Geometry of 7D cross product...?

Higher-dimensional geometry (previously "Polyshapes").

Geometry of 7D cross product...?

Postby Paul » Sat Jan 08, 2005 5:42 pm

Hello all,

I'm having trouble understanding the geometry of the 7D cross product... particularly, as it relates to this passage below from pp. 96-97 in Lounesto's Clifford Algebras and Spinors:

Image


In particular, does Lounesto mean that there will be other 1-vectors in all other 3D-spaces in 7D-space with the same direction as a x b...?

Am I correct that there won't be any 1-vector in the same 3D-space as a x b since once you take the dual of the given bivector (two 1-vectors wedged together) through the volume element of a 3D-space defined by that 3D volume element, i.e. perhaps e<sub>124</sub> for instance, that there won't be any other 1-vectors in this 3D-space so defined by this 3D volume element... just like with the cross product of our 3D-space...?

Does my geometric understanding make sense...?

Can someone explain about the geometry here...?
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Postby Paul » Tue Jan 11, 2005 12:49 am

Hello all,

For all who might be interested...

I wanted to point out that expression of the octonion product

Image

in terms of the Clifford product here for <Cl<sub>0,7</sub>><sub>0,1</sub> (the paravectors of Cl<sub>0,7</sub>) is not as complicated as it may look...

What happens is that the Clifford product produces scalars, 1-vectors, and bivectors... The '1' in (1 - v) catches the scalar and vector components of the Clifford product and passes them over to the octonion product unchanged. The components of the non-simple trivector v of R<sup>0,7</sup> catches all the bivector portions of the Clifford product and duals them into their appropriate octonion product 1-vectors.

Scalars multiplied by v produce trivectors... 1-vectors multiplied by v produce bivectors and 4-vectors... and all of these are excluded since they are not of grades 0 or 1.

In actual computations, only one of the simple trivectors of v will dual with any bivector generated by the Clifford product... so, when you do this by hand, it's not quite as involved as it might at first look... Since you know that only the 1, or one element of v will multiply and contribute the resulting octonion product from the input Clifford product on the left. And it's pretty obvious which term will be the appropriate one for each term of the Clifford product...

I don't know about any of you,... but, for me, v looked pretty scary at first. It took me awhile to see that this is all that Lounesto means for us to do.

Of course, I think it's still easier to figure out an octonion product by hand by using the heptagon and the simple rule illustrated in my post about seeking an alternate basis for the octonions.


Also... how about forming a basis for the quaternions in <Cl<sub>0,3</sub>><sub>0,1</sub> utilizing the product rule

Image ...?

One could probably also adapt how Lounesto forms octonions and their product in <Cl<sub>0,8</sub>><sub>1</sub> similarly for forming quaternions and their product in <Cl<sub>0,4</sub>><sub>1</sub>...?

Likewise, one could also probably make a similar formulations for the complex numbers in <Cl<sub>0,1</sub>><sub>0,1</sub> and <Cl<sub>0,2</sub>><sub>1</sub>...?

What I think is neat is that it appears one can get all the normed division algebras (the reals are kinda a trivial case...) into one (or two) formulations...?


Another question... definitions. Are these two different types of representations the <i>Pinor</i> and <i>Spinor</i> representations? Is this what those terms mean?
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Postby Paul » Tue Jan 11, 2005 4:38 pm

Hello again all,

It occurred to me that this doesn't seem right...

One could probably also adapt how Lounesto forms octonions and their product in <Cl<sub>0,8</sub>><sub>1</sub> similarly for forming quaternions and their product in <Cl<sub>0,4</sub>><sub>1</sub>...?

Likewise, one could also probably make a similar formulations for the complex numbers in <Cl<sub>0,1</sub>><sub>0,1</sub> and <Cl<sub>0,2</sub>><sub>1</sub>...?


