Diagrams for rotations of vectors as double reflections
Would these be useful anywhere? M&and;Ŝc2ħεИτlk 15:36, 20 August 2013 (UTC)
 * The problem is they don't clearly show that it's a rotation. This does, as the objects being transformed are more complex than vectors and show the orientation better. It's also simpler as it has less information, yet it gets across the key relationship between the reflections and rotation, in particular that the rotation angle is twice the angle between the vectors. It certainly could be improved as it's hard to read at the default size, while yours are much clearer and better presented.-- JohnBlackburne wordsdeeds 16:01, 20 August 2013 (UTC)



Thank you for the timely response and good feedback and sorry for the delay...

About mine: It's a fair point they don't clearly show orientation and hence the rotation, a shape does, so replaced the vectors with flags since the reflection and rotation are very clear and the shape is compact.

As for File:Simx2=rotOK.svg, I created a (hopefully) clearer version. --->

The labels of points A,B,C, on the shape were not included to reduce clutter, but shall add them if desired. (The angles may give an impression of clutter, but they are more geometrically significant to showing why there is a factor of 2).

I don't mind which we use, but the reason for two diagrams in each reflection case was to show all cases pictorially.

Thanks, M&and;Ŝc2ħεИτlk 07:58, 24 August 2013 (UTC)


 * Usually the letter "F" does the trick. Teply (talk) 10:56, 24 August 2013 (UTC)


 * Yes, but I think the crescents/flags show the rotations/reflections clearly enough. Is this a suggestion to replace the crescents with the letter F? M&and;Ŝc2ħεИτlk 07:31, 25 August 2013 (UTC)


 * I see this thread went quiet, and I seem to have failed to get involved. The flags work for me better than any of the others, including an 'F' or the crescents. Any unneeded detail is only going to suggest that there is might be more to it, which we would want to avoid. —Quondum 15:19, 17 May 2014 (UTC)

The outer product of vectors can not be associative
In section 1.1 it is written (bold emphasis is mine):


 * The outer product is naturally extended as a completely antisymmetric, associative operator between any number of vectors


 * $$a_1\wedge a_2\wedge\dots\wedge a_r = \frac{1}{r!}\sum_{\sigma\in\mathfrak{S}_r} \operatorname{sgn}(\sigma) a_{\sigma(1)}a_{\sigma(1)} \dots a_{\sigma(r)},$$

Defined as such, the outer product can not be associative, because that would imply that the outer product of two outer products is defined. It is not. I can see it is defined later in the article and that in the end in is indeed associative, but at this point in the article we don't know what it is.--Grondilu (talk) 15:16, 25 August 2014 (UTC)


 * This is a fair point. As it stands, an r-ary operator antisymmetric operator is being defined, so associativity does not enter into it. It is not clear to me whether to move this down to where the various extensions are dealt with. The article does not actually develop the definition of the algebra formally or completely. For now we could simply delete the emphasized words. —Quondum 02:42, 26 August 2014 (UTC)


 * I think it should be removed indeed, as it is not only confusing, but wrong. Mentioning the arity seems like a good idea as well.--Grondilu (talk) 21:21, 26 August 2014 (UTC)


 * You seem to be suggesting deletion of the two words, not the entire thing (though I may be misunderstanding you). On further thought, it seems like a bad idea to introduce an operator with arity not 1 or 2 at this point, even if its arity is clearly defined such as you suggest. This is a new concept for the article that is not used later and in general is not often used in GA. If desired, the outer product expression, considered as a sequence of applications of an associative binary operator, could be stated to be antisymmetric in the original vector arguments.
 * That way the r-ary operator never comes into it, pretty as it might be. So should we drop the entire bit? —Quondum 22:20, 26 August 2014 (UTC)
 * That's precisely the point I'm not satisfied with: the outer product at this point can not be considered as a sequence of applications of an associative binary operator, for $$a\wedge b\wedge c$$ is neither $$(a\wedge b)\wedge c$$ nor $$a\wedge (b\wedge c)$$.  Even once you have defined $$a\wedge b\wedge c$$ as the signed sum of permuted geometric products, the two expressions $$(a\wedge b)\wedge c$$ and $$a\wedge (b\wedge c)$$ are still undefined, for at this point in the article the exterior product is only defined for vectors and neither $$a\wedge b$$ nor $$b\wedge c$$ are vectors.
 * I'm OK with not mentioning the r-arity as it is indeed not necessary but I do object on the associative term.--Grondilu (talk) 19:08, 2 September 2014 (UTC)
 * You inserted your response inside a single signed section of mine, so I have moved it to the end of the post that you were responding to. It is not generally acceptable to break another's comment, thus leaving an effectively unsigned post for the first part. In this instance, your intent might have been to respond to a specific portion of my comment, but in this case, it does not even seem to aid that goal.
 * I agree with your objections. I think we should take a stab at addressing them. —Quondum 20:39, 2 September 2014 (UTC)

how about some concrete?
The article would be more meaningful and useful to me, at least, if it gave a smidgen of attention to arithmetic. Supposing I have two vectors a = (0,1,2,3) and b = (4,5,6,7), what is ab in numbers rather than abstract symbols? —Tamfang (talk) 08:39, 17 May 2014 (UTC)


 * Having more concrete illustrations is not a bad idea, but finding something that builds intuition may be tricky. Use of tuple notation may be confusing (in 4 dimensions, there are 24=16 components), which is why we use explicit basis vectors. So we'd put
 * $a = e_{2} + 2e_{3} + 3e_{4}$ and $b = 4e_{1} + 5e_{2} + 6e_{3} + 7e_{4}$,
 * giving
 * Here we see a multivector with scalar (the 38) and bivector (the remainder) parts. However, I do not see an example of mechanical algebraic manipulation like this being of value or being less abstract, unless we can use it to do something intuitively concrete with say physics. The kind of example that comes to mind for me would be illustrating torque as a bivector in two dimensions (albeit this uses the exterior/wedge product rather than the geometric product) in which bivectors have only one component. Three dimensions tend to confuse because of the familiarity of the cross product, and the coincidence of the number of components that vectors and bivectors have. Four dimensions will lose the audience. If we are to develop one intuition in the article, I'd like to see the idea that torque generalizes to any number of dimensions under the wedge product, whereas the cross product is limited to three dimensions. Another example may be calculation of a specific rotation.
 * Given all this, what would make sense to you? —Quondum 15:05, 17 May 2014 (UTC)
 * Here we see a multivector with scalar (the 38) and bivector (the remainder) parts. However, I do not see an example of mechanical algebraic manipulation like this being of value or being less abstract, unless we can use it to do something intuitively concrete with say physics. The kind of example that comes to mind for me would be illustrating torque as a bivector in two dimensions (albeit this uses the exterior/wedge product rather than the geometric product) in which bivectors have only one component. Three dimensions tend to confuse because of the familiarity of the cross product, and the coincidence of the number of components that vectors and bivectors have. Four dimensions will lose the audience. If we are to develop one intuition in the article, I'd like to see the idea that torque generalizes to any number of dimensions under the wedge product, whereas the cross product is limited to three dimensions. Another example may be calculation of a specific rotation.
 * Given all this, what would make sense to you? —Quondum 15:05, 17 May 2014 (UTC)
 * Here we see a multivector with scalar (the 38) and bivector (the remainder) parts. However, I do not see an example of mechanical algebraic manipulation like this being of value or being less abstract, unless we can use it to do something intuitively concrete with say physics. The kind of example that comes to mind for me would be illustrating torque as a bivector in two dimensions (albeit this uses the exterior/wedge product rather than the geometric product) in which bivectors have only one component. Three dimensions tend to confuse because of the familiarity of the cross product, and the coincidence of the number of components that vectors and bivectors have. Four dimensions will lose the audience. If we are to develop one intuition in the article, I'd like to see the idea that torque generalizes to any number of dimensions under the wedge product, whereas the cross product is limited to three dimensions. Another example may be calculation of a specific rotation.
 * Given all this, what would make sense to you? —Quondum 15:05, 17 May 2014 (UTC)


 * A quick description of what happens when basis vectors (and polyvectors, and polyvectors of different ranks) are multiplied, pointing out the difference from the familiar cross product, should suffice; sorry if it's already there and my eye skipped over it. My grievance is that I've heard several times that GA can simplify (for example) things that I want to do in higher geometry, but every presentation I've seen consists mostly of a list of identity theorems which doesn't help me write code. —Tamfang (talk) 19:23, 17 May 2014 (UTC)


