Geometric Algebra Fundamentals

[ObsessiveMathsFreak]

I've been studying geometric algebra of the form promoted by David Hestenes, but I'm having trouble with the very basics.

Most GA books, in fact, all GA books, begin as follows.

For two vectors $$\mathbf{a}\mathbf{b}$$, they define the symmetrical inner product $$\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos{\theta}$$ as normal, then they define the anti-symmetrical outer product $$\mathbf{a}\wedge\mathbf{b}=|\mathbf{a}||\mathbf{b}|\sin{\theta}\mathbf{I}$$, i.e. the area of the parallelogram spanned by the two vectors times $$\mathbf{I}$$ which is the unit bivector giving the correct orientation.

Then they define the geometric product $$\mathbf{a}\mathbf{b} = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\wedge\mathbf{b}$$, and this is fine if you simply consider a larger vector space with scalars and bivectors as axes.

They then simply have $$\mathbf{a}\cdot\mathbf{b}=\frac{1}{2}(\mathbf{a}\mathbf{b}+\mathbf{b}\mathbf{a})$$ and $$\mathbf{a}\wedge\mathbf{b}=\frac{1}{2}(\mathbf{a}\mathbf{b}-\mathbf{b}\mathbf{a})$$
By symmetry and anti-symmetry. This is all fine too.

But here is where they lose me. They then demand/require/axiomatise/assume? that the geometric product is associative, $$\mathbf{a}(\mathbf{b}\mathbf{c})=(\mathbf{a}\mathbf{b})\mathbf{c}=\mathbf{a}\mathbf{b}\mathbf{c}$$ and go on to "prove" a swath of results involving the inner and outer products of vector and bivectors, etc, etc.

But, it is not clear, at least to me, that this new associative geometric product is in fact the same as the original geometric product that was discussed. In paticular, I do not see why the symmetric part of this associative product should coincide with the standard dot product, or similarly for the outer product. In short, what is the connection between this new associative operator and sines and cosines?

This may seem like a very silly question, but I cannot for the life of me see why we should just be allowed to assume associativity. Is there some kind of proof of this using Clifford Algebra?

 

[ObsessiveMathsFreak]

The thought just occurs to me. Perhaps what the authors meant to say was not that the geometric product automatically has the property of being associative, but that we define the operation of the geometric product to be associative. In other words, since we don't yet have an operation for the product of a vector with a "geometric product" $$\mathbf{a}(\mathbf{b}\mathbf{c})$$, we say;

$$Let\ \ \ \mathbf{a}(\mathbf{b}\mathbf{c})=(\mathbf{a}\mathbf{b})\mathbf{c}$$

And we denote it by just $$\mathbf{a}\mathbf{b}\mathbf{c}$$. And using this definition we can show that this defines a unique(whew!) operator in the space of $$\mathbb{R}^{2n}$$ multivectors, as the authors go on to do.

Maybe I was just to shortsighted, but it seems to me that there is a subtle yet important distinction between declaring that an operator has the associative property, and defining the operator to have the associative property. Most of the authors simply stated that associativity was a given property of the operator. As I read them again, one or two did in fact define the operator to be associative instead, but I was too busy trying to find a proof of associativity to spot that.

Of course there is still always the constructivist, or perhaps even circular argument, objections to this definition of the geometric product, in that we assume the properties without first constructing the operator. But I suppose that since the definition does define a unique operator, we could view it as being something akin to the definition of $$e=\lim_{n \mapsto \infty} (1+\frac{1}{n})^n$$, a number which, though we can never initially construct it, does in fact exist and is unique according to the definition. We just extend this argument from numbers to functions.

But I've babbled away to myself for long enough, so I'll leave it at that.

Edit:
Hmmmmm... But what do people think about the constructivist objection? Can we simply define an operation by specifying its properties? Are there any good examples of this being done?

 

[Hurkyl]

Hmmmmm... But what do people think about the constructivist objection? Can we simply define an operation by specifying its properties? Are there any good examples of this being done?

It's not much of an objection; we define things by properties all the time. e.g. groups, rings, vector spaces, topological spaces, the natural numbers...

In the end, it really doesn't matter what you call a definition and what you call a theorem, so it only really matters for pedagogical purposes. I think there is merit in axiomatically defining things; people sometimes get overly attached to the definition, so it's helpful to get them attached to the more important stuff.

