5 ms·

This post was brought to you by the Geometric Algebra gang. Join us at: https://www.youtube.com/watch?v=60z_hpEAtD8 https://www.youtube.com/watch?v=60z_hpEAtD

This post was brought to you by the Geometric Algebra gang. Join us at: https://www.youtube.com/watch?v=60z_hpEAtD8 https://www.youtube.com/watch?v=60z_hpEAtD8 https://bivector.net/ https://bivector.net/ https://enkimute.github.io/ganja.js/examples/coffeeshop.html#pga2d_points_and_lines https://enkimute.github.io/ganja.js/examples/coffeeshop.html...

contravariant 3y ago
I never did quite 'get' geometric algebra. I mean the exterior algebra gives a couple of clear generalisations, and for the ones that require a metric you can typically use the Hodge star to find a generalisation. Geometric algebra then blends all of this together, and I'm still not entirely convinced that this improves things. Is it actually ever handy to have to deal with mixed degree multivectors?
- adrian_b 3y ago
  Even without using mixed-degree multivectors, the fact that the existence and the properties of all the different kinds of multivectors results from a small set of simple axioms is very satisfying in itself. Before all these concepts were unified by geometric algebra, the system of physical quantities as taught in most places was a huge mess of many different kinds of quantities, scalars, polar vectors, axial vectors, pseudoscalars, tensors, pseudotensors, complex numbers, quaternions, spinors and so on. It was not at all obvious why there are so many kinds of quantities, which are the relationships between them, are there any other kinds of quantities besides those already studied, etc. Geometric algebra has brought order in this chaos and it has enabled a much deeper and more complete understanding of physics, by reducing a long list of seemingly arbitrary rules to a much smaller set of axioms, and by deriving all the many kinds of physical quantities from the vectors in the strict sense, i.e. from the translations of the space, which are themselves derived from the points of the affine space (as equivalence classes of point pairs). (Both logically and historically, in physics the vectors are more fundamental than the scalars. The scalars are obtained by dividing collinear vectors, i.e. they are equivalence classes of pairs of collinear vectors. This division operation is a.k.a. measurement and what are now named as "real numbers" were named as "measures" in the past, for more than two millennia. For any Archimedean group it is possible to define a division operation using the Axiom of Archimedes, generating a set of scalars over which the original group is a vector space).
  - contravariant 3y ago
    Yeah but like all of that is true about the exterior algebra as well. Except exterior algebra is a bit more explicit about where you use the metric, which is really convenient when the metric starts to become important.
- BenoitP 3y ago
  I'm not versed at all into the Hodge star operator but it feels like Maxwell's equations expressed in both systems[1] might locate the answer. With Hodge star: dF = 0 d * F = J Expressed in GA: ∇F = J The eyeball-difference of which (dF = 0) would be your "how GA then blends all of this together", if I understand correctly. I'd guesstimate that dF = 0 to be akin to Gauss' law; and that maybe GA somewhat incorporates that the curl of a gradient is the zero field. [1] https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field https://en.wikipedia.org/wiki/Mathematical_descriptions_of_t...
  - nyssos 3y ago
    These are not quite equivalent, since using the hodge dual explicitly incorporates the geometry of space in a way left implicit by your geometric algebra formulation. The appropriate exterior algebra analogue here is simply `F = dA`. F is the electromagnetic field tensor, d is the exterior derivative, A is the 4-potential.