 * A "quick description" (in the sense of an interpretation) of the general case of a product would be challenging. Perhaps in the section §Geometric_interpretation, I would like to see the geometric interpretation and computations related enough that it should not be difficult to translate operations into code (though inverses can be tricky). The article already tries to address the geometric interpretation of blades (it at least has a diagram to that end), as well as an interpretation of the wedge product of blades. In this context, the geometric product is more useful for its algebraic properties, and hence as a building block for the geometric operations such as projection, reflection, rotation etc. Since many of these are already covered (e.g. in 3 dimensions, the cross product is related to the wedge product in §Rotating systems). The article is not organized too well, and could do with some insights on cleaning up, which is where more specific suggestions from you would be helpful (when you have newly gained an insight, you can still see what's missing). Perhaps we can work through some specific examples of what you find obscure and from that glean where the article needs updating. —Quondum 21:11, 17 May 2014 (UTC)


 * I don't think the product on its own is that interesting: the applications and so the geometric interpretations arise when you do things with it, i.e. with the product and things derived from it. One example is the reflection in a plane, which is given at Reflection (mathematics). It's especially interesting as it's closely related to rotations - a simple rotation can be constructed from two of these operations, and relates to the simpler operation both algebraically and geometrically. Another simple operation is how you get a rotor from a bivector via the exponential map. I can see both of these in the article but buried deep within it; if I didn't know them already I might not recognise them.-- JohnBlackburne wordsdeeds 21:31, 17 May 2014 (UTC)


 * Is it possible to clearly define what ab means? The above talk was really useful to me and if its not 'in' some other wikipedia article (?) then is should be defined here. I get that a ≡ a1e1 + a2e2+... and I get that ab ≡ a1b1e1² + ... and that e1² ≡ 1 (here). What is missing is what is eiej for i≠ j. The fact that the above explicit example doesn't contract (that may not be the right word?) it to one of the four ei's suggests either that it can't be contracted, or the author didn't bother spelling out what the result of the matrix multiplication was either because it 'should be obvious' or because it depends on the (variable) 'metric' of the particular problem. That's three possible interpretations! Please note that a Λ b and a· b are defined in terms of ab and ab is defined in terms of them! Circular definitions should, whenever possible, be avoided like Ebola! ThanksAbitslow (talk) 01:08, 11 January 2015 (UTC)


 * Hi, I'm not an expert on this, but I think the definition of ab via the given rules to be OK. eiej for i ≠ j is just a bi-vector, an element contained in the algebra. Only for i = j there is a "contraction" provided a priori by the quadratic form, applicable to simple vectors and resulting in a "scalar". Furthermore, I think that there is no "definition" of ab in terms of a Λ b and a·b, but just a derivation of a specific representation of the geometric product by the sum of a symmetric 'inner' and an antisymmetric 'exterior product', defined by the quadratic form belonging to the geometric algebra). Purgy (talk) 09:33, 15 January 2015 (UTC)


 * The method here is completely consistent (and not circular). The idea is to describe the algebra in terms of the generators $$e_1,...,e_n$$ constrained by the relation $$aa=g(a,a)$$. This is analogous to forming a polynomial ring in n variables and then taking the quotient by some elements to force relations upon the ring. Rschwieb (talk) 20:15, 15 January 2015 (UTC)


 * What might be missing from the description is the idea that the geometric product is the freest product for an R-linear (associative unital) algebra subject to this relation (though I may still have missed a point or two). This basically guarantees that scalars, vectors, bivectors etc. constitute linearly independent subspaces. It'd be nice to be able to capture this less abstractly, but this may be a challenge. Also, it would make sense to rewrite the section so that non-circularity is clear: defining the dot and wedge products in terms of the geometric product only after defining the geometric product more fully (including for the entire space, not only vectors), and making it clear that the expression giving the geometric product of vectors in terms of the others is merely a result. —Quondum 00:48, 16 January 2015 (UTC)


 * Since this article already refers to Clifford algebra where the universal property and other subtleties of the product are dealt with and I myself did like this here article, I do not feel urged very much to oppress it with my suboptimal English edits. Purgy (talk) 17:07, 16 January 2015 (UTC)

proposed new sections: Reversion, Conjugation
Instead of trying to just add these sections, I thought I better have some peer-review/talk about them first. Very likely the formatting/TeX is objectionable to some. Possibily there are some subtle mathematical issues also. Please have a look and suggest/edit it as appropriate. If any good, maybe later an edit of this can be put on the main page.

In the literature on geometric algebra, I feel there is some confusion on the notation and definitions given to reversion, conjugate, and inverse. I hope this contribution, if it holds up, helps to clarify these basic operations.

Reversion
Reversion is an operation that reverses the order of products. Reversion is a primitive operation that has many uses in the construction of other more complicated operations.

Given an $$r$$-blade
 * $$\begin{array}{rcl}

\mathbf{A}_r & = & \bigwedge_{i = 1}^r \mathbf{a}_i =\mathbf{a}_1 \wedge \ldots \wedge \mathbf{a}_{r - 1} \wedge \mathbf{a}_r\end{array} $$ the reversion, or reverse, of $$\mathbf{A}_r$$ is
 * $$\begin{array}{rcl}

\tilde{\mathbf{A}}_r & = & \bigwedge_{i = 1}^r \mathbf{a}_{r - i + 1} =\mathbf{a}_r \wedge \mathbf{a}_{r - 1} \wedge \ldots \wedge \mathbf{a}_1\\ & = & (- 1)^{r (r - 1) / 2} \mathbf{A}_r .\end{array} $$ The reverse of an $$r$$-vector $$V_{\langle r \rangle}$$ , which is a sum of $$r$$-blades, is
 * $$\begin{array}{rcl}

\widetilde{V_{\langle r \rangle}} & = & (- 1)^{r (r - 1) / 2} V_{\langle r \rangle},\end{array} $$ the sum of the reverses of the $$r$$-blade terms.

More generally, given a multivector of grade $$r$$
 * $$\begin{array}{rcl}

A_r & = & \prod_{i = 1}^r \mathbf{a}_i =\mathbf{a}_1 \ldots \mathbf{a}_{r - 1} \mathbf{a}_r\end{array} $$ the reverse of $$A_r$$ is
 * $$\begin{array}{rcl}

\widetilde{A_r} & = & \prod_{i = 1}^r \mathbf{a}_{r - i + 1} =\mathbf{a}_r \mathbf{a}_{r - 1} \ldots \mathbf{a}_1\\ & = & \sum_{i = 0}^r \widetilde{\langle A_r \rangle_i}\end{array} $$ which is the sum of the reverses of the $$i$$-vector terms.

Properties of the reverse, for any multivectors $$A$$ and $$B$$:
 * $$\begin{array}{rcl}

(A B)^{\sim} & = & \tilde{B} \tilde{A}\\ (\tilde{A})^{\sim} & = & A.\end{array} $$

Conjugation of blades
The conjugation of blades $$\mathbf{a}_1 \wedge \ldots \wedge \mathbf{a}_{r - 1} \wedge \mathbf{a}_r$$ in a geometric algebra generalizes the conjugations of the imaginary unit $$i$$ of complex numbers and the pure quaternion vectors $$x\mathbf{i}+ y\mathbf{j}+ z\mathbf{k}$$ of quaternions. Blades are a generalization of imaginaries to higher dimensions or grades.

In a $$\mathcal{G}_{p, q}$$ geometric algebra, there are $$p + q = n$$ orthogonal unit vectors where the first $$p$$ unit vectors have positive signature and the last $$q$$ unit vectors have negative signature


 * $$\begin{array}{rcl}

\mathbf{e}_i \cdot \mathbf{e}_j & = & \left\{ \begin{array}{lll} + 1 & | & i = j ; 1 \leq i \leq p\\ - 1 & | & i = j ; p + 1 \leq i \leq n\\ 0 & | & i \neq j. \end{array} \right.\end{array} $$

An $$r$$-blade $$\mathbf{A}_r$$ in this algebra is a scalar $$\alpha$$ multiple of the outer product of some combination of $$r$$ elements taken from the set of $$n$$ vector units $$\mathbf{e}_i $$


 * $$\begin{array}{rcl}

\mathbf{A}_r & = & \alpha \bigwedge_{i = 1}^r \mathbf{e}_{\sigma (i)} = \alpha \mathbf{e}_{\sigma (1)} \wedge \ldots \wedge \mathbf{e}_{\sigma (r - 1)} \wedge \mathbf{e}_{\sigma (r)}\end{array} $$

where $$\sigma (i) \in \sigma \{ 1, \ldots, p, p + 1, \ldots, n \}$$ is the $$i$$th element from some permutation of the ordered indices.

The number, or grade, of negative-signature vector units $$\mathbf{e}_{\sigma (i)}$$ in $$\mathbf{A}_r$$ is denoted $${\operatorname{gr}}_- (\mathbf{A}_r)$$, and the number of positive-signature vector units $$\mathbf{e}_{\sigma (i)}$$ in $$\mathbf{A}_r$$ is denoted $${\operatorname{gr}}_+ (\mathbf{A}_r)$$.