 

[cornfall]

It's not much of an objection...so it only
really matters for pedagogical purposes.

It also suggests to me a feeling, a way to invent or learn. I first approached this associative "geometric product" reading, http://www.mrao.cam.ac.uk/%7Eclifford/publications/abstracts/imag_numbs.html". Following this paper; "to see what is going on" and to make it feel satisfying and natural start off with a grade-2 space and work through a detailed example. So instead of exploring associativity with vectors a, b and c, use multivectors. These are arbitrary linear sums of the the basis elements. Keep in mind the peculiar containment of the "objects" and the "operators" within the same algebra. This is a key feature of geometric algebra.

The geometric quantities you can make using the the inner and outer products are scalars (observables or magnitudes), vectors (lines or position locators), and bivectors (planes or orientation rotators). With these, assemble by hand in tedius or fun! detail a Clifford algebra of three multivectors A, B and C. I got a feeling for (AB)C = A(BC) in this way.

 

[ObsessiveMathsFreak]

Keep in mind the peculiar containment of the "objects" and the "operators" within the same algebra. This is a key feature of geometric algebra.

I think this is the most offputting aspect of Geometric Algebra. Especially viewing things like a^b as an operator and a directed line segment. I think it may be better to view every object as an operator.

 

[cornfall]

viewing is seeing

I think this is the most offputting aspect of Geometric Algebra. ... I think it may be better to view every object as an operator.

Your view is more precise than mine, I'm more coarse. Working on a ferris wheel in the rain puts me real $$\textit{e}$$ off. An old carny needs some help with this question. A convex polygon in a plane is specified by the ordered set of points $$\{x_{0},x_{1},\,.\,.\,.\,,x_{n}\}$$. Prove that the directed area of the polygon is given by $$A = \frac{1}{2} (x_0 \wedge x_1 \, + \, x_1 \wedge x_2 \, + \, . \, . \, . \, + \, x_n \wedge x_0)$$

 

[mathwonk]

i always thought the canonical, i.e. hands down best possible, book on geometric algebra, is the one by emil artin, entitled "geometric algebra". have you looked at that one?it is the source of such famous quotes as "in doing linear algebra, on the whole the work is made longer and harder by the introduction of matrices. in general matrices should always be ignored in favor of linear maps. sometimes this cannot be done, a determinant must be computed. in those cases the matrix should be introduced temporarily, the determinant calculated, and the matrices thrown out again."(quoted roughly from memory. the original is much more appealing, and even stronger in tone.)

 

[complexPHILOSOPHY]

i always thought the canonical, i.e. hands down best possible, book on geometric algebra, is the one by emil artin, entitled "geometric algebra". have you looked at that one?it is the source of such famous quotes as "in doing linear algebra, on the whole the work is made longer and harder by the introduction of matrices. in general matrices should always be ignored in favor of linear maps. sometimes this cannot be done, a determinant must be computed. in those cases the matrix should be introduced temporarily, the determinant calculated, and the matrices thrown out again."


(quoted roughly from memory. the original is much more appealing, and even stronger in tone.)

Sorry to interject, however, I have a question regarding your comment. Is it more beneficial to the student to learn a style of LA that uses the least amount of matric theory, or is it beneficial to learn LA using matric theory and then expand your understanding of LA so that you no longer require the use of matric theory?

I have a desire to do mathematical physics (quantum mechanics especially) which I have been told, is primarily linear operators in hilbert spaces and that matric theory is not at all useful in understanding this.

I am a naive maths student in search of guidance.

Thanks Mathwonk,

-cP

 

[mathwonk]

i think the import of artins comment was that one should learn via linear maps, and consider matrices only an occasional computational aid. one should know how to use matrices but keep them in their place, i.e. purely for computational use.

this preference for concepts over computational devices is of course also true for tensors, geometric algebra, clifford algebras, exterior algebra, differential calculus, etc...

 

[complexPHILOSOPHY]

i think the import of artins comment was that one should learn via linear maps, and consider matrices only an occasional computational aid. one should know how to use matrices but keep them in their place, i.e. purely for computational use.

this preference for concepts over computational devices is of course also true for tensors, geometric algebra, clifford algebras, exterior algebra, differential calculus, etc...

That is an interesting perspective, which I appreciate and embrace. I am really enjoying my recent exposure to abstract mathematics. I used to think mathematics was purely computational but now I have definitely started to understand the elegance and beauty that is contained in the concepts of mathematics and the properties and relationships that emerge out of them.