- nyssos 3y ago
  Clifford (i.e. "geometric") algebras are of some mathematical interest, but for physics the only real value I've seen from them is in offering a fairly nice presentation of spin groups, which is ultimately where the Dirac matrices come from. The geometric algebra advocacy, so far as I can tell, comes entirely from engineers who were never given a proper account of tensors in the first place: it's certainly an improvement over whatever godawful basis-dependent stuff gets taught there.
  - itishappy 3y ago
    I'd be interested to know your opinion on space-time algebras. They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts, but that may be blind optimism as your last sentence seems to have been written to describe me specifically...
    - nyssos 3y ago
      Spacetime algebra is just the Clifford algebra Cl_1_3(R), so all of the above applies. > They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts Yes and no: the right way to unify rotations and boosts is to consider them as the orientation-preserving elements of the Lorentz group (sometimes called the proper Lorentz group). You can construct this from the corresponding Clifford algebra, but it's somewhat technical and not physically well-motivated until you start dealing with spinors. It's also the group of symmetries of spacetime that leave the origin unchanged, which is far, far more natural.
  - paholg 3y ago
    I found it a much nicer way to think about physical concepts. Torque and magnetism, for example, make much more sense as bivectors. And never having to do the right-hand rule is a nice plus.
- smaddox 3y ago
  Yes. In physics. E.g. Maxwell's equation reduces to a single equation in geometric Algebra: https://peeterjoot.com/archives/geometric-algebra/maxwells_ga.pdf https://peeterjoot.com/archives/geometric-algebra/maxwells_g...
captainclam 3y ago
Thank you so much for the links! I just happen to be endeavoring to seriously learn this stuff at the moment, driven by interests in physics, differential geometry, and gamedev. I'm seriously jazzed to check these out, thank you.
uxp8u61q 3y ago
Somehow, in your links, there isn't a single explanation anywhere (that's not in the form of a video that I don't have time or inclination to watch) about what "geometric algebra" is supposed to mean. I'm trying to gather what this is supposed to be from the catchphrases on the front page, and honestly, this just looks like linear algebra. Can you explain what "geometric algebra" is supposed to be? Or just link to written explanations? I'm a mathematician (algebraic topology), so don't be afraid to get technical. How is this different from linear algebra? The only things I can see are some tensor products and exterior products, and a few couple of division algebra structures. Can you enlighten me? Is there any actually new math behind all this?
- nyssos 3y ago
  Geometric algebras = Clifford algebras over R.
  - uxp8u61q 3y ago
    And they made dozens of hour long videos about that? That's insane. And the arrogance of calling it "geometric algebra" like they invented a new field of math?!
    - BenoitP 3y ago
      One could argue it is more about physics than math. GA has simpler expressions for a lot of physical topics.
      - uxp8u61q 3y ago
        Perhaps, but it's not new math (Clifford algebras are almost 200 years old). Shrouding it in a veil of buzzwords and pizzaz makes you look more like con artists than scientists. A sentence like this one is just absurd: > Clifford's Geometric Algebra enables a unified, intuitive and fresh perspective on vector spaces, giving elements of arbitrary dimensionality a natural home.
- ngcc_hk 3y ago
WD40forRust 3y ago
Based.
- meindnoch 3y ago
  Based and Grassmann-pilled?

