Talk:Geometric algebra

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Have a look at the recent article on Clifford algebras, largely written by Richard Borcherds. This is some of the most clearly written mathematics I have seen on wikipedia (assuming others have not overedited it). Readers should compare this article with the one under discussion. -Jenny Harrison, June 2006

Here are notes I haven't had time to add TEX to: [edits, anyone?]

Geometric algebra is Clifford algebra given a geometric interpretation which makes it useful in an exceptionally wide range of physics problems, particularly those that involve rotations, phases or imaginary numbers. Proponents of geometric algebra say that it more compactly and intuitively describes classical mechanics, quantum mechanics, electromagnetic theory and relativity than standard methods do.

The elements of geometric algebra are called blades. In two dimensional geometric algebra there are scalars (grade-0 blades), vectors (directed line segments, grade-1 blades), and bivectors (directed areas, grade-2 blades). The direction of an area is defined by the direction taken around its perimeter, usually in a right-handed sense so that counterclockwise is considered positive. Similarly, in higher dimensional algebras there are directed 3D volumes called trivectors. The highest grade blade in an algebra is called a pseudoscalar.

Just as real and imaginary parts are combined to make a single complex number, so all different grades of blades in an algebra are combined to make a single entity called a multivector. In an algebra describing a geometric space of dimension d, there will be d^2 independent blades in a single multivector. For example a basis for a 2D geometric algebra has a scalar, 2 orthogonal vectors and 1 bivector. A 3D space has a basis of 1 scalar, 3 orthogonal vectors, 3 orthogonal bivectors (planes of rotation) and 1 trivector. Algebras of dimension n have the number blades of each grade given by the binomial coefficients in the nth row of Pascal's triangle.

Addition of multivectors is performed by simply adding elements with corresponding blades - scalar parts to scalar parts, vectors in a given direction to vectors in the same direction, and so on.

Multiplication of vectors a and b is defined by the geometric product, ab = a.b + a^b . The inner (dot) product part is the product of the lengths of the two vectors scaled by the extent to which the vectors are going in the same direction. It is a scalar given by the multiplying the lengths of a and b and scaling by the cosine of the angle between them. It is commutative: a.b = b.a .

The outer (wedge) product of two vectors on the other hand is a bivector with a magnitude given by the product of the lengths of the two vectors scaled by the extent to which they are perpendicular to each other. (|a||b|sin(theta)) The outer product is anticommutative: a^b = -b^a.

If one pictures the two vectors tail to tail, the area of the bivector a^b is given by the parallelogram swept out by sliding a down b. If on the other hand one slides b down a the direction indicated by the vectors around the perimeter of the area will go in the opposite direction.

The outer product is analogous to the vector cross product in ordinary vector algebra, but while the cross product only works in 3D, the outer product works in any dimension. This is because the cross product gives an "axial vector" perpendicular to a plane of rotation (which is not unique in dimensions higher than 3), while the outer product gives the plane of rotation itself.

In general the inner product of two blades yields a result one grade lower than the lower of the original two, while the outer product yields a result one grade higher than the higher of the original two.

unwritten ideas: [More about bivector multiplication yielding rotations, unit bivector square = -1, similarities an differences between bivectors, (& pseudoscalars in odd dimensions) and imaginary numbers. Rotors. Compositon of succesive rotations by embedded rotor expressions. Brief EM discussion / relativistic Maxwell's Eqns condensed to "delF = J" - spacetime vector derivative of the EM field strength equals the spacetime current. 4D with ---+ signature -> boosts as hyperbolic rotations . Derivation of quaternions and octonions from the mother algebra, deficiencies of q and o. Descriptions of links.]

<rant> Some previous comments miss the point of physics, as do most theoretical types and mathemeticians - it's not mere math where you can assert nonsense like i with no physical interpretation, its a description of reality which has to explain, and has value only insofar as its application can be understood physically, geometrically. Also it's no good having a different mathematical dialect for every little subspecialty or using unnecesarily general and abstract structures such as matrices for physics with constraints not naturally modelled by such general methods. GA goes a long way to solving all these problems.

Mathemeticians who want to explain the obvious in terms of the obscure have the Clifford algebra page to do their thing on - GA is for physics. Physical intuitions were the basis for geometry, set theory, calculus and so forth - most mathematical symbol shuffling is just imitation physics underneath the bafflegab. It won't do to mess up a good physical description like GA with the obscurantist pseudorigor of mathematicians' style. </rant>

Not that I have anything against Hestenes-worshippers (I met Hestenes once and he's a nice guy, if a little bit ..... well, let me put it this way: I think he said Clifford algebras will settle all questions of physics, or something like that) .... OK, where was I? Oh: Well doesn't Emil Artin also warrant some attention on this page? -- Mike Hardy

I've never heard of a geometric algebra before, but your remark about Grassmann algebras giving a more natural treatment of physics without complex numbers piqued my interest, although I don't quite see how. Would you care to clarify what you meant? Phys 17:33, 20 Aug 2003 (UTC)

Hmm, I just read a bit of it, but I still don't see anything new geometric algebra has to say that can't be said already in the language of differential geometry and "dot products", linear representations, etc.. Phys 10:31, 21 Aug 2003 (UTC)
Also, I'm a bit suspicious of defining the wedge product as 1/2(ab-ba) because unless both a and b have an odd grading, a\wedge b+b\wedge a\neq 0 in general. Phys 11:01, 21 Aug 2003 (UTC)
Though it wasn't stated on the article, small case refer to vectors, so that definition is for vectors only. I added a little wider definition of both inner and outer product with higher grade elements. In any case, even with mixed grade elements, the wedge product is 0 when the elements involved are "parallel" or "containded" or "linearly dependent". With vectors it is clear that wedge product cancels if the vectors are linearly dependent (don't know if the concept can be expanded to multivectors).

