General formulas for inner and outer products...?

Higher-dimensional geometry (previously "Polyshapes").

General formulas for inner and outer products...?

Postby Paul » Fri Nov 26, 2004 11:09 pm

Hello Pat and all,

First, I tried to find the inner and outer products that seemed to make this work out well...

I'm much more interested in what you think about the 'justifications' of why I did what I did below...

Image

As you can see, I believe I've taken the definitions that David Hestenes chose. I believe the defintion of inner product I've chosen is what the author of this webpage, http://www.iancgbell.clara.net/maths/geoalg1.htm#A18, calls the semi-commutative inner product, and the definition of outer product appears to be what the author of this webpage calls the "thin" outer product.

Going with these definitions of inner and outer products, I'm wondering if I can take the associated concept of parallelness with the inner product, and the associated concept of perpendicularly with the outer product, and generalize what I've done here to general multivectors.

Image

Are my geometric and algebraic formulations correct? Is what I saying here essentially what Hestenes is saying... except perhaps more so in my own words...?

In what ways perhaps yes, and in what ways perhaps no...?
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Postby Paul » Fri Nov 26, 2004 11:44 pm

I think I just found a silly error... :oops:

Everywhere you see the stuff on the left side of the arrow, substitute the stuff on the right...

Image
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Postby pat » Sat Nov 27, 2004 7:19 am

And, I'm going to have read more about the semi-commutative inner product and such. I've never tried to dot-product things that weren't pure vectors (1-vectors).

But, you lost me a little bit right out of the gate in the second image there where you define:
coef(A<sub><a<sub>j</sub>></sub>)

There is more than one blade in A<sub><a<sub>j</sub>></sub>. So, there's more than one coefficient. I believe the gist of your argument is fine... but it looks (notationally) like you're thinking there will only be one blade of a given grade. Or, am I reading the line above it incorrectly where you define blades?
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Postby Paul » Sat Nov 27, 2004 5:17 pm

Hello Pat,

I think the formulation is likely... all messed up.

Also... I don't think I really want the "thin" outer product... I think I want the "regular" outer product. However, I want the outer product of two scalars to be zero, but I don't want the outer product of a scalar and a nonscalar to be zero. So, I'm still a bit confused about all this...

I'm pretty sure an outer product where a ^ b, where a, b are scalars would make sense, but that's not what the operation table for the outer product on the linked to page above would indicate for the outer product... anyway, I'm still pretty sure that such an outer product is what I want.

I'll show a good reason why I tend to believe that later...

There is more than one blade in A<sub>a<sub>j</sub></sub>.


Yeah... I think 'blade' is the wrong word. I think I just mean every term...

Also... just noticed something else. I think for the outer product we don't need the intersection between the two permutations to be null... I think what we want is something like [perm(A) - perm(B)] U [perm(B) - perm(A)] not null... or perhaps [perm(A) U perm(B)] - [perm(A) int. perm(B)].

I've got to run out now...
Paul
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Postby Paul » Sat Nov 27, 2004 10:51 pm

Hello Pat and all,

I decided to work out the Clifford formalism for the quaternion product to see if that reveals some more insights...

First, recall quaternion multiplication:

Image

The quaternion product does appear given the definitions of inner and outer product as previously defined:

Image

There appears to be many symmetries that stand out. One thing I noticed is that all the possible products of terms are entirely partitioned into inner and outer products... that is, a term product is either one or the other, not both, and all are one, or the other. What's more if lays out all the possible products in a matrix as above, those on the main diagonal are all the inner products, and all the others are wedge products.

So, perhaps another way of indicating which products are wedge products is just to say that all those products, of all the possible term products, which are NOT inner products are wedge products. That is, all term products such that perm(A) != +/-perm(B).

Generally, I believe the concept of associating parallelness with the inner product and perpendicularly with the outer product is a helpful way of viewing them.

However, these concepts are somewhat more obscure when one's dealing with scalars.

I think I can probably clean up these general formulas for the inner and outer product.


Questions are lurking around here concerning the permutation of the basis vectors. This determines sense(s), handedness, and chirality... yes?

How precisely does one differientiate between sense, handedness, and chirality?
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Postby pat » Sat Nov 27, 2004 11:50 pm

Paul wrote:Questions are lurking around here concerning the permutation of the basis vectors. This determines sense(s), handedness, and chirality... yes?

How precisely does one differientiate between sense, handedness, and chirality?


I've not really messed much with permuting the basis vectors yet myself. But, as I understand it ``handedness'' and ``chirality'' are synonymous. ``Chiral'' is just the Greek word.
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