 

[mathwonk]

ideally the two points of view should enhance each other.

 

[ObsessiveMathsFreak]

That area proof involves summing the areas of the signed triangular areas generated by the differences betwen the point vectors if I remember correctly.

Actually, the Geometric Algebra I was referring to is the "new" Geometric Algebra as evangelised by David Hestenes, not the "old" Geometric Algebra of Artin. Actually, I'm not too sure if there is any difference between them. I think I've come across Artin's book though. If I remember correctly, it had very few pictures, which I found a bit pathological so I didn't look into it too much, so I'm not too sure if these are the same or different subjects.

I'm looking into Hestenes' Geometric Algebra at the moment to see if there is an alternative to differential forms, which are driving me to distraction. Hestenes claims so, a nd apparently one can prove the generalised Stokes theorem using Geometric Calculus, which follows on from his Geometric Algebra.. It's a little odd at the moment, but it does seem interesting. Let's hope it holds promise.

I'll let you know how it goes.

 

[mathwonk]

the difference is artin is a more famous expert in mathematics.

 

[robphy]

the difference is no one has ever heard of hestenes, whereas artin is a world famous expert in all areas of mathematics. so choose your own poison.

Well... "no one" is a little strong. He is known among the Physics Education folks and among some of the older Relativity folks... which I should point out since we are in a physicsforum.

Back on topic: I've been curious about Hestenes' Geometric Algebra/Calculus... but I'm not ready to dive into it yet. Right now, I'm more interested in differential forms and how useful they may be for understanding physics.

 

[mathwonk]

correction noted. along the same lines for differential forms, there is a book by the famous mathematican henri cartan in paperback and cheap.

 

[mathwonk]

here are some excerpts from wikipedia:

Emil Artin:
He was one of the leading algebraists of the century, with an influence larger than might be guessed from the one volume of his Collected Papers edited by Serge Lang and John Tate. He worked in algebraic number theory, contributing largely to class field theory and a new construction of L-function. He also contributed to the pure theories of rings, groups and fields. He developed the theory of braids as a branch of algebraic topology.
He was also an important expositor of Galois theory, and of the group cohomology approach to class field theory (with John Tate), to mention two theories where his formulations became standard. The influential treatment of abstract algebra by van der Waerden is said to derive in part from Artin's ideas, as well as those of Emmy Noether. He wrote a book on geometric algebra that gave rise to the contemporary use of the term, reviving it from the work of W. K. Clifford.


David Hestenes:

Ph.D. (born 1933) is a physicist. For more than 30 years, he was employed in the Department of Physics and Astronomy of Arizona State University (ASU), where he retired with the rank of Research Professor and is now emeritus.
Hestenes has worked in mathematical and theoretical physics, geometric calculus, geometric algebra, neural networks, and cognitive research in science education. He is the prime mover behind the contemporary resurgence of interest in geometric algebras and in other offshoots of Clifford algebras, as ways of formalizing theoretical physics.
From 1976 to 1979, he was an Editorial Advisory Board Member (formerly called Associate Editor) of the American Journal of Physics. He is currently on the editorial board of the journal Foundations of Physics.
In 2002, the American Association of Physics Teachers awarded him its Oersted Medal for his notable contributions to the teaching of physics. He has been a Principal Investigator for NSF grants seeking to model instruction at both the high school and university levels.

Henri Cartan:
Henri Cartan (born July 8, 1904) is a son of Élie Cartan, and is, as his father was, a distinguished and influential French mathematician.
Born in Nancy, France. He studied at the Lycée Hoche in Versailles, then at the ENS. He held academic positions at a number of French universities, spending the bulk of his working life in Paris.
Henri Cartan is known for work in algebraic topology, in particular on cohomology operations, killing homotopy groups and group cohomology. His seminar in Paris in the years after 1945 covered ground on several complex variables, sheaf theory, spectral sequences and homological algebra, in a way that deeply influenced Jean-Pierre Serre, Armand Borel, Alexander Grothendieck and Frank Adams, amongst others of the leading lights of the younger generation. The number of his official students was small, but includes Roger Godement, Max Karoubi, Jean-Pierre Serre and René Thom.
Cartan also was a founding member of the Bourbaki group and one of its most active participants. His book with Samuel Eilenberg Homological Algebra (1956) was an important text, treating the subject with a moderate level of abstraction and category theory.
Henri Cartan received numerous honours and awards. He was a foreign member of the Royal Danish Academy of Sciences and Letters, Royal Society of London, Russian Academy of Sciences, Royal Swedish Academy of Sciences, United States National Academy of Sciences, and other academies and societies.

and a link for an interview with Cartan:
http://www.ams.org/notices/199907/fea-cartan.pdf

 

[mathwonk]

I am just trying to remind learners there is a big difference between learning from those of us, however skilled at exposition, who try to interpret the masters, and learning from the masters themselves.

