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Introduction

To a mathematician, an algebra is a very precise thing. The definition they provide is tight and insightful, but it also goes over the heads of most beginning students. Fortunately, it is possible to describe geometric algebras in a manner that builds on a basic understanding of geometry most people acquire in their every-day lives. This informal description leads to the use of the same objects as the mathematicians define, but it does leave out some of the powerful insights that automatically apply if one knew the formal definition. Students are encouraged to learn the formal definition when they feel prepared to do so.

In the most basic sense, a geometric algebra is a set of rules governing objects that permit us to treat geometric things with the tools we learned to use with algebraic tasks. Most students who take both an algebra and geometry class realize that the two subjects are taught differently. In an algebra class, we are taught about functions and rules that enable solving for unknown variables. We are taught to draw graphs that help us find numerical solutions to our problems. In a geometry class, we are taught about line segments, shapes, and a theorem/proof system we use to know that we actually know something. The two column proof encourages deductive thinking while helping to build a geometric intuition.

Algebra is a subject that concerns itself with numbers. Geometry is a subject that concerns itself with magnitudes. In the historical sense, these are distinct concepts even if we blur them in today's classrooms. Whether we are able to link a number to the length to a line segment or the span of an angle doesn't impact purely geometric theorems. Constructing the bisector of an angle requires no numbers. Whether we know that the unknown variable in the formula for the parabolic arc a rock takes when thrown up and forward does not influence the fact that we can solve the equation for the peak height and range on impact.

Numbers are the stuff of algebra. They are the familiar integers augmented with rational and irrational numbers to make up the whole 'real' number line. They are the complex numbers that make use of imaginary numbers if one requires that all polynomials have roots.

Magnitudes are the stuff of geometry. They are the line segments, areas, volumes, and angles of various types. These magnitudes can't be numbers since even the simplest example shows that magnitudes contain more information. Imagine a line drawn at a certain length in the north-south direction. Draw another one at right angles to the first such that they both share one of their end points. These line segments are different magnitudes if one sticks to the historical definition of the term even though they are of the same length. Today we might be tempted to use a different word like vector, but the difference between the two line segments still stands.

The marriage of algebra with geometry creates a powerful union with many unexpected offspring. Trigonometry is one such well-known offspring. Knowing numeric values for certain magnitudes associated with right triangles allows us to find numeric values for other magnitudes and construct the triangle if we wish. Less well-known offspring include studies of tessellation (tiling) and perspective.

The offspring of most interest here, though, is the subject of geometric algebra described by William Kingdon Clifford in 1878 and many others since then. Geometric algebras provide for a mechanism linking magnitudes and numbers and lend themselves neatly to the representation of physical problems and of reality as we know it.

____________________________________

Section 1: A 3-D Euclidean space

Let us start with the geometry of a three dimensional Euclidean space. This is the space most of us already have a well-built intuition to handle problems of shape, size, and location. This is the space a child learns when they play with wooden blocks or those toys with beads on curvy wires. Human brains are wired for it.

----The Objects----
Describing objects in three dimensions requires both numbers and geometric magnitudes. For our numbers, we will restrict ourselves to real numbers unless otherwise stated. This is the number line we all learned as kids including integers, rationals (one integer/another integer), and irrationals (π, square root of 2, etc). For our geometric magnitudes, we use the symbol e followed by zero to three subscripts and then carefully define what we mean by them.

The first geometric magnitude to define is e with no subscripts. This object is a multiplicative identity. That means it acts like the real number 1 when multiplied by any other geometric magnitude. The reader might occasionally see it written as 1 in other books and articles, but they should not confuse it for the real number 1. To avoid this confusion, e will be used here.

The next geometric magnitudes to define are e1, e2, and e3. These are the magnitudes used as basis vectors along each of the three spatial directions. In a purely geometric sense, they are oriented line segments where one end point defines the start and the other the end.

The next three geometric magnitudes to define are e12, e13, and e23. These represent the basis planes and can be thought of as oriented plane segments. Plane segments require three non-collinear points to define, so oriented plane segments have a start, middle, and end point. Anyone who has programmed with 3-D surfaces for games has already encountered oriented planes as triangles in a surface mesh. These surfaces have a front and backside because they have an orientation.

The last geometric magnitude in the space is e123. This one is the basis volume segment. Only one of these magnitudes is needed here since there is one way to make a volume in a three dimensional space. This object does have a sense of orientation, though. Volume segments require four non-coplanar points in their definition, so there have a sense of order among those points if the segment is oriented.

To summarize our list of geometric magnitudes, we have eight objects representing various ranks of geometry and they all have a sense of orientation except the first. Here is the list.

{ e, e1, e2, e3, e12, e13, e23, e123}
A generic member of this list will be labeled Ei where 'i' can range from one to eight inclusive.
----What to do with them----
The next thing to do is define what one may do with these eight geometric magnitudes. The first operation is called scalar multiplication, which is a little unfortunate since the operation looks more like scaling than scalars. The second operation is addition and the third is multiplication. With these three primary operations, most other that people can use can be defined.

Scalar multiplication is an operation that provides a sense of scale to our objects. Take any geometric magnitude and multiply it by a number to get a different sized magnitude. Multiply a 2 against one of the oriented line segments and one moves the end point twice as far from teh start point as it used to be while also keeping the whole arrangement pointing in the same direction. Actually knowing that direction is a little more complex and will be saved for later. One useful thing to remember about scaling is that it is commutative (aA = Aa) and associative (a(bA)=(ab)A) where lower case letters are numbers and upper case letters are geometric magnitudes.

Addition works as most people would expect it. Apples may be added to apples, but not to oranges. So e1 + e1 may be thought of as 2 e1 while e1 + e2 cannot be further simplified. Addition is commutative (A+B = B+A) and associative ((A+(B+C)) = ((A+B)+C)). Addition is also distributive with respect to scalar multiplication (a(A+B) = aA+aB).

Example 1: Add 5e123 and 2e13 and -2e1 and 14e2 and -e123 and 2e23.
= + 5e123 + 2e13 - 2e1 + 14e2 -e123 + 2e23
= + 5e123 + 2e13 - 2e1 - e123 + 14e2 + 2e23 {Because Addition commutes}
= + 5e123 + 2e13 - e123 - 2e1 + 14e2 + 2e23 {Because Addition commutes}
= + 5e123 - e123 + 2e13 - 2e1 + 14e2 + 2e23 {Because Addition commutes}
= + 4e123 + 2e13 - 2e1 + 14e2 + 2e23 {Because Addition and scalar multiplication are distributive}
No further simplification is possible, but most people will regroup the terms to bring objects with similar rank together and pull out common coefficients. Apply commutativity a few more times and distributivity once to get the following.
= + 2( 2e123 + e13 + e23 - e1 + 7e2)
----Technical Note----
With addition and scalar multiplication, we can form linear combinations and span our related vector space. Any reader with more technical knowledge will know what that means, but it isn't very important here.

Multiplication is the last of the three operations. It's definition is best described with a multiplication table. The table must have eight rows and columns to cover all possible Ei's. After a bit of use, the reader will probably wind up accidentally memorizing large parts of the table. It isn't any harder to do than the real number multiplication tables we learned as kids.

