# I Is Geometric Algebra inconsistent/circular?

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1. Jul 8, 2016

### malawi_glenn

I am trying to learn Geometric Algebra from the textbook by Doran and Lasenby.

They claim in chapter 4 that the geometric product $ab$ between two vectors $a$ and $b$ is defined according to the axioms
i) associativity: $(ab)c = a(bc) = abc$
ii) distributive over addition: $a(b+c) = ab+ac$
iii) The square of any vector is a real scalar

Then they claim that the inner and outer product are defined as
$a \cdot b = \frac{1}{2} (ab+ba)$
$a \wedge b = \frac{1}{2} (ab-ba)$
so that
$a b = a \cdot b + a \wedge b$

My problem is that if you are given two vectors, say
$a = 1e_1 + 3 e_2 - 2e_3$
$b = 5e_1 -2 e_2 + 1e_3$

How to actually compute $ab$?

I mean, you then have to specify how either $a \cdot b$ and $a \wedge b$ works, or how (in detail) $a b$ are to be performed.

This is in my view, circular.

From another point of view, say that you start backwards by defining the inner and outer product, and then define the geometric product as $a b = a \cdot b + a \wedge b$

Then how to show that the geometric product is associative? The usual definition of the outer product is associative, but the usual definition of the inner (dot) product is NOT associative. So how to show that the geometric product is associative if you take the inner and outer product as starting point for the geometric product?

Thank you very much in advance for any kind of feedback

2. Jul 8, 2016

### micromass

Staff Emeritus
This definitely does not define a unique product. So if that is the definition your book gives, then that is incorrect. It is certainly possible to define the geometric product axiomatically, but I think the easiest way is to do as you noticed and define the geometric product "backwards". But for information, the full list of axioms takes a bilinear form $Q$ (usually taken to be the inner product) and defined the geometric product
1) Associative
2) Distributive
3) $a^2 = Q(a,a)$
4) The geometric product on scalars agrees with the usual product

The problem here is that taking the geometric product of two vectors does not give you a vector anymore! It gives you a combination of a scalar and a 2-form. So the definition $xy = x\cdot y + x\wedge y$ is not good enough anymore since it only holds true for scalars. You'll need a definition that holds true for all $n$-forms if you wish to show associativity.

3. Jul 8, 2016

### malawi_glenn

This is the book I am using http://www.cambridge.org/se/academi...hysics/geometric-algebra-physicists?format=PB
My impression was that the authors are considered (?) to be world leading experts on this matter, that is why I choose to study this book

As you point out, $ab$ is no longer a vector but instead a scalar plus a bivector. This means that in order to compute $(ab)c$ one needs to define the geometric product of scalars-vectors and bivectors-vectors (and so forth).

Then my question is, how do you define such geometric products? My impression is that if you can define the geometric product between some "basic elements", then you can recursively work out the geometric product between any "multivector"

4. Jul 8, 2016

### micromass

Staff Emeritus
It's a shame if the book doesn't tell you how to compute the geometric product for multivectors! In any case, you can see a fully rigorous approach here http://arxiv.org/pdf/1205.5935v1.pdf

5. Jul 8, 2016

### micromass

Staff Emeritus
Last edited by a moderator: May 8, 2017
6. Jul 8, 2016

### malawi_glenn

Thank you, I don't know if the book tells me this or not, I am just starting at chapter 4. Perhaps it becomes more clear and complete in the latter chapters. I just felt that I needed some feedback.

The book by Dorst was also something I considered. I had a brief examination on google books and my impression was that it was too detailed (left contraction, right contraction etc) but perhaps this is the way to do it rigorously?

7. Jul 8, 2016

### micromass

Staff Emeritus
Well, it is exactly this left/right contraction that you use to define the geometric product in general!

8. Jul 8, 2016

### malawi_glenn

I guess I am more into physics-applications (I have PhD degree in theoretical particle physics), that was also a reason for why I didn't want to spend 70\$ on a book inclined towards computer graphics. Perhaps I should just "bite the bullet" and use the book by Dorst to straighten out my fundamental issues and then move back to a more "physics application approach" text after that? What do you suggest?

9. Jul 8, 2016

### micromass

Staff Emeritus
I guess it depends on how much you want to take on faith. I don't think this left/right contraction stuff is actually very important in physics, so it might not be very useful for you to learn it. I think you should try to read the Arxiv link I provided above to see how things are done rigorously. I think that should be enough. You should only get Dorst if you want an intuition behind the myriad of different operations in geometric algebra. But I don't think it's necessary to actually get the book if you're into physics. Although it is a really good one.

10. Jul 8, 2016

### malawi_glenn

What is your impression on the books by Hestenes?

11. Jul 8, 2016

### micromass

Staff Emeritus
I'm not a fan. There's definitely a lot of information in them. But they don't give enough intuition to be useful to me.

12. Jul 8, 2016

### malawi_glenn

Ok, thanks!

My goal of learning this is just to have some fun :)

13. Jul 8, 2016

### micromass

Staff Emeritus
Last edited by a moderator: May 8, 2017
14. Jul 8, 2016

### malawi_glenn

as long as things are done in logical and pedagogical consistent manners, I don't care if the level is "too easy" :)

15. Jul 8, 2016

### malawi_glenn

In the ArXiV paper you sent me, it says:

"I begin with a formal product of vectors uv that obeys the usual rules for multiplication; for example,
it’s associative and distributive over addition."

now this directly confuses me, what IS the "usual" rules for multiplication? I often think that multiplication is commutative, but the geometric product is not commutative in general

16. Jul 8, 2016

### mnb96

I think in that context the "usual" rules for multiplication are exactly those that the author means by "usual", i.e. associativity and distributivity over addition. Commutativity holds for multiplications between real scalars but it is certainly not an "usual" property of matrix multiplication, for instance.

I would not focus too much attention on the choice of word "usual" and I think that you can mentally replace the string "the usual rules" with "the following rules".

When the author said "the usual rules for multiplication", he probably had in mind the rules of multiplication in an arbitrary ring (not necessarily in the ring of reals), where multiplication is not necessarily commutative.

17. Jul 8, 2016

### Staff: Mentor

One could also object that "usual" implies a $1$ and an inverse. This shows how personal the term is. If you're used to mappings, rings and algebras then usual becomes something more general. Personally, I even regard associativity as a special condition and would be satisfied with the distributivity.

18. Jul 8, 2016

### malawi_glenn

ok so usual in the ring kind of sense?

19. Jul 8, 2016

### Staff: Mentor

More in the sense of an algebra (= a vector space with multiplication) since you still have vectors and a scalar field like ℝ or ℂ that allows to stretch the vectors.

20. Jul 8, 2016

### micromass

Staff Emeritus
That is the introduction. The introduction is for motivation only. You should start reading at section 2 page 10.