 

[cornfall]

Judge Judgement

the difference is artin is a more famous expert in mathematics.

I understand the comparison with Artin, but I'm not able to judge
your judgement. Anyway, who is the most famous; Hamilton or
Hestenes? Fame is fickle so let's get some criteria down to inform
judgement, all the while expecting these to run like sand between
our fingers.

 

[MaribuS]

I recommend that you see the following page:

https://www.amazon.com/dp/9027725616/?tag=pfamazon01-20

Click at the LOOK INSIDE figure and peek at the axioms of GA.

These axioms may give you an idea of what is better to do in order to construct GA from nothing...

I don't know how mathematical you would like to be, but... Just understand that defining a.b and a^b... and then defining ab in terms of a.b and a^b is not a good idea...

What is happening is that you are reading introductory texts only... They are not rigorous...

You have to define, first, the geometric multiplication (with axioms)... For multivectors in general, not only for vectors...

The geometric multiplication is associate by axiom (for multivectors in general, not only for vectors)

MaribuS.

 

[Doodle Bob]

But, it is not clear, at least to me, that this new associative geometric product is in fact the same as the original geometric product that was discussed. In paticular, I do not see why the symmetric part of this associative product should coincide with the standard dot product, or similarly for the outer product. In short, what is the connection between this new associative operator and sines and cosines?

This may seem like a very silly question, but I cannot for the life of me see why we should just be allowed to assume associativity. Is there some kind of proof of this using Clifford Algebra?

the new product has been *defined* to be: $ab=|a| |b| (cos(\theta) + sin(\theta) I)$. Due to the symmetry properties of cosine and sine, we can split ab into the "wedge" and "symm. pdt." parts as indicated. Since a specific product has been given, any desired properties from here on have to be proven. They can't be assumed. Associativity is not very hard to prove in this case: note that the symmetric part is always "real" and the skew-symm. part is not.

i'm sorry to hear that you're still having problems with forms. Have you tried "Advanced Calculus" by Loomis and Sternberg, or Spivak's book yet?

 

[mathwonk]

i admit that i first got over my fear of differential forms by perusing a little chapter by harley flanders, where he just calculated with them. when i saw how easy it was to use them correctly, i lost my fear of remembering all the definitions and theorems abut them.

i think it was in a little book from the AMS on differential geometry edited by chern. now long out of print of course but available in libraries. also the little course tom mattson followed here, from dave bachman, should provide almost the same hands on ease of introduction.i.e. the first thing to do is just multiply dx + dy times dx - dy + 2dz and see what you get. then take d of ((x^2 - y^3)dx + (xydy) and see what you get.

then crank up a teeny bit and take d of the angle form

-ydx/[x^2+y^2] + xdy/[x^2+y^2] ?? is that it?

change dtheta to x and y coords to be sure, via theta = arctan(y/x). i.e. compute d(arctan(y/x)) then take d of that. see if you get zero! then you are on your way.

 

[mathwonk]

i.e. check that d of a function is its gradient (please don't correct my terminology),

that d of a one form Pdx +Qdy, is its curl,

and d of a two form Pdxdy + Qdydz +R dzdx, is its (what?) oh yes divergence?

then check grens theorem in a rectangle, that the integral of d(Pdx +Qdy) over the rectangle equals the integral of Pdx +Qdy over the boundary.

then you are already way ahead of the game. that's more than most people know and all most people need to know.

 

[Doodle Bob]

i think it was in a little book from the AMS on differential geometry edited by chern. now long out of print of course but available in libraries.

Let me give an amen on that one! An excellent little book.

Isn't it currently being published by the MAA?

