4 ms·

The answer depends on the kind of Clifford/Geometric Algebra being used, the structure for the vector bases and the selected result of the quadratic forms. Thos

by pjbk 6y ago

The answer depends on the kind of Clifford/Geometric Algebra being used, the structure for the vector bases and the selected result of the quadratic forms. Those choices are mostly driven by the end application where the algebra will be used, with its similarities (obviously, they are all algebras) and particularities. * "A brief introduction to Clifford algebra" - https://www.researchgate.net/profile/Giorgio_Vassallo/publication/228955605_A_brief_introduction_to_Clifford_algebra/links/0c9605363ce335ce3d000000/A-brief-introduction-to-Clifford-algebra.pdf https://www.researchgate.net/profile/Giorgio_Vassallo/public... * "Doing euclidean plane geometry using projective geometric algebra" - https://arxiv.org/pdf/1501.06511.pdf https://arxiv.org/pdf/1501.06511.pdf * "Projective geometric algebra: A new framework for doing euclidean geometry" - https://arxiv.org/pdf/1901.05873.pdf https://arxiv.org/pdf/1901.05873.pdf * "3D Motion from structures of points, lines and planes" - https://tubaro.faculty.polimi.it/Journals/Journal_2008_DA.pdf https://tubaro.faculty.polimi.it/Journals/Journal_2008_DA.pd... * "The Construction of 3D Conformal Motions" - https://link.springer.com/content/pdf/10.1007%2Fs11786-016-0250-8.pdf https://link.springer.com/content/pdf/10.1007%2Fs11786-016-0...

alquemist 6y ago
Thanks. Appreciate taking the effort to answer. What I'm trying to do is to gain an intuition of what kind of objects GA manipulates, then stash it for whenever I'll need that kind of manipulation, which, sadly, can be never. Within that context, just grokking R^2 or R^3 would be great progress. To be fair, the article does present the fundamental identity, uv = u.v + u /\ v, where u.v is the usual dot product (scalar denoting relative projective size) and u /\ v = |u||v|sinθi (something related to area and the plane induced by u and v). Even if the article does exemplify with R^2 by defining i = xy, and then dropping the ball: "i is interpreted to mean the plane generated by x and y". I can't grok this part :( * This looks like (almost) a circular definition, defining uv in terms of the specific xy, where x and y are the two orthonormal R^2 vectors forming the basis. What is xy? Is it a scalar? A vector? A normal vector? How does a normal vector work in R^2?. A tensor of some dimensions? For example, assuming x = (1,0) and y = (0,1), how does one compute i = xy? * Plane generated how? a u + b v, given any two scalars a and b? * When moving to R^3, i'd think uv / xy define a plane. What does uvw / xyz mean? * Minor quibble, is it sin(θ * i) or sin(θ) * i?
- creata 6y ago
  * The space of multivectors in Rⁿ is actually also a vector space, and it's 2ⁿ-dimensional. The standard basis of this vector space (in 2 dimensions) is {1, x, y, x ∧ y = xy}, and that's why we care about representing uv in terms of xy. Hopefully that explains why asking how to compute "i = xy" is a meaningless question. By analogy, `(3, 5) = 3x + 5y` in the same way `xy = (0, 0, 0, 1)`, but keeping track of the order of an 2ⁿ-element basis gets annoying really fast, so we just write `xy`. * Yes, span{a, b}. * u ∧ v ∧ w would represent a volume — the one and only 3D subspace of Rⁿ, together with a sign (i.e. an orientation) and a magnitude. * It's sin(θ)*i.
  - alquemist 6y ago
    Thank you kind stranger. The technical machinery makes much more sense now. But the intuition still lacks. An attempt. To map an R^2 vector into G^2 is trivial, (a,b) => a x + b y = (1, a, b, 0). The reverse is hard. s 1 + a x + b y + d xy = (s, a, b, o) => (?, ?). Or perhaps G^2 is a superstructure of R^2, and most elements of G^2 can only be projected (somehow? perhaps trivially?) to R^2, but with loss of information. Which leaves the grok question: what does a vector in G^2 actually represent? Perhaps scale + x position + y position + orientation angle, as the use of 'i' suggests? If so, then in R^3 xy, xz any yz represent rotations in the respective planes? Which leaves a bonus puzzle: what does (multiplying with) xyz represent? FWIW, at this point it's likely I can compute the answers using the code in the OP and well crafted multivectors. Time permitting :)
    - creata 6y ago
      Minor correction: the inclusion of the vectors is (a, b) ⇒ (0, a, b, 0), not (1, a, b, 0). Also, you have an inclusion of the scalars, which is x ⇒ (x, 0, 0, 0). And yeah, G^2 can't be projected back onto R^2 without losing a lot of information. > what does a vector in G^2 actually represent? If you can write a multivector x as the wedge product of k independent vectors (or equivalently, the geometric product of k orthogonal vectors), then x is called a k-blade. In this case, x has a very clear geometric interpretation: * the magnitude of x is the volume of the parallelotope spanned by the vectors. * the "direction" of x corresponds to the space spanned by the vectors — in concrete terms, that space is the set of vectors in R^2 that the map (v ⇒ v ∧ x) takes to zero. * the "direction" also gives an orientation of the subspace. Orientations are defined indirectly: if you have two blades representing the same space, then they must be scalar multiples of one another. If that scaling factor is positive, then they have the same orientation, and if it is negative, then they have opposing orientations. These three pieces of information are enough to uniquely identify x. Unfortunately, I'm not sure how to go from there to interpreting other multivectors, which are sums of blades. > If so, then in R^3 xy, xz any yz represent rotations in the respective planes? Yes, xy represents rotation on the xy plane, as long as you apply it as described in the article. > Which leaves a bonus puzzle: what does (multiplying with) xyz represent? This one should be fairly easy to prove: it represents a kind of dualization, in which the space represented by mi = mxyz is the orthogonal complement of the space represented by m.