It seems I'm forgetting that for the pure vector representations we need one basis vector to square to +1. So, the above should read:

One could probably also adapt how Lounesto forms octonions and their product in <Cl<sub>1,7</sub>><sub>1</sub> similarly for forming quaternions and their product in <Cl<sub>1,3</sub>><sub>1</sub>...?

Likewise, one could also probably make a similar formulations for the complex numbers in <Cl<sub>0,1</sub>><sub>0,1</sub> and <Cl<sub>1,1</sub>><sub>1</sub>...?
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Re: Geometry of 7D cross product...?

Postby pat » Wed Jan 12, 2005 6:50 pm

Paul wrote:In particular, does Lounesto mean that there will be other 1-vectors in all other 3D-spaces in 7D-space with the same direction as a x b...?


There are two different things going on in the passage, from what I can see. It's kind of obscured by some of his language at the beginning of the subsection about how to define the 7-d cross-product.

The rule: e<sub>i</sub> x e<sub>i+1</sub> = e<sub>i+3</sub> is only one of the ways you could define a 7-d cross-product that obeys the orthogonality rule and the Pythagorean rule. That way corresponds to the rule he gives at the bottom if you take v to be the 3-vector that he gives at the top of the next page.

I am sure that if you toggle some of the signs of the blades in v (and, there may even be 3-vectors made up of other blades altogether... maybe so long as each index appears in exactly three blades and doesn't share more than one blade with any other index?), you will also come up with a valid cross-product---it will obey the orthogonality and the Pythagorean requirements. But, you won't have something so nice as e<sub>i</sub> x e<sub>i+1</sub> = e<sub>i+3</sub>.

That's what he means when he says "depends on a 3-vector defining the cross product". You can change that 3-vector a bit and still get a cross product.

Now, there's another thing going on, too, which you highlighted in yellow. In 3-D, if you know that a x b = c x d (and not zero), then you know that both c and d are in the plane spanned by a and b and that a and b are in the plane spanned by c and d. He's saying, in the yellow section, that with the 7-D cross-product, a x b = c x d is not a guarantee that c and d are in the plane spanned by a and b (or vice-versa).

I tried, for a bit last night to come up with two vectors c and d such that
e<sub>1</sub> x e<sub>2</sub> = c x d = e<sub>4</sub>. But, my first few attempts fizzled. And, it got messy pretty fast trying to work something out with 14 variables (two 7-d vectors). So, maybe I'll write some code to find some.

Am I correct that there won't be any 1-vector in the same 3D-space as a x b


I don't understand what you're asking here. Certainly, a, b, and a x b are three mutually perpendicular 1-vectors. Thus, the form a 3D space. And, for example, a+b is in that space as is a+b+(a x b) and any linear combination of the three.
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Re: Geometry of 7D cross product...?

Postby pat » Wed Jan 12, 2005 7:02 pm

pat wrote:The rule: e<sub>i</sub> x e<sub>i+1</sub> = e<sub>i+3</sub> is only one of the ways you could define a 7-d cross-product that obeys the orthogonality rule and the Pythagorean rule.


And, now that I look at it again, you can read that rule straight off the indexes in the blades of v. e<sub>1</sub> x e<sub>2</sub> = e<sub>4</sub> is precisely because of the e<sub>124</sub> term.

So, for example, I contend that the rule at the bottom of the left page there would also satisfy the orthogonality and Pythagorean restrictions if v were: e<sub>126</sub> + e<sub>237</sub> + e<sub>341</sub> + e<sub>452</sub> + e<sub>563</sub> + e<sub>674</sub> + e<sub>715</sub>. This corresponds to the case where e<sub>i</sub> x e<sub>i+1</sub> = e<sub>i+5</sub>.
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Postby pat » Wed Jan 12, 2005 7:20 pm

And, here's another. Let v be: e<sub>123</sub> + e<sub>145</sub> + e<sub>167</sub> + e<sub>246</sub> + e<sub>257</sub> + e<sub>347</sub> + e<sub>356</sub>.

