Eebster the Great wrote:I mean the geometric product. According to that page, a2 is a scalar. But the geometric product ab for a ≠ b must not be a scalar, because if it were, a∧b = ½(ab − ba) would clearly be a scalar too.
Their product lives in the exterior algebra of the vector space (or whatever the word for the algebra of 0-vectors, 1-vectors, 2-vectors, etc. is); so the product could, depending on circumstance, take a value as either a scalar, or a formal sum of scalars, vectors, bivectors (and more in higher dimension, but not on the plane). Since the wedge product a∧a is 0, in the case of aa, only scalars are present, so you can identify that with the scalar ||a||
2. In particular, ab is a scalar exactly when a and b are parallel. (I have to say that that definition really threw me for the same reasons; it's just bad writing, really - a scalar is not the same as a formal sum involving 1).
phlip wrote:As long as we're working in two dimensions, their "outer product" is still a scalar. They're calling it a "bivector", but it's still just a number. They say it has "direction" in the article, but really it only has two directions: clockwise and anticlockwise, which map to positive and negative. Adding it to the dot product isn't necessarily something it makes sense to do, but it's not a type error like adding a scalar and a vector is... it's like, in 3D, adding a vector and a pseudovector... "a + a×b"... the types are compatible, the answer is just usually meaningless. Their "bivector" is basically the 2D equivalent of what the pseudovector is in 3D.
Just because the space of bivectors is only one dimensional doesn't mean it's just a number, unless you a particular basis in mind (which we don't). And they manage to show a sum of a scalar and bivector means something (a combination of a rotation & scaling) - which is reasonable enough.
The problem is that, they seem to implicitly assume that the algebra operators on any formal sum of a scalar, vector, and bivector. However, by taking the geometric product of an even number of vectors, you always get the sum of a bivector and scalar, and of an odd number of vectors yields a vector. So, it's reasonable to say that the system actually discusses two different domains - that of rotation/scaling transformation (bivector + scalar) and that of vectors. Obviously, transformations are composed when multiplied, and a transformation * a vector is the application of the transformation to the vector, but a vector * transformation is the conjugate of the transformation applied to the vector (i.e. the rotation the other way) - which is just how a particular subgroup of matrices work. (Also, the product on the space of transformations is
actually the complex numbers with multiplication; they point the similarity out, but somehow don't seem to make the connection that they're isomorphic). The only particularly interesting thing is that ba is the transformation which takes a to b|a| (when premultiplied).
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.