3 ms·

I’m telling you that after a few years working with the geometric product a whole bunch, the wedge product really doesn’t cut it. > I still have basically no i

I’m telling you that after a few years working with the geometric product a whole bunch, the wedge product really doesn’t cut it. > I still have basically no idea what 'AB' means when both are arbitrary-grade multivectors Any mathematical language (or any natural language) can express a bunch of nonsensical and useless things. What is the sine of the square root of the logarithm of the tangent of some polynomial applied to a scalar? Who knows?! You can certainly make up nonsensical expressions using matrices, differential forms, etc. In a concrete problem, sometimes you have to think a bit to figure out what the geometrical meaning is of a particular multivector, but if the problem came out of a physical situation, I haven’t yet found a case where I couldn’t explain it. Your multivector might be an electromagnetic field. It might be a rotor. It might be a representation of a sphere in the conformal model, ...

ajkjk 8y ago
To be clear, I don't think the wedge product is sufficient either; I just think it's the real reason most people end up finding GA to be appealing. I think there's work to do to clean the whole space up. Particularly, the wedge product is a 'join' or 'union' operator (amusingly, it has the wrong symbol). The 'meet' operator is not widely known but should probably be equally prominent. I haven't figured out for myself where the other holes are yet.
- jacobolus 8y ago
  You might enjoy https://arxiv.org/abs/1205.5935 https://arxiv.org/abs/1205.5935
  - improbable22 8y ago
    As someone unable to fathom the enthusiasm for geometric algebra... at least this defines this product: It's the sum of an inner product and a wedge product. Acting on two vectors these give a scalar and a 2-form. Why is adding these different objects a good idea? Don't they have different units (like, a pure number vs meters squared)?
    - happy4crazy 8y ago
      It's analogous to the real and imaginary parts of a complex number. Does it makes sense to add a real number and a purely imaginary number? Aren't they different kinds of things? Yes, and yes!
      - twic 8y ago
        It would really helpful if the article had some worked examples of this arithmetic, with actual numbers, the kind with digits and decimal points in. Then, things like the structure of a geometric product would be absolutely clear. Let me have a go. I'll use an asciified version of the symbols, with * to mean multiplication of two scalars,, to mean raising one scalar to the power of another, and _ to mean taking a component of a vector. ab = a.b + a^b a.b = a_x * b_x + a_y * b_y + a_z * b_z a^b = (a_x * b_y - b_x * a_y) (x^y) + (a_x * b_z - b_x * a_z) (x^z) + (a_y * b_z - b_y * a_z) (y^z) ab = (a_x * b_x + a_y * b_y + a_z * b_z) + (a_x * b_y - b_x * a_y) (x^y) + (a_x * b_z - b_x * a_z) (x^z) + (a_y * b_z - b_y * a_z) (y^z) So if a = (1, 2, 3) and b = (4, 5, 6): ab = (1 * 4 + 2 * 5 + 3 * 6) + (1 * 5 - 4 * 2) (x^y) + (1 * 6 - 4 * 3) (x^z) + (2 * 6 - 5 * 3) (y^z) = 32 + -3 (x^y) + -6 (x^z) + -3 (y^z) The dot product makes a scalar, the wedge product makes a bivector, and the geometric product makes a scalar plus a bivector You will note that the scalar part is much bigger than the coefficients of the bivector part. That's because the input vectors are actually quite similar - pointing z-by-y-z, with a little bit of x. Hence, their projection onto each other is large, whereas the parallelogram they form is quite small (long and thin). The dot product measures the former, the wedge product the latter. Have i got that right? EDIT And to clarify this: > For any basis vector, such as the x axis, the result [of taking the geometric product with itself] is 1 That '1' isn't the scalar number 1, it's the scalar-plus-bivector 1 + 0 (x^y) + 0 (x^z) + 0 (y^z).
      - improbable22 8y ago
        Sure, you've defined a map V⊗V → R⊕T, where T means 2-forms. But I still don't see why this is useful, apart from being able to extract from it both the wedge product and the inner product which you started with.
  - 0-_-0 8y ago
    Thanks for that, it's a really good explanation!
cygx 8y ago
> the wedge product really doesn’t cut it What about the wedge product together with the Hodge star, ie the language that tends to be used in Riemannian geometry?

ajkjk 8y ago

To be clear, I don't think the wedge product is sufficient either; I just think it's the real reason most people end up finding GA to be appealing. I think there's work to do to clean the whole space up. Particularly, the wedge product is a 'join' or 'union' operator (amusingly, it has the wrong symbol). The 'meet' operator is not widely known but should probably be equally prominent. I haven't figured out for myself where the other holes are yet.

jacobolus 8y ago
You might enjoy https://arxiv.org/abs/1205.5935 https://arxiv.org/abs/1205.5935
- improbable22 8y ago
  As someone unable to fathom the enthusiasm for geometric algebra... at least this defines this product: It's the sum of an inner product and a wedge product. Acting on two vectors these give a scalar and a 2-form. Why is adding these different objects a good idea? Don't they have different units (like, a pure number vs meters squared)?
  - happy4crazy 8y ago
    It's analogous to the real and imaginary parts of a complex number. Does it makes sense to add a real number and a purely imaginary number? Aren't they different kinds of things? Yes, and yes!
    - twic 8y ago
      It would really helpful if the article had some worked examples of this arithmetic, with actual numbers, the kind with digits and decimal points in. Then, things like the structure of a geometric product would be absolutely clear. Let me have a go. I'll use an asciified version of the symbols, with * to mean multiplication of two scalars,, to mean raising one scalar to the power of another, and _ to mean taking a component of a vector. ab = a.b + a^b a.b = a_x * b_x + a_y * b_y + a_z * b_z a^b = (a_x * b_y - b_x * a_y) (x^y) + (a_x * b_z - b_x * a_z) (x^z) + (a_y * b_z - b_y * a_z) (y^z) ab = (a_x * b_x + a_y * b_y + a_z * b_z) + (a_x * b_y - b_x * a_y) (x^y) + (a_x * b_z - b_x * a_z) (x^z) + (a_y * b_z - b_y * a_z) (y^z) So if a = (1, 2, 3) and b = (4, 5, 6): ab = (1 * 4 + 2 * 5 + 3 * 6) + (1 * 5 - 4 * 2) (x^y) + (1 * 6 - 4 * 3) (x^z) + (2 * 6 - 5 * 3) (y^z) = 32 + -3 (x^y) + -6 (x^z) + -3 (y^z) The dot product makes a scalar, the wedge product makes a bivector, and the geometric product makes a scalar plus a bivector You will note that the scalar part is much bigger than the coefficients of the bivector part. That's because the input vectors are actually quite similar - pointing z-by-y-z, with a little bit of x. Hence, their projection onto each other is large, whereas the parallelogram they form is quite small (long and thin). The dot product measures the former, the wedge product the latter. Have i got that right? EDIT And to clarify this: > For any basis vector, such as the x axis, the result [of taking the geometric product with itself] is 1 That '1' isn't the scalar number 1, it's the scalar-plus-bivector 1 + 0 (x^y) + 0 (x^z) + 0 (y^z).
    - improbable22 8y ago
      Sure, you've defined a map V⊗V → R⊕T, where T means 2-forms. But I still don't see why this is useful, apart from being able to extract from it both the wedge product and the inner product which you started with.
- 0-_-0 8y ago
  Thanks for that, it's a really good explanation!

cygx 8y ago

> the wedge product really doesn’t cut it What about the wedge product together with the Hodge star, ie the language that tends to be used in Riemannian geometry?