I do think that there is too much fear over these things called Differential Forms. they (and tensors, in general) have been given a stigma that they don't really deserve. I personally have always thought that one reason why they cause so much trouble is that they are fundamental geometric objects for which there is no obvious picture to think of. Vectors (which are really directional derivatives) can be thought of as directed line segments. You can even imagine transformations (e.g. rotations) through the use of "before and after" pictures. But forms are something else: they lurk in the unseen world of geometric structures. Which means of course that they are imminently cooler than all of the other things.

 

[mathwonk]

good point! here is a true comparison, diferential forms are MUCH easier than determinants, and understanding them magically allows one to also understand determinants!

 

[fopc]

For those interested in working through Flanders, the book edited by Chern can be found here:

https://www.amazon.com/dp/0883851296/?tag=pfamazon01-20

 

[redrzewski]

I really like "A course in mathematics for students of physics" by Bamberg and Sternberg. It's a companion book to Sternberg's "Advanced Calculus". It tones down the mathematical rigor compared to "Advanced Calculus", and has many more worked out examples of differential forms, the chain rule, etc.

 

[robphy]

I really like "A course in mathematics for students of physics" by Bamberg and Sternberg. It's a companion book to Sternberg's "Advanced Calculus". It tones down the mathematical rigor compared to "Advanced Calculus", and has many more worked out examples of differential forms, the chain rule, etc.

In addition, it [and Bill Burke's "Applied Differential Geometry" and Schutz's "Geometrical Methods of Mathematical Physics"] provides interesting physics applications: [geometrical] mechanics, optics, electromagnetism and [contact-geometric] thermodynamics.

 

[mathwonk]

bamberg was a famous outstanding TA in physics (the only good one?) at harvard in the 60's.

 

[Excal]

More Questions on the Fundamentals

Since this thread has pretty much gone inactive, I was wondering if I could get some help, after reading the posts in this thread, as a context. I'm trying to understand Hestenes' GA, but I notice that some see a difference in his work and Artin's, for instance. This intrigues me and I was hoping to get help with understanding Hestenes' contribution, but not with a mathematics career in mind, only as food for thought about the nature of numbers and geometry, or algebra, as a science of numbers, and geometry, as a science of spaces, or multi-dimensional distances, or spatial magnitudes.

I would like to solicit help and comments, using a link I found in this thread, pointing to the Cambridge group's Introduction to GA,

http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node6.html"

as a roadmap, asking some pretty unadvanced and unsophisticated questions about their statements, to try to clarify some things in my mind, if possible.

With your indulgence, the first question (please don't laugh! is this: Given that the authors begin with 2D space, stating that 1D space has "insufficient geometric structure," I was wondering if we could back up a bit. The Clifford algebras are the algebras of the binomial expansion (BE), if I understand correctly.

Therefore, the algebra of the 2D plane starts, as they show implicitly, with the third line of the BE, having the form of 2^2 = 1+2+1 = 4. My question is, do we assume that 2^0 = 1 is the linear space of numbers, corresponding to the geometric algebra of points? Likewise, then, do we assume that 2^1 = 1 + 1 = 2 are the linear spaces of numbers, corresponding to the geometric algebra of points and lines?

If so, then it seems that the reason that the BE is so useful in this regard is that it reflects the duality of the direction property of physical magnitudes, which grows in complexity, as the dimension property increments.

Thus, at dimension 2^0 = 1, there is one type of linear space of numbers, with no directions. This space of numbers, corresponds to the geometric space of points, correct? Incrementing dimensions of the BE from 0 to 1, gives us a new linear space, in addition to the space of points, the new space is the 1D space of numbers, corresponding to the geometric space of lines.

Whereas, points have no dimensions, and thus no directions, lines have one dimension, and thus two directions, correct? If this correspondence between geometric magnitudes and the BE is correct, then the following should hold:

At 2^2 = 1+2+1 = 4 linear spaces, the first space of numbers is the space, corresponding to the space of geometric points, the second and third are the two subspaces of a superspace of numbers, corresponding to two, orthogonal (i.e. independent), geometric lines, and the fourth space of numbers is the space, corresponding to the outer (wedge) product of these two 1D subspaces, forming a new space of numbers, corresponding to the geometric space of area, with four directions, correct?

Hence, in the algebra of the plane, the first space of numbers is dimensionless, with no directions. The second and third subspaces of numbers form a 2D space with two, independent, 1D possibilities, each with two directions. Finally, the fourth space is one space of 2D numbers, with no subspaces, with the four directions of 2D area.