contravariant 3y ago

I never did quite 'get' geometric algebra. I mean the exterior algebra gives a couple of clear generalisations, and for the ones that require a metric you can typically use the Hodge star to find a generalisation. Geometric algebra then blends all of this together, and I'm still not entirely convinced that this improves things. Is it actually ever handy to have to deal with mixed degree multivectors?

adrian_b 3y ago
Even without using mixed-degree multivectors, the fact that the existence and the properties of all the different kinds of multivectors results from a small set of simple axioms is very satisfying in itself. Before all these concepts were unified by geometric algebra, the system of physical quantities as taught in most places was a huge mess of many different kinds of quantities, scalars, polar vectors, axial vectors, pseudoscalars, tensors, pseudotensors, complex numbers, quaternions, spinors and so on. It was not at all obvious why there are so many kinds of quantities, which are the relationships between them, are there any other kinds of quantities besides those already studied, etc. Geometric algebra has brought order in this chaos and it has enabled a much deeper and more complete understanding of physics, by reducing a long list of seemingly arbitrary rules to a much smaller set of axioms, and by deriving all the many kinds of physical quantities from the vectors in the strict sense, i.e. from the translations of the space, which are themselves derived from the points of the affine space (as equivalence classes of point pairs). (Both logically and historically, in physics the vectors are more fundamental than the scalars. The scalars are obtained by dividing collinear vectors, i.e. they are equivalence classes of pairs of collinear vectors. This division operation is a.k.a. measurement and what are now named as "real numbers" were named as "measures" in the past, for more than two millennia. For any Archimedean group it is possible to define a division operation using the Axiom of Archimedes, generating a set of scalars over which the original group is a vector space).
- contravariant 3y ago
  Yeah but like all of that is true about the exterior algebra as well. Except exterior algebra is a bit more explicit about where you use the metric, which is really convenient when the metric starts to become important.
BenoitP 3y ago
I'm not versed at all into the Hodge star operator but it feels like Maxwell's equations expressed in both systems[1] might locate the answer. With Hodge star: dF = 0 d * F = J Expressed in GA: ∇F = J The eyeball-difference of which (dF = 0) would be your "how GA then blends all of this together", if I understand correctly. I'd guesstimate that dF = 0 to be akin to Gauss' law; and that maybe GA somewhat incorporates that the curl of a gradient is the zero field. [1] https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field https://en.wikipedia.org/wiki/Mathematical_descriptions_of_t...
- nyssos 3y ago
  These are not quite equivalent, since using the hodge dual explicitly incorporates the geometry of space in a way left implicit by your geometric algebra formulation. The appropriate exterior algebra analogue here is simply `F = dA`. F is the electromagnetic field tensor, d is the exterior derivative, A is the 4-potential.
nyssos 3y ago
Clifford (i.e. "geometric") algebras are of some mathematical interest, but for physics the only real value I've seen from them is in offering a fairly nice presentation of spin groups, which is ultimately where the Dirac matrices come from. The geometric algebra advocacy, so far as I can tell, comes entirely from engineers who were never given a proper account of tensors in the first place: it's certainly an improvement over whatever godawful basis-dependent stuff gets taught there.
- itishappy 3y ago
  I'd be interested to know your opinion on space-time algebras. They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts, but that may be blind optimism as your last sentence seems to have been written to describe me specifically...
  - nyssos 3y ago
    Spacetime algebra is just the Clifford algebra Cl_1_3(R), so all of the above applies. > They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts Yes and no: the right way to unify rotations and boosts is to consider them as the orientation-preserving elements of the Lorentz group (sometimes called the proper Lorentz group). You can construct this from the corresponding Clifford algebra, but it's somewhat technical and not physically well-motivated until you start dealing with spinors. It's also the group of symmetries of spacetime that leave the origin unchanged, which is far, far more natural.
- paholg 3y ago
  I found it a much nicer way to think about physical concepts. Torque and magnetism, for example, make much more sense as bivectors. And never having to do the right-hand rule is a nice plus.
smaddox 3y ago
Yes. In physics. E.g. Maxwell's equation reduces to a single equation in geometric Algebra: https://peeterjoot.com/archives/geometric-algebra/maxwells_ga.pdf https://peeterjoot.com/archives/geometric-algebra/maxwells_g...

captainclam 3y ago

Thank you so much for the links! I just happen to be endeavoring to seriously learn this stuff at the moment, driven by interests in physics, differential geometry, and gamedev. I'm seriously jazzed to check these out, thank you.

uxp8u61q 3y ago

Somehow, in your links, there isn't a single explanation anywhere (that's not in the form of a video that I don't have time or inclination to watch) about what "geometric algebra" is supposed to mean. I'm trying to gather what this is supposed to be from the catchphrases on the front page, and honestly, this just looks like linear algebra. Can you explain what "geometric algebra" is supposed to be? Or just link to written explanations? I'm a mathematician (algebraic topology), so don't be afraid to get technical. How is this different from linear algebra? The only things I can see are some tensor products and exterior products, and a few couple of division algebra structures. Can you enlighten me? Is there any actually new math behind all this?

nyssos 3y ago
Geometric algebras = Clifford algebras over R.
- uxp8u61q 3y ago
  And they made dozens of hour long videos about that? That's insane. And the arrogance of calling it "geometric algebra" like they invented a new field of math?!
  - BenoitP 3y ago
    One could argue it is more about physics than math. GA has simpler expressions for a lot of physical topics.
    - uxp8u61q 3y ago
      Perhaps, but it's not new math (Clifford algebras are almost 200 years old). Shrouding it in a veil of buzzwords and pizzaz makes you look more like con artists than scientists. A sentence like this one is just absurd: > Clifford's Geometric Algebra enables a unified, intuitive and fresh perspective on vector spaces, giving elements of arbitrary dimensionality a natural home.
ngcc_hk 3y ago

WD40forRust 3y ago

Based.

meindnoch 3y ago
Based and Grassmann-pilled?