An $$r$$-blade $$\mathbf{A}_r$$ can also be written as the outer product of $$r$$ linearly independent vectors $$\mathbf{a}_i$$ as


 * $$\begin{array}{rcl}

\mathbf{A}_r & = & \bigwedge_{i = 1}^r \mathbf{a}_i =\mathbf{a}_1 \wedge \ldots \wedge \mathbf{a}_{r - 1} \wedge \mathbf{a}_r .\end{array} $$

An $$r$$-blade $$\mathbf{A}_r$$ is called a  null r-blade   $$\mathbf{A}_r \in \mathcal{G}^{\circ r}_{p, q}$$ if $$\mathbf{A}_r \mathbf{A}_r=0$$. A null blade  does not   have an inverse $$\mathbf{A}^{- 1}_r$$ but   does have   a pseudoinverse $$\mathbf{A}^+_r$$ with respect to the inner product.

The conjugate of any $$r$$-blade $$\mathbf{A}_r$$ is defined as


 * $$\begin{array}{rcl}

\mathbf{A}^{\dagger}_r & = & (- 1)^{{\operatorname{gr}}_- (\mathbf{A}_r)} \tilde{\mathbf{A}}_r\\ & = & (- 1)^{{\operatorname{gr}}_- (\mathbf{A}_r) + r (r - 1) / 2} \mathbf{A}_r .\end{array} $$

In any $$\mathcal{G}_n$$ geometric algebra of $$n$$ positive-signature vector units,


 * $$\begin{array}{rcl}

\mathbf{A}^{\dagger}_r & = & \tilde{\mathbf{A}}_r .\end{array} $$

Properties of the conjugate of a  non-null   $$r$$-blade $$\mathbf{A}_r \in \mathcal{G}^{\varnothing r}_{p, q}$$:


 * $$\begin{array}{rcl}

\mathbf{A}^{\dagger}_r \mathbf{A}_r & = & \mathbf{A}_r \mathbf{A}^{\dagger}_r\\ & = & | \mathbf{A}_r |^2\\ & = & \mathbf{A}^{\dagger}_r \cdot \mathbf{A}_r\\ \mathbf{A}^{- 1}_r & = & \frac{\mathbf{A}^{\dagger}_r}{\mathbf{A}^{\dagger}_r \mathbf{A}_r} = \frac{\tilde{\mathbf{A}}_r}{\tilde{\mathbf{A}}_r \mathbf{A}_r} .\end{array} $$

The inverse $$\mathbf{A}^{- 1}_r$$ of a non-null $$r$$-blade $$\mathbf{A}_r \in \mathcal{G}^{\varnothing r}_{p, q}$$ is with respect to either the geometric product or inner product, as $$\mathbf{A}^{- 1}_r \mathbf{A}=\mathbf{A}^{- 1}_r \cdot \mathbf{A}_r = + 1$$.

Properties of the conjugate of a  null   $$r$$-blade $$\mathbf{A}_r \in \mathcal{G}^{\circ r}_{p, q}$$:


 * $$\begin{array}{rcl}

\mathbf{A}^{\dagger}_r \cdot \mathbf{A}_r & = & \mathbf{A}_r \cdot \mathbf{A}^{\dagger}_r\\ & = & | \mathbf{A}_r |^2\\ & \neq & \mathbf{A}^{\dagger}_r \mathbf{A}_r\\ \mathbf{A}^+_r & = & \frac{\mathbf{A}^{\dagger}_r}{\mathbf{A}^{\dagger}_r \cdot \mathbf{A}_r}\end{array} $$

A null $$r$$-blade $$\mathbf{A}_r \in \mathcal{G}^{\circ r}_{p, q}$$ has a  pseudoinverse   $$\mathbf{A}^+_r$$ with respect to the inner product $$\mathbf{A}^+_r \cdot \mathbf{A}_r = + 1$$, but does not have an inverse $$\mathbf{A}^{- 1}_r$$ with respect to the geometric product.

For a non-null blade, $$\mathbf{A}^+_r =\mathbf{A}^{- 1}_r$$. For any $$r$$-blade, $$\mathbf{A}^+_r \cdot \mathbf{A}_r = + 1$$ such that the pseudoinverse of an $$r$$-blade is always an inverse with respect to the inner product.

If $$\mathbf{E}_i$$ is the $$i$$th basis blade in a ordered set of canonical bases for any $$\mathcal{G}_{p, q}$$ geometric algebra, then $$\mathbf{E}^{\dagger}_i =\mathbf{E}^{- 1}_i$$.

The conjugation of blades is also called  spatial reversion   and can also be identified with   Hermitian conjugation   of complex matrices.

responses
The concepts of reversion and conjugation do appear in most of the texts on GA that I've read, and it would be fair to define them. There are a few points I'd make: —Quondum 18:46, 11 March 2015 (UTC)
 * These involutions are two of several involutions, and would be best presented (as I've seen done) within a framework of involutions. The important ones are these two and their composite (plus of course the identity). There are, however, others.
 * The statement of being "primitive" is relative to the starting point. The simplest definition is in terms of the grade projection operation, rather than via reversal of a product of vectors.
 * The properties of reversal, expressed as the reverse of a product of vectors, applies regardless of whether the product used is the geometric product or the exterior (or outer) product.
 * The "norm" of a general multivector has not been defined in the article, and is usually used in restricted contexts (i.e. when not dealing with general multivectors). We'd have to consider whether introducing this would not detract from the article.
 * The conjugation that I'm familiar with is independent of the signature. What you have presented is some operator that I've never taken note of, and I'm not sure what its utility might be; there is a simpler signature-independent version that I'd prefer (negating all the odd-numbered grades). What you've presented might not even be basis-independent, but I'd have to check.

I made some changes to the Conjugation, because I noticed that it only applies to blades. It is "Conjugation of blades." It can't apply to multivectors in the way I originally proposed. There was also an issue where null vectors that square zero do not have an inverse. I cut out any statements that seemed to have some doubts. There are likely more issues! Twy2008 (talk) 22:26, 12 March 2015 (UTC)


 * You don't seem to have taken note of what I've said. We can't just expand on every possible interpretation of an operation. Mike Adler (2009) "4P8/Whatever Geometric Algebra: An Introduction to Clifford Algebras" says "we can define conjugation of h by g by sending h to ghg−1". This seemed to tie in with your previous description of the conjugate, (using a special element γ0). He also refers to reversion, rather than conjugation, as being a generalization of complex conjugation. Conjugation as you've defined it now should extend by linearity to general multivectors, but I'm pretty sure it doesn't. There are other involutions that have the properties that you seem to want (in relation to the inverse etc.) that have no dependence on the signature and are better-behaved. Are you indulging in OR here? —Quondum 14:25, 13 March 2015 (UTC)

Not original research, OR. A reference on the "Conjugate" of blades is page 59 in the book "Geometric Algebra with Applications in Engineering" by Christian Perwass (2009). The book "New Foundations in Mathematics, The Geometric Concept of Number" by Sobczyk (2013) seems to have a chapter 10 that is related to this. We can delete/ignore these proposed sections if they represent too much specific/technical detail for a wiki article. I am beginning to understand more clearly that this site is mostly interested in small abstracts with large lists of references. In this way, the site is mainly of interest to book publishers as a book promotion platform via references sections. Actual useful content in wikis isn't really desired. OR, and other rules of this site, are in place to discourage people from writing articles that contain actual content beyond abstracts and pointers to book stores. Twy2008 (talk) 16:52, 13 March 2015 (UTC)


 * I have to admit that I partially agree with your comments on the conservatory habits around W and the partly hilarious calls for citations, having dramatically increased lately in my perception, but I'd like to ask why to independantly expand on this article instead of integrating (not literally) it in the concept of Clifford algebrae. Regards. Purgy (talk) 09:58, 14 March 2015 (UTC)


 * I hear your exasperation, Twy2008. I had forgotten about Perwass's definition, and yes, I see it in his book. I do not have the Sobczyk reference. I'll make the following observation: the reversal is pretty much universally defined by all authors of GA texts, but the conjugate as defined by Perwass does not seem to be widely used. For example Dorst, Fontijne and Mann in "Geometric Algebra for Computer Science", a fairly thorough text on the subject, does not appear to mention it (though uses something similar, as already mentioned in a footnote in this article). There are many other more notable "conjugates", such as the Clifford conjugate. I was jumping to conclusions about OR, but that was prompted by the algebraic ugliness of the operation.  My concern with Perwass's conjugate is that it seems to be ill-defined: it depends on the choice of basis, and is thus simply a calculation convenience, not an operator of the algebra at all. We also need to remember that an encyclopaedia article is not a textbook. On the reversal, it is already mentioned, but not defined. We could define it and give its properties in about three lines; if we devoted this much space to each of the common operators, this article would become a monster. —Quondum 18:47, 14 March 2015 (UTC)