by pjbk 6y ago

alquemist 6y ago

Thanks. Appreciate taking the effort to answer. What I'm trying to do is to gain an intuition of what kind of objects GA manipulates, then stash it for whenever I'll need that kind of manipulation, which, sadly, can be never. Within that context, just grokking R^2 or R^3 would be great progress. To be fair, the article does present the fundamental identity, uv = u.v + u /\ v, where u.v is the usual dot product (scalar denoting relative projective size) and u /\ v = |u||v|sinθi (something related to area and the plane induced by u and v). Even if the article does exemplify with R^2 by defining i = xy, and then dropping the ball: "i is interpreted to mean the plane generated by x and y". I can't grok this part :( * This looks like (almost) a circular definition, defining uv in terms of the specific xy, where x and y are the two orthonormal R^2 vectors forming the basis. What is xy? Is it a scalar? A vector? A normal vector? How does a normal vector work in R^2?. A tensor of some dimensions? For example, assuming x = (1,0) and y = (0,1), how does one compute i = xy? * Plane generated how? a u + b v, given any two scalars a and b? * When moving to R^3, i'd think uv / xy define a plane. What does uvw / xyz mean? * Minor quibble, is it sin(θ * i) or sin(θ) * i?

creata 6y ago
* The space of multivectors in Rⁿ is actually also a vector space, and it's 2ⁿ-dimensional. The standard basis of this vector space (in 2 dimensions) is {1, x, y, x ∧ y = xy}, and that's why we care about representing uv in terms of xy. Hopefully that explains why asking how to compute "i = xy" is a meaningless question. By analogy, `(3, 5) = 3x + 5y` in the same way `xy = (0, 0, 0, 1)`, but keeping track of the order of an 2ⁿ-element basis gets annoying really fast, so we just write `xy`. * Yes, span{a, b}. * u ∧ v ∧ w would represent a volume — the one and only 3D subspace of Rⁿ, together with a sign (i.e. an orientation) and a magnitude. * It's sin(θ)*i.
- alquemist 6y ago
  Thank you kind stranger. The technical machinery makes much more sense now. But the intuition still lacks. An attempt. To map an R^2 vector into G^2 is trivial, (a,b) => a x + b y = (1, a, b, 0). The reverse is hard. s 1 + a x + b y + d xy = (s, a, b, o) => (?, ?). Or perhaps G^2 is a superstructure of R^2, and most elements of G^2 can only be projected (somehow? perhaps trivially?) to R^2, but with loss of information. Which leaves the grok question: what does a vector in G^2 actually represent? Perhaps scale + x position + y position + orientation angle, as the use of 'i' suggests? If so, then in R^3 xy, xz any yz represent rotations in the respective planes? Which leaves a bonus puzzle: what does (multiplying with) xyz represent? FWIW, at this point it's likely I can compute the answers using the code in the OP and well crafted multivectors. Time permitting :)
  - creata 6y ago
    Minor correction: the inclusion of the vectors is (a, b) ⇒ (0, a, b, 0), not (1, a, b, 0). Also, you have an inclusion of the scalars, which is x ⇒ (x, 0, 0, 0). And yeah, G^2 can't be projected back onto R^2 without losing a lot of information. > what does a vector in G^2 actually represent? If you can write a multivector x as the wedge product of k independent vectors (or equivalently, the geometric product of k orthogonal vectors), then x is called a k-blade. In this case, x has a very clear geometric interpretation: * the magnitude of x is the volume of the parallelotope spanned by the vectors. * the "direction" of x corresponds to the space spanned by the vectors — in concrete terms, that space is the set of vectors in R^2 that the map (v ⇒ v ∧ x) takes to zero. * the "direction" also gives an orientation of the subspace. Orientations are defined indirectly: if you have two blades representing the same space, then they must be scalar multiples of one another. If that scaling factor is positive, then they have the same orientation, and if it is negative, then they have opposing orientations. These three pieces of information are enough to uniquely identify x. Unfortunately, I'm not sure how to go from there to interpreting other multivectors, which are sums of blades. > If so, then in R^3 xy, xz any yz represent rotations in the respective planes? Yes, xy represents rotation on the xy plane, as long as you apply it as described in the article. > Which leaves a bonus puzzle: what does (multiplying with) xyz represent? This one should be fairly easy to prove: it represents a kind of dualization, in which the space represented by mi = mxyz is the orthogonal complement of the space represented by m.