----Some geometric meanings----
We could stop our description of multiplication by encouraging the reader to look up table entries and not worry about why the entry is what it is. We won't though. The table entries are worth thinking about because they make some sense after a fashion.

If one looks down the main diagonal of the table from the top-left to the bottom-right, one will see that any geometric magnitude multiplied by itself results in +e or -e. This shows that each of the geometric magnitudes has an inverse, so it is possible to divide by them. Many of the objects we encounter later will not have inverses, so the reader should not assume division can always be done to undo multiplication by geometric magnitudes.

The main diagonal also shows what we mean by the size of a geometric magnitude. The three basis line segments all have squares of +e, so they are said to have a squared length of +1. The three basis plane segments have squares of -e, so they are said to have squared lengths of -1. Objects with negative squared lengths behave somewhat like the imaginary magnitude i if one ignores the extra geometric meaning. The fact that these 'imaginaries' are present makes some researchers wonder if we needed to invent i in the first place or if there is deeper physical meaning to the many equations that make strong use of complex numbers to solve physical problems. Those who work with the subject area known as 'Clifford Analysis' are doing more work on this issue.

The last general note about the table to be made here will be about the shading of some cells. The cells that are shaded yellow signify that the two geometric magnitudes happen to commute. Those that do not commute happen to anticommute. In general, then AB = ±BA if A and B are chosen from the list of eight geometric magnitudes. When two of them anti-commute, they are said to be perpendicular. When two of them commute, they are said to be parallel. Both meanings are used most often when A and B are of the same geometric rank since they don't make a lot of intuitional sense otherwise.

----How multiplication really works----
There is an algorithm for determining the entry in each cell of the multiplication table. There are a couple of index rules and a few entries to memorize in the brute force fashion. After memorizing three entries and the two rules, the reader should be able to reproduce the entire multiplication table if they wish to do so.

Multiplication starts with making a long list of all the indices involved with both objects. Make sure to keep the indices from the left side object on the left of the list of indices from the right side object. The memorization part requires the reader to remember that the three geometric magnitudes with one index have squares of +e. The first index rule is one that states that two neighboring indices can have their order swapped if the sign of the cell entry is switched. This means that e12 = -e21. The second index rule is one that states two neighboring indices may be removed if they are the same and the cell entry does not have to switch signs. This means that e112 = e2.

Example 2: Multiply e123 and e13.
= + e12313 {Just write long list of indices in row-column order}
= - e12133 {Swapping uses index rule 1}
= - e121 {Eliminating pair uses index rule 2}
= + e112 {Swapping uses index rule 1}
= + e2 {Eliminating pair uses index rule 2}
(See problems 1 and 2.)
If the reader works with these rules a bit, they should be able to figure out the entries in the multiplication table. With a little bit of thought, they should also see that knowing how multiplication works for e1, e2, and e3 is enough to know how multiplication works for all other objects. These three objects will be referred to as 'generators' because one may start with them and generate all other objects in the R(3,0) geometric algebra through multiplication, addition, and the scaling operations.
----Technical Note----
These generators are the geometric equivalent of a basis in a vector space. The actual vector space contained within R(3,0) happens to be eight dimensional in the sense of the meaning of 'span'. In the geometric sense, though, there are really only three degrees of freedom and the generators represent them. Multiplication ensures the other five dimensions linear algebra students might expect are not available as degrees of freedom since the other geometric magnitudes are products of the three generators.

The last example for this section shows how two line segments multiplied together work in general. This result will be used later in section 2.

Example 3: Multiply two line segments M and N and show the general result.
M N= ( M ) ( Npar + Nperp )
where N is decomposed into pieces parallel and perpendicular to M

=( M · direction of M ) ( N cos(θ) · unit direction of M
+ N sin(θ) · a unit direction perpendicular to M )
where M and N are the numeric lengths of M and N and θ is the angle between them.

=( M N ) [cos(θ) · (direction of M)2
+sin(θ) · (unit direction of M)(a unit direction perpendicular to M )]

=( M N ) [cos(θ) · e + sin(θ) · unit plane containing M and then N]

In general, two line segments multiplied together produce a sum of a scalar and a plane segment. If the line segments happen to be parallel (θ=0 or 180) the plane segment vanishes and we are left with the scalar. This is the inner product some students may find familiar from classes involving vectors. If the line segments happen to be perpendicular (θ=±90), the scalar part vanishes and we are left with the plane segment. Some students know this product as a wedge or outer product which is related to the vector cross product.

Summary

In this section, the basic meanings of the objects within a geometric algebra were explained. The eight geometric magnitudes combined with numbers to make sums give us a way to represent our geometric objects with algebraic expressions. Constructing all the objects is done through a combination of addition, multiplication, and scalar multiplication.

With a bit of practice, the basic operations can be added to a student's intuitive toolbox. Simplification of these operations will be as simple as the addition and multiplication we learn as young children. With a bit more practice, the student will discover new things real numbers can't do, though. There are objects whose squares are themselves and they aren't e. There are even objects whose squares are zero and they aren't zero. With enough practice, the tools of geometric algebra will be available to students even when they are employing a different mathematical formalism because these tools bring geometric magnitudes into the powerful language of algebra.

----Problems for Section 1----

1: Try multiplying both objects from example 2 in the opposite order.

2: What would happen to the multiplication table if the square of e1 were -e?

3: Add (3e1 + 5e3) and (4e2 + 7e3 -2e23).

4: Multiply (3e1 + 5e3) and (4e2 + 7e3 -2e23).

5: Multiply (4e2 + 7e3 -2e23) and (3e1 + 5e3)

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 Introduction to Geometric Algebra | 85 comments (74 topical, 11 editorial, 0 hidden)
 If my (none / 0) (#4) by tzigane on Sat Sep 07, 2002 at 02:05:46 AM EST

 college math professors had been this straightforward I would have enjoyed math. You should be a teacher! Will these articles be another book? . Knit on with confidence and hope through all crises. E. Zimmermann
 book (5.00 / 1) (#5) by adiffer on Sat Sep 07, 2002 at 02:14:55 AM EST

 I am putting together a book. I also used to be a teacher.  I got a little tired of being poor and semi-employed.  I still love it, though. -Dream Big. --Grow Up.[ Parent ]
 Get thee (5.00 / 1) (#6) by tzigane on Sat Sep 07, 2002 at 02:29:48 AM EST

 back into the classroom if you can manage it; you are needed there. I want my copy of your book autographed  ;-) . Knit on with confidence and hope through all crises. E. Zimmermann[ Parent ]
 make do (5.00 / 3) (#8) by adiffer on Sat Sep 07, 2002 at 02:52:43 AM EST

 Thanks for the support, but you all will have to be the classroom for now if you want it.  The schools can't afford to pay what IT folks can make even with this downturn. The book will come in two varieties.  One will be open licensed and free to print.  The other will be bound if I can find someone to do it.  The book is intended to be a bit revolutionary and adress how geometric algebras can be put to work in introductory and advanced physics topics.  I'm working on a friend of mine to see if he is ready to go through with what we had in mind 10 years ago.  He would be a co-author. -Dream Big. --Grow Up.[ Parent ]
 Book (5.00 / 1) (#9) by tzigane on Sat Sep 07, 2002 at 03:04:47 AM EST