Well. This conjugate is a bit complicated. I've made another revision to "Conjugate of blades" and it now includes the null-blade and pseudoinverse cases. I still hold some hope that with some more revisions this info can make it into the article. These are basic operations that would be quite handy on the page as a quick reference/cheat sheet. As maybe you can tell, a clear idea of even these basic operations requires careful digging in the literature. Twy2008 (talk) 01:06, 15 March 2015 (UTC)

Quondum, I do hear what you are saying about "define conjugation of h by g by sending h to ghg−1". That conjugation is nice algebraically, and it is the versor conjugation of h by g, rotating vector h around vector g by 180 degrees... general conjugation makes something like a+b -> a-b, but should for r-vectors also have a squared magnitude for the r-vector times its conjugate. That rotation makes h parallel to g plus h perpendicular to g -> h parallel to g minus h perpendicular to g. With a good choice of g, you might be able to get the conjugate you want in a nice algebraic way. In the spacetime algebra y0By0 happens to give the conjugate of blade B in that nice way. I would like to generalize that type of conjugation by choosing a certain subspace pseudoscalar as the versor g. For each algebra G(p,q), maybe there is a certain versor g that produce conjugates? Kinda researchy. A rough idea is (conjugate of h) = -(h.I)h(h.I), where I is the pseudoscalar of the space in which h is and relative to which you want its conjugate. Not sure if this works or not. Maybe I will drop this whole thing until I can find something better in the literature. Twy2008 (talk) 14:07, 15 March 2015 (UTC)


 * The terminology "conjugation by an element" presumably comes from the identical operation in group theory. You will see that it is essentially the versor operation already in the article. There is nothing mysterious about it, it is algebraically nice, and it gets used a lot, although there is some variation between whether the inverse or reversal is used (essentially they're equivalent if the versor is normalized). Hestenes and others seem to have deemed it appropriate to re-invent the terminology (or rather IMO confusingly repurpose old terms). In this article conjugation by an element is simply called a versor operation.  When the versor (the conjugating element) is a vector, we simply get reflection on a vector.  So you'll see that it is not a 180° rotation, but a reflection.  Calling it "the conjugate" would be a misnomer, as there is no "good choice of g": pretty much any non-null vector will do. In the spacetime algebra, the vector is chosen to produce effectively a P (all three spatial directions) or T (time direction) reflection, depending on sign, but this only makes sense with respect to an observer's frame of reference.  The γ0 is simply the observer's worldline tangent vector, and there are as many of these as there are timelike vectors. —Quondum 18:05, 15 March 2015 (UTC)

merge Unit pseudoscalars into relationship to other formalisms
The section relationship to other formalisms also contains the examples of the complex numbers and the quaternions but shorter than the section unit pseudoscalars. Merge unit pseudoscalar into the other section. — Preceding unsigned comment added by 2A02:A03F:24D5:D300:C18E:EB07:DAF8:588D (talk) 18:07, 11 October 2015 (UTC)


 * I think, don't do it. There are quite different topics addressed in the mentioned sections. Purgy (talk) 06:24, 12 October 2015 (UTC)

External links modified
Hello fellow Wikipedians,

I have just added archive links to 3 one external links on Geometric algebra. Please take a moment to review my edit. If necessary, add after the link to keep me from modifying it. Alternatively, you can add to keep me off the page altogether. I made the following changes:
 * Added archive https://web.archive.org/20141129023605/http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/chris_thesis.html to http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/chris_thesis.html
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incorrect extensions of the inner and outer products?
A recent edit by an IP user added some formulas for the inner and outer products of k-vectors that appear to be incorrect. It claims that the formulas are common practice, but gives no references to support that claim. I think that a basic check of the algebra will verify that these formulas are wrong in general, and are just the commutator and anticommutator products, which are not generally equal to inner and outer products. Please confirm and remove these edits, if appropriate. Thanks. — Preceding unsigned comment added by 125.25.10.75 (talk) 22:33, 10 August 2016 (UTC)
 * I don’t think they were entirely wrong, but they were unnecessary: the section already gives definitions of the various generalisations of the inner product immediately below, with sources. I have therefore removed them.-- JohnBlackburne wordsdeeds 23:18, 10 August 2016 (UTC)

Typesetting
It turned out to me that I am disturbed by the different font of "a" and "$$a$$" denoting the same object in one paragraph or even in one line.

Please, may I ask if this is my singular sensitivity, or if -on occasion- one should try to unify the typesetting for the math-objects at least throughout a paragraph, and what the requirements by the MOS are. Purgy (talk) 13:43, 3 February 2017 (UTC)


 * I agree with you. We should strive to have consistent formatting within an article.  For the time being, formatting is considered separately for inline formulae and standalone formulae.  My feeling is that inline formulae should be consistently formatted (one of three: HTML, math, or &lt;math>, and further, that they should not result in significantly increased line spacing.  The question becomes which one to choose.  In this article, my preference is for one of the latter two.  —Quondum 19:02, 4 February 2017 (UTC)


 * I gave the first section's intro a shot with the &lt;math>-method. Please, have a look at it, and especially at the denotation for the Cliffford Algebra. Purgy (talk) 13:24, 5 February 2017 (UTC)


 * Mixing HTML and &lt;math> won't work. The options I see: $$\operatorname{Cl}(V,g)$$, $$Cl(V,g)$$, $$\operatorname{C\ell}(V,g)$$, $$C\ell(V,g)$$.  I think the first fits in best with the conventions used in WP, though it is it not common in this context (yet?).  —Quondum 16:00, 5 February 2017 (UTC)

Overemphasis on the geometric product?
Following on from the comment by above, I would like to see whether there are others who would like to comment on the emphasis on the geometric product in this article. I, for one, would like to see the exterior product being presented as more core to geometry than the geometric product, since it embodies more geometric information than the geometric product does, applies in more general contexts, and is more amenable to (general, rather than metric) geometric manipulations. It is also more directly geometrically interpretable. —Quondum 20:11, 23 February 2017 (UTC)
 * I tend to think of it the other way around: the geometric product is core to GA, more fundamental algebraically and geometrically. In simple operations you can think of interchangeably with the exterior and interior product, but for anything more complex you have to use the geometric product. E.g. reflection given by $ab = (e_{2} + 2e_{3} + 3e_{4})(4e_{1} + 5e_{2} + 6e_{3} + 7e_{4})$ for vector a being reflected in a plane with normal n. Sure you could break that down into parallel and perpendicular components, using interior and exterior products, but that stops being practical once you replace vector a with a bivector or something more complex, or replace n with two or more vectors representing multiple reflections/rotations.-- JohnBlackburne wordsdeeds 20:30, 23 February 2017 (UTC)
 * I guess that I should expand on what I said, and see how you respond. The geometric product (as a ring with addition and multiplication) does not identify the vector subspace of the algebra; it does not even uniquely determine the signature of the bilinear form.  To get these, you have to add structure not contained in the geometric product: the identification of the vector subspace.  Until you have this, you cannot define geometric (1-)vectors, lengths, areas, angles, or even of versors.  It also singularly useless at nonmetric geometries as pioneered by Grassmann, as exemplified by its inability to express a general outermorphism, or to deal with projective geometries.  In contrast, all of the following naturally follow from the exterior product (even if you "forget" the vector subspace): the grading of the algebra, the blade representation of oriented k-shapes and subspaces, general linear transforms of vectors and their associated outermorphisms (of which versors are simply the subset that preserve orthogonality).  It gives all this geometric content before a bilinear form is added.  Add only a bilinear form, and the geometric product (and the entire geometric algebra) is a consequence.  One has to go "outside the algebra (of the geometric product)" to achieve most geometric constructions (ever tried to express the oft-used exterior product or a contraction in terms of the geometric product without the additional structure of grade projection?).  —Quondum 21:39, 23 February 2017 (UTC)
 * I don’t have any problems with that – The algebra is an algebra over a vector space, so there is the vector space, or subspace. There are many products you could choose that satisfy the properties of geometric algebra, so you also need to specify a signature. Once you do that there is only one algebra, up to isomorphism. And there are only a handful of distinct signatures, especially if you disallow ones with zero terms. I am not familiar with Grassmann’s approach, except as as precursor to Clifford algebra and GA. But you can do different things with them, they are different algebras. As I noted you can do reflections and rotations in any dimension. You can do calculus with geometric calculus. You can do projective geometry with conformal geometric algebra.-- JohnBlackburne wordsdeeds 22:02, 23 February 2017 (UTC)
 * Combining isomorphism and definition in one argument feels circular to me. If one starts from a structure (such as a geometric algebra), one considers only its properties, not how one got there (e.g. that it is an algebra over a vector space, or that one has specified a signature).  One also has to take care to specify the type of isomorphism.  If we are referring to a ring isomorphism, what you say gets confused by isomorphism of distinct signatures.  If one means a "Clifford algebra isomorphism" (by which I mean a mapping that preserves not only the ring structure, but also the identification of the vector subspace), it makes more sense.  Anyhow, from a structural perspective, I don't see that the geometric product stands out as any more "fundamental" than any others; from a definitional perspective there are diverse ways to reach the same result.  If one wishes to consider it in the context of more general structures (such as differential geometry, and geometric algebra applies as long as the manifold is pseudo-Riemannian), one has to abandon the geometric product approach and start afresh.  Approaches that do not generalize to their own natural and well-studied generalizations are ... a little limited (less "fundamental"?).
 * However, I am not trying to trash the geometric product. It is a highly useful product, along with several others, particularly the exterior product.  Either suffices for defining a GA.  However, the opening "A geometric algebra (GA) is a unital associative algebra on a vector space over the field of real numbers endowed with a quadratic form" neither reflects its history nor the only way of characterizing a GA.  Besides, the same description fits when one is talking about the algebra with the exterior product, only it would continue: "[...], i. e. a real exterior algebra with a quadratic form."  I am only suggesting that we could open the description without appealing to the abstractness of a Clifford algebra, and try for something more concrete.  Maybe I should do a little homework and try to craft an alternate introductory sentence before going further.   —Quondum 23:49, 23 February 2017 (UTC)