That one has no simple shorthand rule.
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Postby Paul » Wed Jan 12, 2005 7:52 pm

Hello Pat,

Thanks for your reponse.

So, for example, I contend that the rule at the bottom of the left page there would also satisfy the orthogonality and Pythagorean restrictions if v were: e<sub>126</sub> + e<sub>237</sub> + e<sub>341</sub> + e<sub>452</sub> + e<sub>563</sub> + e<sub>674</sub> + e<sub>715</sub>. This corresponds to the case where e<sub>i</sub> x e<sub>i+1</sub> = e<sub>i+5</sub>.


Interesting... In fact, that's the rule that Geoffrey Dixon, Division Algebras: Octonions, Quaternions, Complex Numbers and the Algebraic Design of Physics prefers.

Am I correct that there won't be any 1-vector in the same 3D-space as a x b


I don't understand what you're asking here. Certainly, a, b, and a x b are three mutually perpendicular 1-vectors. Thus, the form a 3D space. And, for example, a+b is in that space as is a+b+(a x b) and any linear combination of the three.


Probably what I wrote didn't make sense. What I mean is... what Lounesto is saying is true of the 3D cross product... wouldn't all these things also be true of a cross product in a 3D subspace of a higher space?

That is, the 3D subspace of 7D or 8D space is isomorphic to a 3D space... correct? So, differences that arise between the 3D cross product and the 7D cross product of 2 1-vectors must arise out of the fact that the 7D cross product is embedded in a 7D space...?

So, if Lounesto says that the direction of the 3D cross product is unique (up to sign, or two directions)... then, doesn't it follow that the 7D cross product seen only in a 3D subspace of 7D space will also be unique up to sign, or two directions...?

So, this makes me think that Lounesto is saying that the other 1-vectors having the same direction as a x b (besides the two of sign)... lie outside this 3D subspace of 7D space...

I kinda get the impression that he's talking about dualing the simple bivector of the 2 1-vector cross product with a simple trivector... in 3D space, of course, there's only one of these. But, in 7D space, there's 35... 7 of which form quaternion subalgebras of the octonion algebra.

Of course, I'm still confused... because any simple bivector is only going to dual into a 1-vector with one of the 7 which form quaternion subalgebras...

But, is it possible that for Lounesto's purposes, perhaps one, or more, of the other 28 simple trivectors could also serve as a dual to a 1-vector for a simple bivector of 7D space...? Perhaps this wouldn't be what one might want to do with octonions, but perhaps it's valid, in general...?

Is this what Lounesto's talking about... or, perhaps I have some misunderstanding otherwise... concerning dimensionality issues would seem the most probable... that interfering with my understanding what Lounesto's saying...?
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Postby pat » Thu Jan 13, 2005 7:48 am

Hmmm... I will have to think more about your uniqueness thing. I can see your point. And, I agree.

I have a side question for you though. I'm confused about Lounesto's stuff about contractions on page 46. He says that if x and y are pure vectors, then x ⌋ y = x ⋅ y. It looks to me, like he's got the two equations at the top of page 46 backwards.

For example, let x = y = e<sub>1</sub>. Then x ⌋ y = ( e<sub>1</sub>e<sub>1</sub> - e<sub>1</sub>e<sub>1</sub> ) / 2 = 0, but x ⋅ y = 1.

The opposite goes is you let x = e<sub>1</sub> and y = e<sub>2</sub>.

Am I confused?

