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Jun 4, 2018, 3:53:18 PM6/4/18

to Geometric_Algebra

Hi -- I have a beginner question that I feel has probably been answered many times before, yet I've not been able to find the answer.

Give a Clifford Algebra with a specific signature (i.e. Cl(p,q,r)), how does one go about determining possible geometric interpretations of the objects in the algebra,

e.g. wedging four vectors from Cl(4,1,0) can be interpreted as a sphere through four points, etc

Best, nehal

Jun 4, 2018, 7:03:07 PM6/4/18

to geometri...@googlegroups.com

Nehal,

Great question; I'll be eager to see any responses. Thanks for asking it.

Are you aware of any a basic, entry-level senior undergraduate level intro (text, article, slide deck) to Cl algebra entities like the one you just asked about?

I know a bit about the history, and a rough smattering of the areas of application in physics (EM, e.g.) but would like to see a survey, "Clifford for Poets" coverage of the overall interpretations and usages of the the different signatures you mention.

Thanks again for your question,

Best Wishes,

Ross

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Jun 5, 2018, 8:29:39 AM6/5/18

to Geometric_Algebra

i think if you are speaking of geometrical interpretations you want to use the term geometric algebra. check out sec 2 of this ,

if you know things about the object, like a sphere is defined by 4 points, it follows such ang such intersection rules, etc, then these things will guide your search. but, does your question assume the CGA map is being used? if not, then it sounds to me like you are asking; how did people 'figure out' conformal or projective geometry? i dont think this is an easy question with a single answer.

On Mon, Jun 4, 2018 at 12:17 PM, Nehal Patel <nehal...@gmail.com> wrote:

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Jun 5, 2018, 8:47:01 AM6/5/18

to geometri...@googlegroups.com

Yes sir, Alex, that's exactly correct....and i will check out sec #2 provided..

.... and I deliberately did not use the term <GA> because of the proliferation of books out there with titles like "geometric algebra for XXX"

I'm not looking for so much interpretations of the CL signatures in "R^3" like applications --computer vision, astronautic navigation, EM re-reformulations, etc.-- but rather what the abstract guys are using them for across-the-board of all Clifford signatures. ( maybe something like a survey article?)

I acknowledge in advance the "Clifford algebra for poets" tone of this email and thank you in advance for any pointers provided! 🙏

Ross

On Tue, Jun 5, 2018, 8:29 AM alexander arsenovic <al...@810lab.com> wrote:

i think if you are speaking of geometrical interpretations you want to use the term geometric algebra. check out sec 2 of this ,if you know things about the object, like a sphere is defined by 4 points, it follows such ang such intersection rules, etc, then these things will guide your search. but, does your question assume the CGA map is being used? if not, then it sounds to me like you are asking; how did people 'figure out' conformal or projective geometry? i dont think this is an easy question with a single answer.

<GA>

On Mon, Jun 4, 2018 at 12:17 PM, Nehal Patel <nehal...@gmail.com> wrote:

Hi -- I have a beginner question that I feel has probably been answered many times before, yet I've not been able to find the answer.Give a Clifford Algebra with a specific signature (i.e. Cl(p,q,r)), how does one go about determining possible geometric interpretations of the objects in the algebra,e.g. wedging four vectors from Cl(4,1,0) can be interpreted as a sphere through four points, etcBest, nehal

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Jun 5, 2018, 9:41:52 AM6/5/18

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Chapter 4 of "Geometric Algebra with Applications in Engineering" by Christian Perwass describes algebras with various signatures, and how various objects in them can be associated with various geometric entities. The types of signatures/intepretations he considers include those that could be described as "euclidean", "projective", "conformal", and "conic" . His presentation is satisfying in the sense that he "works out the algebra" for various expressions to demonstrate that indeed what was claimed to be a sphere is indeed a sphere (in the sense that it is being represented by the zeros of a function)

On Monday, June 4, 2018 at 7:03:07 PM UTC-4, Ross wrote:

Nehal,Great question; I'll be eager to see any responses. Thanks for asking it.Are you aware of any a basic, entry-level senior undergraduate level intro (text, article, slide deck) to Cl algebra entities like the one you just asked about?I know a bit about the history, and a rough smattering of the areas of application in physics (EM, e.g.) but would like to see a survey, "Clifford for Poets" coverage of the overall interpretations and usages of the the different signatures you mention.Thanks again for your question,Best Wishes,Ross