 I'll buy a bound copy if you get it printed and I will want an autograph. ;-) It is criminal that teachers are not paid enough. We entrust them with our future and pay them a pittance. Is there anything more valuable to us than our children?   +1FP this is good writing about a tough subject. . Knit on with confidence and hope through all crises. E. Zimmermann[ Parent ]
 book publishing (none / 0) (#34) by adiffer on Sat Sep 07, 2002 at 11:28:52 PM EST

 I spent some time talking to Robert today about book publishing.  I don't think I'll have any trouble pulling it off now.  8) -Dream Big. --Grow Up.[ Parent ]
 Great! (none / 0) (#39) by tzigane on Sun Sep 08, 2002 at 12:49:30 AM EST

 I still want an autograph ;-) Let me know if you need text formatting stuff done, I'll help if you need it. . Knit on with confidence and hope through all crises. E. Zimmermann[ Parent ]
 well (none / 0) (#57) by Hillman on Mon Sep 09, 2002 at 10:30:56 PM EST

 It is criminal that teachers are not paid enough Well, at least you can be sure they teachers aren't there for the money. [ Parent ]
 Sounds WONDERFUL (none / 0) (#82) by mcherm on Thu Sep 12, 2002 at 01:50:49 PM EST

 Will you please add my name to the list you will begin to collect of people who asked to be notified when this book is complete? I can be reached via email using "mcherm" "mcherm.com". -- Michael Chermside[ Parent ]
 Ignorance Hurts (4.00 / 1) (#12) by bugmaster on Sat Sep 07, 2002 at 04:03:46 AM EST

 Forgive the stupid question, but where does Geometric Algebra come from ? I have never heard of it until now, and it seems like I should have, since it looks pretty neat. Was it invented/discovered for some specific purpose (like Calculus) ? What problems can I solve with GA that I can't solve with regular Linear Algebra ? >|<*:=
 history (5.00 / 1) (#15) by adiffer on Sat Sep 07, 2002 at 04:34:48 AM EST

 It is 19th century stuff that didn't quite win out in the debate over what formalism to use when teaching physics.  It's loss had little to do with its drawbacks and much to do with the fact that physicists often don't understand the tools the mathemeticians craft for us.  It's not surprising, though, when you realize we aren't mathemeticians anymore than members of the general public are. If you look at the subject of Clifford Algebras the way the math people write about it up, it will probably be unrecognizable if you expect it to compare to what I've written here.  I've kept the focus on one algebra and avoided the higher  power vocabulary.  Therefore, if you have been exposed to the topic, you might still not recognize it here. If you are really curious, look up the link to Clifford.  Look for other people like Hamilton and Grassman too.  Hamilton's quaternions are the generators of a two-dimensional geometric algebra.  Grassman's work led to forms and an outer product that is easily recognizable to any user of geometric algebras.  There is also a neat little chart here that might help lay things out a bit.  Remember that Gibbs was a physicist, you it's not hard to understand why we (and the engineers) were taught the way we were.  8) -Dream Big. --Grow Up.[ Parent ]
 Thankyou so much! (5.00 / 1) (#14) by Miscreant on Sat Sep 07, 2002 at 04:26:10 AM EST

 I do not study maths of any sort at university, but I sometimes wish i did. Especially after reading articles like this. I have given this the once over already, and after taking it all in i plan to reread your "Usage of Geometric Algebra as a representational tool" article (which i had bookmarked a while ago, but never got round to finding my way through it properly). Your writing is concise and well written, and allows for a pretty easy understanding, after a few thorough readings. Please keep writing. I for one love this style of article, and think kuroshin could do with far more educational writing......Its a positive thing for all involved. Thankyou.... and +1FP :)
 Please solve: (2.00 / 1) (#19) by egg troll on Sat Sep 07, 2002 at 12:57:44 PM EST

 Solve for t using logarithms: P=P0at Show your work. He's a bondage fan, a gastronome, a sensualist Unparalleled for sinister lasciviousness.
 hmm? (none / 0) (#22) by adiffer on Sat Sep 07, 2002 at 02:57:31 PM EST

 P = P0 at implies ln(P/P0) = t ln(a) if ln(P), ln(P0) and ln(a) exist. -Dream Big. --Grow Up.[ Parent ]
 What about t? (none / 0) (#23) by egg troll on Sat Sep 07, 2002 at 03:28:38 PM EST

 ln(P/P0) = t ln(a) You still haven't solved for t. What would the next step be? He's a bondage fan, a gastronome, a sensualist Unparalleled for sinister lasciviousness. [ Parent ]
 No, Adiffer, No! (5.00 / 1) (#24) by bjlhct on Sat Sep 07, 2002 at 04:49:35 PM EST

 Don't do his homework for him! * Beware, gentle knight - the greatest monster of them all is reason. -Cervantes[ Parent ]
 If he doesn't know how to do it (none / 0) (#29) by Rk on Sat Sep 07, 2002 at 05:35:08 PM EST

 He isn't going to learn by just staring at a textbook. Better to explain it once, so he understands (though there isn't really a lot to understand there), and there are no doubt plenty more excercises where that came from, if his maths textbook looks anything like mine (which it probably doesn't). Alternative, he may just be trolling. Doesn't bother me greatly, I wasn't doing anything worthwhile anyhow. [ Parent ]
 show work (5.00 / 1) (#33) by adiffer on Sat Sep 07, 2002 at 10:32:56 PM EST

 I'm pretty sure it wouldn't be useful since my typical weakness is in showing the work.  I jumped at least a half dozen steps with my word 'implies'.   I used to annoy my professors when I did that in grad school.  My jumps were wrong often enough that they couldn't help correct my work and thinking because I hadn't shown my thinking.  It cost me a whole letter grade on the final exam for my second quarter quantum class.  I got one particular problem exactly right, but I jumped so many steps that the professor had a hard time believing I understood how to do the problem.  It turned out that I used a completely different technique than he had intended, but we didn't uncover that until after a long discussion about whether I had been looking over someones shoulder.  I was so happy to have avoided flunking the test that I let it go. So... I'm probably the worst person to ask 'show the work' problems.  If anyone took my solutions to their teachers, they would probably get asked which alien species taught them to do the problem.  8) -Dream Big. --Grow Up.[ Parent ]
 Well..... (5.00 / 1) (#35) by bjlhct on Sat Sep 07, 2002 at 11:29:17 PM EST

 It was supposed to be a joke. Anyhoo, I have the same problem. Actually, my brain is used to writing what I write, but used to reading what others read. So when I look at something I did from long enough ago that I've forgotten what I was thinking, it looks like it was written by .... (P-|) (*-}) (:: - }) )~ take your pick, something strange. * Beware, gentle knight - the greatest monster of them all is reason. -Cervantes[ Parent ]
 That's so easy that even I could do it... (none / 0) (#28) by Rk on Sat Sep 07, 2002 at 05:29:53 PM EST

