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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. Ac

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.
marctenbosch 8y ago
I think it is a good idea because both components contain useful complementary information (the cosine/sine of the angle and the plane they form), and the algebra corresponds to geometric operations. Having an algebra that corresponds to geometric operations is good because you get a high-level framework for how to manipulate quantities instead of having to do things case by case.

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.

marctenbosch 8y ago

I think it is a good idea because both components contain useful complementary information (the cosine/sine of the angle and the plane they form), and the algebra corresponds to geometric operations. Having an algebra that corresponds to geometric operations is good because you get a high-level framework for how to manipulate quantities instead of having to do things case by case.