On Mon, Jun 4, 2018 at 12:17 PM, Nehal Patel <nehal...@gmail.com> wrote:

Hi -- I have a beginner question that I feel has probably been answered many times before, yet I've not been able to find the answer.Give a Clifford Algebra with a specific signature (i.e. Cl(p,q,r)), how does one go about determining possible geometric interpretations of the objects in the algebra,e.g. wedging four vectors from Cl(4,1,0) can be interpreted as a sphere through four points, etcBest, nehal

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Jun 5, 2018, 10:12:23 AM6/5/18

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On Tuesday, June 5, 2018 at 8:29:39 AM UTC-4, alexander arsenovic wrote:

i think if you are speaking of geometrical interpretations you want to use the term geometric algebra. check out sec 2 of this ,

Thanks for the link to the Hestenes retrospective. It's somewhat helpful.

how did people 'figure out' conformal or projective geometry?

In the sense that projective and conformal geometry were being studied well before clifford, this is not really my question. (I'm not asking why did 18th/19th century mathematicians choose to formulate and study these notions)

Restated a different way:

The abstract algebraic objects that mathematicians call "Clifford Algebras" seem to be easily interpreted in concrete geometric terms. The two questions then are 1) why is does this happen in general, but also how might one work out the details for a specific Clifford Algebra (allowing that there could be more than one useful interpretation)

An analogy might be that Coxeter Groups can be defined formally in terms of their presentations. Yet, it turns out that these groups often model the symmetries of concrete geometric objects -- so again there are two natural questions 1) why does this seem to happen in general, and 2) for a specific Coxeter group how would you go about working out the specific geometric object whose symmetries it models e.g. a hexagon in 2D euclidean space.

In both cases, question 1) is about "intuition" and the answer is probably not so easily communicated, but still can be attempted (e.g. the generators of Coexter groups are "mirrors", which is the foundation fo geometric symmetry, etc), but question 2) is seems for directly answerable in terms providing a list of concrete techniques, theorems, etc.

Fuzzy questions for sure, but (to me) they feel natural as a beginner.

Jun 5, 2018, 10:24:57 AM6/5/18

to geometri...@googlegroups.com

Gold.

Thanks!!

Jun 5, 2018, 10:37:41 AM6/5/18

to Geometric_Algebra

i think the reason the abstract objects of clifford algebras are easily interpreted in concrete geometric terms is because the algebra was intentionally created for that purpose. like; hey , addition of translations should be associative so we will keep that rule. but you could abandon the associative axiom and develop a different an algebra.

but i like your questions 2, and i think its natural. one you model a single concept in a GA, then you have the rest of the GA to 'interpret'. i wish i knew a systematic method for this!

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Jun 5, 2018, 1:09:53 PM6/5/18

to Geometric_Algebra

As Alexander suggests, you do need to presuppose a particular representation (e.g. CGA); just saying "what does an element of this algebra represent geometrically?" is too open-ended.

For example: the exterior (i.e. wedge) product of three *null* vectors in Cl(3,1,0) can represent a circle in two-dimensional Euclidean space if it is being used as a CGA representation, and of any three vectors can represent a 3-d subspace in a Lorentzian 4-d vector space. A (null) vector can represent a point (2-d Euclidean CGA), an vector a 4-vector (Lorentzian vector space), or a positive-square vector a point of a hyperbolic geometry (3-d hyperbolic geometry). It may help to start with thinking about what the vector space can represent before looking at further multivectors.

Jun 5, 2018, 4:04:53 PM6/5/18

to Geometric_Algebra

Thanks Manfred,

the exterior (i.e. wedge) product of three *null* vectors in Cl(3,1,0)

The hint emphasizing *null* vectors for CGA was helpful -- that clarified a point of confusion/sloppiness on my end. (Though it makes me even more curious how in the world some one thought up this representation...)

cheers, nehal

Jun 5, 2018, 6:23:39 PM6/5/18

to Geometric_Algebra

Maybe a similar one that led me to a GA of projective geometry in arbitrary dimension that expresses all projective transformations within the algebra (a strong claim, given that the GA community still seems to be trying to get beyond three dimensions). One gets an insight when working on something peripherally related, and notices how this applies in GA.