 Divide by ln(a): ln(P / P0) / ln(a) = t NB: The use of ln isn't actually necessary. You can use log (decimal logarithim) as well, or any other logarithmic base, though most calculators only support 10 and e as a base. That wasn't even difficult. It's only basic algebra. If you can't understand that, maybe you should talk to your teacher and ask him/her to explain it in detail. [ Parent ]
 Your "NB" and properties of logarithms (none / 0) (#30) by Kalani on Sat Sep 07, 2002 at 05:43:33 PM EST

 ln(P / P0) / ln(a) = t NB: The use of ln isn't actually necessary. You can use log (decimal logarithim) as well, or any other logarithmic base, though most calculators only support 10 and e as a base. Actually, if you want to simplify it as much as possible in terms of minimal symbols, you'd write log-a(P / P0). In general log(V) / log(B) = log-B(V) where B does not equal 1. PS: What does "NB" mean? -----"Images containing sufficiently large skin-colored groups of possible limbs are reported as potentially containing naked people." -- [ Parent ]
 NB = Nota bene (5.00 / 2) (#31) by Rk on Sat Sep 07, 2002 at 07:17:28 PM EST

 Latin for "Note well". [ Parent ]
 yes!!!!!111111 (none / 0) (#36) by rev ine on Sun Sep 08, 2002 at 12:27:55 AM EST

 :DDD [ Parent ]
 Non-base 10/e logs (5.00 / 1) (#80) by bil on Wed Sep 11, 2002 at 09:13:37 AM EST

 NB: The use of ln isn't actually necessary. You can use log (decimal logarithim) as well, or any other logarithmic base, though most calculators only support 10 and e as a base. But you can get other logs by knowing that logXN=(log10N)/(log10X) so log2128 = (log10128)/(log102) =7 Oh and isn't log(P/P0) = log(p) - log(P0) bil bil Where you stand depends on where you sit...[ Parent ]
 An obvious question (5.00 / 2) (#20) by Shimmer on Sat Sep 07, 2002 at 02:24:35 PM EST

 This is all well and good, but what does it really mean to multiply a line segment by a plane? Because you provide no motivation for the purpose of these operations, the topic seems completely artificial to me. -- Brian Wizard needs food badly.
 reasonable question (5.00 / 1) (#21) by adiffer on Sat Sep 07, 2002 at 02:52:53 PM EST

 Multiplication of different rank objects is relatively easy to picture.  I'll have to beef up or add a motivation section I guess.  The harder thing to explain is addition of different rank objects. When you multiply two line segments, you get a scalar or a plane segment or both.    If the two have parts that are parallel, you get the scalar.  If the two have parts that are perpendicular, you get the parallelogram they define as the plane segment.  If you know cross products, think about how we get the legth of the resulting axial vector from the area of the parallelogram the first two vectors define. When you multiply a line segment against a plane segment, you get a line segment in the plane perpendicular to the line segment or a volume segment or both.  If the line segment lies in the plane, you get a line segment in that plane but perpendicular to the first one.  If the line segment is perpendicular to the plane, you get a volume segment that looks like a parallelopiped. Try a few by hand and I think you will see it.  The multiplication table is structured so you get geometric magnitudes that match up with the numeric magnitudes for lengths, areas, and volumes.  If you multiply 2e1 and 2e2 and 2e3, you get a box named 8e123. -Dream Big. --Grow Up.[ Parent ]
 Thanks, that's helpful (none / 0) (#38) by Shimmer on Sun Sep 08, 2002 at 12:32:14 AM EST

 Examples like that help me develop an intuitive understanding of what "multiplication" means. -- Brian Wizard needs food badly.[ Parent ]
 Suggestion (5.00 / 1) (#25) by Kalani on Sat Sep 07, 2002 at 04:55:46 PM EST

 First of all, great job! This certainly is (as I read your lower comment) perfect book material. So, having said that, I think that you ought to talk to the editors about getting images in your next story (or even inserted after-the-fact into this one if you'd like). Some basic diagrams go a long way toward demonstrating what the different members of e map onto in the geometric sense (as well as demonstrating what you're doing with, for instance, Example 3). They don't have to be really fancy ... take a look at www.geometryalgorithms.com (a site for computational geometry and associated subjects) for some examples of very simple diagrams that help the descriptions/demonstrations very much. Otherwise, this submission is absolutely fantastic. -----"Images containing sufficiently large skin-colored groups of possible limbs are reported as potentially containing naked people." --
 :DD (2.50 / 2) (#37) by rev ine on Sun Sep 08, 2002 at 12:28:03 AM EST

 [n/t]! !!!!!111 [ Parent ]
 Thought about that (none / 0) (#40) by adiffer on Sun Sep 08, 2002 at 02:58:23 AM EST

 I decided not to ask about images, though.  Can you imagine what some in this community would submit.  8)  On top of that, the storage requirements for the site would probably jump through the roof.  Maybe when K5 has its own server farm and a staff to maintain the site, they will be able to improve collaborative media by adding other channels besides text. What I will do is reproduce my articles as I package them for the book on my own site.  That version will have the images and a richer rendering technique made available.  Until then, I'm thankful people are willing to read the text and comment. I will look at the site you mentioned.  I think I'll go do that right now.  Thanks. -Dream Big. --Grow Up.[ Parent ]
 Well (none / 0) (#42) by Kalani on Sun Sep 08, 2002 at 04:13:27 AM EST

 They've done it for some other submissions. I don't remember hearing anything about it screwing up k5 very badly. Anyway, if you do it right (small gif files with only a couple of colors) the actual image files will be about as large as a couple of comments. I don't think that they have any problem with putting the images in at your request (and it's set up this way to keep "unsavory" images from being submitted). Anyway, the site that I mentioned kind of looks like just another junk FrontPage website. However, the descriptions of basic computational geometry algorithms are pretty decent, and the "basic linear algebra" tutorial they have (in pdf form) goes over a few of the same topics (although your geometric algebra description generalizes some of them of course). The diagrams are, I think, perfect for the accompanying descriptions. They're compact and they feature just enough visual information to convey whatever situation they're describing. I thought that in your Example 3 above you might show a few intermediate solutions for theta = 0 to 90 degrees, which I think would be a very dramatic demonstration of the generality of the formalism you describe over traditional (messy) notation. In any case, it'd just be icing on the cake. The core of what you've got here is great. -----"Images containing sufficiently large skin-colored groups of possible limbs are reported as potentially containing naked people." -- [ Parent ]
 Burn All Math Textbooks! (5.00 / 1) (#26) by bjlhct on Sat Sep 07, 2002 at 04:56:40 PM EST

 Or maybe put  thetm in Yucca Mountain. Use The Higher Mathematics, What Is Mathematics? and Adiffer's Book instead. Good job, adiffer. This is clear and concise. The only thing I can think of that you could add is examples of what you're doing; "Multiply 1 meter to the south with one meter to the west, and you get 1 square meter...." like Shimmer sez. - * Beware, gentle knight - the greatest monster of them all is reason. -Cervantes
 *The Higher Arithmetic (en sea) (5.00 / 1) (#55) by bjlhct on Mon Sep 09, 2002 at 07:07:05 PM EST