--Xavier 18:48, 2005 Mar 30 (UTC)

Replacing vector spaces and algebras over the complex numbers with algebras over real Clifford algebras achieves just exactly what? Sure, any quantity whose square is -1 and commutes with everything can be thought of as for all intents and purposes as i, but choosing the n-vector for an n-dimensional space as i doesn't really make any difference. It doesn't really matter what i "really" is. Insisting it's a certain element of a Clifford algebra doesn't really matter. It's already well-known that in many fields like quantum mechanics, we could simply deal with real algebras and real vector spaces provided we define an element in the center whose square is -1. Phys 14:00, 21 Aug 2003 (UTC)
in my humble opinion, and as a non-mathemathitian, one of the strong points of geometric algebra is that is connected to geometry (something more or less real), and so it is much more easy to understand/learn. Things like "polar" vectors got sense if you see them as bivectors; or to me it's easy to see that the scare of a unitary bivector is -1, while the imaginary unit always was kind of too imaginary. --Xavier 18:48, 2005 Mar 30 (UTC)


I'm not too sure, but it sure seems like the treatment of Maxwell's equations using real Clifford algebras sure does look quite a bit like the stuff Reddi and some other contributors are adding everywhere about quarternions... Phys 15:28, 21 Aug 2003 (UTC)

This article is extremely POV, to say the least. David Hestenes should be mentioned, maybe after two or three long paragraphs, and Emil Artin should have higher billing than Hestenes. Michael Hardy 22:20, 13 Jan 2004 (UTC)

The history of GA including the people involved should be moved down to a separate section, and an explanation be at the top, yes.

Who should have higher billing may depend on your POV -- Hestenes' wrote for people with a mathematical background like engineers and natural scientists. As far as I can see, he was the first to do so. Presenting a subject in an accessible way is a highly valuable thing to do (especially in this case, as far as I am concerned). Artin on the other hand seems to have written for mathematicians. Quote from the description of Emil Artin's book "Geometric Algebra" (supposedly from the publisher):

Exposition is centered on the foundations of affine geometry, the geometry of quadratic forms, and the structure of the general linear group. Context is broadened by the inclusion of projective and symplectic geometry and the structure of symplectic and orthogonal groups.

It would be interesting to know who contributed what to the development, understanding and application of GA.

RainerBlome 21:58, 19 Sep 2004 (UTC)

The article is not POV. Emil Artin has nothing to do with geometric algebra except that he wrote a book by the same name. Putting Emil Artin in this article would absolutely make no sense. Geometric Algebra is a book written by Artin. But it is also a mathematical system proposed by Hestenes. However the book and the mathematical system having nothing to do with each other and discussing them in the same article would only confuse things. If you want to write about Emil Artin's book then it should be written about in a seperate article.

The mathematical system that Hestenes proposed is however a special kind of Clifford algebra. So the people who contributed to it are exactly the people who formulated clifford algebra. The main advantage of Hestenes work is that it is much easier to understand than clifford algebras and it has a very nice geometric interpretation. Some might say that Hestenes has not come up with anything new because all of the formalism is already present in Clifford algebras. However the geometric interpretation is more important than the formalisms IMHO because people understand visual ideas but they don't understand formalisms.

This article lacks a definition. Is a geometric algebra nothing but a Clifford algebra over the reals? If so, which quadratic form is being used to define the Clifford algebra? Is there one geometric algebra for every quadratic form on the reals?

Also, the relation to exterior powers remains unclear. Are the geometric algebra and the exterior power the same as vector spaces? AxelBoldt 01:04, 28 Sep 2004 (UTC)

The definition is where the three axioms for the geometric product are given. Maybe it should be more prominent. One axiom may be missing, along 1A = A. And only an implicit definition of what a multivector is is given. I have read elsewhere that a geometric algebra is indeed a Clifford algebra over the reals. There is a degree of freedom with regard to the choice of contraction. That may be equivalent to freedom in choice of quadratic form, but I don't know.
The 1A = A axiom was missing. --Xavier 18:48, 2005 Mar 30 (UTC)
What do you mean by "exterior power"? The exterior algebra? A geometric algebra is a vector space, but the converse is not true. Same for an exterior algebra. For example, with GA, you get inverses for vectors. RainerBlome 19:31, 22 Oct 2004 (UTC)

It is not clear from the article what is the geometric product of multivectors.Serg3d2 20:28, 29 November 2005 (UTC)

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