Jun 5, 2018, 8:45:29 PM6/5/18

to geometri...@googlegroups.com

<< a GA of projective geometry in arbitrary dimension >>

I am intrigued, though sceptical: the projective group dimension in
d-space equals (d+1)^2 - 1 , whereas Clifford algebra represents

transformation groups of dimension k(k+1)/2 , where k is roughly

equal to the vector dimension.

Fred Lunnon

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Jun 5, 2018, 10:12:03 PM6/5/18

to Geometric_Algebra

Well-observed; we need k=2(d+1). It seems quite expensive, considering that the dimension of the Clifford algebra is exponential in k, though this is hardly a reason to abandon it as a direction of development. I found the implication that the projective group is a subgroup of an orthogonal group somewhat counterintuitive, though easily proven.

I am intrigued, though sceptical: the projective group dimension in

d-space equals (d+1)^2 - 1 , whereas Clifford algebra represents

transformation groups of dimension k(k+1)/2 , where k is roughly

equal to the vector dimension.

Fred Lunnon

Aug 7, 2018, 8:15:12 PM8/7/18

to geometri...@googlegroups.com

In connection with representations of the projective group by

Clifford algebras, I may not have mentioned the special case

Cl(3, 3) representing lines in 3-space, which fails to generalise

to other dimensions in any obvious fashion. Among other nice

properties, it incorporates 3-space projectivities and quadrics;

although sadly complex components are required for ellipsoids,

and there seems no representation available for quadric cones.

I have just come across a couple of relatively recent papers

discussing these matters in detail:

Daniel Klawitter (2014)

"Null Polarities as Generators of the Projective Group"

https://arxiv.org/pdf/1406.0278

Daniel Klawitter (2014)

"A Clifford algebraic Approach to Line Geometry"

https://arxiv.org/pdf/1311.0131.pdf

Fred Lunnon

Clifford algebras, I may not have mentioned the special case

Cl(3, 3) representing lines in 3-space, which fails to generalise

to other dimensions in any obvious fashion. Among other nice

properties, it incorporates 3-space projectivities and quadrics;

although sadly complex components are required for ellipsoids,

and there seems no representation available for quadric cones.

I have just come across a couple of relatively recent papers

discussing these matters in detail:

Daniel Klawitter (2014)

"Null Polarities as Generators of the Projective Group"

https://arxiv.org/pdf/1406.0278

Daniel Klawitter (2014)

"A Clifford algebraic Approach to Line Geometry"

https://arxiv.org/pdf/1311.0131.pdf

Fred Lunnon

Aug 16, 2018, 7:32:30 PM8/16/18

to Geometric_Algebra

Sorry, I meant to respond more quickly. Yes, the use of Cl(3,3) to represent the projective transformations of a 3-dimensional space appears to rely on on an exceptional group isomorphism, which would explain why it would not generalize directly to other dimensionalities. However, Cl(4,4) achieves the same thing, and is part of a particularly natural and easy-to-construct general family for any finite number of dimensions. I am not particularly convinced that it is less efficient, either, if computations exploit known restrictions (in particular, knowing which specific components of a multivector are necessarily always zero in the context).

Aug 17, 2018, 7:07:38 PM8/17/18

to geometri...@googlegroups.com

For dynamical applications a natural representation of quadrics (particularly

ellipsoids) is of interest. Have you considered this question in relation to

Cl(n+1,n+1) ?

WFL

ellipsoids) is of interest. Have you considered this question in relation to

Cl(n+1,n+1) ?

WFL

Aug 17, 2018, 8:47:55 PM8/17/18

to Geometric_Algebra

Yes, I have considered these, though thusfar only speculating that at least some types of quadric would take the form of the solutions x to xZx = 0 in a specific (n+1)-dimensional subspace of the 1-vectors, where Z is a rotor representing the quadric. This is equivalent to x.x=0 (the null vectors) in a Cl(p,q) implied by Z, where p+q=n+1.

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