 * Beware, gentle knight - the greatest monster of them all is reason. -Cervantes[ Parent ]
 solution to #2 (5.00 / 1) (#27) by bitmask on Sat Sep 07, 2002 at 04:59:20 PM EST

 What I had thought naively, was that the new e1 would switch places with e12 or e13, so that the multiplication table would be the same but with 2 rows and columns switched around. I don't know if that makes sense, but anyway, it's not what happens (and can't happen - the number of negative squares is wrong), so it probably doesn't make sense. What does happen, I found after a bit of poking around on clados.sourceforge.net, is that e1 becomes timelike; the multiplication table becomes isomorphic to that of R(2,1). The elements that commute are the same as for R(3,0), and all of the elements in the e2, e3 and e23 columns and rows are unchanged in sign. All the other signs are flipped. Looking back on it, this should have been obvious. Thanks for a good problem.
 very good (none / 0) (#32) by adiffer on Sat Sep 07, 2002 at 10:16:38 PM EST

 That's it.  You wind up with an algebra useful for describing a universe with two spatial directions and one time axis.  It is possible to write some amusing line segments that are made of equal parts space and time directions that have zero squares.  These 'nilpotents' are really useful if you want to represent light waves. The other version of this problem that I didn't include involved adding a fourth generator (e0) and giving it a negative square.  The multiplication table widens out to 16x16 and the algebra becomes R(3,1).  This is the algebra I typically use to understand special relativity.  Some of the weirder results of relativity are pretty obvious if you work with objects represented in this algebra.  Of course, R(1,3) works just as well. Keep what you've learned in mind for section two of this intro.  The rotation operation becomes something neat if you decide to change one of the spatial directions into a timelike direction. -Dream Big. --Grow Up.[ Parent ]
 Pehaps you can help me (none / 0) (#43) by epepke on Sun Sep 08, 2002 at 05:08:48 PM EST

 I'm a geometry kind of guy. I've been beating the bongos for quaternions for fifteen years and like them so much I put them as elementary types in my LISP system. I'm comfortable with taking the products of tori and projective planes in topology. Hyperbolic manifolds and even orbifolds don't faze me a bit. I did this in work for scientific visualization for a long time, and I'm now getting back into it for making multidimensional environments for games. I'm always looking for better terminology. However, I've read this several times, and my reaction is still "eh, so nu?" I don't see some special magic in this notation that really does something useful, unlike, for example, the several kinds of magic that you obviously get out of quaternions. Is there some magic that you haven't told us about, or am I just being dimwitted? OK, yeah, you can add and multiply these things and you have an identity that you call e instead of I for some reason, but what of it? The truth may be out there, but lies are inside your head.--Terry Pratchett
 Weird isn't it? (none / 0) (#46) by Humuhumunukunukuapuaa on Mon Sep 09, 2002 at 12:19:17 PM EST

 I have this problem too. Quaternions do have a kind of magic to them. They give a great representation of SO(3) that really is different from the usual one and is much simpler to deal with. But Geometric Algebra is different. It's more of a religious thing - some people prefer to use it. Not that Clifford Algebras aren't interesting. There's a great wealth of interesting material: the representation theory of rotation groups, spinors and quantum mechanics and other fun stuff. But as a replacement for the usual vector notation: I just don't see the point. -- (&()*&^#@!!&_(\$&)!&\$(*#\$(!\$&_(!\$*&&!\$@[ Parent ]
 Three things (none / 0) (#56) by Pseudonym on Mon Sep 09, 2002 at 08:19:58 PM EST

 First, the 3D geometric algebra contains the quaternions, which is actually a fairly deep result. Second, you can add a fourth dimension with an antisymmetric generator and get Minkowsti space-time, which is great for physicists. Third... remember this is only an introduction. I suspect that adiffer has more stuff waiting, like deriving algorithms in computation geometry from their GA representation. sub f{(\$f)=@_;print"\$f(q{\$f});";}f(q{sub f{(\$f)=@_;print"\$f(q{\$f});";}f});[ Parent ]
 a little magic (none / 0) (#58) by adiffer on Tue Sep 10, 2002 at 12:04:32 AM EST

 There is a little magic coming, but I can't do it a the introductory level.  Most people have an intuitive feel for points, lines, planes and so on.  Few have a good feel for spinors.  It is in the effort to write a spinor that leads to the first stark difference.  I'm thinking hard about spinors and how to introduce them, but I don't think they will go over well in the introduction. If you are such a big fan of quaternions, try writing out the multiplication table for R(0,2).  If you play around with the objects, you will find it is the quaternion algebra.  If I remember my names correctly, R(2,0) is the dihedral algebra. From a physics perspective, some algebras may be more interesting than others.  We use which ever ones seem to be useful as solutions spaces.  There is no doubt that R(0,2) is useful.  My angle for why R(3,0) would be useful is that it is easier to teach that the combination of formalisms we wind up using in early physics classes.  I honestly feel that a student getting used to R(3,0) could avoid many of the pitfalls they would otherwise hit when they get to rotational kinematics, rotating frames of reference, and magnetic fields. -Dream Big. --Grow Up.[ Parent ]
 R (3,0) notation (none / 0) (#44) by hugh mcd on Sun Sep 08, 2002 at 06:33:18 PM EST

 If the multiplication table is written with e31 (= -e13) it is more obvious that {e, e12, e23, e31} are a basis for the quaternions, just as {e, e123} are a basis for the complex numbers.
 agreed (none / 0) (#61) by adiffer on Tue Sep 10, 2002 at 12:22:43 AM EST

 The even ranked objects are closed under multiplication and form a subalgebra.  This one happens to be the quaternions if you make the choices you made. If you look at all the smaller algebras, they can all be shown to be products of a few simple ones like the quaternion, dihedral, and complex algebras.  If I remember correctly, there are a few others for the list, but I've never worked in an algebra with more than eight generators, so I rarely run into them.  The more mathematically oriented researchers have this all figured out, though. I believe Clifford used the term 'biquaternion algebra' for what we know as R(3,1).  It's quite interesting how it is possible to build larger algebras from smaller ones, but I wanted to avoid that for now.  There is little motivation for a physicist to learn that until quite late in the game. -Dream Big. --Grow Up.[ Parent ]
 Editorial (none / 0) (#45) by p3d0 on Mon Sep 09, 2002 at 10:56:50 AM EST

 Hmm... Whether we know that the unknown variable in the formula for the parabolic arc a rock takes when thrown up and forward does not influence the fact that we can solve the equation for the peak height and range on impact. Woah. This sentence needs some serious help. I think it's (barely) parsable if you remove the first "that", but even then it's hard to tell what's being said. Is it something like this? We can determine the peak hight and range on impact of a thrown rock, even if we don't know the indepent variable in the equation for the rock's parabolic motion. Even this doesn't make a lot of sense as-is, but maybe it's a step in the right direction. -- Patrick Doyle My comments do not reflect the opinions of my employer.
 agreed (none / 0) (#62) by adiffer on Tue Sep 10, 2002 at 12:31:43 AM EST