To me a geometric algebra is an algebra with the geometric product. You can approach it a number of ways, and it’s common especially when first encountering GA to start with the inner and exterior product. But in this context they are only learning aids – you start with something familiar and use it to build something new. Once you have the geometric product you effectively have all you need. You can then use it to (re-)define the inner and exterior products, but the geometric product is fundamental. One reason is it generalises, to products of any two objects in the algebra. Generalising the inner and exterior product though is much harder, as seen in geometric algebra there are multiple ways to do it. The geometric product has no such issues.

I do agree the current definition in the first sentence is somewhat lacking. It’s correct but is not really useful for understanding the algebra, and the lead section should be all about helping people understand what it is about, with more technical definitions appearing later on. It would instead be better to have something of the definition based on, and the relationship to, inner and exterior products. Not the whole of geometric algebra but the gist of it, to draw in readers familiar with the products from e.g. vector algebra.-- JohnBlackburne wordsdeeds 09:20, 24 February 2017 (UTC)


 * Okay, so I have a simple answer to my first implied question: it does not make sense to focus on any change in emphasis in the article. But we seem to have agreement that the theoretical terminology in the lead could be toned down, even if I might not be fully with you on the detail.  Let me see what I can do, and see whether it is then generally agreed to.  —Quondum 21:09, 24 February 2017 (UTC)

When I first read this after having taken only high school mathematics and physics, I saw the geometric intuition of an oriented plane for bivectors, but was utterly and completely confused by the definition of a geometric product. Such eluded me until I took elementary abstract algebra, learned the concept of "free" constructions (like free groups), and began to understand geometric algebra as a quotient algebra. Even then, I don't think the current article captures much of the original motivation of this definition, and I feel that abstract algebra shouldn't be needed to get at least a basic understanding of this concept.--Jasper Deng (talk) 02:50, 25 February 2017 (UTC)
 * , I'm with you on this. Finding an approach that provides an understandable, direct description is not easy.  I find the usual pedagogical approach of trying to build the geometric product up from the dot product unhelpful, circular and not appropriate to an encyclopaedia, and starting from a Clifford algebra is a non-starter in this respect.
 * , a niggle about your comment on representing a projective geometry using GA: CGA does not cut it. To quote from Dorst et al (2010), Geometric Algebra for Computer Science: An Object-Oriented Approach to Geometry, p. 499: "Projectve Geometry.  Unfortunately, there is not yet an operational model for projective geometry.  That would have projective transformations as versors (rather than as linear transformations, as in the homogeneous coordinate approach).  ...  Its blades would naturally represent the conic sections.  Initial attempts [...] do not quite have this structure, but we hope [...]."  Though I have my own hopes and ideas on how to achieve this, it still seems to be a gaping hole from the literature.  —Quondum 17:37, 26 February 2017 (UTC)
 * I think what's missing is the notion of a "free" construction here. When I first read the article, I specifically was confused by the seemingly-circular definition: the geometric product of two vectors is a bivector, and in turn a (simple) bivector is the geometric product of vectors. I could not see how one could define an operation without specifying the domain and codomain first, and could not grasp the notion that the geometric product generates bivectors and other multivectors from vectors. Along with a discussion of the motivation of geometric algebra, I think explaining this could help clear things up. Ideally, we'd get an applied textbook (e.g. in physics) that provides a great explanation of these as a source.--Jasper Deng (talk) 20:06, 26 February 2017 (UTC)
 * I'm having a little confusion following you, but I have a feeling that we share the same feeling. It is definitely confusing if one is told that one can define a simple bivector as the geometric product of two vectors, but then only if they happen to be orthogonal, because if they're not, we have something that's not even a pure bivector.  I will assume that by "free" you mean "noncircular".  To take your example, construction of a bivector from two vectors, I think one can explain the construction of a codomain via the exterior product: consider the example of torque in physics.  This clearly acts like an element of a vector space in both two and three dimensions: additive with scalar multiplication, only the dimension of this space is obviously 0 in one dimension, 1 in two dimensions and 3 in three dimensions.  This generalizes to higher dimensions, and we can already reasonably assume that there is no natural identification with the initial vector space (and if one needs any convincing, then just try to match a general basis transform of three dimensions with the codomain!).  In all, the geometric construction to produce the exterior algebra seems straightforward (understandable) and "free" (if you mean it in the way I'm taking it) if done along these lines, and is undeniably geometric.  Building reflections much as it is done in the article would then allow us to "discover" the geometric product lurking underneath.  I would love finding a textbook that takes such an approach, but so far I remember seeing only abstract algebraic and the more applied "let's assume the geometric product cos we know it works and see how we can define/teach it" approaches.  —Quondum 21:30, 26 February 2017 (UTC)
 * Oh what I meant by "free" is the following: the tensor algebra is the freest algebra over our original vector space, and the geometric algebra is only "less free" than that because of us imposing $$\mathbf{v} \otimes \mathbf{v} - Q(v, v)1 = 0$$ by taking the quotient. Basically, subject to that restriction, we get the geometric algebra by freely applying the geometric product among all the vectors of the original vector space. The resulting objects aren't something that we defined before defining these operations. Rather, these objects come into being to "flesh out" the codomain of the product - instead of first defining the domain and codomain's members and then the operation, we are (in some sense at least) first defining a domain, then the operation, and only finally the codomain, before extending further. In other words, I was very confused by the statements "a bivector is the geometric product of (orthogonal) vectors" and "the geometric product produces bivectors when it takes in (orthogonal) vectors" when considered definitions: the former statement begs the question of "what is the geometric product?" and the latter statement begs the question of "what is a bivector?". This is what I saw as circular. Now I understand that the elements of the geometric algebra are really equivalence classes of tensors under our quotient relation $$\mathbf{v} \otimes \mathbf{v} - Q(v, v)1 = 0$$, since I now have a grasp of tensor products and taking quotients.
 * Perhaps more to the point, though, why are we taking the quotient? There are tons of other quotients I could take of the tensor algebra, what makes this one particularly special? As a reader I'd like to see that discussed.--Jasper Deng (talk) 22:09, 26 February 2017 (UTC)
 * Ah, that "free". I'd hoped not, because then all we would have to do is to replace the article with a redirect to Clifford algebra.  Or, we could turn it into an article "Adaptations of Clifford algebras for use in physics".  Whatever, there would be no point in repeating the construction of a Clifford algebra as a quotient of a tensor algebra; that would be linked.  Conversely, quotients are the only motivation I've seen for the construction of a tensor algebra as a single vector space, so maybe we're facing multiple "why?"s.  I don't think there is any answer to the "why this particular quotient?" other than that this is a rigorous and simple construction that yields the mathematical object that we seek.  The construction is not unique.  (It also has the rather nasty side affect that what are perfectly natural tensors inside the algebra, such as the stress–energy tensor, no longer have a natural representation inside the algebra!)  Construction via the tensor algebra simply gives us confidence that we have not missed something.  I also think that this is the wrong approach for introducing a perfectly reasonable high-school physics-related topic for people who will never use quotients, even though, once you are a math/physics under/postgrad, revisiting it using the new tools at your disposal might make it look so simple and neat.  —Quondum 22:47, 26 February 2017 (UTC)
 * Exactly - I would like a presentation free of abstract algebra (no pun intended). But how would we define bivectors without appeal to tensor math? What are the applied motivations for seeking these objects in the first place? I like the part that summarizes the geometric product in terms of four axioms (which easily result from the quotient algebra construction), but it still begs the question of the result of the geometric product. In physics, it would be nice to say that a bivector is a quantity with magnitude and two directions, just like an ordinary vector is a quantity with magnitude and direction, but that only applies to simple bivectors.--Jasper Deng (talk) 23:52, 26 February 2017 (UTC)