I was trying to debug some code to do the 7-d cross-product using his formula at the bottom of page 96, when I ran into this. I got exactly the terms that I had expected to cancel out instead of the term that I expected not to. Erf.
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Postby pat » Thu Jan 13, 2005 7:58 am

Oops... I skimmed over some of page 46... particularly the (-1)<sup>k</sup> in the third set of equations.
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Postby pat » Thu Jan 13, 2005 8:09 am

Hmmm... so, I guess I know now how to do the contraction of a 1-vector and a k-vector. But, I don't know how to do the contraction of a j-vector and a k-vector. In particular, I'm having trouble with e<sub>12</sub> contracted by e<sub>346</sub>. I believe the answer should be 0 e<sub>12346</sub>, but I'm getting 1 e<sub>12346</sub>.
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Postby pat » Thu Jan 13, 2005 9:56 pm

Eh, I dodged the major issue by using the formula on the middle of page 6 that u contracted with v is: [ u ∧ ( v e<sub>1234567</sub> ) ] e<sub>1234567</sub><sup>-1</sup>.

So, now I can calculate the 7-D cross-product using my Geometric-Algebra library and this snippet of code.
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Postby Paul » Fri Jan 14, 2005 6:51 am

Hello Pat,

I was going to say... ask... isn't the (left) contraction just the same as the Hestenes semi-commutative inner product, except that multiplication on the left nulls the product.

I think here it's all multiplication on the right... isn't it? That is, v, the nonsimple trivector which is multiplied by a^b is multiplied on the right, and it's always the larger of the two operands...

If that's the case, then here, at least, you I think(!) the (left) contraction is represented by this venn diagram where A<sub>i</sub>, B<sub>i</sub> are the sets of basis indices in any product pair:

Image

If you look at the operation tables for the (left) contraction, and the Hestenes inner product, as found at Ian Bell's Multivector Methods website:

Image
Image


Both look the same besides the part on both matrices above the main diagonal... AND, the contraction includes product with scalars, whereas the Hestenes inner product does not. (However again,... this really isn't important here since I don't think there's going to be any scalars in this computation)

So, I think(!) as far as this particular computation is concerned, the Hestenes inner product and the (left) contraction yield the same answer... is this correct?
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Postby pat » Fri Jan 14, 2005 4:57 pm

That looks reasonable to me for this calculation. I think. I'm a bit worried about the difference in symmetry though. Inner products have to be symmetric, and I thought that contraction was either symmetric or antisymmetric depending on the ranks of the factors.... Hmmm...
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Postby Paul » Fri Jan 14, 2005 6:56 pm

Hello Pat,

Do you mean symmetric like ab = ba...?

I thought with bivectors it was AB = -BA for the inner product... anti-symmetric. I kinda thought the inner product was only symmetric for odd k-vectors.

I think once all the inner products get into k-vectors greater than 1, this kind of symmetric is lost.
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Postby pat » Fri Jan 14, 2005 7:15 pm

I meant ab = ba. Check out rule #3 here: http://mathworld.wolfram.com/InnerProduct.html

But, I suppose equation 4 later down that page would be applicable here. Just the conjugate of a k-blade would be (-1)<sup>k</sup> times the blade. (I think.)
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Postby Paul » Sat Jan 15, 2005 1:49 am

Hello Pat,

I think I see what you mean...

That operation table for the (left) contractive inner product isn't symmetric even for 1-vectors...?
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Postby wendy » Wed Jan 19, 2005 4:51 am

I read this topic.

When i read the excert from the book on octonions.

Let us first note that the requirement for this cross product is that the axies be ordered. That is, there is a way of meaningly deriving e4 from e1 and e2 respectively. This is not the nature of the horozettix (E7-space), or even the horoyottix (E8-space). [One distinguishes the difference between En (euclidean n-space), and Rn (the expansion of n real axies, without interrelation)]

What we are then dealing with is that octonional horolatrix (OE1 = octonion-euclidean-line) is a subspace of E8, because it adds extra restrictions onto E8 (and hence E7), that these spaces do not have.

When i think in terms of cross-poducts of vectors, i tend to think of the operation of a tensor of the form n^n. ( ie n*n*n... n terms).

While i too am trying to understand octonions, i am trying to do it the easy way (ie by feeling what happens to the eight-dimensional space), rather than the hard way (ie by algebra). At the moment i am having limited success....

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