 That sentance basically sucks.  That's what I get for being convoluted during the wee hours of the morning. It must go away.   'thrown up'...  What was I thinking?  8) How about something like: We can determine the peak height and impact range for a projectile even if we don't realize one of the equations for the flight describes a parabola. It would have to be fit into the paragraph of course. -Dream Big. --Grow Up.[ Parent ]
 Aha (none / 0) (#65) by p3d0 on Tue Sep 10, 2002 at 09:10:30 AM EST

 Now I see what you're saying. That sentence is much better. -- Patrick Doyle My comments do not reflect the opinions of my employer.[ Parent ]
 The one useful thing (2.50 / 2) (#47) by Humuhumunukunukuapuaa on Mon Sep 09, 2002 at 12:42:02 PM EST

 There are even objects whose squares are zero and they aren't zero The one useful thing in everything you say and you just skip over it as if it's nothing but a curiosity. -- (&()*&^#@!!&_(\$&)!&\$(*#\$(!\$&_(!\$*&&!\$@
 non-introductory direction (none / 0) (#59) by adiffer on Tue Sep 10, 2002 at 12:07:04 AM EST

 This is important stuff, but the water gets deep very quickly. When I get around to spinors, that little line/comment will be developed more fully. -Dream Big. --Grow Up.[ Parent ]
 Actually I'm probably thinking of something... (none / 0) (#69) by Humuhumunukunukuapuaa on Tue Sep 10, 2002 at 12:38:04 PM EST

 ...different to you. Nilpotent elements have cool uses in computing that seem to have been overlooked by lots of people. For example if e^2=0 then f(x+e)=f(x)+f'(x)e. This gives a nice way to compute derivatives without using symbolic algebra (ridiculously heavy for just computing derivatives) or numerical derivatives (which are expensive to compute and just plain wrong). There are many computational problems that can be solved very elegantly and efficiently this way but for some reason nobody ever thinks of doing it. Elements that have the property e^n=0 can be used to find (n-1)-th derivatives. -- (&()*&^#@!!&_(\$&)!&\$(*#\$(!\$&_(!\$*&&!\$@[ Parent ]
 An example? (none / 0) (#84) by mysta on Wed Dec 11, 2002 at 05:49:30 PM EST

 That f(x+e) = f(x)+f'(x)e when e^2=0 is a consequence of the Taylor expansion of f(x+e) about x, right? This would also explain why you can get nth derivatives from elements e such that e^n = 0. I agree with you that this would be a nice way of calculating derviatives but I can't for the life of me think of any nilpotent elements that could be used in the computation. The reals don't have any so I'm guessing you are thinking of using functions over matricies. f(x) becomes f(xI) and you can use, say [ 1 0 ] [ 0 0 ] or something similar as the nilpotent element. Is this even close to what you had in mind when you were talking about nilpotents? I've not seen this used in computing derivatives so I'd like to understand it. This is now way off topic. Oh well... --- Are we not drawn onward, we few, drawn onward to new era?[ Parent ]
 BRING ON SECTION 2!!! (none / 0) (#48) by Rock Joe on Mon Sep 09, 2002 at 12:59:02 PM EST

 I attended the course on geometric algebra at siggraph last year and when the day was over, the only impression I had was that I didn't understand a thing, but that I was missing out on ALOT! Now here you come, able to explain this stuff in pseudo-laymen's terms and I'm eating this stuff up! I haven't finished all the problems and I'm very impatiently waiting for section 2. Bring it on! :o) Signatures are for losers! --Rock Joe
 Not too sure about number four. (none / 0) (#49) by Rock Joe on Mon Sep 09, 2002 at 01:19:49 PM EST

 To multiply the two terms, I merely distributed one into the other and solved, giving me 35e + 10e2 + 12e12 + 21e13 -2e23 - 6e123 (I'm too lazy to learn how to write the indices the way they're supposed to be written, but you get the idea.) The thing is I'm not sure if I was allowed to do that. If not, can you gimme a hint? :o) Please thank you. P.S:I'm sure that the answer to question five will only switch the signs of afew of those terms, so I'm not gonna do it. Signatures are for losers! --Rock Joe[ Parent ]
 My results.... (none / 0) (#50) by BlaisePascal on Mon Sep 09, 2002 at 02:35:57 PM EST

 I got the same thing, but with -20e23 instead of -2e23. Since I had +2e32 running through most of my calculations before I realized I dropped a 0, I think you made a similar error and didn't catch it. [ Parent ]
 It's not me! It's YOU! [nt] (none / 0) (#51) by Rock Joe on Mon Sep 09, 2002 at 03:09:24 PM EST

 EVERYONE knows that 5*4 = 2!!! :o) Signatures are for losers! --Rock Joe[ Parent ]
 distribution (none / 0) (#64) by adiffer on Tue Sep 10, 2002 at 01:23:21 AM EST

 Distribution is legal as long as you don't get the right/left order mixed up. Problem 5 is there just to make sure people don't stay in the habit of commuting operands without paying for it.  8) -Dream Big. --Grow Up.[ Parent ]
 Thanx, but... (none / 0) (#66) by Rock Joe on Tue Sep 10, 2002 at 10:22:39 AM EST

 I'm not sure I get the answer to problem 2. What do the two numbers in parenthesis mean to begin with? I'm talking about things like R(3, 0) and R(2, 1) and the like... And how far into this topic do you plan on bringing us? Signatures are for losers! --Rock Joe[ Parent ]
 My take on R(3,0), R(2,1), etc... (none / 0) (#67) by BlaisePascal on Tue Sep 10, 2002 at 11:39:03 AM EST

 The comments "spacelike", "timelike" generators has been bandied around, and I believe that the parenthetical numbers refer to the number of spacelike and timelike generators. Spacelike generators follow the formula e = ei2, while timelike generators follow the formula e = -ej2.    Personally, it makes more sense to me to see them as "real" and "imaginary" generators, since they act (I think) as if the scaling coefficients of the "timelike" generators are imaginary. As an example of several geometries: R(0,0) would have only a scalar basis e.  It would act like regular real numbers, I think. R(1,0) would have two orthogonal bases e and e1, with e = e12.   This would act like polynomials of the form a+bx with the simplifying rule of x2 = 1. R(0,1) would likewise have two orthogonal bases, but in this case e = -e12.  This would act like complex numbers. R(2,0) would have four orthogonal bases e, e1, e2, e12, with the two generators each squaring to e.  The two generators will anti-commute (e1e2 = -e2e1), and unlike the R(3,0) case we've been studying, the "all generators" basis (Argh, I hate doing maths in HTML...  TeX notation from now on....), i.e., \$e_{12}\$ (or \$e_{123}\$ in the R(3,0) case), anticommutes with everything but \$e\$ and \$e_{12}\$.  \$e_{12}^2 = e\$, as well. R(1,1) would also have four orthogonal bases, \$e\$, \$e_1\$, \$e_2\$, with \$e_2^2 = -e\$.  I should be working now, not figuring out how that works... etc. Is that right?  Does that help? [ Parent ]
 majority (none / 0) (#70) by adiffer on Tue Sep 10, 2002 at 04:18:01 PM EST