The best elementary definition of a bivector is an area-element with magnitude and orientation, just as a vector can be defined as a line segment with magnitude and direction. As with vectors this is a signed orientation, so bivectors B and -B are parallel but oriented in opposite ways, just like vectors v and -v. You can scale bivectors the obvious way, and add two bivectors just like two vectors. In physics they have a number of applications: angular momentum, angular velocity and torque are all better described with bivectors than vectors (otherwise you have to create a new class of vectors, the pseudovectors, to account for their odd behaviour). There is no need to define all that here though, just link to the article on bivector.-- JohnBlackburne wordsdeeds 00:11, 27 February 2017 (UTC)
 * How about this then: we introduce the geometric product as a (bilinear) operation taking pairs of vectors to sums of bivectors (as defined by you) and scalars. This definition suffices in 2 dimensions where one only has bivectors, vectors, and scalars. Then we can generalize to three dimensions and speak of trivectors. Only after these special cases do we then introduce the Clifford algebra connection. This would be a lot more accessible to high-school physics students; technical-minded readers can read the main Clifford algebra article.--Jasper Deng (talk) 00:47, 27 February 2017 (UTC)
 * This suffices in any number of dimensions right from the start, with no need to define higher-dimensional k-vectors at first, except that it is the exterior product, not the geometric product. Two dimensions is a good place to start the illustration of bivectors.  And my feeling is that one should not introduce the Clifford/geometric product until all higher k-vectors and metric aspects are introduced, for example how rotation preserves a quadratic form.  I expect that the geometric product can be derived at that point, no need to appeal to (what is at least to many readers) a mythical/abstract/unknown Clifford product.  It could end by saying "Oh, look: this happens to be precisely the real Clifford algebras."  —Quondum 01:41, 27 February 2017 (UTC)
 * Let me see if I can find such a presentation of geometric algebra and its motivations for use in my university's library (I am already planning a trip there to answer my recent question at RDMA about left null spaces and cokernels).--Jasper Deng (talk) 21:00, 27 February 2017 (UTC)
 * I've been re-reading a bit of An Introduction to Clifford Algebras and Spinors by Vaz and da Rocha. They define what they call a Grassmann algebra (which they use to mean an exterior algebra with an extension of a bilinear form) and a geometric (aka Clifford) algebra.  Identifying them via isomorphism as vector spaces, each introduces its own multiplication (exterior and geometric respectively) to the space.  Even when talking of a geometric algebra, they say (p. 83): "To summarise, in order to define the multivector structure in a basis-independent way, it is best to use the exterior product."  Which supports the idea that the exterior product may be the best way to introduce a geometric algebra.  A subtlety supports JohnBlackburne, too, however: from a definitional perspective, the multiplication operation of a geometric algebra is the geometric product, period.  This suggests that we should not treat the exterior product as a defined part of a geometric algebra (notwithstanding its derivation and clear geometric interpretations), but that it can be defined and would be useful to "build" the algebra.  —Quondum 03:51, 1 March 2017 (UTC)

Rewording the lead
, and others, I wrote the following as a possible replacement of the first paragraph and first sentence of the second paragraph. Do you think that this adequately introduces it in a way that tells the average reader about what to expect in the article rather than trying to define the topic?
 * The geometric algebra of a vector space is an algebraic structure, noted for its multiplication operation called the geometric product on a space of elements called multivectors, which is a superset of both the scalars $$\mathbf{R}$$ and the vector space $$V$$. Mathematically, a geometric algebra is defined as the Clifford algebra of a real vector space with a quadratic form.  The vector space may be the vector space of interest (as with vector algebra, but for any number of dimensions and with any quadratic form), or a larger space, as with conformal geometric algebra.  In practice, several derived operations are generally defined, and together these allow a correspondence of elements, subspaces and operations of the algebra with physical interpretations.  The algebra may be graded under  admits a derived operation, the exterior product, which defines an exterior algebra on the same space and provides many of the geometric interpretations of its elements, as well as demonstrating that a GA is basis-independent.
 * Elements of the algebra are usually assigned interpretations. The scalars and vectors may have their usual interpretation, and are elements of distinct subspaces of a GA.  Bivectors provide a more natural representation of classically pseudovector quantities such as oriented area, torque, angular momentum, electromagnetic field, the Poynting vector and oriented angle of rotation.  A trivector can represent an oriented volume, and so on.  A blade can represent any subspace of $$V$$.  Operations such as rotations and reflections have a particularly simple algebraic are represented as elements of the algebra called versors , and orthogonal projections onto a subspace as a blade .  Unlike vector algebra, a GA naturally accommodates any number of dimensions and any quadratic form such as used in relativity.  Not all physical and geometric operations find direct representation within the algebra, including general linear transforms of $$V$$, which are represented as a class of linear function named outermorphisms, as well as several other quantities such as the stress–energy tensor or Ricci curvature, in contrast with the more general  universal tensor algebra.

All comments welcome. —Quondum 00:31, 6 March 2017 (UTC)
 * One thing is that we shouldn't limit ourselves to just geometric motivations, but also physical ones, since physics students will be reading this. This would entail mentioning how many pseudovectors in rotational motion (such as angular momentum, torque) and elsewhere (e.g. the Poynting vector) are more naturally interpreted as bivectors.--Jasper Deng (talk) 01:13, 6 March 2017 (UTC)
 * Thanks – good comment. I've copyedited the text somewhat to incorporate this.  If others don't object in the next few days, this makes me feel comfortable enough to use this, so that further input can be via direct editing.  —Quondum 03:00, 6 March 2017 (UTC)
 * I think it can be put in as-is. I wish to edit it further, but would rather do it directly.--Jasper Deng (talk) 08:59, 6 March 2017 (UTC)


 * As an occasional passer by I want to state (on invitation ;) ) that I definitely prefer this suggestion to the status quo. Especially, I highly estimate the neatening of pseudovectors and other ideosyncrasies from inherited physicists' "sloppiness". I am just lightly unsure about the formulation "may be the vector space of interest" and perhaps miss some emphasis in the lede itself on the reasons for coining this term (referring to Selfstudier!) at all, when it is mathematically a real Clifford algebra. I am not sufficiently competent to contribute on the thoughts about alternative constructions; avoiding the quotient of a free algebra in the lede, is certainly along my lines of thought.


 * In any case I feel indebted to Quondum for implementing the whole task of reformatting the math. Thanks, I enjoyed rereading! :) Purgy (talk) 11:42, 6 March 2017 (UTC)


 * To let everyone directly edit it, and since I've given a day for comment, I've replaced the beginning of the lead with my suggestion above, with minor copyediting. I've simply removed the clumsy part about "the vector space of interest".  —Quondum 04:28, 7 March 2017 (UTC)

First shot
Isn't generalizing to arbitrary fields dangerous to Clifford algebras for their possible characteristic 2? :) I do not know about a complex geometry (complex quadrics, conics, varieties, ...), but neither about outstanding rigor in articles promoting GA (I always read the wrong ones). Purgy (talk) 07:30, 2 September 2017 (UTC)


 * No, I don't believe so, if you mean specifically Clifford algebras, but I'd say also for GAs. It seems that Clifford algebras generalize to arbitrary free modules over commutative (unital) rings, including where 2$−1$ does not exist (though the definition in terms of a quadratic form remains unaffected, except that I guess that it may be necessary to a restrict the quadratic form in infinite dimensions).  Care is indeed needed in the working when 2$−1$ does not exist.  The generalization of CAs to characteristic 2 seems to me to be standard (I do not recollect ever seeing an exclusion from the definition of CA that purports to be general).  In particular, see Clifford algebra, where no such exclusion is made from the definition, though it does restrict itself to fields.
 * Here it is necessary to distinguish, at least semantically, a CA from a GA. Cortzen does so by requiring the structure of the exterior product on the CA, which disqualifies some CAs in characteristic 2 from being what he defines as GAs (and I believe I've captured this requirement).  This still allows characteristic 2 even in a GA.  Because Cortzen is probably not the final word on this, the exact definition of a GA is still uncertain (and indeed I have counterexamples that he excludes but I think should be regarded as GAs).  —Quondum 12:10, 2 September 2017 (UTC)

Statement about limitations of GA
I think that it is worth making the point that the algebraic operations have limitations. For example, the outermorphism of a homothety other than the identity cannot be expressed algebraically without a helper function such as grade projection. This relates to the text
 * "However, a direct representation within the algebra has not been found for all physical and geometric operations, including general linear transforms of $$V$$, which are represented as a class of linear function named outermorphisms, as well as other quantities such as the stress–energy tensor or Ricci curvature, in contrast with the more universal tensor algebra."

