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From your description, you have not experienced [what I consider] a breakthrough with the cross product. In 3D there is a one-to-one mapping between vectors and

From your description, you have not experienced [what I consider] a breakthrough with the cross product. In 3D there is a one-to-one mapping between vectors and oriented areas (feel free to think of it as a plane along with scalar denoting an area of the plane). When I learned about vector calculus, this one-to-one mapping was used liberally and implicitly, even though the vector a×b behaves differently from the vector a and the vector b. Even if you take the magnitude of a×b to get a scalar |a×b|, that scalar does not behave the same as e.g. the scalar a ⋅ b. The SI system cannot distinguish between a scalar torque Newton-meters and scalar work Newton-meters, though they are different. Geometric Algebra makes the difference explicit. For me, it's not about giving a meaning to a×b after the fact, but rather to ensure it has meaning by construction. I think it's valid to assume the operation a×b and then figure out what it means (or doesn't mean, or call it a weird tool). But Geometric Algebra gives one way to think about it more clearly. It's not about fundamental beauty for me. It's about having a consistent and clear system to operate within.

andi999 6y ago
The cross product is just an infinitesimal rotation. If you rotate b around the axis a, then the derivative of the new vector is axb. So the cross product is the product in the Lie Algebra of rotations. (Thats why you have the Jacobi identity). I find this gives much more insight than labeling it a bivector. Edit: also this view can generalize. You can ask what other reprensentation of the rotation Lie algebra exists and find e.g su(2), this leads you to spin-bundles then.
- gugagore 6y ago
  This is not very clear to me. Can you be more precise? If we're talking about the Lie algebra of space rotations, then the elements of that are infinitesimal rotations. Those elements form a vector space, and there's also the Lie bracket, which is the cross product as you say. I don't see the sense in saying "rotate b about a" when both a and b are to represent infinitesimal rotations. If we're taking b to be a position in space, and a to be an infinitesimal rotation, then your explanation makes sense (though in a backwards way from how I think of it. a x b is the vector you rotate a about in order to get the derivative of the tip to be b). How do you reconcile that with the Lie Algebra? More importantly, what if your vectors don't model rotations to begin with? Then you still are left with giving a meaning to the cross product. How can you clarify anything by taking your vectors to be infinitesimal rotations, whose cross product gives a third infinitesimal rotation, which carries information about the non-commutativity of the [non-infinitesimal] rotations? It is one thing that the cross product does, but I don't see how it explains, for example, what the cross product has to do with calculating volumes of parallelepipeds, or torques.
benrbray 6y ago
Geometric Algebra is certainly not the only theory to make this distinction. Most treatments of tensor algebra will cover Hodge duality and emphasize the importance of ensuring all quantities have well-behaved "types". Category theory helps to organize everything a bit more neatly, too. Some good examples are [1] and [2]. [1] https://ccom.ucsd.edu/~ctiee/notes/tensors.pdf https://ccom.ucsd.edu/~ctiee/notes/tensors.pdf [2] https://www.cs.cmu.edu/~kmcrane/Projects/DDG/ https://www.cs.cmu.edu/~kmcrane/Projects/DDG/

andi999 6y ago

The cross product is just an infinitesimal rotation. If you rotate b around the axis a, then the derivative of the new vector is axb. So the cross product is the product in the Lie Algebra of rotations. (Thats why you have the Jacobi identity). I find this gives much more insight than labeling it a bivector. Edit: also this view can generalize. You can ask what other reprensentation of the rotation Lie algebra exists and find e.g su(2), this leads you to spin-bundles then.

gugagore 6y ago
This is not very clear to me. Can you be more precise? If we're talking about the Lie algebra of space rotations, then the elements of that are infinitesimal rotations. Those elements form a vector space, and there's also the Lie bracket, which is the cross product as you say. I don't see the sense in saying "rotate b about a" when both a and b are to represent infinitesimal rotations. If we're taking b to be a position in space, and a to be an infinitesimal rotation, then your explanation makes sense (though in a backwards way from how I think of it. a x b is the vector you rotate a about in order to get the derivative of the tip to be b). How do you reconcile that with the Lie Algebra? More importantly, what if your vectors don't model rotations to begin with? Then you still are left with giving a meaning to the cross product. How can you clarify anything by taking your vectors to be infinitesimal rotations, whose cross product gives a third infinitesimal rotation, which carries information about the non-commutativity of the [non-infinitesimal] rotations? It is one thing that the cross product does, but I don't see how it explains, for example, what the cross product has to do with calculating volumes of parallelepipeds, or torques.

benrbray 6y ago

Geometric Algebra is certainly not the only theory to make this distinction. Most treatments of tensor algebra will cover Hodge duality and emphasize the importance of ensuring all quantities have well-behaved "types". Category theory helps to organize everything a bit more neatly, too. Some good examples are [1] and [2]. [1] https://ccom.ucsd.edu/~ctiee/notes/tensors.pdf https://ccom.ucsd.edu/~ctiee/notes/tensors.pdf [2] https://www.cs.cmu.edu/~kmcrane/Projects/DDG/ https://www.cs.cmu.edu/~kmcrane/Projects/DDG/