 There are quite a few people who think of the generators as you do.  The physicists are actually the minority, so our description of space-like and time-like is not real common.  There are even differences among us for labelling the algebras too. Someday, we may adopt the sensible approach offered by the mathematicians.  We will see.  8) -Dream Big. --Grow Up.[ Parent ]
 What is next? (1.00 / 3) (#52) by darthaya on Mon Sep 09, 2002 at 03:56:39 PM EST

 Introduction to geometry? Since when k5 had become a website of rudimental education? And people voted this crap to the front page?!?!
 Nice troll. (none / 0) (#53) by upsilon on Mon Sep 09, 2002 at 05:23:05 PM EST

 It's geometric algebra, not geometry. So should I be commending you on a quality troll, or on not even reading the title of the article, let alone the body? -- Once, I was the King of Spain.[ Parent ]
 Why shouldn't it be? (none / 0) (#54) by Spendocrat on Mon Sep 09, 2002 at 07:02:14 PM EST

 You're complaining about it, but I don't see you stating any reasons why this sort of "crap" shouldn't be voted up. [ Parent ]
 poll (none / 0) (#60) by adiffer on Tue Sep 10, 2002 at 12:09:02 AM EST

 That's why I put the poll in.  I really was curious to see if the K5 community felt that educational articles were appropriate.  The numbers strongly suggest a landslide against your POV. -Dream Big. --Grow Up.[ Parent ]
 solutions (none / 0) (#63) by adiffer on Tue Sep 10, 2002 at 01:17:10 AM EST

 1: Try multiplying both objects from example 2 in the opposite order. e13 and e123 = e13123 = -e11323 = -e323 = e233 = e2 There is no change because e123 happens to commute with everything in our algebra. 2: What would happen to the multiplication table if the square of e1 were -e? As has been pointed out by 'bitmask', the generator becomes time-like. A few negative signs get added to the table. The multiplication table for R(2,1) is similar. If you renumber the generators from 0, 1, 2 to 1, 2, 3 you are done. 3: Add (3e1 + 5e3) and (4e2 + 7e3 -2e23). = (3e1 + 5e3) + (4e2 + 7e3 -2e23) = 3e1 +4e2 + 5e3 + + 7e3 -2e23 = 3e1 +4e2 + 12e3 -2e23 4: Multiply (3e1 + 5e3) and (4e2 + 7e3 -2e23). = (3e1 + 5e3) (4e2 + 7e3 -2e23) =12e12 + 21e13 -6e123 +20e32 + 35e33 -10e323 =12e12 + 21e13 -6e123 -20e23 + 35e +10e2 =+ 35e +10e2 +12e12 + 21e13 -20e23 -6e123 5: Multiply (4e2 + 7e3 -2e23) and (3e1 + 5e3) = (4e2 + 7e3-2e23) (3e1 + 5e3) = 12e21 + 21e31 -6e231 +20e23 + 35e33 -10e233 = + 35e -10e2 -12e12 -21e13 +20e23 -6e123 -Dream Big. --Grow Up.
 Huzzah! ... and applications. (none / 0) (#68) by spcmanspiff on Tue Sep 10, 2002 at 12:36:16 PM EST

 First of all, thanks for posting this; it's interesting, easy to read, and right at my level. I've got a few questions about how this could be used, stealing things that I need to do often with linear algebra. Finding the perpendicular to a plane: I'm not sure how to represent the plane.... I guess as the product of two vectors (Ae1 + Be2 + Ce3)(De1 + Ee2 + Fe3) = (AD+BE+CF) + (AE-BD)e12 + (AF-CD)e13 + (BF-CE)e23 (and it's even a nice product of the right basises! bases?) The geometric product of a plane and its perpendicular would result a unit volume segment (extrapolating blindly from result w/ two perp. line segments) ... it's a straightforwardish matter of solving for the perpendicular from (K+ Ae12 + B13 + C23)(Xe1 + Ye2 + Z3) = Me123 Distributing that out, I get: (KX+AY+BZ)e1 + (KY-AX+CZ)e2 + (KZ-BX-CY)e3 + (AZ + BY + CX)e123 = Me123 Which turns into a nice easy linear system of: KX+AY+BZ=0 KY-AX+CZ=0 KZ-BX-CY=0 Very convenient, assuming my blind extrapolation was right. :) Also, unlike cross products, it's easy to extend to higher dimensions! I guess, after going through all that, I don't have any specific questions. Just, am I doing this right? Do I get it? Thanks much for taking the time to write this! I'm looking forward to upcoming chapters / sections.
 looks good (none / 0) (#71) by adiffer on Tue Sep 10, 2002 at 04:28:59 PM EST

 Your linear system leads to solutions that force the line segment to be perpendicular to the plane.  If what you want is a volume segment, you have the right approach.  In general, you get a line segment and a volume segment as you've seen.  It's all right to have a sum like this.  What you are running into is the distinction we will make later between the inner and outer products. Here is an easy way to find a perpendicular to a plane that works in three dimensions.  If your plane is ∑ Mij eij then multiply by e123.  The result is a vector perpendicular to the plane. -Dream Big. --Grow Up.[ Parent ]
 Details.... (none / 0) (#74) by spcmanspiff on Tue Sep 10, 2002 at 09:19:59 PM EST

 Just out of curiosity, I tried to derive that and got stuck: I started with: plane =  (Ae12 + Be13 + Ce23) vector = (Xe1 + Ye2 + Ze3) -- solving for this! (plane)(vector) = Me123 Taking the product of both sides by e123, then (vector), I get: (e123)(plane)(|vector|2) = (-M)(vector) I can divide by the length2 of the unknown vector, but I don't see how to get rid of the scalar terms. Is it safe to just ignore them?? (e.g., assume a unit volume and a unit vector?) If so, why is that okay? If not, then where did I screw up? Gracias!   [ Parent ]
 like this (none / 0) (#75) by adiffer on Tue Sep 10, 2002 at 09:46:01 PM EST

 Plane = Ae12 + Be13 + Ce23 Multiply the plane by e123 and see what happens. Plane e123 = Ae12123 + Be13123 + Ce23123 = -Ae3 + Be2 - Ce1 which is a vector. If you were trying to solve for Xe1 + Ye2 + Ze3  then X = -C, Y = B, Z = -A -Dream Big. --Grow Up.[ Parent ]
 But ... (none / 0) (#78) by spcmanspiff on Tue Sep 10, 2002 at 11:00:19 PM EST