If this is correct, it seems to me that we should then be able to identify the second and third subspaces, in the middle space of this line of the BE, with the two bases, sigma 1 and sigma 2, as the two inner products s1^2 = 1 and s2^2 = 1, but this is not noted by the authors in the tutorial.

It may be that the authors assume that everyone understands this, or they don't feel it worthwhile to point out, or they don't see it that way. It's really hard for me to tell, which it is, but I would really like to know.

Taking this approach, it's possible to show, as the authors do, that only the non-zero elements remain in the algebra's multivectors, all the zero elements dropping out, leaving us with a set of spaces of numbers, described in terms of algebraic operations, corresponding to geometric magnitudes, up to 2D area magnitudes, without having to state that "we begin with two dimensions, taking two orthononormal basis vectors," correct?

If this is a correct view of things, it is very satisfying to finally have filled in this gap in my knowledge that evidently everyone else in the "class" already knew. The square of the inner product seems to imply an orthonormal basis, meaning that the square of sigma 1 is independent of the square of sigma 2, but both have to inhabit this space of numbers, given the BE is a dimensional expansion of nature's fundamental duality, which I guess we could express as "every direction in nature, from a point, has an (equal?) and opposite direction."

Putting it another way, "In the algebra of the plane, any direction, from a point, can be rotated to an opposite direction, by means of one of two rotation operations: A unit direction can be rotated (transformed) into its opposite direction (|-a|*|a| = 1), where it joins with its opposite, or else it is transformed from its opposite (joined) direction, back to its original direction, where it opposes the opposite (One of these operations contitutes an inner transformation (contracting the 180 degree angle between the two directions in a given dimension, from pi to 0), while the other operation constitutes an outer transformation (expanding the angle between them, from 0 to pi)."

In other words, the only logical rotations of these opposite angles, in the plane, is isomorphic to opening and closing a jack knife: The 180 degree angle can either be opened (expanded) or closed (contracted), in the plane of rotation. The inner product is the closing operation, while the outer product is the opening operation.

This last part, of course, is deduced from the assumptions in the BE, since any given dimension has two, and only two directions. However, the authors don't follow this reasoning, but instead identify the outer rotation with a rotation from pi/2 to 0, and the inner rotation from 0 to pi/2, mapping them to the trig functions, the sine and cosine, which enables/requires them to resort to a geometric product, consisting of a combined operation, containing both the inner and outer products, which can be mapped to magnitude and direction in the plane.

This "departure" from the directly obtainable conclusion, where the pi rotation, representing the inner and outer product, or transformation of opposite directions in two, orthogonal, dimensions, is replaced by the pi/2 rotation, in GA, representing the inner and outer product, or transformation of two, orthogonal, directions in two dimensions, not all four directions in two, orthogonal, dimensions, must have some fundamental implications, but I can't see, yet, what they would be.

It may be significant, given what http://links.jstor.org/sici?sici=0025-570X(198912)62%3A5%3C291%3AHRATQS%3E2.0.CO%3B2-8". Maybe, we could discuss it in more detail later?

 

[Chris Hillman]

Criterion for necropost? One Year? Six Months?

Post #28 dated 02-15-2007, 08:24 PM
Post #29 dated 11-20-2007, 06:49 PM

Excal, reviving a thread after a 10 month hiatus is disorientating to frequent posters and seems to be frowned upon at PF. Better to start a new thread and provide a link to the old one!

About your question, I understand this in terms of writing down an expression like
$$x_0 + x_1 \, \vec{e}_1 + \dots x_n \, \vec{e}_n$$
and formally squaring it (outer product!), then setting each $\vec{e}_j \vee \vec{e}_j = \varepsilon_j$ where $\varepsilon_j \in \{ -1, \, 0, \, 1 \}$. This then leads to a real algebra of dimension $2^n$ which decomposes (orthogonal direct sum) into subspaces having dimensions given by the binomial coefficients $\left( \begin{smallmatrix} n \\ m \end{smallmatrix} \right)$ and spanned by "outer products" $\vee$ of various $\vec{e}_j$. The case where all the $\varepsilon_j=0$ gives the exterior algebra on R^n. Then from $\left( \vec{e}_1 + \vec{e}_2 \right)^2 = \dots$ we deduce...

The first chapter of the book by Ahlfors has a good short explanation.