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The formula $A \cdot (B \times C) = \textrm{Det}(A,B,C)$ shows this the cross product can be thought of as the transpose of the linear map $\textrm{Det}(\cdot,B,C)$. Using the notation of riemannian geometry (hodge star, sharps, and flats) another way to say this is that $A \times B=\star(A^\flat \wedge B^\flat)^\sharp$. This is the connection between the ...

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I think I was just being overly cautious here. The starting inner product $\langle x|y\rangle$ on $M$ is positive definite, i.e. of signature $(n,0)$, simply by the definition of "inner product". For each $\sigma\in \Gamma_{kl}$, the form $\langle \sigma(x)|\sigma(y)\rangle$ on $M$ is positive definite, again by simple observation: for any $x\in M$ ...

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Doran and Lasenby's book Geometric algebra for physicists does an OK job of presenting those three topics. Normally the way physicists write does not make me very happy, but their exposition for these particular topics was very helpful to me.

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I see no reason for that inner product to be equivalent to the/an Euclidean inner product on M. It seems as if you are interpreting "orthogonal" to be "Euclidean," which is probably the problem. The modifier "orthogonal" is used to describe any geometry which preserves a symmetric bilinear form. So what I think is going on is that they are claiming ...

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Scroll to the bottom of Clifford Algebra: A Visual Introduction, right under Clifford the Big Red Algebra and you will find two links: The story continues with Geometric Algebra: Projective Geometry. The final chapter is Geometric Algebra: Conformal Geometry. If you like this intuitive graphical presentation of Clifford Algebra you might also be ...

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For the sake of completeness, I'll just post up the full answer, which is relatively simple after ahmetselcuk's response. Since $v^2 = v_0^2$, then we can describe $v$ purely by means of rotations. That is, for some unit vector $u$, we have: $$v = v_0RuR^\dagger$$ where all of the information of motion is purely on the rotor $R$. Then, we must have ...

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The restriction comes from the fact that $v$ is on a curve. If you look at the Rotating frames section in the same chapter you will find the answer.

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As much as I love Clifford / geometric algebra, I don't think it is of use here. CA allows you to deal in a basis-free way with multidimensional subspaces as algebraic objects, making various complicated derivations more elegant and transparent. In the case of neural nets, however, we really are dealing with a plain vector of parameters. If \$\theta \in ...

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