 How do I know that vector is a perpendicular to the plane, as opposed to just any old vector? I was trying to go about it from the (plane)(vector) = (scalar)(e123) form that I had in my first comment, since a vector that satisfies this will be perpendicular to the plane, but couldn't get rid of scalar coefficients. Of course, since direction is the important property here and not magnitude, I could just drop the scalars and arrive at (plane)(e123) = (vector), but I'm not sure that's the best thing to do. Leaving them in, I can get at: (plane)(e123) = (-M)(1/L2)(vector) where M is the scalar coefficient of the unit volume (e123) and L the length of (vector). Do these have any meaning? Or should they be dropped? I'm beginning to think that "dropped" is the correct answer.... Actually, solving for a unit-length vector makes sense, so that term can disappear. I don't know about the -M, though, especially the -1 part. Suggestions? (I guess if a unit plane * a perp. unit vector = a -1 unit volume, it will drop too if the plane is normal... the multiplication table hints that this is the case....)   [ Parent ]
 my confusion (none / 0) (#79) by adiffer on Wed Sep 11, 2002 at 12:11:09 AM EST

 oh.  The general rule would go like this. Multiply a plane and a vector.  If you get a vector piece in the result, part of the first vector lies in the plane.  If you get a volume piece in the result, part of the first vector is perpendicular to the plane. The test for perpendicularity between a plane and a vector is that their product must be a volume and only a volume.  Your approach with  (plane)(vector) = (scalar)(e123) and the coefficients would produce a system of equations that solve for vectors perpendicular to the plane.   I would think it would be easier to pull the length of the vector out ||v|| and be left with a unit vector in the product.  If you're solving it with a computer, though, it won't care. -Dream Big. --Grow Up.[ Parent ]
 So, to summarize, (none / 0) (#81) by spcmanspiff on Wed Sep 11, 2002 at 12:27:49 PM EST

 (any plane)(unit volume) = (mystery coefficient)(unit vector perp to plane)  = (non-unit vector perp to plane) If the plane is a unit plane, too, then the mystery coefficient will be e. Otherwise, it'll be some weird number. :) Then, just for fun... in higher dimensions (n), (any plane)(/n/-dimentional basis) = (mystery coefficient)(n-2-dimensional object with all components perpendicular to the plane) That makes sense, I think. I'm off to read your next article; sorry for bothering you with questions! but it's pretty neat stuff.   [ Parent ]
 Error, and complaint (none / 0) (#72) by Sacrifice on Tue Sep 10, 2002 at 04:47:16 PM EST

 http://clados.sourceforge.net/r30.html: "An 'e' by itself is the additive identity." I was somewhat distressed by this article.  I thought there was something wrong with me when (having no prior knowledge of the topic) I did not understand it.  Then I figured out I am supposed to fill in the blanks by analogy to high school algebra and 3-dimensional Euclidean space. The author should say what he means, and not just allude to it. If the generators are "used as basis vectors" (so they would consist of a length and a direction), then why would they require two points (a start and end) to define them?  Wouldn't one point suffice? Furthermore, I assume that the product of two line segments is in fact an oriented parallelogram, not a triangle (in terms of interpreting multiplication and magnitudes by analogy to the usual area). I am also supposed to guess (nowhere is this specified) that multiplication distributes over the components.  (so that (a,b)*(c,d) = a*c + a*d + b*c + b*d). I am completely floored by the non sequitur introduction of the cosine of the angle between two line segments (I assume a line segment is a linear combination of the generators).  Are we to assume now that the generator are in fact orthogonal basis vectors of a Euclidean space?  And that "perpendicular" and "parallel" have the usual geometric meaning instead of the "commute/anti-commute" meaning given earlier?
 answers (none / 0) (#76) by adiffer on Tue Sep 10, 2002 at 09:54:07 PM EST

 My prerequisites statement was buried in an editorial comment.  I should have moved it into the body of the article. Vectors as directed line segments do rely upon two points.  One of them is usually implied, though, as the origin.  Think geometrically and remember that two points define a line. I didn't think anyone would think of the multiplication in terms of triangles.  I shall have to give some thought to that.  You are right about the parallelograms, though. Distribution is first shown in example one.  I debated listing the rules one is allowed to use instead of just showing them through examples.  If I list them, I tick off the math neophytes.  If I don't, I tick off the more experienced readers.  I decided to go with a more example based approach to 'uncover' the rules.  Most people will try to use distribution without being told it is OK.  Math teachers shouldn't take this approach, but I'll try to get away with it and claim I'm a physicist.  8) You are right about example three.  I like it less and less every day. -Dream Big. --Grow Up.[ Parent ]
 Why the obsession with perpenduculars? (none / 0) (#73) by expro on Tue Sep 10, 2002 at 06:42:58 PM EST

 Some feel that perpendiculars are unnecessary. According to Bucky our space has four natural directions not three orthogonal ones. There are many other good choices as well. You can do most math more convienently without the peculiar perpendiculars everywhere and you can produce more meaningful exact answers. I never studied it in school, but I find much of geometric algebra unnecessarily arbitrary and fragile because of reliance on such perpendicular axes.
 getting there (none / 0) (#77) by adiffer on Tue Sep 10, 2002 at 09:59:10 PM EST

 By the time we get to relativity, there are four directions and they are only locally perpendicular in the rest frame.  Other than that, I will have to admit that there are a lot of things 'Bucky' said that I don't understand. The reliance on the axes you don't like is a representation choice.  There is no reason you couldn't make a different choice.  What I've shown is just one way to do it. -Dream Big. --Grow Up.[ Parent ]
 Unfortunately, I am not a mathematician. (none / 0) (#83) by expro on Fri Sep 13, 2002 at 11:00:27 AM EST

 By now I am a bit off topic. I pounced on this because it is something I can talk about and perhaps some day contribute. But I have experimented a bit with non-perpendicular axes, starting with Bucky's tetrahedral coordinates and doing the math to actually operate on such a space going all the way up to icosahedral coordinates (there are some interesting results here as well, I think) and looking at the more general problems in isolation from the perpendicular assumptions. I've redesigned the trig functions to match, tried to look at the more-traditional geometric products, explored what that does to area, how area relates to volume, examined the value of normalization and how having a non-normalized representation can be a great benefit, examined how spaces described by different coordinate systems are open or closed under rational operations in different situations, and more. I think there are some results here that are not so obvious, and perpendicularity is built in in very strange ways because everyone has been making these assumptions since Pythagoras, which have been apparent in Hamilton's document and all those I have read which followed. There is some real beauty in nature, and perpendicularity does turn out to be an artificial case, unstable, and hiding quite a bit of natural beauty. I've written up some of it with graphics and references, but I am ashamed to even show it until I figure out how to shed some of the additional orthogonalisms that have been bestowed upon me during my short education, because it seems I have just revealed the tip of the iceberg and need to rewrite it with much more. But I can't find any reputable mathematician who has explored this area, which frustrates me. In short, I think that the choice of "convenient" perpenducular coordinates may hide more than mathematicians think, and be less convenient than they think. [ Parent ]
 Alternatives to orthogonal coordinate systems (none / 0) (#85) by pdx4d on Sat Jan 25, 2003 at 01:18:55 PM EST

 You're not alone in investigating coordinate systems not built around perpendicular relationships. I invite you to visit my Introduction to Quadray Coordinates and explore the links that you find there. I especially bring your attention to Ray Whitmer's paper (listed in For Further Reading). I'd be interested in learning more about your results. Also, it may be that geometric algebra has alternative visual representations which do not effect the algebra e.g. e1e2e3 might relate to a tetrahedron. [ Parent ]
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