Macdonald merely gives examples of outermorphisms that can be expressed simply in terms of the basic algebraic operations, which does not contradict the position. The question is: what is the actual point that is worth making? For example, look at Dorst, Fontijne, Mann (2007) p. 499: "Unfortunately, there is not yet an operational model for projective geometry. That would have projective transformations as versors (rather than as linear transformations, as in the homogeneous coordinate approach)." They were trying to make a point. The point that the algebra does not really do everything needed from it is not to be ignored. General linear transformations are needed. These functions are defined as primitives in GA to get around the problem. —Quondum 01:12, 30 August 2017 (UTC)
 * https://link.springer.com/content/pdf/10.1007/s00006-015-0625-y.pdf (Dorst) Selfstudier (talk) 09:26, 30 August 2017 (UTC) and
 * https://arxiv.org/abs/1507.06634v1Selfstudier (talk) 09:32, 30 August 2017 (UTC) and
 * https://link.springer.com/article/10.1007/s00006-017-0798-7 Selfstudier (talk) 09:50, 30 August 2017 (UTC)
 * , if you are trying to make the point that one or two specific groups of transformations have been represented, I am aware of this. In particular, a representation in terms of the geometric product of GL(R$3$) and a few others have been found.  The deficit from a general representation of GL transformations (equivalently general 2nd-order tensors) in any space is a gap in the literature, not in the algebraic possibilities (hence only "has not been found"; a better description would be "appears not to have been published yet").  The same is true for general outermorphisms.  The fact of this gap seems worth mentioning (Dorst, Fontijne, Mann (2007) and Perwass (2009) appear to lament this deficit).  However, it is evident that this can be done only in the algebra of a higher dimension.
 * The point remains that the operations that are routinely applied in a GA (grade projection, the main involution, reversion, general outermorphisms, exterior product, contractions) cannot be expressed within the algebra itself (an OR point, but nevertheless true). This gap in the capability of the algebra is again worth understanding.  Overall, GA routinely defines and uses many operations/functions that lie outside the algebra (not to be confused with representation through embedding in a larger algebra).  Do you think there is no statement to be made, or just that it should be revised?  —Quondum 11:53, 30 August 2017 (UTC)
 * I'm interested in what can be done, not what can't, you choose. Good explanation of "vector space of interest" tho.Selfstudier (talk) 12:05, 30 August 2017 (UTC)
 * Btw, do you know this book? https://books.google.es/books?id=QP1oDQAAQBAJ&pg=PA9&redir_esc=y#v=onepage&q&f=false Selfstudier (talk) 12:36, 30 August 2017 (UTC)
 * Interested in what can be done? That general GL and hence PG has not been published yet intrigues me, especially since there appears to remain active interest and it is not difficult.  What cannot be done is interesting because it guides searches away from futile directions.
 * I'm not sure what you're referring to on the "vector space of interest" – that was trimmed from the lead. However, it seems to be a useful concept, essentially the space you wish to construct the algebra around to allow its manipulation.
 * No, I did not know that book, but it looks good. Though its table of contents suggests that it does not cover GL transformations.  —Quondum 13:01, 30 August 2017 (UTC)

Did I not do it right? I thought I set it to come up at page 9 and 10? What do you make of his explanation of the difference between GA and CA? Selfstudier (talk) 14:16, 30 August 2017 (UTC)
 * It does come up on page 9. However, what I get about the difference between GA and CA is a difference of emphasis, not a mathematical one, including the association of geometric meaning with elements and operations of the algebra.  I disagree with some statements, e.g. "Multiplication and division [...] are the only two geometrically meaningful operations upon algebraic expressions having geometric meaning"; as I've said, one cannot algebraically synthesize many of the standard operations from them.  It would help if you were more specific, e.g. what specific text you mean, and what point you are trying to make.  —Quondum 17:02, 30 August 2017 (UTC)
 * He's still around, y'know...https://link.springer.com/content/pdf/10.1007/s00006-016-0664-z.pdf The Genesis of Geometric Algebra:A Personal Retrospective David Hestenes Start reading at "Do not confuse Geometric Algebra (GA) with Clifford Algebra (CA)!" Lol.Selfstudier (talk) 18:08, 30 August 2017 (UTC)
 * Good grief :S Rschwieb (talk) 18:18, 30 August 2017 (UTC)
 * Idk whether Dorst is still working with his "multiplicative principle" I can sympathize with it, tho, in a computing sense (same idea, blades and versors, no arbitrary additions so you know the result is always geometric)Selfstudier (talk) 18:49, 30 August 2017 (UTC)
 * Interesting. Not sure what to make of the paper.  But the difference emphasized in "Though they stand on common ground, they differ profoundly in culture, content and praxis" does not contradict my position.  On the multiplicative principle, I'm with Dorst, at least for projective transformations in arbitrary dimension: these can be expressed as rotors.  —Quondum 19:27, 30 August 2017 (UTC)
 * Bit more at bottom p372 and 373 "An explanation for this can be found in the conceptual difference that underlies the superficially synonymous terms ‘Geometric Algebra’ and ‘Clifford Algebra'.. Selfstudier (talk) 21:37, 30 August 2017 (UTC)
 * K, that's 3 RS secondaries for GA not equals (equivalent?) CA, must be worth a para somewhere, haha.(Maybe I should edit the CA page) Selfstudier (talk) 21:43, 30 August 2017 (UTC)
 * Under History? Not unless we devote space to Hestenes's personal perspectives.  —Quondum 22:02, 30 August 2017 (UTC)


 * Replying to the original proposal, I don’t see why that needs adding. Yes, you cannot represent all of physics in GA. You cannot represent all of physics in the Natural number system. But there is no such disclaimer in that article, and none is needed here either. Simply say what the algebra can do. If there is any claim that GA can be used for the whole of physics in the article it should be removed. There is no point adding a disclaimer, when no such claim is being made.-- JohnBlackburne wordsdeeds 21:45, 30 August 2017 (UTC)
 * Unfortunately, there is some confusion about what constitutes the algebra. If limited to the CA operations (i.e. a true algebra), then as I've been saying, many of the operations being discussed fall outside the algebra.  This is not about all of physics, just what is in the article.  I think it is worth clarifying such a confusion.  —Quondum 22:02, 30 August 2017 (UTC)
 * I don’t see where there is confusion in the article. It can be confusing at times as e.g. different authors use the names CA and GA in different ways, but I don’t think this article suffers from this problem. And even if there is such confusion, I don’t see how it justifies the disclaimer you proposed: nowhere does the article say "GA can be used for all of physics"-- JohnBlackburne wordsdeeds 22:36, 30 August 2017 (UTC)
 * My original proposal intended no such a claim, but I guess the wording could be interpreted that way. (Keep in mind: I was not proposing re-insertion of the removed text quoted above.)  Let's see if I can reword it: To consider the operations in the article (such as ⟨A⟩$r$, A ⌋ B, A ∧ B) as "part" of the algebra, we need to add structure to a CA: the ring operations (+, ×, ÷) of the algebra are insufficient to generate these operations; it is not possible to map an element of the algebra to the corresponding basis coefficient scalar using only the ring operations and inverses.  A GA must be defined as a CA with added structure if the exterior product etc. are considered to be within the algebra.  This is a purely mathematical argument.  The question arises: are these extra very natural ubiquitous geometric operations just functions that cannot be expressed algebraically, or are they considered to be "part" of a GA?  —Quondum 23:54, 30 August 2017 (UTC)
 * In my perception there are sentiments, which oppose to mentioning any disclaimers about capabilities of GA, as regards their applications in e.g. physics. Well, I never noticed any claim about the exhaustive and superior capabilities of integers in their general application in physics &mdash;disregarding (halved) quantum numbers. In stark contrast, I noticed GA as promulgated as a general savior for the problems of handling newer physical quandaries (spin, GTR, ...). So imho, a well reasoned disclaimer about the intrinsic deficiencies in strict GA in this article, if not even a hint to more complete "tools", is a recommendable service to the readers. Purgy (talk) 08:07, 31 August 2017 (UTC)
 * I want to see secondary RS making these arguments (that's Wikipedia, your and my opinion don't count, we need sources)I think I need a cn tag for GA is a CA in the lead...:)Selfstudier (talk) 08:49, 31 August 2017 (UTC)
 * Here we need to distinguish between excessive claims about GA and attempts to define GA. The first might be best ignored rather than disclaimed (e.g. the claims that GA is a "universal mathematical language" does not seem to be echoed, only ignored), though problems that it does solve can be listed (e.g. that a GA/CA provides a natural spinor representation, but claims of "significant advances in GR" probably have little secondary support).  Selfstudier, you make a good point, which I read as follows: we are lacking a clear cited definition of a GA as a mathematical object, and we should seek such a definition from a secondary source.  We currently have an attempted definition, which deserves a  tag.  A suitable definition should address my concern about prominent and geometrically significant operations (e.g. ∧) that do not seem to be included in this definition, but which are in general treated as part of a GA.  —Quondum 12:00, 31 August 2017 (UTC)

