Comment by jefftk | Hacker News Reader

Comment by jefftk | Hacker News Reader

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> The similarities are so striking that we might think of them as "pseudovpseudovectors". But I won't write them this way because I think that obscures their true nature. Written this way it looks like a bivector only encapsulates three degrees of freedom!

> Instead, I will use: ... Because it forces us to remember what those coefficients are attached to. Knowing that a bivector contains five degrees of freedom, can you figure out what the other two describe?

I'm confused here and don't understand why they keep saying a bivector has five degrees of freedom. If you can uniquely identify one with three scalar coefficients, doesn't it only have three degrees of freedom?

Yes, only three. As defined, two bivectors are equal if their areas are equal and if their oriented planes are equal. Therefore two more degrees of freedom are absorbed by taking rotations of the two vectors in the plane.Along with the rescaling the author noted, we're down to three from six.

That makes complete sense to me. But then later on they say "The output is a Geometric with a scalar component s and a bivector component ⇒c, which has 1 + 5 = 6 degrees of freedom so this system is not lossy! It should permit an unambiguous inversion operation!" If a bivector only has 3 degrees of freedom then the total is 4, which seems like it would be lossy?

I was also wondering this. But note that x^y is always perpendicular to x, so really only has two degrees of freedom while you need three to recover y (knowing x). Add in the dot product part to make up for it.

Interesting, I was confused about the same thing. So the author is not correct when they say a bivector has 5 degrees of freedom?

Yes, a bivector has 3 degrees of freedom. I was mistaken when I wrote 5. I'll fix the text tonight.

Author here: I was wrong about the 5 degrees of freedom thing. A bivector has 3 degrees of freedom. I'll correct the text tonight.

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