https://groups.google.com/forum/#!topic/geometric_algebra/-t8wIdUHvOQ (from back when I was trying to do this page up years ago :) so https://arxiv.org/pdf/1707.02338.pdf might be helpful. Selfstudier (talk) 15:45, 31 August 2017 (UTC)
 * I like the definition in :
 * If the linear space of the Clifford algebra W also is given a Grassmann structure by the zero bilinear form, we get a double algebra W$B,0$ = W. A geometric algebra over B, denoted Λ(V, B), is a double algebra isomorphic to W$B,0$ with an isomorphism fixing V.
 * The key here seems to be that it is defined as a double algebra, which has two products, the geometric and exterior products. Any field can be used, but the bilinear form must be diagonalizable, or equivalently, there must exist an orthogonal basis of the vector space V for the bilinear form.  I think that despite the isolated reference, this is better than what we currently have (notwithstanding statements like Perwass's equating GA and CA: "... we will talk about Clifford algebra instead of Geometric algebra. Recall that these are just two names for the same thing. The only difference is that when we talk about Geometric algebra we would like to emphasize the geometric interpretation of the elements of that algebra."  Close reading suggests that he really implicitly means a double algebra, but falls into the trap of assuming that this is inherent.)  If we define a GA as a double algebra, then my reservations about inadequate algebraic underpinnings disappear.  (A personal perspective: it addresses the algebraic issue, but the orthogonal basis requirement excludes some particularly interesting cases.)  —Quondum 01:35, 1 September 2017 (UTC)
 * It's a usage thing mainly, I think, Clifford fans say a GA is a CA because it contains one, people using the "Galgebra" for something geometrical/physical would call it a GA more often than not and some algebras referred to as geometric by CA people would not be called that by most GA people. There are often interesting papers on GA "hidden" in the CA pile.Selfstudier (talk) 11:12, 1 September 2017 (UTC)
 * Yes and no. From a classification of structures perspective, we say that every ring is a commutative group but not vice versa, or equivalently that a ring is a group with the further structure of multiplication.  Similarly, every GA is a CA with some additional structure: the structure that gives us the exterior and contractive products, at least on the subsets of interest (e.g. in CGA, I think we are interested in these products on elements of the null cone).  Perwass glosses over this, Cortzen defines the subset of interest as the whole algebra, and I define this subset in a more restricted way.  In every case, we assume the existence of these two operations with their specific geometric meaning on the subsets of interest.  I think the simplest would be to say something along the lines that a GA is a CA with (further) products that are ascribed their geometric meanings, with references to the disparate descriptions.  Then there are the sociological aspects, as you say.  We cannot simply say that a GA is defined as a CA.  —Quondum 12:22, 1 September 2017 (UTC)
 * I would also avoid the use of the word "double", this is used a bit inconsistently and will likely add to rather than subtract from the general confusion--Grassmann–Cayley algebra Selfstudier (talk) 12:52, 1 September 2017 (UTC):::
 * Cortzen, new and improved version http://ctz.dk/wp-content/uploads/2016/10/Direct-Construction-of-Clifford-and-Geometric-Algebras-and-their-basic-algebraic-Structures-v.2.0.pdf Selfstudier (talk) 18:55, 1 September 2017 (UTC)
 * Hmm. IMO, his construction is lacking in one or two spots.  Such as in the definition of a Clifford algebra.  If one is to define a Clifford algebra for an arbitrary commutative ring, then restriction to diagonalizable bilinear forms is probably nonstandard and unusually restrictive.  One is also interested in Clifford algebras of quadratic forms for which there does not exist an orthogonal basis.  Especially in geometry.
 * Anyhow, I've included the idea that a GA is defined such that the exterior product is part of the definition; if this can be done with Grassmann–Cayley algebra, it makes sense to do so here too. I think I should pause now while everyone shoots me down ;-)  —Quondum 02:04, 2 September 2017 (UTC)
 * Reals...from Clifford Algebra page "The geometric interpretation of real Clifford algebras is known as geometric algebra." together with a link out to this page. Selfstudier (talk) 08:47, 2 September 2017 (UTC)
 * And who is guilty of adding that unsourced statement? .  Probably best removed.  —Quondum 13:27, 2 September 2017 (UTC)
 * Orthogonal basis, forget about tensors.Selfstudier (talk) 08:54, 2 September 2017 (UTC)
 * I like a construction in terms of a basis, and which does not require a quotient of the tensor algebra to define. The orthogonality requirement on the basis is too strict, however.  It redefines CAs in a nonstandard way that excludes some very relevant CAs.  The good news is that the orthogonality requirement can be dropped, recovering all the excluded CAs.  —Quondum 13:41, 2 September 2017 (UTC)
 * Nothing wrong with restricting the usual CA definition which is way too general for GA purposes. What excluded CA's are you referring to? And how they are used by GA people? Selfstudier (talk) 15:07, 2 September 2017 (UTC)
 * Cortzen redefines Clifford algebras in a nonstandard, inequivalent way. This is not okay.  This is separate from any consideration of GAs.  Who cares about "GA people"?  I care about what is geometrically meaningful.  The CA of a quadratic form over a field of characteristic 2 $$q\left(\sum_{j=0}^{2n-1} x_i e_i\right)=\sum_{i=0}^{n-1} x_{2i} x_{2i+1}$$ is geometrically meaningful, but is excluded by Cortzen because no orthogonal basis exists.  I think a more sophisticated definition of GA will be needed if a definition of GA is to encompass cases such as this.  At the least, no quadratic form should be excluded.  —Quondum 15:48, 2 September 2017 (UTC)

https://www.physicsforums.com/insights/interview-mathematician-david-hestenes/ Shoulda asked him to define a GA, lol."Virtually anyone who prefers the term “Clifford algebra” to “Geometric algebra” is unaware of the power and broader range of applications given to GA by its various geometric interpretations. That is abundantly clear in the GA references already cited." Selfstudier (talk) 12:35, 2 September 2017 (UTC)
 * H has a great intuition and deserves a lot of credit for increasing awareness of GA, but lacks rigour. He also makes denigratory statements and grandiose claims without being able to pinpoint what he means.  He should defer to mathematicians to help find and restate what he means.  Vector algebra fails to adequately distinguish vectors from pseudovectors; GA, in its current form, makes a similar blunder, which I expect is partly responsible for its slow uptake.  —Quondum 13:55, 2 September 2017 (UTC)

Lead paragraph digression
Maybe we should avoid a digression to describe a related but different algebra in the very first paragraph of the lead? Also, what is the focus – Hestenes's vision of what GA is? Or should it be describing what is known mathematically about GAs? The overlap of the two has not total, IMO. Also, to describe Hestenes's approach as either original or axiomatic might not go down well with an average modern pure mathematician. But maybe I should avoid pushing my own biases too brashly. —Quondum 14:32, 4 September 2017 (UTC)
 * In a sense it's the same algebra, just a different product on the same space like the exterior algebra is. This is part of the problem here, we start off (not unreasonably) trying to answer the question "What is a GA?" One answer is it's a Clifford algebra, tho if you want to know what is a Clifford algebra, you could just go to that page, right? So what's this page for? Its what's a GA is used for (GA is a practical subject) that mainly determines the (signature of the) algebra you use (real, relatively low dimensional). Although we could (for motivation, perhaps), we wouldn't define a Clifford algebra in order to investigate the complex numbers, we can and do define all sorts of algebras for different purposes. The ones I have put answer the question "What is a GA" in a way that emphasizes the G in GA and provides the connection to CGA.(I am not entirely happy with this, it should also address the poor ol physicists a bit more directly than it does at the moment, Lie, Spin, Pin an so on). As for Hestenes, if you have references that critique him or his approach or indeed GA in general, I have no objection their being included here, in fact, if they exist then they should be included here if we want to be encyclopaedic. Selfstudier (talk) 15:39, 4 September 2017 (UTC)
 * I am not objecting to the mention of Grassmann–Cayley algebra in the article; it seems it may be quite a beautiful structure and is particularly relevant to the geometric aspects, though I have not looked into it yet. But I'm trying to say that to insert that amount of detail in the first paragraph is counter to the Wikipedia style.  There are also several points of inaccuracy in what has been added.  I see you've removed my addition the exterior product as a defined operation, thus considering the GA to me merely an applied CA (and effectively disagreeing with Cortzen's view, though this point is unclear anyway).  I have no objection to addressing physicists, but I would really prefer to keep the mathematics straight.  —Quondum 21:17, 4 September 2017 (UTC)
 * If you want to shift some of it into the body, expand it there and then summarize in the lead, that's fine by me, the body may need quite a bit of rearranging to accommodate it correctly.If you want to add Cortzen's latest in there somewhere as a ref, that's also fine with me.I have no problem with the mathematics and that doesn't have anything to do with the physicists so I don't really understand your point there.You can say its an applied CA if that floats your boat,whether that would actually add any value to the article I couldn't really say.Selfstudier (talk) 21:52, 4 September 2017 (UTC)